{ "source_file": "Gr11_Mathematics_Learner_Eng.txt", "title": "Grade 11 Mathematics", "table_of_contents": [ { "section_id": "1", "title": "Exponents and surds" }, { "section_id": "1.1", "title": "Revision" }, { "section_id": "1.2", "title": "Rational exponents and surds" }, { "section_id": "1.3", "title": "Solving surd equations" }, { "section_id": "1.4", "title": "Applications of exponentials" }, { "section_id": "1.5", "title": "Summary" }, { "section_id": "7", "title": "Equations and inequalities" }, { "section_id": "2.1", "title": "Revision" }, { "section_id": "2.2", "title": "Completing the square" }, { "section_id": "2.3", "title": "Quadratic formula" }, { "section_id": "2.4", "title": "Substitution" }, { "section_id": "2.5", "title": "Finding the equation" }, { "section_id": "2.6", "title": "Nature of roots" }, { "section_id": "2.7", "title": "Quadratic inequalities" }, { "section_id": "2.8", "title": "Simultaneous equations" }, { "section_id": "2.9", "title": "Word problems" }, { "section_id": "2.10", "title": "Summary" }, { "section_id": "18", "title": "Number patterns" }, { "section_id": "3.1", "title": "Revision" }, { "section_id": "3.2", "title": "Quadratic sequences" }, { "section_id": "3.3", "title": "Summary" }, { "section_id": "22", "title": "Analytical geometry" }, { "section_id": "4.1", "title": "Revision" }, { "section_id": "4.2", "title": "Equation of a line" }, { "section_id": "4.3", "title": "Inclination of a line" }, { "section_id": "4.4", "title": "Parallel lines" }, { "section_id": "4.5", "title": "Perpendicular lines" }, { "section_id": "4.6", "title": "Summary" }, { "section_id": "29", "title": "Functions" }, { "section_id": "5.1", "title": "Quadratic functions" }, { "section_id": "5.2", "title": "Average gradient" }, { "section_id": "5.3", "title": "Hyperbolic functions" }, { "section_id": "5.4", "title": "Exponential functions" }, { "section_id": "5.5", "title": "The sine function" }, { "section_id": "5.6", "title": "The cosine function" }, { "section_id": "5.7", "title": "The tangent function" }, { "section_id": "5.8", "title": "Summary" }, { "section_id": "38", "title": "Trigonometry" }, { "section_id": "6.1", "title": "Revision" }, { "section_id": "6.2", "title": "Trigonometric identities" }, { "section_id": "6.3", "title": "Reduction formula" }, { "section_id": "6.4", "title": "Trigonometric equations" }, { "section_id": "6.5", "title": "Area, sine, and cosine rules" }, { "section_id": "6.6", "title": "Summary" }, { "section_id": "45", "title": "Measurement" }, { "section_id": "7.1", "title": "Area of a polygon" }, { "section_id": "7.2", "title": "Right prisms and cylinders" }, { "section_id": "7.3", "title": "Right pyramids, right cones and spheres" }, { "section_id": "7.4", "title": "Multiplying a dimension by a constant factor" }, { "section_id": "7.5", "title": "Summary" }, { "section_id": "51", "title": "Euclidean geometry" }, { "section_id": "8.1", "title": "Revision" }, { "section_id": "8.2", "title": "Circle geometry" }, { "section_id": "8.3", "title": "Summary" }, { "section_id": "55", "title": "Finance, growth and decay" }, { "section_id": "9.1", "title": "Revision" }, { "section_id": "9.2", "title": "Simple and compound depreciation" }, { "section_id": "9.3", "title": "Timelines" }, { "section_id": "9.4", "title": "Nominal and effective interest rates" }, { "section_id": "9.5", "title": "Summary" }, { "section_id": "10", "title": "Probability" }, { "section_id": "10.1", "title": "Revision" }, { "section_id": "10.2", "title": "Dependent and independent events" }, { "section_id": "10.3", "title": "More Venn diagrams" }, { "section_id": "10.4", "title": "Tree diagrams" }, { "section_id": "10.5", "title": "Contingency tables" }, { "section_id": "10.6", "title": "Summary" }, { "section_id": "11", "title": "Statistics" }, { "section_id": "11.1", "title": "Revision" }, { "section_id": "11.2", "title": "Histograms" }, { "section_id": "11.3", "title": "Ogives" }, { "section_id": "11.4", "title": "Variance and standard deviation" }, { "section_id": "11.5", "title": "Symmetric and skewed data" }, { "section_id": "11.6", "title": "Identification of outliers" }, { "section_id": "11.7", "title": "Summary" }, { "section_id": "12", "title": "Linear programming" }, { "section_id": "12.1", "title": "Introduction" }, { "section_id": "78", "title": "Solutions to exercises" } ], "front_matter": "EVERYTHING MATHS\nGRADE 11 MATHEMATICS\nVERSION 1 CAPS\nWRITTEN BY VOLUNTEERS\n\nCOPYRIGHT NOTICE\nYou are allowed and encouraged to copy any of the Everything Maths and Everything Science \ntextbooks. 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Practise the exercises from this text-\nbook, additional exercises and questions from past exam papers on m.everythingmaths.\nco.za and m.everythingscience.co.za and Mxit Reach. \nPRACTISE INTELLIGENTLY\nCHECK YOUR ANSWERS ON YOUR PHONE\nm.everythingmaths.co.za and m.everythingscience.co.za \nPRACTISE FOR TESTS AND EXAMS ON YOUR PHONE\nm.everythingmaths.co.za and m.everythingscience.co.za\n\nIf you complete you practice homework and test questions at m.everythingmaths.co.za \nor m.everythingscience.co.za, you can track of your work. Your dashboard will show you \nyour progress and mastery for every topic in the book and help you to manage your stud-\nies. You can use your dashboard to show your teachers, parents, universities or bursary \ninstitutions what you have done during the year.\nMANAGE YOUR STUDIES\nYOUR DASHBOARD\n\nEVERYTHING MATHS\nMathematics is commonly thought of as being about numbers but mathematics is actu-\nally a language! Mathematics is the language that nature speaks to us in. As we learn to \nunderstand and speak this language, we can discover many of nature’s secrets. Just as \nunderstanding someone’s language is necessary to learn more about them, mathemat-\nics is required to learn about all aspects of the world – whether it is physical sciences, life \nsciences or even finance and economics.\nThe great writers and poets of the world have the ability to draw on words and put them \ntogether in ways that can tell beautiful or inspiring stories. In a similar way, one can draw \non mathematics to explain and create new things. Many of the modern technologies that \nhave enriched our lives are greatly dependent on mathematics. DVDs, Google searches, \nbank cards with PIN numbers are just some examples. And just as words were not created \nspecifically to tell a story but their existence enabled stories to be told, so the mathemat-\nics used to create these technologies was not developed for its own sake, but was avail-\nable to be drawn on when the time for its application was right.\nThere is in fact not an area of life that is not affected by mathematics. Many of the most \nsought after careers depend on the use of mathematics. Civil engineers use mathematics \nto determine how to best design new structures; economists use mathematics to describe \nand predict how the economy will react to certain changes; investors use mathematics to \nprice certain types of shares or calculate how risky particular investments are; software \ndevelopers use mathematics for many of the algorithms (such as Google searches and \ndata security) that make programmes useful.\nBut, even in our daily lives mathematics is everywhere – in our use of distance, time and \nmoney. Mathematics is even present in art, design and music as it informs proportions \nand musical tones. The greater our ability to understand mathematics, the greater our \nability to appreciate beauty and everything in nature. Far from being just a cold and ab-\nstract discipline, mathematics embodies logic, symmetry, harmony and technological \nprogress. More than any other language, mathematics is everywhere and universal in its \napplication.\n\nContents\n1\nExponents and surds\n4\n1.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n4\n1.2\nRational exponents and surds . . . . . . . . . . . . . . . . . . . . . . .\n8\n1.3\nSolving surd equations\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n19\n1.4\nApplications of exponentials . . . . . . . . . . . . . . . . . . . . . . .\n23\n1.5\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n25\n2\nEquations and inequalities\n30\n2.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n30\n2.2\nCompleting the square\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n38\n2.3\nQuadratic formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n44\n2.4\nSubstitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n48\n2.5\nFinding the equation\n. . . . . . . . . . . . . . . . . . . . . . . . . . .\n50\n2.6\nNature of roots\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n52\n2.7\nQuadratic inequalities\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n60\n2.8\nSimultaneous equations . . . . . . . . . . . . . . . . . . . . . . . . . .\n67\n2.9\nWord problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n74\n2.10 Summary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n80\n3\nNumber patterns\n86\n3.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n86\n3.2\nQuadratic sequences . . . . . . . . . . . . . . . . . . . . . . . . . . .\n90\n3.3\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n99\n4\nAnalytical geometry\n104\n4.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n104\n4.2\nEquation of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n113\n4.3\nInclination of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n124\n4.4\nParallel lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n132\n4.5\nPerpendicular lines . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n136\n4.6\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n142\n5\nFunctions\n146\n5.1\nQuadratic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n146\n5.2\nAverage gradient\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n164\n5.3\nHyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . .\n170\n5.4\nExponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . .\n184\n5.5\nThe sine function . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n197\n5.6\nThe cosine function . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n209\n5.7\nThe tangent function\n. . . . . . . . . . . . . . . . . . . . . . . . . . .\n222\n5.8\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n235\n6\nTrigonometry\n240\n6.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n240\n6.2\nTrigonometric identities . . . . . . . . . . . . . . . . . . . . . . . . . .\n247\n6.3\nReduction formula\n. . . . . . . . . . . . . . . . . . . . . . . . . . . .\n253\n\n6.4\nTrigonometric equations\n. . . . . . . . . . . . . . . . . . . . . . . . .\n266\n6.5\nArea, sine, and cosine rules . . . . . . . . . . . . . . . . . . . . . . . .\n280\n6.6\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n301\n7\nMeasurement\n308\n7.1\nArea of a polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n308\n7.2\nRight prisms and cylinders . . . . . . . . . . . . . . . . . . . . . . . .\n311\n7.3\nRight pyramids, right cones and spheres . . . . . . . . . . . . . . . . .\n318\n7.4\nMultiplying a dimension by a constant factor . . . . . . . . . . . . . .\n322\n7.5\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n326\n8\nEuclidean geometry\n332\n8.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n332\n8.2\nCircle geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n333\n8.3\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n363\n9\nFinance, growth and decay\n374\n9.1\nRevision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n374\n9.2\nSimple and compound depreciation . . . . . . . . . . . . . . . . . . .\n377\n9.3\nTimelines\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n388\n9.4\nNominal and effective interest rates\n. . . . . . . . . . . . . . . . . . .\n394\n9.5\nSummary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n398\n10 Probability\n402\n10.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n402\n10.2 Dependent and independent events . . . . . . . . . . . . . . . . . . .\n411\n10.3 More Venn diagrams\n. . . . . . . . . . . . . . . . . . . . . . . . . . .\n419\n10.4 Tree diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n426\n10.5 Contingency tables . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n431\n10.6 Summary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n435\n11 Statistics\n440\n11.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n440\n11.2 Histograms\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n444\n11.3 Ogives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n451\n11.4 Variance and standard deviation . . . . . . . . . . . . . . . . . . . . .\n455\n11.5 Symmetric and skewed data\n. . . . . . . . . . . . . . . . . . . . . . .\n461\n11.6 Identification of outliers . . . . . . . . . . . . . . . . . . . . . . . . . .\n464\n11.7 Summary\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n467\n12 Linear programming\n472\n12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n472\nSolutions to exercises\n483\n2\nContents", "chapters": [ { "title": "Exponents and surds", "content": "Chapter 1.\nExponents and surds\n\nExamples:\n1. 2 × 2 × 2 × 2 = 24\n2. 0,71 × 0,71 × 0,71 = (0,71)3\n3. (501)2 = 501 × 501\n4. k6 = k × k × k × k × k × k\nFor x2, we say x is squared and for y3, we say that y is cubed. In the last example we\nhave k6; we say that k is raised to the sixth power.\nWe also have the following definitions for exponents. It is important to remember that\nwe always write the final answer with a positive exponent.\n• a0 = 1\n(a ̸= 0 because 00 is undefined)\n• a−n =\n1\nan\n(a ̸= 0 because 1\n0 is undefined)\nExamples:\n1. 5−2 = 1\n52 = 1\n25\n2. (−36)0x = (1)x = x\n3. 7p−1\nq3t−2 = 7t2\npq3\nWe use the following laws for working with exponents:\n• am × an = am+n\n•\nam\nan = am−n\n• (ab)n = anbn\n•\n\u0000 a\nb\n\u0001n = an\nbn\n• (am)n = amn\nwhere a > 0, b > 0 and m, n ∈Z.\nWorked example 1: Laws of exponents\nQUESTION\nSimplify the following:\n1. 5(m2t)p × 2(m3p)t\n2. 8k3x2\n(xk)2\n6\n1.1.\nRevision\n\n3. 22 × 3 × 74\n(7 × 2)4\n4. 3(3b)a\nSOLUTION\n1. 5(m2t)p × 2(m3p)t = 10m2pt+3pt = 10m5pt\n2. 8k3x2\n(xk)2 = 8k3x2\nx2k2 = 8k(3−2)x(2−2) = 8k1x0 = 8k\n3. 22 × 3 × 74\n(7 × 2)4\n= 22 × 3 × 74\n74 × 24\n= 2(2−4) × 3 × 7(4−4) = 2−2 × 3 = 3\n4\n4. 3(3b)a = 3 × 3ab = 3ab+1\nWorked example 2: Laws of exponents\nQUESTION\nSimplify: 3m −3m+1\n4 × 3m −3m\nSOLUTION\nStep 1: Simplify to a form that can be factorised\n3m −3m+1\n4 × 3m −3m = 3m −(3m × 3)\n4 × 3m −3m\nStep 2: Take out a common factor\n= 3m(1 −3)\n3m(4 −1)\nStep 3: Cancel the common factor and simplify\n= 1 −3\n4 −1\n= −2\n3\n7\nChapter 1.\nExponents and surds\n\nExercise 1 – 2: Laws of exponents\nSimplify the following:\n1. 4 × 42a × 42 × 4a\n2.\n32\n2−3\n3. (3p5)2\n4. k2k3x−4\nkx\n5. (5z−1)2 + 5z\n6. (1\n4)0\n7. (x2)5\n8.\n\u0000 a\nb\n\u0001−2\n9. (m + n)−1\n10. 2(pt)s\n11.\n1\n\u0000 1\na\n\u0001−1\n12.\nk0\nk−1\n13.\n−2\n−2−a\n14.\n−h\n(−h)−3\n15.\n\u0012a2b3\nc3d\n\u00132\n16. 107(70) × 10−6(−6)0 −6\n17. m3n2 ÷ nm2 × mn\n2\n18. (2−2 −5−1)−2\n19. (y2)−3 ÷\n\u0010\nx2\ny3\n\u0011−1\n20. 2c−5\n2c−8\n21. 29a × 46a × 22\n85a\n22. 20t5p10\n10t4p9\n23.\n\u0012\n9q−2s\nq−3sy−4a−1\n\u00132\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 222R\n2. 222S\n3. 222T\n4. 222V\n5. 222W\n6. 222X\n7. 222Y\n8. 222Z\n9. 2232\n10. 2233\n11. 2234\n12. 2235\n13. 2236\n14. 2237\n15. 2238\n16. 2239\n17. 223B\n18. 223C\n19. 223D\n20. 223F\n21. 223G\n22. 223H\n23. 223J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 222Q at www.everythingmaths.co.za\n1.2\nRational exponents and surds\nEMBF5\nThe laws of exponents can also be extended to include the rational numbers. A rational\nnumber is any number that can be written as a fraction with an integer in the numerator\nand in the denominator.\nWe also have the following definitions for working with\nrational exponents.\n8\n1.2.\nRational exponents and surds\n\n• If rn = a, then r =\nn√a\n(n ≥2)\n• a\n1\nn =\nn√a\n• a−1\nn = (a−1)\n1\nn =\nnq\n1\na\n• a\nm\nn = (am)\n1\nn =\nn√\nam\nwhere a > 0, r > 0 and m, n ∈Z, n ̸= 0.\nFor\n√\n25 = 5, we say that 5 is the square root of 25 and for\n3√\n8 = 2, we say that 2 is\nthe cube root of 8. For\n5√\n32 = 2, we say that 2 is the fifth root of 32.\nWhen dealing with exponents, a root refers to a number that is repeatedly multiplied\nby itself a certain number of times to get another number. A radical refers to a number\nwritten as shown below.\nradical sign\nn√a\ndegree\nradicand\nradical\n}\nSee video: 223K at www.everythingmaths.co.za\nThe radical symbol and degree show which root is being determined. The radicand is\nthe number under the radical symbol.\n• If n is an even natural number, then the radicand must be positive, otherwise the\nroots are not real. For example,\n4√\n16 = 2 since 2 × 2 × 2 × 2 = 16, but the roots\nof\n4√−16 are not real since (−2) × (−2) × (−2) × (−2) ̸= −16.\n• If n is an odd natural number, then the radicand can be positive or negative. For\nexample,\n3√\n27 = 3 since 3 × 3 × 3 = 27 and we can also determine\n3√−27 = −3\nsince (−3) × (−3) × (−3) = −27.\nIt is also possible for there to be more than one nth root of a number. For example,\n(−2)2 = 4 and 22 = 4, so both −2 and 2 are square roots of 4.\nA surd is a radical which results in an irrational number. Irrational numbers are num-\nbers that cannot be written as a fraction with the numerator and the denominator as\nintegers. For example,\n√\n12,\n3√\n100,\n5√\n25 are surds.\nWorked example 3: Rational exponents\nQUESTION\nWrite each of the following as a radical and simplify where possible:\n1. 18\n1\n2\n2. (−125)−1\n3\n9\nChapter 1.\nExponents and surds\n\n3. 4\n3\n2\n4. (−81)\n1\n2\n5. (0,008)\n1\n3\nSOLUTION\n1. 18\n1\n2 =\n√\n18\n2. (−125)−1\n3 =\n3p\n(−125)−1 =\n3\nr\n1\n−125 =\n3\nr\n1\n(−5)3 = −1\n5\n3. 4\n3\n2 = (43)\n1\n2 =\n√\n43 =\n√\n64 = 8\n4. (−81)\n1\n2 = √−81 = not real\n5. (0,008)\n1\n3 =\n3\nr\n8\n1000 =\n3\nr\n23\n103 = 2\n10 = 1\n5\nSee video: 223M at www.everythingmaths.co.za\nWorked example 4: Rational exponents\nQUESTION\nSimplify without using a calculator:\n\u0012\n5\n4−1 −9−1\n\u0013 1\n2\nSOLUTION\nStep 1: Write the fraction with positive exponents in the denominator\n \n5\n1\n4 −1\n9\n! 1\n2\nStep 2: Simplify the denominator\n10\n1.2.\nRational exponents and surds\n\n=\n \n5\n9−4\n36\n! 1\n2\n=\n \n5\n5\n36\n! 1\n2\n=\n\u0012\n5 ÷ 5\n36\n\u0013 1\n2\n=\n\u0012\n5 × 36\n5\n\u0013 1\n2\n= (36)\n1\n2\nStep 3: Take the square root\n=\n√\n36\n= 6\nExercise 1 – 3: Rational exponents and surds\n1. Simplify the following and write answers with positive exponents:\na)\n√\n49\nb)\n√\n36−1\nc)\n3√\n6−2\nd)\n3\nr\n−64\n27\ne)\n4p\n(16x4)3\n2. Simplify:\na) s\n1\n2 ÷ s\n1\n3\nb)\n\u000064m6\u0001 2\n3\nc)\n12m\n7\n9\n8m−11\n9\nd) (5x)0 + 5x0 −(0,25)−0,5 + 8\n2\n3\n3. Use the laws to re-write the following expression as a power of x:\nx\nr\nx\nq\nx\np\nx√x\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 223P\n1b. 223Q\n1c. 223R\n1d. 223S\n1e. 223T\n2a. 223V\n2b. 223W\n2c. 223X\n2d. 223Y\n3. 223Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11\nChapter 1.\nExponents and surds\n\nSimplification of surds\nEMBF6\nWe have seen in previous examples and exercises that rational exponents are closely\nrelated to surds. It is often useful to write a surd in exponential notation as it allows us\nto use the exponential laws.\nThe additional laws listed below make simplifying surds easier:\n•\nn√a\nn√\nb =\nn√\nab\n•\nn\nra\nb =\nn√a\nn√\nb\n•\nmp\nn√a =\nmn√a\n•\nn√\nam = a\nm\nn\n• ( n√a)m = a\nm\nn\nSee video: 223N at www.everythingmaths.co.za\nWorked example 5: Simplifying surds\nQUESTION\nShow that:\n1.\nn√a ×\nn√\nb =\nn√\nab\n2.\nn\nra\nb =\nn√a\nn√\nb\nSOLUTION\n1.\nn√a ×\nn√\nb = a\n1\nn × b\n1\nn\n= (ab)\n1\nn\n=\nn√\nab\n2.\nn\nra\nb =\n\u0010a\nb\n\u0011 1\nn\n= a\n1\nn\nb\n1\nn\n=\nn√a\nn√\nb\n12\n1.2.\nRational exponents and surds\n\nExamples:\n1.\n√\n2 ×\n√\n32 = √2 × 32 =\n√\n64 = 8\n2.\n3√\n24\n3√\n3 =\n3\nr\n24\n3 =\n3√\n8 = 2\n3.\np√\n81 =\n4√\n81 =\n4√\n34 = 3\nLike and unlike surds\nEMBF7\nTwo surds\nm√a and\nn√\nb are like surds if m = n, otherwise they are called unlike surds.\nFor example,\nq\n1\n3 and −\n√\n61 are like surds because m = n = 2. Examples of unlike\nsurds are\n3√\n5 and\n5p\n7y3 since m ̸= n.\nSimplest surd form\nEMBF8\nWe can sometimes simplify surds by writing the radicand as a product of factors that\ncan be further simplified using\nn√\nab =\nn√a ×\nn√\nb.\nSee video: 2242 at www.everythingmaths.co.za\nWorked example 6: Simplest surd form\nQUESTION\nWrite the following in simplest surd form:\n√\n50\nSOLUTION\nStep 1: Write the radicand as a product of prime factors\n√\n50 =\n√\n5 × 5 × 2\n=\np\n52 × 2\nStep 2: Simplify using\nn√\nab =\nn√a ×\nn√\nb\n=\n√\n52 ×\n√\n2\n= 5 ×\n√\n2\n= 5\n√\n2\n13\nChapter 1.\nExponents and surds\n\nSometimes a surd cannot be simplified. For example,\n√\n6,\n3√\n30 and\n4√\n42 are already in\ntheir simplest form.\nWorked example 7: Simplest surd form\nQUESTION\nWrite the following in simplest surd form:\n3√\n54\nSOLUTION\nStep 1: Write the radicand as a product of prime factors\n3√\n54 =\n3√\n3 × 3 × 3 × 2\n=\n3p\n33 × 2\nStep 2: Simplify using\nn√\nab =\nn√a ×\nn√\nb\n=\n3√\n33 ×\n3√\n2\n= 3 ×\n3√\n2\n= 3\n3√\n2\nWorked example 8: Simplest surd form\nQUESTION\nSimplify:\n√\n147 +\n√\n108\nSOLUTION\nStep 1: Write the radicands as a product of prime factors\n√\n147 +\n√\n108 =\n√\n49 × 3 +\n√\n36 × 3\n=\np\n72 × 3 +\np\n62 × 3\nStep 2: Simplify using\nn√\nab =\nn√a ×\nn√\nb\n14\n1.2.\nRational exponents and surds\n\n=\n\u0010√\n72 ×\n√\n3\n\u0011\n+\n\u0010√\n62 ×\n√\n3\n\u0011\n=\n\u0010\n7 ×\n√\n3\n\u0011\n+\n\u0010\n6 ×\n√\n3\n\u0011\n= 7\n√\n3 + 6\n√\n3\nStep 3: Simplify and write the final answer\n13\n√\n3\nWorked example 9: Simplest surd form\nQUESTION\nSimplify:\n\u0000√\n20 −\n√\n5\n\u00012\nSOLUTION\nStep 1: Factorise the radicands were possible\n\u0010√\n20 −\n√\n5\n\u00112\n=\n\u0010√\n4 × 5 −\n√\n5\n\u00112\nStep 2: Simplify using\nn√\nab =\nn√a ×\nn√\nb\n=\n\u0010√\n4 ×\n√\n5 −\n√\n5\n\u00112\n=\n\u0010\n2 ×\n√\n5 −\n√\n5\n\u00112\n=\n\u0010\n2\n√\n5 −\n√\n5\n\u00112\nStep 3: Simplify and write the final answer\n=\n\u0010√\n5\n\u00112\n= 5\n15\nChapter 1.\nExponents and surds\n\nWorked example 10: Simplest surd form with fractions\nQUESTION\nWrite in simplest surd form:\n√\n75 ×\n3p\n(48)−1\nSOLUTION\nStep 1: Factorise the radicands were possible\n√\n75 ×\n3p\n(48)−1 =\n√\n25 × 3 ×\n3\nr\n1\n48\n=\n√\n25 × 3 ×\n1\n3√8 × 6\nStep 2: Simplify using\nn√\nab =\nn√a ×\nn√\nb\n=\n√\n25 ×\n√\n3 ×\n1\n3√\n8 ×\n3√\n6\n= 5 ×\n√\n3 ×\n1\n2 ×\n3√\n6\nStep 3: Simplify and write the final answer\n= 5\n√\n3 ×\n1\n2\n3√\n6\n= 5\n√\n3\n2\n3√\n6\nExercise 1 – 4: Simplification of surds\n1. Simplify the following and write answers with positive exponents:\na)\n3√\n16 ×\n3√\n4\nb)\n√\na2b3 ×\n√\nb5c4\nc)\n√\n12\n√\n3\nd)\np\nx2y13 ÷\np\ny5\n16\n1.2.\nRational exponents and surds\n\n2. Simplify the following:\na)\n\u00121\na −1\nb\n\u0013−1\nb)\nb −a\na\n1\n2 −b\n1\n2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2243\n1b. 2244\n1c. 2245\n1d. 2246\n2a. 2247\n2b. 2248\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nRationalising denominators\nEMBF9\nIt is often easier to work with fractions that have rational denominators instead of surd\ndenominators. By rationalising the denominator, we convert a fraction with a surd in\nthe denominator to a fraction that has a rational denominator.\nWorked example 11: Rationalising the denominator\nQUESTION\nRationalise the denominator:\n5x −16\n√x\nSOLUTION\nStep 1: Multiply the fraction by\n√x\n√x\nNotice that\n√x\n√x = 1, so the value of the fraction has not been changed.\n5x −16\n√x\n×\n√x\n√x =\n√x(5x −16)\n√x × √x\nStep 2: Simplify the denominator\n=\n√x(5x −16)\n(√x)2\n=\n√x(5x −16)\nx\nThe term in the denominator has changed from a surd to a rational number. Expressing\nthe surd in the numerator is the preferred way of writing expressions.\n17\nChapter 1.\nExponents and surds\n\nWorked example 12: Rationalising the denominator\nQUESTION\nWrite the following with a rational denominator:\ny −25\n√y + 5\nSOLUTION\nStep 1: Multiply the fraction by\n√y−5\n√y−5\nTo eliminate the surd from the denominator, we must multiply the fraction by an ex-\npression that will result in a difference of two squares in the denominator.\ny −25\n√y + 5 ×\n√y −5\n√y −5\nStep 2: Simplify the denominator\n= (y −25)(√y −5)\n(√y + 5)(√y −5)\n= (y −25)(√y −5)\n(√y)2 −25\n= (y −25)(√y −5)\ny −25\n= √y −5\nSee video: 2249 at www.everythingmaths.co.za\nExercise 1 – 5: Rationalising the denominator\nRationalise the denominator in each of the following:\n1. 10\n√\n5\n2.\n3\n√\n6\n3.\n2\n√\n3 ÷\n√\n2\n3\n4.\n3\n√\n5 −1\n5.\nx\n√y\n6.\n√\n3 +\n√\n7\n√\n2\n7. 3√p −4\n√p\n8.\nt −4\n√\nt + 2\n9. (1 + √m)−1\n10. a\n\u0010√a ÷\n√\nb\n\u0011−1\n18\n1.2.\nRational exponents and surds\n\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 224B\n2. 224C\n3. 224D\n4. 224F\n5. 224G\n6. 224H\n7. 224J\n8. 224K\n9. 224M\n10. 224N\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n1.3\nSolving surd equations\nEMBFB\nWe also need to be able to solve equations that involve surds.\nSee video: 224P at www.everythingmaths.co.za\nWorked example 13: Surd equations\nQUESTION\nSolve for x: 5\n3√\nx4 = 405\nSOLUTION\nStep 1: Write in exponential notation\n5\n\u0000x4\u0001 1\n3 = 405\n5x\n4\n3 = 405\nStep 2: Divide both sides of the equation by 5 and simplify\n5x\n4\n3\n5\n= 405\n5\nx\n4\n3 = 81\nx\n4\n3 = 34\nStep 3: Simplify the exponents\n\u0010\nx\n4\n3\n\u0011 3\n4 =\n\u000034\u0001 3\n4\nx = 33\nx = 27\n19\nChapter 1.\nExponents and surds\n\nStep 4: Check the solution by substituting the answer back into the original equation\nLHS = 5\n3√\nx4\n= 5(27)\n4\n3\n= 5(33)\n4\n3\n= 5(34)\n= 405\n= RHS\nWorked example 14: Surd equations\nQUESTION\nSolve for z: z −4√z + 3 = 0\nSOLUTION\nStep 1: Factorise\nz −4√z + 3 = 0\nz −4z\n1\n2 + 3 = 0\n(z\n1\n2 −3)(z\n1\n2 −1) = 0\nStep 2: Solve for both factors\nThe zero law states: if a × b = 0, then a = 0 or b = 0.\n∴(z\n1\n2 −3) = 0 or (z\n1\n2 −1) = 0\nTherefore\nz\n1\n2 −3 = 0\nz\n1\n2 = 3\n\u0010\nz\n1\n2\n\u00112\n= 32\nz = 9\n20\n1.3.\nSolving surd equations\n\nor\nz\n1\n2 −1 = 0\nz\n1\n2 = 1\n\u0010\nz\n1\n2\n\u00112\n= 12\nz = 1\nStep 3: Check the solution by substituting both answers back into the original equa-\ntion\nIf z = 9:\nLHS = z −4√z + 3\n= 9 −4\n√\n9 + 3\n= 12 −12\n= 0\n= RHS\nIf z = 1:\nLHS = z −4√z + 3\n= 1 −4\n√\n1 + 3\n= 4 −4\n= 0\n= RHS\nStep 4: Write the final answer\nThe solution to z −4√z + 3 = 0 is z = 9 or z = 1.\nWorked example 15: Surd equations\nQUESTION\nSolve for p: √p −2 −3 = 0\nSOLUTION\nStep 1: Write the equation with only the square root on the left hand side\nUse the additive inverse to get all other terms on the right hand side and only the\n21\nChapter 1.\nExponents and surds\n\nsquare root on the left hand side.\np\np −2 = 3\nStep 2: Square both sides of the equation\n\u0010p\np −2\n\u00112\n= 32\np −2 = 9\np = 11\nStep 3: Check the solution by substituting the answer back into the original equation\nIf p = 11:\nLHS =\np\np −2 −3\n=\n√\n11 −2 −3\n=\n√\n9 −3\n= 3 −3\n= 0\n= RHS\nStep 4: Write the final answer\nThe solution to √p −2 −3 = 0 is p = 11.\nExercise 1 – 6: Solving surd equations\nSolve for the unknown variable (remember to check that the solution is valid):\n1. 2x+1 −32 = 0\n2. 125 (3p) = 27 (5p)\n3. 2y\n1\n2 −3y\n1\n4 + 1 = 0\n4. t −1 = √7 −t\n5. 2z −7√z + 3 = 0\n6. x\n1\n3 (x\n1\n3 + 1) = 6\n7. 24n −\n1\n4√\n16 = 0\n8.\n√\n31 −10d = 4 −d\n9. y −10√y + 9 = 0\n10. f = 2 + √19 −2f\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 224Q\n2. 224R\n3. 224S\n4. 224T\n5. 224V\n6. 224W\n7. 224X\n8. 224Y\n9. 224Z\n10. 2252\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n22\n1.3.\nSolving surd equations\n\n1.4\nApplications of exponentials\nEMBFC\nThere are many real world applications that require exponents. For example, expo-\nnentials are used to determine population growth and they are also used in finance to\ncalculate different types of interest.\nWorked example 16: Applications of exponentials\nQUESTION\nA type of bacteria has a very high exponential growth rate at 80% every hour. If there\nare 10 bacteria, determine how many there will be in five hours, in one day and in\none week?\nSOLUTION\nStep 1: Exponential formula\nfinal population = initial population × (1 + growth percentage)time period in hours\nTherefore, in this case:\nfinal population = 10 (1,8)n\nwhere n = number of hours.\nStep 2: In 5 hours\nfinal population = 10 (1,8)5 ≈189\nStep 3: In 1 day = 24 hours\nfinal population = 10 (1,8)24 ≈13 382 588\nStep 4: In 1 week = 168 hours\nfinal population = 10 (1,8)168 ≈7,687 × 1043\nNote this answer is given in scientific notation as it is a very big number.\n23\nChapter 1.\nExponents and surds\n\nWorked example 17: Applications of exponentials\nQUESTION\nA species of extremely rare deep water fish has a very long lifespan and rarely has\noffspring. If there are a total of 821 of this type of fish and their growth rate is 2% each\nmonth, how many will there be in half of a year? What will the population be in ten\nyears and in one hundred years?\nSOLUTION\nStep 1: Exponential formula\nfinal population = initial population × (1 + growth percentage)time period in months\nTherefore, in this case:\nfinal population = 821(1,02)n\nwhere n = number of months.\nStep 2: In half a year = 6 months\nfinal population = 821(1,02)6 ≈925\nStep 3: In 10 years = 120 months\nfinal population = 821(1,02)120 ≈8838\nStep 4: In 100 years = 1200 months\nfinal population = 821(1,02)1200 ≈1,716 × 1013\nNote this answer is also given in scientific notation as it is a very big number.\nExercise 1 – 7: Applications of exponentials\n1. Nqobani invests R 5530 into an account which pays out a lump sum at the end\nof 6 years. If he gets R 9622,20 at the end of the period, what compound interest\nrate did the bank offer him? Give answer correct to one decimal place.\n2. The current population of Johannesburg is 3 885 840 and the average rate of\npopulation growth in South Africa is 0,7% p.a. What can city planners expect\nthe population of Johannesburg to be in 13 years time?\n3. Abiona places 3 books in a stack on her desk. The next day she counts the\nbooks in the stack and then adds the same number of books to the top of the\nstack. After how many days will she have a stack of 192 books?\n24\n1.4.\nApplications of exponentials\n\n4. A type of mould has a very high exponential growth rate of 40% every hour. If\nthere are initially 45 individual mould cells in the population, determine how\nmany there will be in 19 hours.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2253\n2. 2254\n3. 2255\n4. 2256\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n1.5\nSummary\nEMBFD\nSee presentation: 2257 at www.everythingmaths.co.za\n1. The number system:\n• N: natural numbers are {1; 2; 3; . . .}\n• N0: whole numbers are {0; 1; 2; 3; . . .}\n• Z: integers are {. . . ; −3; −2; −1; 0; 1; 2; 3; . . .}\n• Q: rational numbers are numbers which can be written as a\nb where a and b\nare integers and b ̸= 0, or as a terminating or recurring decimal number.\n• Q′: irrational numbers are numbers that cannot be written as a fraction\nwith the numerator and denominator as integers. Irrational numbers also\ninclude decimal numbers that neither terminate nor recur.\n• R: real numbers include all rational and irrational numbers.\n• R′: non-real numbers or imaginary numbers are numbers that are not real.\n2. Definitions:\n• an = a × a × a × · · · × a (n times)\n(a ∈R, n ∈N)\n• a0 = 1\n(a ̸= 0 because 00 is undefined)\n• a−n =\n1\nan\n(a ̸= 0 because 1\n0 is undefined)\n3. Laws of exponents:\n• am × an = am+n\n• am\nan = am−n\n• (ab)n = anbn\n•\n\u0000 a\nb\n\u0001n = an\nbn\n• (am)n = amn\nwhere a > 0, b > 0 and m, n ∈Z.\n25\nChapter 1.\nExponents and surds\n\n4. Rational exponents and surds:\n• If rn = a, then r =\nn√a\n(n ≥2)\n• a\n1\nn =\nn√a\n• a−1\nn = (a−1)\n1\nn =\nn\nr\n1\na\n• a\nm\nn = (am)\n1\nn =\nn√\nam\nwhere a > 0, r > 0 and m, n ∈Z, n ̸= 0.\n5. Simplification of surds:\n•\nn√a\nn√\nb =\nn√\nab\n•\nn\nra\nb =\nn√a\nn√\nb\n•\nmp\nn√a =\nmn√a\nExercise 1 – 8: End of chapter exercises\n1. Simplify as far as possible:\na) 8−2\n3\nb)\n√\n16 + 8−2\n3\n2. Simplify:\na)\n\u0000x3\u0001 4\n3\nb)\n\u0000s2\u0001 1\n2\nc)\n\u0000m5\u0001 5\n3\nd)\n\u0000−m2\u0001 4\n3\ne) −\n\u0000m2\u0001 4\n3\nf)\n\u0010\n3y\n4\n3\n\u00114\n3. Simplify the following:\na) 3a−2b15c−5\n(a−4b3c)\n−5\n2\nb)\n\u00009a6b4\u0001 1\n2\nc)\n\u0010\na\n3\n2 b\n3\n4\n\u001116\nd) x3√x\ne)\n3√\nx4b5\n4. Re-write the following expression as a power of x:\nx\nr\nx\nq\nx\np\nx√x\nx2\n5. Expand:\n\u0000√x −\n√\n2\n\u0001 \u0000√x +\n√\n2\n\u0001\n6. Rationalise the denominator:\n10\n√x −1\nx\n26\n1.5.\nSummary\n\n7. Write as a single term with a rational denominator:\n3\n2√x + √x\n8. Write in simplest surd form:\na)\n√\n72\nb)\n√\n45 +\n√\n80\nc)\n√\n48\n√\n12\nd)\n√\n18 ÷\n√\n72\n√\n8\ne)\n4\n\u0000√\n8 ÷\n√\n2\n\u0001\nf)\n16\n\u0000√\n20 ÷\n√\n12\n\u0001\n9. Expand and simplify:\na)\n\u00002 +\n√\n2\n\u00012\nb)\n\u00002 +\n√\n2\n\u0001 \u00001 +\n√\n8\n\u0001\nc)\n\u00001 +\n√\n3\n\u0001 \u00001 +\n√\n8 +\n√\n3\n\u0001\n10. Simplify, without use of a calculator:\na)\n√\n5\n\u0000√\n45 + 2\n√\n80\n\u0001\nb)\n√\n98 −\n√\n8\n√\n50\n11. Simplify:\n√\n98x6 +\n√\n128x6\n12. Rationalise the denominator:\na)\n√\n5 + 2\n√\n5\nb)\ny −4\n√y −2\nc)\n2x −20\n√x −\n√\n10\n13. Evaluate without using a calculator:\n \n2 −\n√\n7\n2\n! 1\n2\n×\n \n2 +\n√\n7\n2\n! 1\n2\n14. Prove (without the use of a calculator):\nr\n8\n3 + 5\nr\n5\n3 −\nr\n1\n6 = 10\n√\n15 + 3\n√\n6\n6\n15. Simplify completely by showing all your steps (do not use a calculator):\n3−1\n2\n\"\n√\n12 +\n3\nr\u0010\n3\n√\n3\n\u0011#\n16. Fill in the blank surd-form number on the right hand side of the equal sign which\nwill make the following a true statement: −3\n√\n6 × −2\n√\n24 = −\n√\n18 × ...\n27\nChapter 1.\nExponents and surds\n\n17. Solve for the unknown variable:\na) 3x−1 −27 = 0\nb) 8x −\n1\n3√\n8 = 0\nc) 27(4x) = (64)3x\nd) √2x −5 = 2 −x\ne) 2x\n2\n3 + 3x\n1\n3 −2 = 0\n18.\na) Show that\nr\n3x+1 −3x\n3x−1\n+ 3 is equal to 3\nb) Hence solve\nr\n3x+1 −3x\n3x−1\n+ 3 =\n\u00121\n3\n\u0013x−2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2258\n1b. 2259\n2a. 225B\n2b. 225C\n2c. 225D\n2d. 225F\n2e. 225G\n2f. 225H\n3a. 225J\n3b. 225K\n3c. 225M\n3d. 225N\n3e. 225P\n4. 225Q\n5. 225R\n6. 225S\n7. 225T\n8a. 225V\n8b. 225W\n8c. 225X\n8d. 225Y\n8e. 225Z\nIf. 2262\n9a. 2263\n9b. 2264\n9c. 2265\n10a. 2266\n10b. 2267\n11. 2268\n12a. 2269\n12b. 226B\n12c. 226C\n13. 226D\n14. 226F\n15. 226G\n16. 226H\n17a. 226J\n17b. 226K\n17c. 226M\n17d. 226N\n17e. 226P\n18. 226Q\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n28\n1.5.\nSummary\n\nCHAPTER\n2\nEquations and inequalities", "chapter_id": "1" }, { "title": "Revision", "content": "2.1\nRevision\n30\n2.2\nCompleting the square\n38\n2.3\nQuadratic formula\n44\n2.4\nSubstitution\n48\n2.5\nFinding the equation\n50\n2.6\nNature of roots\n52\n2.7\nQuadratic inequalities\n60\n2.8\nSimultaneous equations\n67\n2.9\nWord problems\n74\n2.10\nSummary\n80\n\n2\nEquations and inequalities\n2.1\nRevision\nEMBFF\nSolving quadratic equations using factorisation\nEMBFG\nTerminology:\nExpression\nAn expression is a term or group of terms consist-\ning of numbers, variables and the basic operators\n(+, −, ×, ÷, xn).\nEquation\nA mathematical statement that asserts that two ex-\npressions are equal.\nInequality\nAn inequality states the relation between two ex-\npressions (>, <, ≥, ≤).\nSolution\nA value or set of values that satisfy the original prob-\nlem statement.\nRoot\nA root of an equation is the value of x such that\nf(x) = 0.\nA quadratic equation is an equation of the second degree; the exponent of one variable\nis 2.\nThe following are examples of quadratic equations:\n2x2 −5x = 12\na(a −3) −10 = 0\n3b\nb + 2 + 1 =\n4\nb + 1\nA quadratic equation has at most two solutions, also referred to as roots. There are\nsome situations, however, in which a quadratic equation has either one solution or no\nsolutions.\n30\n2.1.\nRevision\n\nx\ny\ny = x2 −4\n0\n−2\n2\n−4\nx\ny\ny = x2\n0\nx\ny\ny = x2 + x + 1\n0\ny = (x −2)(x + 2)\n= x2 −4\nGraph of a quadratic\nequation with two roots:\nx = −2 and x = 2.\ny = x2\nGraph of a quadratic\nequation with one root:\nx = 0.\ny = x2 + x + 1\nGraph of a quadratic\nequation with no real\nroots.\nOne method for solving quadratic equations is factorisation. The standard form of a\nquadratic equation is ax2 + bx + c = 0 and it is the starting point for solving any\nequation by factorisation.\nIt is very important to note that one side of the equation must be equal to zero.\nInvestigation: Zero product law\nSolve the following equations:\n1. 6 × 0 = ?\n2. −25 × 0 = ?\n3. 0 × 0,69 = ?\n4. 7 × ? = 0\nNow solve for the variable in each of the following:\n1. 6 × m = 0\n2. 32 × x × 2 = 0\n3. 11(z −3) = 0\n4. (k + 3)(k −4) = 0\nTo obtain the two roots we use the fact that if a × b = 0, then a = 0 and/or b = 0. This\nis called the zero product law.\n31\nChapter 2.\nEquations and inequalities\n\nMethod for solving quadratic equations\nEMBFH\n1. Rewrite the equation in the standard form ax2 + bx + c = 0.\n2. Divide the entire equation by any common factor of the coefficients to obtain a\nsimpler equation of the form ax2+bx+c = 0, where a, b and c have no common\nfactors.\n3. Factorise ax2 + bx + c = 0 to be of the form (rx + s) (ux + v) = 0.\n4. The two solutions are\n(rx + s) = 0\n(ux + v) = 0\nSo x = −s\nr\nSo x = −v\nu\n5. Always check the solution by substituting the answer back into the original equa-\ntion.\nSee video: 226R at www.everythingmaths.co.za\nWorked example 1: Solving quadratic equations using factorisation\nQUESTION\nSolve for x: x (x −3) = 10\nSOLUTION\nStep 1: Rewrite the equation in the form ax2 + bx + c = 0\nExpand the brackets and subtract 10 from both sides of the equation x2 −3x −10 = 0\nStep 2: Factorise\n(x + 2) (x −5) = 0\nStep 3: Solve for both factors\nx + 2 = 0\nx = −2\nor\nx −5 = 0\nx = 5\n32\n2.1.\nRevision\n\nThe graph shows the roots of the equation x = −2 or x = 5. This graph does not\nform part of the answer as the question did not ask for a sketch. It is shown here for\nillustration purposes only.\n2\n4\n−2\n−4\n−6\n−8\n−10\n−12\n2\n4\n6\n−2\n−4\nx\nf(x)\ny = x2 −3x −10\n0\nStep 4: Check the solution by substituting both answers back into the original equa-\ntion\nStep 5: Write the final answer\nTherefore x = −2 or x = 5.\nWorked example 2: Solving quadratic equations using factorisation\nQUESTION\nSolve the equation: 2x2 −5x −12 = 0\nSOLUTION\nStep 1: There are no common factors\nStep 2: The quadratic equation is already in the standard form ax2 + bx + c = 0\nStep 3: Factorise\nWe must determine the combination of factors of 2 and 12 that will give a middle term\ncoefficient of 5.\nWe find that 2 × 1 and 3 × 4 give a middle term coefficient of 5 so we can factorise the\nequation as\n(2x + 3)(x −4) = 0\nStep 4: Solve for both roots\n33\nChapter 2.\nEquations and inequalities\n\nWe have\n2x + 3 = 0\nx = −3\n2\nor\nx −4 = 0\nx = 4\nStep 5: Check the solution by substituting both answers back into the original equa-\ntion\nStep 6: Write the final answer\nTherefore, x = −3\n2 or x = 4.\nWorked example 3: Solving quadratic equations using factorisation\nQUESTION\nSolve for y: y2 −7 = 0\nSOLUTION\nStep 1: Factorise as a difference of two squares\nWe know that\n\u0010√\n7\n\u00112\n= 7\nWe can write the equation as\ny2 −(\n√\n7)2 = 0\nStep 2: Factorise\n(y −\n√\n7)(y +\n√\n7) = 0\nTherefore y =\n√\n7 or y = −\n√\n7\n34\n2.1.\nRevision\n\nEven though the question did not ask for a sketch, it is often very useful to draw the\ngraph. We can let f(y) = y2 −7 and draw a rough sketch of the graph to see where\nthe two roots of the equation lie.\n1\n2\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n3\n−1\n−2\n−3\ny\nf(y)\nf(y) = y2 −7\n0\nStep 3: Check the solution by substituting both answers back into the original equa-\ntion\nStep 4: Write the final answer\nTherefore y = ±\n√\n7.\nWorked example 4: Solving quadratic equations using factorisation\nQUESTION\nSolve for b:\n3b\nb + 2 + 1 =\n4\nb + 1\nSOLUTION\nStep 1: Determine the restrictions\nThe restrictions are the values for b that would result in the denominator being equal\nto 0, which would make the fraction undefined. Therefore b ̸= −2 and b ̸= −1.\nStep 2: Determine the lowest common denominator\nThe lowest common denominator is (b + 2) (b + 1).\nStep 3: Multiply each term in the equation by the lowest common denominator and\nsimplify\n35\nChapter 2.\nEquations and inequalities\n\n3b(b + 2)(b + 1)\nb + 2\n+ (b + 2)(b + 1) = 4(b + 2)(b + 1)\nb + 1\n3b(b + 1) + (b + 2)(b + 1) = 4(b + 2)\n3b2 + 3b + b2 + 3b + 2 = 4b + 8\n4b2 + 2b −6 = 0\n2b2 + b −3 = 0\nStep 4: Factorise and solve the equation\n(2b + 3)(b −1) = 0\n2b + 3 = 0 or b −1 = 0\nb = −3\n2 or b = 1\nStep 5: Check the solution by substituting both answers back into the original equa-\ntion\nStep 6: Write the final answer\nTherefore b = −1 1\n2 or b = 1.\nWorked example 5: Squaring both sides of the equation\nQUESTION\nSolve for m: m + 2 = √7 + 2m\nSOLUTION\nStep 1: Square both sides of the equation\nBefore we square both sides of the equation, we must make sure that the radical is the\nonly term on one side of the equation and all other terms are on the other, otherwise\nsquaring both sides will make the equation more complicated to solve.\n(m + 2)2 =\n\u0000√\n7 + 2m\n\u00012\nStep 2: Expand the brackets and simplify\n36\n2.1.\nRevision\n\n(m + 2)2 =\n\u0000√\n7 + 2m\n\u00012\nm2 + 4m + 4 = 7 + 2m\nm2 + 2m −3 = 0\nStep 3: Factorise and solve for m\n(m −1)(m + 3) = 0\nTherefore m = 1 or m = −3\nStep 4: Check the solution by substituting both answers back into the original equa-\ntion\nTo find the solution we squared both sides of the equation. Squaring an expression\nchanges negative values to positives and can therefore introduce invalid answers into\nthe solution. Therefore it is very important to check that the answers obtained are\nvalid. To test the answers, always substitute back into the original equation.\nIf m = 1:\nRHS =\np\n7 + 2(1)\n=\n√\n9\n= 3\nLHS = 1 + 2\n= 3\nLHS = RHS\nTherefore m = 1 is valid.\nIf m = −3:\nRHS =\np\n7 + 2(−3)\n=\n√\n1\n= 1\nLHS = −3 + 2\n= −1\nLHS ̸= RHS\nTherefore m = −3 is not valid.\nStep 5: Write the final answer\nTherefore m = 1.\nSee video: 226S at www.everythingmaths.co.za\n37\nChapter 2.\nEquations and inequalities\n\nExercise 2 – 1: Solution by factorisation\nSolve the following quadratic equations by factorisation. Answers may be left in surd\nform, where applicable.\n1. 7t2 + 14t = 0\n2. 12y2 + 24y + 12 = 0\n3. 16s2 = 400\n4. y2 −5y + 6 = 0\n5. y2 + 5y −36 = 0\n6. 4 + p = √p + 6\n7. −y2 −11y −24 = 0\n8. 13y −42 = y2\n9. (x −1)(x + 10) = −24\n10. y2 −5ky + 4k2 = 0\n11. 2y2 −61 = 101\n12. 2y2 −10 = 0\n13. −8 + h2 = 28\n14. y2 −4 = 10\n15. √5 −2p −4 = 1\n2p\n16. y2 + 28 = 100\n17. f (2f + 1) = 15\n18. 2x = √21x −5\n19.\n5y\ny −2 + 3\ny + 2 =\n−6\ny2 −2y\n20.\nx + 9\nx2 −9 +\n1\nx + 3 =\n2\nx −3\n21. y −2\ny + 1 = 2y + 1\ny −7\n22. 1 + t −2\nt −1 =\n5\nt2 −4t + 3 +\n10\n3 −t\n23.\n4\nm + 3 +\n4\n4 −m2 =\n5m −5\nm2 + m −6\n24. 5√5t + 1 −4 = 5t + 1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 226T\n2. 226V\n3. 226W\n4. 226X\n5. 226Y\n6. 226Z\n7. 2272\n8. 2273\n9. 2274\n10. 2275\n11. 2276\n12. 2277\n13. 2278\n14. 2279\n15. 227B\n16. 227C\n17. 227D\n18. 227F\n19. 227G\n20. 227H\n21. 227J\n22. 227K\n23. 227M\n24. 227N\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.2\nCompleting the square\nEMBFJ\nInvestigation: Completing the square\nCan you solve each equation using two different methods?\n1. x2 −4 = 0\n2. x2 −8 = 0\n3. x2 −4x + 4 = 0\n38\n2.2.\nCompleting the square\n\n4. x2 −4x −4 = 0\nFactorising the last equation is quite difficult. Use the previous examples as a hint and\ntry to create a difference of two squares.\nSee video: 227P at www.everythingmaths.co.za\nWe have seen that expressions of the form x2 −b2 are known as differences of squares\nand can be factorised as (x −b)(x + b). This simple factorisation leads to another\ntechnique for solving quadratic equations known as completing the square.\nConsider the equation x2 −2x −1 = 0. We cannot easily factorise this expression.\nWhen we expand the perfect square (x −1)2 and examine the terms we see that\n(x −1)2 = x2 −2x + 1.\nWe compare the two equations and notice that only the constant terms are different.\nWe can create a perfect square by adding and subtracting the same amount to the\noriginal equation.\nx2 −2x −1 = 0\n(x2 −2x + 1) −1 −1 = 0\n(x2 −2x + 1) −2 = 0\n(x −1)2 −2 = 0\nMethod 1: Take square roots on both sides of the equation to solve for x.\n(x −1)2 −2 = 0\n(x −1)2 = 2\np\n(x −1)2 = ±\n√\n2\nx −1 = ±\n√\n2\nx = 1 ±\n√\n2\nTherefore x = 1 +\n√\n2 or x = 1 −\n√\n2\nVery important: Always remember to include both a positive and a negative answer\nwhen taking the square root, since 22 = 4 and (−2)2 = 4.\nMethod 2: Factorise the expression as a difference of two squares using 2 =\n\u0000√\n2\n\u00012.\nWe can write\n(x −1)2 −2 = 0\n(x −1)2 −\n\u0010√\n2\n\u00112\n= 0\n\u0010\n(x −1) +\n√\n2\n\u0011 \u0010\n(x −1) −\n√\n2\n\u0011\n= 0\n39\nChapter 2.\nEquations and inequalities\n\nThe solution is then\n(x −1) +\n√\n2 = 0\nx = 1 −\n√\n2\nor\n(x −1) −\n√\n2 = 0\nx = 1 +\n√\n2\nMethod for solving quadratic equations by completing the square\n1. Write the equation in the standard form ax2 + bx + c = 0.\n2. Make the coefficient of the x2 term equal to 1 by dividing the entire equation by\na.\n3. Take half the coefficient of the x term and square it; then add and subtract it\nfrom the equation so that the equation remains mathematically correct. In the\nexample above, we added 1 to complete the square and then subtracted 1 so\nthat the equation remained true.\n4. Write the left hand side as a difference of two squares.\n5. Factorise the equation in terms of a difference of squares and solve for x.\nSee video: 227Q at www.everythingmaths.co.za\nWorked example 6: Solving quadratic equations by completing the square\nQUESTION\nSolve by completing the square: x2 −10x −11 = 0\nSOLUTION\nStep 1: The equation is already in the form ax2 + bx + c = 0\nStep 2: Make sure the coefficient of the x2 term is equal to 1\nx2 −10x −11 = 0\nStep 3: Take half the coefficient of the x term and square it; then add and subtract it\nfrom the equation\nThe coefficient of the x term is −10. Half of the coefficient of the x term is −5 and\nthe square of it is 25. Therefore x2 −10x + 25 −25 −11 = 0.\nStep 4: Write the trinomial as a perfect square\n40\n2.2.\nCompleting the square\n\n(x2 −10x + 25) −25 −11 = 0\n(x −5)2 −36 = 0\nStep 5: Method 1: Take square roots on both sides of the equation\n(x −5)2 −36 = 0\n(x −5)2 = 36\nx −5 = ±\n√\n36\nImportant: When taking a square root always remember that there is a positive and\nnegative answer, since (6)2 = 36 and (−6)2 = 36.\nx −5 = ±6\nStep 6: Solve for x\nx = −1 or x = 11\nStep 7: Method 2: Factorise equation as a difference of two squares\n(x −5)2 −(6)2 = 0\n[(x −5) + 6] [(x −5) −6] = 0\nStep 8: Simplify and solve for x\n(x + 1)(x −11) = 0\n∴x = −1 or x = 11\nStep 9: Write the final answer\nx = −1 or x = 11\nNotice that both methods produce the same answer. These roots are rational because\n36 is a perfect square.\n41\nChapter 2.\nEquations and inequalities\n\nWorked example 7: Solving quadratic equations by completing the square\nQUESTION\nSolve by completing the square: 2x2 −6x −10 = 0\nSOLUTION\nStep 1: The equation is already in standard form ax2 + bx + c = 0\nStep 2: Make sure that the coefficient of the x2 term is equal to 1\nThe coefficient of the x2 term is 2. Therefore divide the entire equation by 2:\nx2 −3x −5 = 0\nStep 3: Take half the coefficient of the x term, square it; then add and subtract it\nfrom the equation\nThe coefficient of the x term is −3, so then\n\u0012−3\n2\n\u00132\n= 9\n4:\n\u0012\nx2 −3x + 9\n4\n\u0013\n−9\n4 −5 = 0\nStep 4: Write the trinomial as a perfect square\n\u0012\nx −3\n2\n\u00132\n−9\n4 −20\n4 = 0\n\u0012\nx −3\n2\n\u00132\n−29\n4 = 0\nStep 5: Method 1: Take square roots on both sides of the equation\n\u0012\nx −3\n2\n\u00132\n−29\n4 = 0\n\u0012\nx −3\n2\n\u00132\n= 29\n4\nx −3\n2 = ±\nr\n29\n4\nRemember: When taking a square root there is a positive and a negative answer.\nStep 6: Solve for x\n42\n2.2.\nCompleting the square\n\nx −3\n2 = ±\nr\n29\n4\nx = 3\n2 ±\n√\n29\n2\n= 3 ±\n√\n29\n2\nStep 7: Method 2: Factorise equation as a difference of two squares\n\u0012\nx −3\n2\n\u00132\n−29\n4 = 0\n\u0012\nx −3\n2\n\u00132\n−\n r\n29\n4\n!2\n= 0\n \nx −3\n2 −\nr\n29\n4\n! \nx −3\n2 +\nr\n29\n4\n!\n= 0\nStep 8: Solve for x\n \nx −3\n2 −\n√\n29\n2\n! \nx −3\n2 +\n√\n29\n2\n!\n= 0\nTherefore x = 3\n2 +\n√\n29\n2\nor x = 3\n2 −\n√\n29\n2\nNotice that these roots are irrational since 29 is not a perfect square.\nSee video: 227R at www.everythingmaths.co.za\nExercise 2 – 2: Solution by completing the square\n1. Solve the following equations by completing the square:\na) x2 + 10x −2 = 0\nb) x2 + 4x + 3 = 0\nc) p2 −5 = −8p\nd) 2(6x + x2) = −4\ne) x2 + 5x + 9 = 0\nf) t2 + 30 = 2(10 −It)\ng) 3x2 + 6x −2 = 0\nh) z2 + 8z −6 = 0\ni) 2z2 = 11z\nj) 5 + 4z −z2 = 0\n43\nChapter 2.\nEquations and inequalities\n\n2. Solve for k in terms of a: k2 + 6k + a = 0\n3. Solve for y in terms of p, q and r: py2 + qy + r = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 227S\n1b. 227T\n1c. 227V\n1d. 227W\n1e. 227X\n1f. 227Y\n1g. 227Z\n1h. 2282\n1i. 2283\n1j. 2284\n2. 2285\n3. 2286\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.3\nQuadratic formula\nEMBFK\nIt is not always possible to solve a quadratic equation by factorisation and it can take a\nlong time to complete the square. The method of completing the square provides a way\nto derive a formula that can be used to solve any quadratic equation. The quadratic\nformula provides an easy and fast way to solve quadratic equations.\nConsider the standard form of the quadratic equation ax2 + bx + c = 0. Divide both\nsides by a (a ̸= 0) to get\nx2 + bx\na + c\na = 0\nNow using the method of completing the square, we must halve the coefficient of x\nand square it. We then add and subtract\n\u0012 b\n2a\n\u00132\nso that the equation remains true.\nx2 + bx\na + b2\n4a2 −b2\n4a2 + c\na = 0\n\u0012\nx2 + bx\na + b2\n4a2\n\u0013\n−b2\n4a2 + c\na = 0\n\u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a2\n= 0\nWe add the constant to both sides and take the square root of both sides of the equa-\ntion, being careful to include a positive and negative answer.\n44\n2.3.\nQuadratic formula\n\n\u0012\nx + b\n2a\n\u00132\n= b2 −4ac\n4a2\ns\u0012\nx + b\n2a\n\u00132\n= ±\nr\nb2 −4ac\n4a2\nx + b\n2a = ±\n√\nb2 −4ac\n2a\nx = −b\n2a ±\n√\nb2 −4ac\n2a\nx = −b ±\n√\nb2 −4ac\n2a\nTherefore, for any quadratic equation ax2 + bx + c = 0 we can determine two roots\nx = −b +\n√\nb2 −4ac\n2a\nor x = −b −\n√\nb2 −4ac\n2a\nIt is important to notice that the expression b2 −4ac must be greater than or equal to\nzero for the roots of the quadratic to be real. If the expression under the square root\nsign is less than zero, then the roots are non-real (imaginary).\nSee video: 2287 at www.everythingmaths.co.za\nWorked example 8: Using the quadratic formula\nQUESTION\nSolve for x and leave your answer in simplest surd form: 2x2 + 3x = 7\nSOLUTION\nStep 1: Check whether the expression can be factorised\nThe expression cannot be factorised, so the general quadratic formula must be used.\nStep 2: Write the equation in the standard form ax2 + bx + c = 0\n2x2 + 3x −7 = 0\nStep 3: Identify the coefficients to substitute into the formula\na = 2;\nb = 3;\nc = −7\n45\nChapter 2.\nEquations and inequalities\n\nStep 4: Apply the quadratic formula\nAlways write down the formula first and then substitute the values of a, b and c.\nx = −b ±\n√\nb2 −4ac\n2a\n=\n−(3) ±\nq\n(3)2 −4 (2) (−7)\n2 (2)\n= −3 ±\n√\n65\n4\nStep 5: Write the final answer\nThe two roots are x = −3 +\n√\n65\n4\nor x = −3 −\n√\n65\n4\n.\nWorked example 9: Using the quadratic formula\nQUESTION\nFind the roots of the function f(x) = x2 −5x + 8.\nSOLUTION\nStep 1: Finding the roots\nTo determine the roots of f(x), we let x2 −5x + 8 = 0.\nStep 2: Check whether the expression can be factorised\nThe expression cannot be factorised, so the general quadratic formula must be used.\nStep 3: Identify the coefficients to substitute into the formula\na = 1;\nb = −5;\nc = 8\nStep 4: Apply the quadratic formula\nx = −b ±\n√\nb2 −4ac\n2a\n=\n−(−5) ±\nq\n(−5)2 −4 (1) (8)\n2 (1)\n= 5 ± √−7\n2\n46\n2.3.\nQuadratic formula\n\nStep 5: Write the final answer\nThere are no real roots for f(x) = x2 −5x + 8 since the expression under the square\nroot is negative (√−7 is not a real number). This means that the graph of the quadratic\nfunction has no x-intercepts; the entire graph lies above the x-axis.\n2\n4\n6\n8\n10\n2\n4\n6\n−2\n−4\nx\nf(x)\nf(x) = x2 −5x + 8\n0\nSee video: 2288 at www.everythingmaths.co.za\nExercise 2 – 3: Solution by the quadratic formula\nSolve the following using the quadratic formula.\n1. 3t2 + t −4 = 0\n2. x2 −5x −3 = 0\n3. 2t2 + 6t + 5 = 0\n4. 2p(2p + 1) = 2\n5. −3t2 + 5t −8 = 0\n6. 5t2 + 3t −3 = 0\n7. t2 −4t + 2 = 0\n8. 9(k2 −1) = 7k\n9. 3f −2 = −2f2\n10. t2 + t + 1 = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2289\n2. 228B\n3. 228C\n4. 228D\n5. 228F\n6. 228G\n7. 228H\n8. 228J\n9. 228K\n10. 228M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n47\nChapter 2.\nEquations and inequalities\n\n2.4\nSubstitution\nEMBFM\nIt is often useful to make a substitution for a repeated expression in a quadratic equa-\ntion. This makes the equation simpler and much easier to solve.\nWorked example 10: Solving by substitution\nQUESTION\nSolve for x: x2 −2x −\n3\nx2 −2x = 2\nSOLUTION\nStep 1: Determine the restrictions for x\nThe restrictions are the values for x that would result in the denominator being equal\nto 0, which would make the fraction undefined. Therefore x ̸= 0 and x ̸= 2.\nStep 2: Substitute a single variable for the repeated expression\nWe notice that x2 −2x is a repeated expression and we therefore let k = x2 −2x so\nthat the equation becomes\nk −3\nk = 2\nStep 3: Determine the restrictions for k\nThe restrictions are the values for k that would result in the denominator being equal\nto 0, which would make the fraction undefined. Therefore k ̸= 0.\nStep 4: Solve for k\nk −3\nk = 2\nk2 −3 = 2k\nk2 −2k −3 = 0\n(k + 1)(k −3) = 0\nTherefore k = −1 or k = 3\nWe check these two roots against the restrictions for k and confirm that both are valid.\nStep 5: Use values obtained for k to solve for the original variable x\n48\n2.4.\nSubstitution\n\nFor k = −1\nx2 −2x = −1\nx2 −2x + 1 = 0\n(x −1)(x −1) = 0\nTherefore x = 1\nFor k = 3\nx2 −2x = 3\nx2 −2x −3 = 0\n(x + 1)(x −3) = 0\nTherefore x = −1 or x = 3\nWe check these roots against the restrictions for x and confirm that all three values are\nvalid.\nStep 6: Write the final answer\nThe roots of the equation are x = −1, x = 1 and x = 3.\nExercise 2 – 4:\nSolve the following quadratic equations by substitution:\n1. −24 = 10(x2 + 5x) + (x2 + 5x)2\n2. (x2 −2x)2 −8 = 7(x2 −2x)\n3. x2 + 3x −\n56\nx(x + 3) = 26\n4. x2 −18 + x +\n72\nx2 + x = 0\n5. x2 −4x + 10 −7(4x −x2) = −2\n6.\n9\nx2 + 2x −12 = x2 + 2x −12\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 228N\n2. 228P\n3. 228Q\n4. 228R\n5. 228S\n6. 228T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n49\nChapter 2.\nEquations and inequalities\n\n2.5\nFinding the equation\nEMBFN\nWe have seen that the roots are the solutions obtained from solving a quadratic equa-\ntion. Given the roots, we are also able to work backwards to determine the original\nquadratic equation.\nWorked example 11: Finding an equation when the roots are given\nQUESTION\nFind an equation with roots 13 and −5.\nSOLUTION\nStep 1: Assign a variable and write roots as two equations\nx = 13 or x = −5\nUse additive inverses to get zero on the right-hand sides\nx −13 = 0 or x + 5 = 0\nStep 2: Write down as the product of two factors\n(x −13)(x + 5) = 0\nNotice that the signs in the brackets are opposite of the given roots.\nStep 3: Expand the brackets\nx2 −8x −65 = 0\nNote that if each term in the equation is multiplied by a constant then there could be\nother possible equations which would have the same roots. For example,\nMultiply by 2:\n2x2 −16x −130 = 0\nMultiply by −3:\n−3x2 + 24x + 195 = 0\n50\n2.5.\nFinding the equation\n\nWorked example 12: Finding an equation when the roots are fractions\nQUESTION\nFind an equation with roots −3\n2 and 4.\nSOLUTION\nStep 1: Assign a variable and write roots as two equations\nx = 4 or x = −3\n2\nUse additive inverses to get zero on the right-hand sides.\nx −4 = 0 or x + 3\n2 = 0\nMultiply the second equation through by 2 to remove the fraction.\nx −4 = 0 or 2x + 3 = 0\nStep 2: Write down as the product of two factors\n(2x + 3)(x −4) = 0\nStep 3: Expand the brackets\nThe quadratic equation is 2x2 −5x −12 = 0.\nExercise 2 – 5: Finding the equation\n1. Determine a quadratic equation for a graph that has roots 3 and −2.\n2. Find a quadratic equation for a graph that has x-intercepts of (−4; 0) and (4; 0).\n3. Determine a quadratic equation of the form ax2 + bx + c = 0, where a, b and c\nare integers, that has roots −1\n2 and 3.\n4. Determine the value of k and the other root of the quadratic equation kx2 −7x+\n4 = 0 given that one of the roots is x = 1.\n5. One root of the equation 2x2 −3x = p is 21\n2. Find p and the other root.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 228V\n2. 228W\n3. 228X\n4. 228Y\n5. 228Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n51\nChapter 2.\nEquations and inequalities\n\nExercise 2 – 6: Mixed exercises\nSolve the following quadratic equations by either factorisation, using the quadratic\nformula or completing the square:\n• Always try to factorise first, then use the formula if the trinomial cannot be fac-\ntorised.\n• In a test or examination, only use the method of completing the square when\nspecifically asked.\n• Answers can be left in surd or decimal form.\n1. 24y2 + 61y −8 = 0\n2. 8x2 + 16x = 42\n3. 9t2 = 24t −12\n4. −5y2 + 0y + 5 = 0\n5. 3m2 + 12 = 15m\n6. 49y2 + 0y −25 = 0\n7. 72 = 66w −12w2\n8. −40y2 + 58y −12 = 0\n9. 37n + 72 −24n2 = 0\n10. 6y2 + 7y −24 = 0\n11. 3 = x(2x −5)\n12. −18y2 −55y −25 = 0\n13. −25y2 + 25y −4 = 0\n14. 8(1 −4g2) + 24g = 0\n15. 9y2 −13y −10 = 0\n16. (7p −3)(5p + 1) = 0\n17. −81y2 −99y −18 = 0\n18. 14y2 −81y + 81 = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2292\n2. 2293\n3. 2294\n4. 2295\n5. 2296\n6. 2297\n7. 2298\n8. 2299\n9. 229B\n10. 229C\n11. 229D\n12. 229F\n13. 229G\n14. 229H\n15. 229J\n16. 229K\n17. 229M\n18. 229N\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.6\nNature of roots\nEMBFP\nInvestigation:\n1. Use the quadratic formula to determine the roots of the quadratic equations given\nbelow and take special note of:\n• the expression under the square root sign and\n• the type of number for the final answer (rational/irrational/real/imaginary)\na) x2 −6x + 9 = 0\n52\n2.6.\nNature of roots\n\nb) x2 −4x + 3 = 0\nc) x2 −4x −3 = 0\nd) x2 −4x + 7 = 0\n2. Choose the appropriate words from the table to describe the roots obtained for\nthe equations above.\nrational\nunequal\nreal\nimaginary\nnot perfect square\nequal\nperfect square\nirrational\nundefined\n3. The expression under the square root, b2 −4ac, is called the discriminant. Can\nyou make a conjecture about the relationship between the discriminant and the\nroots of quadratic equations?\nThe discriminant\nEMBFQ\nThe discriminant is defined as ∆= b2 −4ac\nThis is the expression under the square root in the quadratic formula. The discriminant\ndetermines the nature of the roots of a quadratic equation. The word ‘nature’ refers to\nthe types of numbers the roots can be — namely real, rational, irrational or imaginary.\n∆is the Greek symbol for the letter D.\nFor a quadratic function f (x) = ax2 + bx + c, the solutions to the equation f (x) = 0\nare given by the formula x = −b ±\n√\nb2 −4ac\n2a\n= −b ±\n√\n∆\n2a\n• If ∆< 0, then roots are imaginary (non-real) and beyond the scope of this book.\n• If ∆≥0, the expression under the square root is non-negative and therefore\nroots are real. For real roots, we have the following further possibilities.\n• If ∆= 0, the roots are equal and we can say that there is only one root.\n• If ∆> 0, the roots are unequal and there are two further possibilities.\n• ∆is the square of a rational number: the roots are rational.\n• ∆is not the square of a rational number: the roots are irrational and can be\nexpressed in decimal or surd form.\n53\nChapter 2.\nEquations and inequalities\n\n∆\n∆< 0: imaginary/non-real roots\n∆≥0: real roots\n∆= 0\nequal roots\n∆> 0\nunequal roots\n∆a squared ratio-\nnal: rational roots\n∆not a squared\nrational: irrational\nroots\nNature of roots\nDiscriminant\na > 0\na < 0\nRoots are non-real\n∆< 0\nRoots are real and\nequal\n∆= 0\nRoots are real and\nunequal:\n• rational\nroots\n• irrational\nroots\n∆> 0\n• ∆\n=\nsquared\nrational\n• ∆\n=\nnot\nsquared ra-\ntional\nSee video: 229P at www.everythingmaths.co.za\nWorked example 13: Nature of roots\nQUESTION\nShow that the roots of x2 −2x −7 = 0 are irrational.\nSOLUTION\nStep 1: Interpret the question\nFor roots to be real and irrational, we need to calculate ∆and show that it is greater\nthan zero and not a perfect square.\nStep 2: Check that the equation is in standard form ax2 + bx + c = 0\n54\n2.6.\nNature of roots\n\nx2 −2x −7 = 0\nStep 3: Identify the coefficients to substitute into the formula for the discriminant\na = 1;\nb = −2;\nc = −7\nStep 4: Write down the formula and substitute values\n∆= b2 −4ac\n= (−2)2 −4(1)(−7)\n= 4 + 28\n= 32\nWe know that 32 > 0 and is not a perfect square.\nThe graph below shows the roots of the equation x2 −2x −7 = 0. Note that the graph\ndoes not form part of the answer and is included for illustration purposes only.\n2\n4\n6\n8\n−2\n−4\n−6\n−8\n2\n4\n6\n−2\n−4\nx\nf(x)\nx2 −2x −7 = 0\n0\nStep 5: Write the final answer\nWe have calculated that ∆> 0 and is not a perfect square, therefore we can conclude\nthat the roots are real, unequal and irrational.\n55\nChapter 2.\nEquations and inequalities\n\nWorked example 14: Nature of roots\nQUESTION\nFor which value(s) of k will the roots of 6x2 + 6 = 4kx be real and equal?\nSOLUTION\nStep 1: Interpret the question\nFor roots to be real and equal, we need to solve for the value(s) of k such that ∆= 0.\nStep 2: Check that the equation is in standard form ax2 + bx + c = 0\n6x2 −4kx + 6 = 0\nStep 3: Identify the coefficients to substitute into the formula for the discriminant\na = 6;\nb = −4k;\nc = 6\nStep 4: Write down the formula and substitute values\n∆= b2 −4ac\n= (−4k)2 −4(6)(6)\n= 16k2 −144\nFor roots to be real and equal, ∆= 0.\n∆= 0\n16k2 −144 = 0\n16(k2 −9) = 0\n(k −3)(k + 3) = 0\nTherefore k = 3 or k = −3.\nStep 5: Check both answers by substituting back into the original equation\n56\n2.6.\nNature of roots\n\nFor k = 3:\n6x2 −4(3)x + 6 = 0\n6x2 −12x + 6 = 0\nx2 −2x + 1 = 0\n(x −1)(x −1) = 0\n(x −1)2 = 0\nTherefore x = 1\nWe see that for k = 3 the quadratic equation has real, equal roots x = 1.\nFor k = −3:\n6x2 −4(−3)x + 6 = 0\n6x2 + 12x + 6 = 0\nx2 + 2x + 1 = 0\n(x + 1)(x + 1) = 0\n(x + 1)2 = 0\nTherefore x = −1\nWe see that for k = −3 the quadratic equation has real, equal roots x = −1.\nStep 6: Write the final answer\nFor the roots of the quadratic equation to be real and equal, k = 3 or k = −3.\nWorked example 15: Nature of roots\nQUESTION\nShow that the roots of (x + h)(x + k) = 4d2 are real for all real values of h, k and d.\nSOLUTION\nStep 1: Interpret the question\nFor roots to be real, we need to calculate ∆and show that ∆≥0 for all real values of\nh, k and d.\nStep 2: Check that the equation is in standard form ax2 + bx + c = 0\nExpand the brackets and gather like terms\n57\nChapter 2.\nEquations and inequalities\n\n(x + h)(x + k) = 4d2\nx2 + hx + kx + hk −4d2 = 0\nx2 + (h + k)x + (hk −4d2) = 0\nStep 3: Identify the coefficients to substitute into the formula for the discriminant\na = 1;\nb = h + k;\nc = hk −4d2\nStep 4: Write down the formula and substitute values\n∆= b2 −4ac\n= (h + k)2 −4(1)(hk −4d2)\n= h2 + 2hk + k2 −4hk + 16d2\n= h2 −2hk + k2 + 16d2\n= (h −k)2 + (4d)2\nFor roots to be real, ∆≥0.\nWe know that (4d)2 ≥0\nand (h −k)2 ≥0\nso then (h −k)2 + (4d)2 ≥0\ntherefore ∆≥0\nStep 5: Write the final answer\nWe have shown that ∆≥0, therefore the roots are real for all real values of h, k and\nd.\nExercise 2 – 7: From past papers\n1. Determine the nature of the roots for each of the following equations:\na) x2 + 3x = −2\nb) x2 + 9 = 6x\nc) 6y2 −6y −1 = 0\nd) 4t2 −19t −5 = 0\ne) z2 = 3\nf) 0 = p2 + 5p + 8\ng) x2 = 36\nh) 4m + m2 = 1\ni) 11 −3x + x2 = 0\nj) y2 + 1\n4 = y\n2. Given: x2 + bx −2 + k\n\u0000x2 + 3x + 2\n\u0001\n= 0, (k ̸= −1)\na) Show that the discriminant is given by: ∆= k2 + 6bk + b2 + 8\n58\n2.6.\nNature of roots\n\nb) If b = 0, discuss the nature of the roots of the equation.\nc) If b = 2, find the value(s) of k for which the roots are equal.\n[IEB, Nov. 2001, HG]\n3. Show that k2x2 + 2 = kx −x2 has non-real roots for all real values for k.\n[IEB, Nov. 2002, HG]\n4. The equation x2 + 12x = 3kx2 + 2 has real roots.\na) Find the greatest value of value k such that k ∈Z.\nb) Find one rational value of k for which the above equation has rational roots.\n[IEB, Nov. 2003, HG]\n5. Consider the equation:\nk = x2 −4\n2x −5\nwhere x ̸= 5\n2.\na) Find a value of k for which the roots are equal.\nb) Find an integer k for which the roots of the equation will be rational and\nunequal.\n[IEB, Nov. 2004, HG]\n6.\na) Prove that the roots of the equation x2 −(a + b) x + ab −p2 = 0 are real for\nall real values of a, b and p.\nb) When will the roots of the equation be equal?\n[IEB, Nov. 2005, HG]\n7. If b and c can take on only the values 1, 2 or 3, determine all pairs (b; c) such\nthat x2 + bx + c = 0 has real roots.\n[IEB, Nov. 2005, HG]\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 229Q\n1b. 229R\n1c. 229S\n1d. 229T\n1e. 229V\n1f. 229W\n1g. 229X\n1h. 229Y\n1i. 229Z\n1j. 22B2\n2. 22B3\n3. 22B4\n4. 22B5\n5. 22B6\n6. 22B7\n7. 22B8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n59\nChapter 2.\nEquations and inequalities\n\n2.7\nQuadratic inequalities\nEMBFR\nQuadratic inequalities can be of the following forms:\nax2 + bx + c > 0\nax2 + bx + c ≥0\nax2 + bx + c < 0\nax2 + bx + c ≤0\nTo solve a quadratic inequality we must determine which part of the graph of a\nquadratic function lies above or below the x-axis. An inequality can therefore be\nsolved graphically using a graph or algebraically using a table of signs to determine\nwhere the function is positive and negative.\nWorked example 16: Solving quadratic inequalities\nQUESTION\nSolve for x: x2 −5x + 6 ≥0\nSOLUTION\nStep 1: Factorise the quadratic\n(x −3)(x −2) ≥0\nStep 2: Determine the critical values of x\nFrom the factorised quadratic we see that the values for which the inequality is equal\nto zero are x = 3 and x = 2. These are called the critical values of the inequality and\nthey are used to complete a table of signs.\nStep 3: Complete a table of signs\nWe must determine where each factor of the inequality is positive and negative on the\nnumber line:\n• to the left (in the negative direction) of the critical value\n• equal to the critical value\n• to the right (in the positive direction) of the critical value\nIn the final row of the table we determine where the inequality is positive and negative\nby finding the product of the factors and their respective signs.\n60\n2.7.\nQuadratic inequalities\n\nCritical values\nx = 2\nx = 3\nx −3\n−\n−\n−\n0\n+\nx −2\n−\n0\n+\n+\n+\nf(x) = (x −3)(x −2)\n+\n0\n−\n0\n+\nFrom the table we see that f(x) is greater than or equal to zero for x ≤2 or x ≥3.\nStep 4: A rough sketch of the graph\nThe graph below does not form part of the answer and is included for illustration pur-\nposes only. A graph of the quadratic helps us determine the answer to the inequality.\nWe can find the answer graphically by seeing where the graph lies above or below the\nx-axis.\n• From the standard form, x2 −5x + 6, a > 0 and therefore the graph is a “smile”\nand has a minimum turning point.\n• From the factorised form, (x −3)(x −2), we know the x-intercepts are (2; 0) and\n(3; 0).\n1\n2\n3\n4\n5\n1\n2\n3\n4\n5\nx\ny\ny = x2 −5x + 6\nThe graph is above or on the x-axis for x ≤2 or x ≥3.\nStep 5: Write the final answer and represent on a number line\nx2 −5x + 6 ≥0 for x ≤2 or x ≥3\n1\n2\n3\n4\nb\nb\n61\nChapter 2.\nEquations and inequalities\n\nWorked example 17: Solving quadratic inequalities\nQUESTION\nSolve for x: 4x2 −4x + 1 ≤0\nSOLUTION\nStep 1: Factorise the quadratic\n(2x −1)(2x −1) ≤0\n(2x −1)2 ≤0\nStep 2: Determine the critical values of x\nFrom the factorised quadratic we see that the value for which the inequality is equal to\nzero is x = 1\n2. We know that a2 > 0 for any real number a, a ̸= 0, so then (2x −1)2\nwill never be negative.\nStep 3: A rough sketch of the graph\nThe graph below does not form part of the answer and is included for illustration\npurposes only.\n• From the standard form, 4x2 −4x + 1, a > 0 and therefore the graph is a “smile”\nand has a minimum turning point.\n• From the factorised form, (2x−1)(2x−1), we know there is only one x-intercept\nat\n\u0000 1\n2; 0\n\u0001\n.\n1\n2\n3\n4\n1\n2\nx\ny\ny = 4x2 −4x + 1\n0\nNotice that no part of the graph lies below the x-axis.\nStep 4: Write the final answer and represent on a number line\n4x2 −4x + 1 ≤0 for x = 1\n2\n−2\n−1\n0\n1\n2\nb\n62\n2.7.\nQuadratic inequalities\n\nWorked example 18: Solving quadratic inequalities\nQUESTION\nSolve for x: −x2 −3x + 5 > 0\nSOLUTION\nStep 1: Examine the form of the inequality\nNotice that the coefficient of the x2 term is −1. Remember that if we multiply or\ndivide an inequality by a negative number, then the inequality sign changes direction.\nSo we can write the same inequality in different ways and still get the same answer, as\nshown below.\n−x2 −3x + 5 > 0\nMultiply by −1 and change direction of the inequality sign\nx2 + 3x −5 < 0\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n−1\n−2\n−3\n−4\nx\ny\n−x2 −3x + 5 > 0\n0\n1\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n−1\n−2\n−3\n−4\nx\ny\nx2 + 3x −5 < 0\n0\nFrom this rough sketch, we can see that both inequalities give the same solution; the\nvalues of x that lie between the two x-intercepts.\nStep 2: Factorise the quadratic\nWe notice that −x2−3x+5 > 0 cannot be easily factorised. So we let −x2−3x+5 = 0\nand use the quadratic formula to determine the roots of the equation.\n−x2 −3x + 5 = 0\nx2 + 3x −5 = 0\n63\nChapter 2.\nEquations and inequalities\n\n∴x =\n−3 ±\nq\n(3)2 −4 (1) (−5)\n2 (1)\n= −3 ±\n√\n29\n2\nx1 = −3 −\n√\n29\n2\n≈−4,2\nx2 = −3 +\n√\n29\n2\n≈1,2\nTherefore we can write, correct to one decimal place,\nx2 + 3x −5 < 0\nas (x −1,2)(x + 4,2) < 0\nStep 3: Determine the critical values of x\nFrom the factorised quadratic we see that the critical values are x = 1,2 and x =\n−4,2.\nStep 4: Complete a table of signs\nCritical values\nx = −4,2\nx = 1,2\nx + 4,2\n−\n0\n+\n+\n+\nx −1,2\n−\n−\n−\n0\n+\nf(x) = (x + 4,2)(x −1,2)\n+\n0\n−\n0\n+\nFrom the table we see that the function is negative for −4,2 < x < 1,2.\nStep 5: A sketch of the graph\n• From the standard form, x2 + 3x −5, a > 0 and therefore the graph is a “smile”\nand has a minimum turning point.\n• From the factorised form, (x −1,2)(x + 4,2), we know the x-intercepts are\n(−4,2; 0) and (1,2; 0).\n1\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n−1\n−2\n−3\n−4\nx\ny\ny = x2 + 3x −5\n0\nFrom the graph we see that the function lies below the x-axis between −4,2 and 1,2.\n64\n2.7.\nQuadratic inequalities\n\nStep 6: Write the final answer and represent on a number line\nx2 + 3x −5 < 0 for −4,2 < x < 1,2\n−5\n−4\n−3\n−2\n−1\n0\n1\n2\n−4,2\n1,2\nImportant: When working with an inequality in which the variable is in the denomi-\nnator, a different approach is needed. Always remember to check for restrictions.\nWorked example 19: Solving quadratic inequalities with fractions\nQUESTION\nSolve for x:\n1.\n2\nx + 3 =\n1\nx −3, x ̸= ±3\n2.\n2\nx + 3 ≤\n1\nx −3, x ̸= ±3\nSOLUTION\nStep 1: Solving the equation\nTo solve this equation we multiply both sides of the equation by (x + 3)(x −3) and\nsimplfy:\n2\nx + 3 × (x + 3)(x −3) =\n1\nx −3 × (x + 3)(x −3)\n2(x −3) = x + 3\n2x −6 = x + 3\nx = 9\nStep 2: Solving the inequality\nIt is very important to recognise that we cannot use the same method as above to solve\nthe inequality. If we multiply or divide an inequality by a negative number, then the\ninequality sign changes direction. We must rather simplify the inequality to have a\nlowest common denominator and use a table of signs to determine the values that\nsatisfy the inequality.\n65\nChapter 2.\nEquations and inequalities\n\nStep 3: Subtract\n1\nx −3 from both sides of the inequality\n2\nx + 3 −\n1\nx −3 ≤0\nStep 4: Determine the lowest common denominator and simplify the fraction\n2(x −3) −(x + 3)\n(x + 3)(x −3)\n≤0\nx −9\n(x + 3)(x −3) ≤0\nKeep the denominator because it affects the final answer.\nStep 5: Determine the critical values of x\nFrom the factorised inequality we see that the critical values are x = −3, x = 3 and\nx = 9.\nStep 6: Complete a table of signs\nCritical values\nx = −3\nx = 3\nx = 9\nx + 3\n−\nundef\n+\n+\n+\n+\n+\nx −3\n−\n−\n−\nundef\n+\n+\n+\nx −9\n−\n−\n−\n−\n−\n0\n+\nf(x) =\nx −9\n(x + 3)(x −3)\n−\nundef\n+\nundef\n−\n0\n+\nFrom the table we see that the function is less than or equal to zero for x < −3 or\n3 < x ≤9. We do not include x = −3 or x = 3 in the solution because of the\nrestrictions on the denominator.\nStep 7: Write the final answer and represent on a number line\nx < −3\nor\n3 < x ≤9\n−3\n−2\n−1\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\nb\n66\n2.7.\nQuadratic inequalities\n\nExercise 2 – 8: Solving quadratic inequalities\n1. Solve the following inequalities and show each answer on a number line:\na) x2 −x < 12\nb) 3x2 > −x + 4\nc) y2 < −y −2\nd) (3 −t)(1 + t) > 0\ne) s2 −4s > −6\nf) 0 ≥7x2 −x + 8\ng) x ≥−4x2\nh) 2x2 + x + 6 ≤0\ni)\nx\nx −3 < 2, x ̸= 3\nj) x2 + 4\nx −7 ≥0, x ̸= 7\nk) x + 2\nx\n−1 ≥0, x ̸= 0\n2. Draw a sketch of the following inequalities and solve for x:\na) 2x2 −18 > 0\nb) 5 −x2 ≤0\nc) x2 < 0\nd) 0 ≥6x2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22B9\n1b. 22BB\n1c. 22BC\n1d. 22BD\n1e. 22BF\n1f. 22BG\n1g. 22BH\n1h. 22BJ\n1i. 22BK\n1j. 22BM\n1k. 22BN\n2a. 22BP\n2b. 22BQ\n2c. 22BR\n2d. 22BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.8\nSimultaneous equations\nEMBFS\nSimultaneous linear equations can be solved using three different methods: substi-\ntution, elimination or using a graph to determine where the two lines intersect. For\nsolving systems of simultaneous equations with linear and non-linear equations, we\nmostly use the substitution method. Graphical solution is useful for showing where\nthe two equations intersect.\nIn general, to solve for the values of n unknown variables requires a system of n\nindependent equations.\nAn example of a system of simultaneous equations with one linear equation and one\nquadratic equation is\ny −2x = −4\nx2 + y = 4\nSolving by substitution\nEMBFT\n• Use the simplest of the two given equations to express one of the variables in\nterms of the other.\n• Substitute into the second equation. By doing this we reduce the number of\nequations and the number of variables by one.\n67\nChapter 2.\nEquations and inequalities\n\n• We now have one equation with one unknown variable which can be solved.\n• Use the solution to substitute back into the first equation to find the value of the\nother unknown variable.\nWorked example 20: Simultaneous equations\nQUESTION\nSolve for x and y:\ny −2x = −4\n. . . (1)\nx2 + y = 4\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of the first equation\ny = 2x −4\nStep 2: Substitute into the second equation and simplify\nx2 + (2x −4) = 4\nx2 + 2x −8 = 0\nStep 3: Factorise the equation\n(x + 4) (x −2) = 0\n∴x = −4 or x = 2\nStep 4: Substitute the values of x back into the first equation to determine the corre-\nsponding y-values\nIf x = −4:\ny = 2(−4) −4\n= −12\nIf x = 2:\ny = 2(2) −4\n= 0\nStep 5: Check that the two points satisfy both original equations\n68\n2.8.\nSimultaneous equations\n\nStep 6: Write the final answer\nThe solution is x = −4 and y = −12 or x = 2 and y = 0. These are the coordinate\npairs for the points of intersection as shown below.\n2\n4\n6\n−2\n−4\n−6\n−8\n−10\n−12\n−14\n2\n4\n6\n−2\n−4\n−6\ny = 2x −4\ny = 4\n−x2\n(−4; −12)\nx\ny\nb\nb\nSolving by elimination\nEMBFV\n• Make one of the variables the subject of both equations.\n• Equate the two equations; by doing this we reduce the number of equations and\nthe number of variables by one.\n• We now have one equation with one unknown variable which can be solved.\n• Use the solution to substitute back into either original equation, to find the cor-\nresponding value of the other unknown variable.\nWorked example 21: Simultaneous equations\nQUESTION\nSolve for x and y:\ny = x2 −6x\n. . . (1)\ny + 1\n2x −3 = 0\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of the second equation\ny + 1\n2x −3 = 0\ny = −1\n2x + 3\n69\nChapter 2.\nEquations and inequalities\n\nStep 2: Equate the two equations and solve for x\nx2 −6x = −1\n2x + 3\nx2 −6x + 1\n2x −3 = 0\n2x2 −12x + x −6 = 0\n2x2 −11x −6 = 0\n(2x + 1)(x −6) = 0\nTherefore x = −1\n2 or x = 6\nStep 3: Substitute the values for x back into the second equation to calculate the\ncorresponding y-values\nIf x = −1\n2:\ny = −1\n2\n\u0012\n−1\n2\n\u0013\n+ 3\n∴y = 31\n4\nThis gives the point\n\u0012\n−1\n2; 31\n4\n\u0013\n.\nIf x = 6:\ny = −1\n2(6) + 3\n= −3 + 3\n∴y = 0\nThis gives the point (6; 0).\nStep 4: Check that the two points satisfy both original equations\nStep 5: Write the final answer\nThe solution is x = −1\n2 and y = 31\n4 or x = 6 and y = 0. These are the coordinate\npairs for the points of intersection as shown below.\n2\n4\n−2\n−4\n−6\n−8\n1\n2\n3\n4\n5\n6\n7\n−1\nx\ny\ny = x2 −6x\ny = 1\n2x + 3\n\u0000−1\n2; 3 1\n4\n\u0001\n70\n2.8.\nSimultaneous equations\n\nWorked example 22: Simultaneous equations\nQUESTION\nSolve for x and y:\ny =\n5\nx −2\n. . . (1)\ny + 1 = 2x\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of the second equation\ny + 1 = 2x\ny = 2x −1\nStep 2: Equate the two equations and solve for x\n2x −1 =\n5\nx −2\n(2x −1)(x −2) = 5\n2x2 −5x + 2 = 5\n2x2 −5x −3 = 0\n(2x + 1)(x −3) = 0\nTherefore x = −1\n2 or x = 3\nStep 3: Substitute the values for x back into the second equation to calculate the\ncorresponding y-values\nIf x −1\n2:\ny = 2(−1\n2) −1\n∴y = −2\nThis gives the point (−1\n2; −2).\nIf x = 3:\ny = 2(3) −1\n= 5\nThis gives the point (3; 5).\nStep 4: Check that the two points satisfy both original equations\n71\nChapter 2.\nEquations and inequalities\n\nStep 5: Write the final answer\nThe solution is x = −1\n2 and y = −2 or x = 3 and y = 5. These are the coordinate\npairs for the points of intersection as shown below.\n2\n4\n−2\n−4\n1\n2\n3\n4\n5\n−1\n−2\nx\ny\ny = 2x −1\ny =\n5\nx−2\nSolving graphically\nEMBFW\n• Make y the subject of each equation.\n• Draw the graph of each equation on the same system of axes.\n• The final solutions to the system of equations are the coordinates of the points\nwhere the two graphs intersect.\nWorked example 23: Simultaneous equations\nQUESTION\nSolve graphically for x and y:\ny + x2 = 1\n. . . (1)\ny −x + 5 = 0\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of both equations\nFor the first equation we have\ny + x2 = 1\ny = −x2 + 1\nand for the second equation\n72\n2.8.\nSimultaneous equations\n\ny −x + 5 = 0\ny = x −5\nStep 2: Draw the straight line graph and parabola on the same system of axes\n2\n−2\n−4\n−6\n−8\n−10\n2\n4\n−2\n−4\nx\ny\ny = −x2 + 1\ny = x −5\n0\nb\nb\nStep 3: Determine where the two graphs intersect\nFrom the diagram we see that the graphs intersect at (−3; −8) and (2; −3).\nStep 4: Check that the two points satisfy both original equations\nStep 5: Write the final answer\nThe solutions to the system of simultaneous equations are (−3; −8) and (2; −3).\nExercise 2 – 9: Solving simultaneous equations\n1. Solve the following systems of equations algebraically. Leave your answer in\nsurd form, where appropriate.\na) y + x = 5\ny −x2 + 3x −5 = 0\nb) y = 6 −5x + x2\ny −x + 1 = 0\nc) y = 2x + 2\n4\ny −2x2 + 3x + 5 = 0\nd) a −2b −3 = 0; a −3b2 + 4 = 0\ne) x2 −y + 2 = 3x\n4x = 8 + y\nf) 2y + x2 + 25 = 7x\n3x = 6y + 96\n73\nChapter 2.\nEquations and inequalities\n\n2. Solve the following systems of equations graphically. Check your solutions by\nalso solving algebraically.\na) x2 −1 −y = 0\ny + x −5 = 0\nb) x + y −10 = 0\nx2 −2 −y = 0\nc) xy = 12\n7 = x + y\nd) 6 −4x −y = 0\n12 −2x2 −y = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22BT\n1b. 22BV\n1c. 22BW\n1d. 22BX\n1e. 22BY\n1f. 22BZ\n2a. 22C2\n2b. 22C3\n2c. 22C4\n2d. 22C5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.9\nWord problems\nEMBFX\nSolving word problems requires using mathematical language to describe real-life con-\ntexts. Problem-solving strategies are often used in the natural sciences and engineering\ndisciplines (such as physics, biology, and electrical engineering) but also in the social\nsciences (such as economics, sociology and political science). To solve word problems\nwe need to write a set of equations that describes the problem mathematically.\nExamples of real-world problem solving applications are:\n• modelling population growth;\n• modelling effects of air pollution;\n• modelling effects of global warming;\n• computer games;\n• in the sciences, to understand how the natural world works;\n• simulators that are used to train people in certain jobs, such as pilots, doctors\nand soldiers;\n• in medicine, to track the progress of a disease.\nSee video: 22C6 at www.everythingmaths.co.za\n74\n2.9.\nWord problems\n\nProblem solving strategy\nEMBFY\n1. Read the problem carefully.\n2. What is the question and what do we need to solve for?\n3. Assign variables to the unknown quantities, for example, x and y.\n4. Translate the words into algebraic expressions by rewriting the given information\nin terms of the variables.\n5. Set up a system of equations.\n6. Solve for the variables using substitution.\n7. Check the solution.\n8. Write the final answer.\nInvestigation: Simple word problems\nWrite an equation that describes the following real-world situations mathematically:\n1. Mohato and Lindiwe both have colds. Mohato sneezes twice for each sneeze of\nLindiwe’s. If Lindiwe sneezes x times, write an equation describing how many\ntimes they both sneezed.\n2. The difference of two numbers is 10 and the sum of their squares is 50. Find the\ntwo numbers.\n3. Liboko builds a rectangular storeroom. If the diagonal of the room is\n√\n1312 m\nand the perimeter is 80 m, determine the dimensions of the room.\n4. It rains half as much in July as it does in December. If it rains y mm in July, write\nan expression relating the rainfall in July and December.\n5. Zane can paint a room in 4 hours. Tlali can paint a room in 2 hours. How long\nwill it take both of them to paint a room together?\n6. 25 years ago, Arthur was 5 years more than a third of Bongani’s age. Today,\nBongani is 26 years less than twice Arthur’s age. How old is Bongani?\n7. The product of two integers is 95. Find the integers if their total is 24.\n75\nChapter 2.\nEquations and inequalities\n\nWorked example 24: Gym membership\nQUESTION\nThe annual gym subscription for a single member is R 1000, while an annual family\nmembership is R 1500. The gym is considering increasing all membership fees by\nthe same amount. If this is done then a single membership would cost 5\n7 of a family\nmembership. Determine the amount of the proposed increase.\nSOLUTION\nStep 1: Identify the unknown quantity and assign a variable\nLet the amount of the proposed increase be x.\nStep 2: Use the given information to complete a table\nnow\nafter increase\nsingle\n1000\n1000 + x\nfamily\n1500\n1500 + x\nStep 3: Set up an equation\n1000 + x = 5\n7(1500 + x)\nStep 4: Solve for x\n7000 + 7x = 7500 + 5x\n2x = 500\nx = 250\nStep 5: Write the final answer\nThe proposed increase is R 250.\nWorked example 25: Corner coffee house\nQUESTION\nErica has decided to treat her friends to coffee at the Corner Coffee House. Erica\npaid R 54,00 for four cups of cappuccino and three cups of filter coffee. If a cup of\ncappuccino costs R 3,00 more than a cup of filter coffee, calculate how much a cup of\neach type of coffee costs?\nSOLUTION\nStep 1: Method 1: identify the unknown quantities and assign two variables\nLet the cost of a cappuccino be x and the cost of a filter coffee be y.\n76\n2.9.\nWord problems\n\nStep 2: Use the given information to set up a system of equations\n4x + 3y = 54\n. . . (1)\nx = y + 3\n. . . (2)\nStep 3: Solve the equations by substituting the second equation into the first equation\n4(y + 3) + 3y = 54\n4y + 12 + 3y = 54\n7y = 42\ny = 6\nIf y = 6, then using the second equation we have\nx = y + 3\n= 6 + 3\n= 9\nStep 4: Check that the solution satisfies both original equations\nStep 5: Write the final answer\nA cup of cappuccino costs R 9 and a cup of filter coffee costs R 6.\nStep 6: Method 2: identify the unknown quantities and assign one variable\nLet the cost of a cappuccino be x and the cost of a filter coffee be x −3.\nStep 7: Use the given information to set up an equation\n4x + 3(x −3) = 54\nStep 8: Solve for x\n4x + 3(x −3) = 54\n4x + 3x −9 = 54\n7x = 63\nx = 9\nStep 9: Write the final answer\nA cup of cappuccino costs R 9 and a cup of filter coffee costs R 6.\n77\nChapter 2.\nEquations and inequalities\n\nWorked example 26: Taps filling a container\nQUESTION\nTwo taps, one more powerful than the other, are used to fill a container. Working\non its own, the less powerful tap takes 2 hours longer than the other tap to fill the\ncontainer. If both taps are opened, it takes 1 hour, 52 minutes and 30 seconds to fill\nthe container. Determine how long it takes the less powerful tap to fill the container\non its own.\nSOLUTION\nStep 1: Identify the unknown quantities and assign variables\nLet the time taken for the less powerful tap to fill the container be x and let the time\ntaken for the more powerful tap be x −2.\nStep 2: Convert all units of time to be the same\nFirst we must convert 1 hour, 52 minutes and 30 seconds to hours:\n1 + 52\n60 +\n30\n(60)2 = 1,875 hours\nStep 3: Use the given information to set up a system of equations\nWrite an equation describing the two taps working together to fill the container:\n1\nx +\n1\nx −2 =\n1\n1,875\nStep 4: Multiply the equation through by the lowest common denominator and sim-\nplify\n1,875(x −2) + 1,875x = x(x −2)\n1,875x −3,75 + 1,875x = x2 −2x\n0 = x2 −5,75x + 3,75\nMultiply the equation through by 4 to make it easier to factorise (or use the quadratic\nformula)\n0 = 4x2 −23x + 15\n0 = (4x −3)(x −5)\nTherefore x = 3\n4 or x = 5.\nWe have calculated that the less powerful tap takes 3\n4 hours or 5 hours to fill the\ncontainer, but we know that when both taps are opened it takes 1,875 hours. We can\ntherefore discard the first solution x = 3\n4 hours.\n78\n2.9.\nWord problems\n\nSo the less powerful tap fills the container in 5 hours and the more powerful tap takes\n3 hours.\nStep 5: Check that the solution satisfies the original equation\nStep 6: Write the final answer\nThe less powerful tap fills the container in 5 hours and the more powerful tap takes 3\nhours.\nExercise 2 – 10:\n1. Mr. Tsilatsila builds a fence around his rectangular vegetable garden of 8 m2.\nIf the length is twice the breadth, determine the dimensions of Mr. Tsilatsila’s\nvegetable garden.\n2. Kevin has played a few games of ten-pin bowling. In the third game, Kevin\nscored 80 more than in the second game. In the first game Kevin scored 110 less\nthan the third game. His total score for the first two games was 208. If he wants\nan average score of 146, what must he score on the fourth game?\n3. When an object is dropped or thrown downward, the distance, d, that it falls in\ntime, t, is described by the following equation:\ns = 5t2 + v0t\nIn this equation, v0 is the initial velocity, in m·s−1. Distance is measured in\nmeters and time is measured in seconds. Use the equation to find how long it\ntakes a tennis ball to reach the ground if it is thrown downward from a hot-air\nballoon that is 500 m high. The tennis ball is thrown at an initial velocity of\n5 m·s−1.\n4. The table below lists the times that Sheila takes to walk the given distances.\ntime (minutes)\n5\n10\n15\n20\n25\n30\ndistance (km)\n1\n2\n3\n4\n5\n6\nPlot the points.\nFind the equation that describes the relationship between time and distance.\nThen use the equation to answer the following questions:\na) How long will it take Sheila to walk 21 km?\nb) How far will Sheila walk in 7 minutes?\nIf Sheila were to walk half as fast as she is currently walking, what would the\ngraph of her distances and times look like?\n5. The power P (in watts) supplied to a circuit by a 12 volt battery is given by the\nformula P = 12I −0,5I2 where I is the current in amperes.\na) Since both power and current must be greater than 0, find the limits of the\ncurrent that can be drawn by the circuit.\n79\nChapter 2.\nEquations and inequalities\n\nb) Draw a graph of P = 12I −0,5I2 and use your answer to the first question\nto define the extent of the graph.\nc) What is the maximum current that can be drawn?\nd) From your graph, read off how much power is supplied to the circuit when\nthe current is 10 A. Use the equation to confirm your answer.\ne) At what value of current will the power supplied be a maximum?\n6. A wooden block is made as shown in the diagram. The ends are right-angled\ntriangles having sides 3x, 4x and 5x. The length of the block is y. The total\nsurface area of the block is 3600 cm2.\ny\n3x\n4x\n5x\nShow that\ny = 300−x2\nx\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22C7\n2. 22C8\n3. 22C9\n4. 22CB\n5. 22CC\n6. 22CD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.10\nSummary\nEMBFZ\nSee presentation: 22CF at www.everythingmaths.co.za\n• Zero product law: if a × b = 0, then a = 0 and/or b = 0.\n• Quadratic formula: x = −b ±\n√\nb2 −4ac\n2a\n• Discriminant: ∆= b2 −4ac\n80\n2.10.\nSummary\n\nNature of roots\nDiscriminant\nRoots are non-real\n∆< 0\nRoots are real and equal\n∆= 0\nRoots are real and unequal:\n– Rational roots\n– Irrational roots\n∆> 0\n– ∆= squared rational number\n– ∆= not squared rational num-\nber\nExercise 2 – 11: End of chapter exercises\n1. Solve: x2 −x −1 = 0. Give your answer correct to two decimal places.\n2. Solve: 16 (x + 1) = x2 (x + 1)\n3. Solve: y2 + 3 +\n12\ny2 + 3 = 7\n4. Solve for x: 2x4 −5x2 −12 = 0\n5. Solve for x:\na) x (x −9) + 14 = 0\nb) x2 −x = 3 (correct to one decimal place)\nc) x + 2 = 6\nx (correct to two decimal places)\nd)\n1\nx + 1 +\n2x\nx −1 = 1\n6. Solve for x in terms of p by completing the square: x2 −px −4 = 0\n7. The equation ax2 + bx + c = 0 has roots x = 2\n3 and x = −4. Find one set of\npossible values for a, b and c.\n8. The two roots of the equation 4x2 + px −9 = 0 differ by 5. Calculate the value\nof p.\n9. An equation of the form x2 +bx+c = 0 is written on the board. Saskia and Sven\ncopy it down incorrectly. Saskia has a mistake in the constant term and obtains\nthe solutions −4 and 2. Sven has a mistake in the coefficient of x and obtains\nthe solutions 1 and −15. Determine the correct equation that was on the board.\n10. For which values of b will the expression b2 −5b + 6\nb + 2\nbe:\na) undefined?\nb) equal to zero?\n11. Given (x2 −6)(2x + 1)\nx + 2\n= 0 solve for x if:\na) x is a real number.\nb) x is a rational number.\nc) x is an irrational number.\nd) x is an integer.\n81\nChapter 2.\nEquations and inequalities\n\n12. Given (x −6)\n1\n2\nx2 + 3 , for which value(s) of x will the expression be:\na) equal to zero?\nb) defined?\n13. Solve for a if\n√8 −2a\na −3\n≥0.\n14. Abdoul stumbled across the following formula to solve the quadratic equation\nax2 + bx + c = 0 in a foreign textbook.\nx =\n2c\n−b ±\n√\nb2 −4ac\na) Use this formula to solve the equation: 2x2 + x −3 = 0.\nb) Solve the equation again, using factorisation, to see if the formula works for\nthis equation.\nc) Trying to derive this formula to prove that it always works, Abdoul got stuck\nalong the way. His attempt is shown below:\nax2 + bx + c = 0\na + b\nx + c\nx2 = 0\nDivided by x2 where x ̸= 0\nc\nx2 + b\nx + a = 0\nRearranged\n1\nx2 + b\ncx + a\nc = 0\nDivided by c where c ̸= 0\n1\nx2 + b\ncx = −a\nc\nSubtracted a\nc from both sides\n∴1\nx2 + b\ncx + . . .\nGot stuck\nComplete his derivation.\n15. Solve for x:\na)\n4\nx −3 ≤1\nb)\n4\n(x −3)2 < 1\nc) 2x −2\nx −3 > 3\nd)\n−3\n(x −3) (x + 1) < 0\ne) (2x −3)2 < 4\nf) 2x ≤15 −x\nx\ng) x2 + 3\n3x −2 ≤0\nh) x −2 ≥3\nx\ni) x2 + 3x −4\n5 + x4\n≤0\nj) x −2\n3 −x ≥1\n82\n2.10.\nSummary\n\n16. Solve the following systems of equations algebraically. Leave your answer in\nsurd form, where appropriate.\na) y −2x = 0\ny −x2 −2x + 3 = 0\nb) a −3b = 0\na −b2 + 4 = 0\nc) y −x2 −5x = 0\n10 = y −2x\nd) p = 2p2 + q −3\np −3q = 1\ne) a −b2 = 0\na −3b + 1 = 0\nf) a −2b + 1 = 0\na −2b2 −12b + 4 = 0\ng) y + 4x −19 = 0\n8y + 5x2 −101 = 0\nh) a + 4b −18 = 0\n2a + 5b2 −57 = 0\n17. Solve the following systems of equations graphically:\na) 2y + x −2 = 0\n8y + x2 −8 = 0\nb) y + 3x −6 = 0\ny = x2 + 4 −4x\n18. A stone is thrown vertically upwards and its height (in metres) above the ground\nat time t (in seconds) is given by:\nh (t) = 35 −5t2 + 30t\nFind its initial height above the ground.\n19. After doing some research, a transport company has determined that the rate at\nwhich petrol is consumed by one of its large carriers, travelling at an average\nspeed of x km per hour, is given by:\nP(x) = 55\n2x + x\n200\nlitres per kilometre\nAssume that the petrol costs R 4,00 per litre and the driver earns R 18,00 per\nhour of travel time. Now deduce that the total cost, C, in Rands, for a 2000 km\ntrip is given by:\nC(x) = 256 000\nx\n+ 40x\n20. Solve the following quadratic equations by either factorisation, completing the\nsquare or by using the quadratic formula:\n• Always try to factorise first, then use the formula if the trinomial cannot be\nfactorised.\n• Solve some of the equations by completing the square.\na) −4y2 −41y −45 = 0\nb) 16x2 + 20x = 36\nc) 42p2 + 104p + 64 = 0\nd) 21y + 3 = 54y2\ne) 36y2 + 44y + 8 = 0\nf) 12y2 −14 = 22y\ng) 16y2 + 0y −81 = 0\nh) 3y2 + 10y −48 = 0\ni) 63 −5y2 = 26y\nj) 2x2 −30 = 2\nk) 2y2 = 98\n83\nChapter 2.\nEquations and inequalities\n\n21. One root of the equation 9y2 + 32 = ky is 8. Determine the value of k and the\nother root.\n22.\na) Solve for x in x2 −x = 6.\nb) Hence, solve for y in (y2 −y)2 −(y2 −y) −6 = 0.\n23. Solve for x: x = √8 −x + 2\n24.\na) Solve for y in −4y2 + 8y −3 = 0.\nb) Hence, solve for p in 4(p −3)2 −8(p −3) + 3 = 0.\n25. Solve for x: 2(x + 3)\n1\n2 = 9\n26.\na) Without solving the equation x + 1\nx = 3, determine the value of x2 + 1\nx2 .\nb) Now solve x + 1\nx = 3 and use the result to assess the answer obtained in\nthe question above.\n27. Solve for y: 5(y −1)2 −5 = 19 −(y −1)2\n28. Solve for t: 2t(t −3\n2) =\n3\n2t2 −3t + 2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22CG\n2. 22CH\n3. 22CJ\n4. 22CK\n5a. 22CM\n5b. 22CN\n5c. 22CP\n5d. 22CQ\n6. 22CR\n7. 22CS\n8. 22CT\n9. 22CV\n10. 22CW\n11. 22CX\n12. 22CY\n13. 22CZ\n14. 22D2\n15a. 22D3\n15b. 22D4\n15c. 22D5\n15d. 22D6\n15e. 22D7\n15f. 22D8\n15g. 22D9\n15h. 22DB\n15i. 22DC\n15j. 22DD\n16a. 22DF\n16b. 22DG\n16c. 22DH\n16d. 22DJ\n16e. 22DK\n16f. 22DM\n16g. 22DN\n16h. 22DP\n17a. 22DQ\n17b. 22DR\n18. 22DS\n19. 22DT\n20a. 22DV\n20b. 22DW\n20c. 22DX\n20d. 22DY\n20e. 22DZ\n20f. 22F2\n20g. 22F3\n20h. 22F4\n20i. 22F5\n20j. 22F6\n20k. 22F7\n21. 22F8\n22. 22F9\n23. 22FB\n24. 22FC\n25. 22FD\n26. 22FF\n27. 22FG\n28. 22FH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n84\n2.10.\nSummary\n\nCHAPTER\n3\nNumber patterns\n3.1\nRevision\n86\n3.2\nQuadratic sequences\n90\n3.3\nSummary\n99\n\n3\nNumber patterns\nIn earlier grades we learned about linear sequences, where the difference between\nconsecutive terms is constant. In this chapter, we will learn about quadratic sequences,\nwhere the difference between consecutive terms is not constant, but follows its own\npattern.\n3.1\nRevision\nEMBG2\nTerminology:\nSequence/pattern\nA sequence or pattern is an ordered set of numbers\nor variables.\nSuccessive/consecutive\nSuccessive or consecutive terms are terms that di-\nrectly follow one after another in a sequence.\nCommon difference\nThe common or constant difference (d) is the differ-\nence between any two consecutive terms in a linear\nsequence.\nGeneral term\nA mathematical expression that describes the se-\nquence and that generates any term in the pattern\nby substituting different values for n.\nConjecture\nA statement, consistent with known data, that has\nnot been proved true nor shown to be false.\nImportant: a series is not the same as a sequence or pattern. Different types of series\nare studied in Grade 12. In Grade 11 we study sequences only.\nSee video: 22FJ at www.everythingmaths.co.za\nDescribing patterns\nEMBG3\nTo describe terms in a pattern we use the following notation:\n• T1 is the first term of a sequence.\n• T4 is the fourth term of a sequence.\n• Tn is the general term and is often expressed as the nth term of a sequence.\nA sequence does not have to follow a pattern but when it does, we can write an\nequation for the general term. The general term can be used to calculate any term in\nthe sequence. For example, consider the following linear sequence: 1; 4; 7; 10; 13; . . .\nThe nth term is given by the equation Tn = 3n −2.\nYou can check this by substituting values for n:\nT1 = 3(1) −2 = 1\nT2 = 3(2) −2 = 4\nT3 = 3(3) −2 = 7\nT4 = 3(4) −2 = 10\nT5 = 3(5) −2 = 13\n86\n3.1.\nRevision\n\nIf we find the relationship between the position of a term and its value, we can describe\nthe pattern and find any term in the sequence.\nSee video: 22FK at www.everythingmaths.co.za\nLinear sequences\nEMBG4\nDEFINITION: Linear sequence\nA sequence of numbers in which there is a common difference (d) between any term\nand the term before it is called a linear sequence.\nImportant: d = T2 −T1, not T1 −T2.\nWorked example 1: Linear sequence\nQUESTION\nDetermine the common difference (d) and the general term for the following sequence:\n10; 7; 4; 1; . . .\nSOLUTION\nStep 1: Determine the common difference\nTo calculate the common difference, we find the difference between any term and the\nprevious term:\nd = Tn −Tn−1\nTherefore d = T2 −T1\n= 7 −10\n= −3\nor d = T3 −T2\n= 4 −7\n= −3\nor d = T4 −T3\n= 1 −4\n= −3\n10\n7\n4\n1\n−3\n−3\n−3\nStep 2: Determine the general term\nTo find the general term Tn, we must identify the relationship between:\n87\nChapter 3.\nNumber patterns\n\n• the value of a number in the pattern and\n• the position of a number in the pattern\nposition\n1\n2\n3\n4\nvalue\n10\n7\n4\n1\nWe start with the value of the first term in the sequence. We need to write an expres-\nsion that includes the value of the common difference (d = −3) and the position of\nthe term (n = 1).\nT1 = 10\n= 10 + (0)(−3)\n= 10 + (1 −1)(−3)\nNow we write a similar expression for the second term.\nT2 = 7\n= 10 + (1)(−3)\n= 10 + (2 −1)(−3)\nWe notice a pattern forming that links the position of a number in the sequence to its\nvalue.\nTn = 10 + (n −1)(−3)\n= 10 −3n + 3\n= −3n + 13\nStep 3: Drawing a graph of the pattern\nWe can also represent this pattern graphically, as shown below.\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n0\n1\n2\n3\n4\n5\n6\nb\nb\nb\nb\nPattern number (n)\nTerm value Tn\nNotice that the position numbers (n) can be positive integers only.\nThis pattern can also be expressed in words: “each term in the sequence can be cal-\nculated by multiplying negative three and the position number, and then adding thir-\nteen.”\n88\n3.1.\nRevision\n\nSee video: 22FM at www.everythingmaths.co.za\nExercise 3 – 1: Linear sequences\n1. Write down the next three terms in each of the following sequences:\n45; 29; 13; −3; . . .\n2. The general term is given for each sequence below. Calculate the missing terms.\na) −4; −9; −14; . . . ; −24\nTn = 1 −5n\nb) 6; . . . ; 24; . . . ; 42\nTn = 9n −3\n3. Find the general formula for the following sequences and then find T10, T15 and\nT30:\na) 13; 16; 19; 22; . . .\nb) 18; 24; 30; 36; . . .\nc) −10; −15; −20; −25; . . .\n4. The seating in a classroom is arranged so that the first row has 20 desks, the\nsecond row has 22 desks, the third row has 24 desks and so on. Calculate how\nmany desks are in the ninth row.\n5.\na) Complete the following:\n13 + 31 = . . .\n24 + 42 = . . .\n38 + 83 = . . .\nb) Look at the numbers on the left-hand side, what do you notice about the\nunit digit and the tens-digit?\nc) Investigate the pattern by trying other examples of 2-digit numbers.\nd) Make a conjecture about the pattern that you notice.\ne) Prove this conjecture.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22FN\n2a. 22FP\n2b. 22FQ\n3a. 22FR\n3b. 22FS\n3c. 22FT\n4. 22FV\n5. 22FW\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n89\nChapter 3.\nNumber patterns\n\n3.2\nQuadratic sequences\nEMBG5\nInvestigation: Quadratic sequences\nb\nb\nb\nb\nb\nb\nb\nb\nb\n1. Study the dotted-tile pattern shown and answer the following questions.\na) Complete the fourth pattern in the diagram.\nb) Complete the table below:\npattern number\n1\n2\n3\n4\n5\n20\nn\ndotted tiles\n1\n3\n5\ndifference (d)\n−\n2\nc) What do you notice about the change in number of dotted tiles?\nd) Describe the pattern in words: “The number of dotted tiles...”.\ne) Write the general term: Tn = . . .\nf) Give the mathematical name for this kind of pattern.\ng) A pattern has 819 dotted tiles. Determine the value of n.\n2. Now study the number of blank tiles (tiles without dots) and answer the following\nquestions:\na) Complete the table below:\npattern number\n1\n2\n3\n4\n5\n10\nblank tiles\n3\n6\n11\nfirst difference\n−\n3\nsecond difference\n−\n−\nb) What do you notice about the change in the number of blank tiles?\nc) Describe the pattern in words: “The number of blank tiles...”.\nd) Write the general term: Tn = . . .\ne) Give the mathematical name for this kind of pattern.\nf) A pattern has 227 blank tiles. Determine the value of n.\ng) A pattern has 79 dotted tiles. Determine the number of blank tiles.\n90\n3.2.\nQuadratic sequences\n\nDEFINITION: Quadratic sequence\nA quadratic sequence is a sequence of numbers in which the second difference be-\ntween any two consecutive terms is constant.\nConsider the following example: 1; 2; 4; 7; 11; . . .\nThe first difference is calculated by finding the difference between consecutive terms:\n1\n2\n4\n7\n11\n+1\n+2\n+3\n+4\nThe second difference is obtained by taking the difference between consecutive first\ndifferences:\n1\n2\n3\n4\n+1\n+1\n+1\nWe notice that the second differences are all equal to 1. Any sequence that has a\ncommon second difference is a quadratic sequence.\nIt is important to note that the first differences of a quadratic sequence form a sequence.\nThis sequence has a constant difference between consecutive terms. In other words, a\nlinear sequence results from taking the first differences of a quadratic sequence.\nGeneral case\nIf the sequence is quadratic, the nth term is of the form Tn = an2 + bn + c.\nn = 1\nn = 2\nn = 3\nn = 4\nTn\na + b + c\n4a + 2b + c\n9a + 3b + c\n16a + 4b + c\n1st difference\n3a + b\n5a + b\n7a + b\n2nd difference\n2a\n2a\nIn each case, the common second difference is a 2a.\nExercise 3 – 2: Quadratic sequences\n1. Determine the second difference between the terms for the following se-\nquences:\na) 5; 20; 45; 80; . . .\nb) 6; 11; 18; 27; . . .\nc) 1; 4; 9; 16; . . .\nd) 3; 0; −5; −12; . . .\ne) 1; 3; 7; 13; . . .\nf) 0; −6; −16; −30; . . .\ng) −1; 2; 9; 20; . . .\nh) 1; −3; −9; −17; . . .\ni) 3a+1; 12a+1; 27a+1; 48a+1 . . .\nj) 2; 10; 24; 44; . . .\nk) t −2; 4t −1; 9t; 16t + 1; . . .\n91\nChapter 3.\nNumber patterns\n\n2. Complete the sequence by filling in the missing term:\na) 11; 21; 35; . . . ; 75\nb) 20; . . . ; 42; 56; 72\nc) . . . ; 37; 65; 101\nd) 3; . . . ; −13; −27; −45\ne) 24; 35; 48; . . . ; 80\nf) . . . ; 11; 26; 47\n3. Use the general term to generate the first four terms in each sequence:\na) Tn = n2 + 3n −1\nb) Tn = −n2 −5\nc) Tn = 3n2 −2n\nd) Tn = −2n2 + n + 1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22FX\n1b. 22FY\n1c. 22FZ\n1d. 22G2\n1e. 22G3\n1f. 22G4\n1g. 22G5\n1h. 22G6\n1i. 22G7\n1j. 22G8\n1k. 22G9\n2a. 22GB\n2b. 22GC\n2c. 22GD\n2d. 22GF\n2e. 22GG\n2f. 22GH\n3a. 22GJ\n3b. 22GK\n3c. 22GM\n3d. 22GN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nWorked example 2: Quadratic sequences\nQUESTION\nWrite down the next two terms and determine an equation for the nth term of the\nsequence 5; 12; 23; 38; . . .\nSOLUTION\nStep 1: Find the first differences between the terms\n5\n12\n23\n38\n+7\n+11\n+15\nStep 2: Find the second differences between the terms\n7\n11\n15\n+4\n+4\nSo there is a common second difference of 4. We can therefore conclude that this is a\nquadratic sequence of the form Tn = an2 + bn + c.\nContinuing the sequence, the next first differences will be:\n...15\n19\n23...\n+4\n+4\n92\n3.2.\nQuadratic sequences\n\nStep 3: Finding the next two terms in the sequence\nThe next two terms will be:\n...38\n57\n80...\n+19\n+23\nSo the sequence will be: 5; 12; 23; 38; 57; 80; . . .\nStep 4: Determine the general term for the sequence\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve a set of simultaneous equations to determine the values of a, b and c\nWe know that T1 = 5, T2 = 12 and T3 = 23\na + b + c = 5\n4a + 2b + c = 12\n9a + 3b + c = 23\nT2 −T1 = 4a + 2b + c −(a + b + c)\n12 −5 = 4a + 2b + c −a −b −c\n7 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n23 −12 = 9a + 3b + c −4a −2b −c\n11 = 5a + b\n. . . (2)\n(2) −(1) = 5a + b −(3a + b)\n11 −7 = 5a + b −3a −b\n4 = 2a\n∴a = 2\nUsing equation (1) :\n3(2) + b = 7\n∴b = 1\nAnd using\na + b + c = 5\n2 + 1 + c = 5\n∴c = 1\nStep 5: Write the general term for the sequence\nTn = 2n2 + n + 2\n93\nChapter 3.\nNumber patterns\n\nWorked example 3: Plotting a graph of terms in a sequence\nQUESTION\nConsider the following sequence:\n3; 6; 10; 15; 21; . . .\n1. Determine the general term (Tn) for the sequence.\n2. Is this a linear or a quadratic sequence?\n3. Plot a graph of Tn vs n.\nSOLUTION\nStep 1: Determine the first and second differences\nn = 1\nn = 2\nn = 3\nn = 4\nTn\n3\n6\n10\n15\n1st difference\n3\n4\n5\n2nd difference\n1\n1\nWe see that the first differences are not constant and form the sequence 3; 4; 5; . . . and\nthat there is a common second difference of 1. Therefore the sequence is quadratic\nand has a general term of the form Tn = an2 + bn + c.\nStep 2: Determine the general term Tn\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve this set of simultaneous equations to determine the values of a, b and c. We\nknow that T1 = 3, T2 = 6 and T3 = 10.\na + b + c = 3\n4a + 2b + c = 6\n9a + 3b + c = 10\nT2 −T1 = 4a + 2b + c −(a + b + c)\n6 −3 = 4a + 2b + c −a −b −c\n3 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n10 −6 = 9a + 3b + c −4a −2b −c\n4 = 5a + b\n. . . (2)\n94\n3.2.\nQuadratic sequences\n\n(2) −(1) = 5a + b −(3a + b)\n4 −3 = 5a + b −3a −b\n1 = 2a\n∴a = 1\n2\nUsing equation (1) :\n3\n\u00121\n2\n\u0013\n+ b = 3\n∴b = 3\n2\nAnd using\na + b + c = 3\n1\n2 + 3\n2 + c = 3\n∴c = 1\nTherefore the general term for the sequence is Tn = 1\n2n2 + 3\n2n + 1.\nStep 3: Plot a graph of Tn vs n\nUse the general term for the sequence, Tn = 1\n2n2 + 3\n2n + 1, to complete the table.\nn\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nTn\n3\n6\n10\n15\n21\n28\n36\n45\n55\n66\nUse the table to plot the graph:\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nTerm value (Tn)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nPosition number (n)\nT1\nT2\nT3\nT4\nT5\nT6\nT7\nT8\nT9\nT10\nIn this case it would not be accurate to join these points, since n indicates the position\nof a term in a sequence and can therefore only be a positive integer. We can, however,\nsee that the plot of the points lies in the shape of a parabola.\n95\nChapter 3.\nNumber patterns\n\nWorked example 4: Olympic Games soccer event\nQUESTION\nIn the first stage of the soccer event at the Olympic Games, there are teams from\nfour different countries in each group. Each country in a group must play every other\ncountry in the group once.\n1. How many matches will be played in each group in the first stage of the event?\n2. How many matches would be played if there are 5 teams in each group?\n3. How many matches would be played if there are 6 teams in each group?\n4. Determine the general formula of the sequence.\nSOLUTION\nStep 1: Determine the number of matches played if there are 4 teams in a group\nLet the teams from four different countries be A, B, C and D.\nteams in a group\nmatches played\nA\nAB, AC, AD\nB\nBC, BD\nC\nCD\nD\n4\n3 + 2 + 1 = 6\nAB means that team A plays team B and BA would be the same match as AB. So if\nthere are four different teams in a group, each group plays 6 matches.\nStep 2: Determine the number of matches played if there are 5 teams in a group\nLet the teams from five different countries be A, B, C, D and E.\nteams in a group\nmatches played\nA\nAB, AC, AD, AE\nB\nBC, BD, BE\nC\nCD, CE\nD\nDE\nE\n5\n4 + 3 + 2 + 1 = 10\nSo if there are five different teams in a group, each group plays 10 matches.\nStep 3: Determine the number of matches played if there are 6 teams in a group\nLet the teams from six different countries be A, B, C, D, E and F.\n96\n3.2.\nQuadratic sequences\n\nteams in a group\nmatches to be played\nA\nAB, AC, AD, AE, AF\nB\nBC, BD, BE, BF\nC\nCD, CE, CF\nD\nDE, DF\nE\nEF\nF\n5\n5 + 4 + 3 + 2 + 1 = 15\nSo if there are six different teams in a group, each group plays 15 matches.\nWe continue to increase the number of teams in a group and find that a group of 7\nteams plays 21 matches and a group of 8 teams plays 28 matches.\nStep 4: Consider the sequence\nWe examine the sequence to determine if it is linear or quadratic:\nn = 1\nn = 2\nn = 3\nn = 4\nn = 5\nTn\n6\n10\n15\n21\n28 . . .\nfirst difference\n4\n5\n6\n7\nsecond difference\n1\n1\n1\nWe see that the first differences are not constant and that there is a common second\ndifference of 1. Therefore the sequence is quadratic and has a general term of the form\nTn = an2 + bn + c.\nStep 5: Determine the general term Tn\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve a set of simultaneous equations to determine the values of a, b and c. We\nknow that T1 = 6, T2 = 10 and T3 = 15\na + b + c = 6\n4a + 2b + c = 10\n9a + 3b + c = 15\nT2 −T1 = 4a + 2b + c −(a + b + c)\n10 −6 = 4a + 2b + c −a −b −c\n4 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n15 −10 = 9a + 3b + c −4a −2b −c\n5 = 5a + b\n. . . (2)\n97\nChapter 3.\nNumber patterns\n\n(2) −(1) = 5a + b −(3a + b)\n5 −4 = 5a + b −3a −b\n1 = 2a\n∴a = 1\n2\nUsing equation (1) :\n3\n\u00121\n2\n\u0013\n+ b = 4\n∴b = 5\n2\nAnd using a + b + c = 6\n1\n2 + 5\n2 + c = 6\n∴c = 3\nTherefore the general term for the sequence is Tn = 1\n2n2 + 5\n2n + 3.\nExercise 3 – 3: Quadratic sequences\n1. Calculate the common second difference for each of the following quadratic\nsequences:\na) 3; 6; 10; 15; 21; ...\nb) 4; 9; 16; 25; 36; ...\nc) 7; 17; 31; 49; 71; ...\nd) 2; 10; 26; 50; 82; ...\ne) 31; 30; 27; 22; 15; ...\n2. Find the first five terms of the quadratic sequence defined by: Tn = 5n2 +3n+4.\n3. Given Tn = 4n2 + 5n + 10, find T9.\n4. Given Tn = 2n2, for which value of n does Tn = 32?\n5.\na) Write down the next two terms of the quadratic sequence: 16; 27; 42; 61; . . .\nb) Find the general formula for the quadratic sequence above.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22GP\n1b. 22GQ\n1c. 22GR\n1d. 22GS\n1e. 22GT\n2. 22GV\n3. 22GW\n4. 22GX\n5. 22GY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n98\n3.2.\nQuadratic sequences\n\n3.3\nSummary\nEMBG6\nSee presentation: 22GZ at www.everythingmaths.co.za\n• Tn is the general term of a sequence.\n• Successive or consecutive terms are terms that follow one after another in a\nsequence.\n• A linear sequence has a common difference (d) between any two successive\nterms.\nd = Tn −Tn−1\n• A quadratic sequence has a common second difference between any two suc-\ncessive terms.\n• The general term for a quadratic sequence is\nTn = an2 + bn + c\n• A general quadratic sequence:\nn = 1\nn = 2\nn = 3\nn = 4\nTn\na + b + c\n4a + 2b + c\n9a + 3b + c\n16a + 4b + c\n1st difference\n3a + b\n5a + b\n7a + b\n2nd difference\n2a\n2a\nExercise 3 – 4: End of chapter exercises\n1. Find the first five terms of the quadratic sequence defined by:\nTn = n2 + 2n + 1\n2. Determine whether each of the following sequences is:\n• a linear sequence,\n• a quadratic sequence,\n• or neither.\na) 6; 9; 14; 21; 30; ...\nb) 1; 7; 17; 31; 49; ...\nc) 8; 17; 32; 53; 80; ...\nd) 9; 26; 51; 84; 125; ...\ne) 2; 20; 50; 92; 146; ...\nf) 5; 19; 41; 71; 109; ...\ng) 2; 6; 10; 14; 18; ...\nh) 3; 9; 15; 21; 27; ...\ni) 1; 2,5; 5; 8,5; 13; ...\nj) 10; 24; 44; 70; 102; ...\nk) 21\n2; 6; 101\n2; 16; 221\n2; . . .\nl) 3p2; 6p2; 9p2; 12p2; 15p2; . . .\nm) 2k; 8k; 18k; 32k; 50k; . . .\n99\nChapter 3.\nNumber patterns\n\n3. Given the pattern: 16; x; 46; . . ., determine the value of x if the pattern is linear.\n4. Given Tn = 2n2, for which value of n does Tn = 242?\n5. Given Tn = 3n2, find T11.\n6. Given Tn = n2 + 4, for which value of n does Tn = 85?\n7. Given Tn = 4n2 + 3n −1, find T5.\n8. Given Tn = 3\n2n2, for which value of n does Tn = 96?\n9. For each of the following patterns, determine:\n• the next term in the pattern,\n• and the general term,\n• the tenth term in the pattern.\na) 3; 7; 11; 15; . . .\nb) 17; 12; 7; 2; . . .\nc)\n1\n2; 1; 11\n2; 2; . . .\nd) a; a + b; a + 2b; a + 3b; . . .\ne) 1; −1; −3; −5; . . .\n10. For each of the following sequences, find the equation for the general term and\nthen use the equation to find T100.\na) 4; 7; 12; 19; 28; ...\nb) 2; 8; 14; 20; 26; ...\nc) 7; 13; 23; 37; 55; ...\nd) 5; 14; 29; 50; 77; ...\nGiven: Tn = 3n −1\n11.\na) Write down the first five terms of the sequence.\nb) What do you notice about the difference between any two consecutive\nterms?\nc) Will this always be the case for a linear sequence?\nGiven the following sequence: −15; −11; −7; . . . ; 173\n12.\na) Determine the equation for the general term.\nb) Calculate how many terms there are in the sequence.\n13. Given 3; 7; 13; 21; 31; . . .\na) Thabang determines that the general term is Tn = 4n −1. Is he correct?\nExplain.\nb) Cristina determines that the general term is Tn = n2 + n + 1. Is she correct?\nExplain.\n100\n3.3.\nSummary\n\n14. Given the following pattern of blocks:\n2\n3\n4\na) Draw pattern 5.\nb) Complete the table below:\npattern number (n)\n2\n3\n4\n5\n10\n250\nn\nnumber of white blocks (w)\n4\n8\nc) Is this a linear or a quadratic sequence?\n15. Cubes of volume 1 cm3 are stacked on top of each other to form a tower:\n1\n2\n3\na) Complete the table for the height of the tower:\ntower number (n)\n1\n2\n3\n4\n10\nn\nheight of tower (h)\n2\nb) What type of sequence is this?\nc) Now consider the number of cubes in each tower and complete the table\nbelow:\ntower number (n)\n1\n2\n3\n4\nnumber of cubes (c)\n3\nd) What type of sequence is this?\ne) Determine the general term for this sequence.\nf) How many cubes are needed for tower number 21?\ng) How high will a tower of 496 cubes be?\n16. A quadratic sequence has a second term equal to 1, a third term equal to −6 and\na fourth term equal to −14.\na) Determine the second difference for this sequence.\nb) Hence, or otherwise, calculate the first term of the pattern.\n101\nChapter 3.\nNumber patterns\n\n17. There are 15 schools competing in the U16 girls hockey championship and every\nteam must play two matches — one home match and one away match.\na) Use the given information to complete the table:\nno. of schools\nno. of matches\n1\n0\n2\n3\n4\n5\nb) Calculate the second difference.\nc) Determine a general term for the sequence.\nd) How many matches will be played if there are 15 schools competing in the\nchampionship?\ne) If 600 matches must be played, how many schools are competing in the\nchampionship?\n18. The first term of a quadratic sequence is 4, the third term is 34 and the common\nsecond difference is 10. Determine the first six terms in the sequence.\n19. Challenge question:\nGiven that the general term for a quadratic sequences is Tn = an2 + bn + c, let\nd be the first difference and D be the second common difference.\na) Show that a = D\n2 .\nb) Show that b = d −3\n2D.\nc) Show that c = T1 −d + D.\nd) Hence, show that Tn = D\n2 n2 +\n\u0012\nd −3\n2D\n\u0013\nn + (T1 −d + D).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22H2\n2a. 22H3\n2b. 22H4\n2c. 22H5\n2d. 22H6\n2e. 22H7\n2f. 22H8\n2g. 22H9\n2h. 22HB\n2i. 22HC\n2j. 22HD\n2k. 22HF\n2l. 22HG\n2m. 22HH\n3. 22HJ\n4. 22HK\n5. 22HM\n6. 22HN\n7. 22HP\n8. 22HQ\n9a. 22HR\n9b. 22HS\n9c. 22HT\n9d. 22HV\n9e. 22HW\n10a. 22HX\n10b. 22HY\n10c. 22HZ\n10d. 22J2\n11. 22J3\n12. 22J4\n13. 22J5\n14. 22J6\n15. 22J7\n16. 22J8\n17. 22J9\n18. 22JB\n19. 22JC\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n102\n3.3.\nSummary\n\nCHAPTER\n4\nAnalytical geometry\n4.1\nRevision\n104\n4.2\nEquation of a line\n113\n4.3\nInclination of a line\n124\n4.4\nParallel lines\n132\n4.5\nPerpendicular lines\n136\n4.6\nSummary\n142\n\n4\nAnalytical geometry\nAnalytical geometry, also referred to as coordinate or Cartesian geometry, is the study\nof geometric properties and relationships between points, lines and angles in the Carte-\nsian plane. Geometrical shapes are defined using a coordinate system and algebraic\nprinciples. In this chapter we deal with the equation of a straight line, parallel and\nperpendicular lines and inclination of a line.\n4.1\nRevision\nEMBG7\nPoints A(x1; y1), B(x2; y2) and C(x2; y1) are shown in the diagram below:\nb\nb\nA(x1; y1)\nC(x2; y1)\nB(x2; y2)\nx\ny\n0\nTheorem of Pythagoras\nAB2 = AC2 + BC2\nDistance formula\nDistance between two points:\nAB =\np\n(x2 −x1)2 + (y2 −y1)2\nNotice that (x1 −x2)2 = (x2 −x1)2.\nSee video: 22JD at www.everythingmaths.co.za\nGradient\nGradient (m) describes the slope or steepness of the line joining two points. The\ngradient of a line is determined by the ratio of vertical change to horizontal change.\nmAB = y2 −y1\nx2 −x1\nor\nmAB = y1 −y2\nx1 −x2\nRemember to be consistent: m ̸= y1 −y2\nx2 −x1\n.\n104\n4.1.\nRevision\n\nHorizontal lines\nx\ny\n0\nm = 0\nVertical lines\nx\ny\n0\nm is undefined\nParallel lines\nθ\nθ\nx\ny\n0\nm1 = m2\nPerpendicular lines\nθ2\nθ1\nx\ny\n0\nm1 × m2 = −1\nMid-point of a line segment\nA(x1; y1)\nM(x; y)\nB(x2; y2)\nx\ny\n0\nThe coordinates of the mid-point M(x; y) of a line between any two points A(x1; y1)\nand B(x2; y2):\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nSee video: 22JF at www.everythingmaths.co.za\n105\nChapter 4.\nAnalytical geometry\n\nPoints on a straight line\nThe diagram shows points P(x1; y1), Q(x2; y2) and R(x; y) on a straight line.\nb\nb\nx\ny\nR(x; y)\n0\nb\nQ(x2; y2)\nP(x1; y1)\nWe know that mPR = mQR = mPQ.\nUsing mPR = mPQ, we obtain the following for any point (x; y) on a straight line\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 1: Revision\nQUESTION\nGiven the points P(−5; −4) and Q(0; 6):\n1. Determine the length of the line segment PQ.\n2. Determine the mid-point T(x; y) of the line segment PQ.\n3. Show that the line passing through R(1; −3\n4) and T(x; y) is perpendicular to the\nline PQ.\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n−2\n−4\n−6\n2\n−2\n−4\n−6\nb\nb\nb\nP(−5; −4)\nT(x; y)\nQ(0; 6)\nx\ny\n0\n106\n4.1.\nRevision\n\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q (x2; y2)\nx1 = −5;\ny1 = −4;\nx2 = 0;\ny2 = 6\nWrite down the distance formula\nPQ =\np\n(x2 −x1)2 + (y2 −y1)2\n=\np\n(0 −(−5))2 + (6 −(−4))2\n=\n√\n25 + 100\n=\n√\n125\n= 5\n√\n5\nThe length of the line segment PQ is 5\n√\n5 units.\nStep 3: Write down the mid-point formula and substitute the values\nT(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nx = x1 + x2\n2\n= −5 + 0\n2\n= −5\n2\ny = y1 + y2\n2\n= −4 + 6\n2\n= 2\n2\n= 1\nThe mid-point of PQ is T(−5\n2; 1).\nStep 4: Determine the gradients of PQ and RT\nm = y2 −y1\nx2 −x1\nmPQ = 6 −(−4)\n0 −(−5)\n= 10\n5\n= 2\n107\nChapter 4.\nAnalytical geometry\n\nmRT = −3\n4 −1\n1 −(−5\n2)\n= −7\n4\n7\n2\n= −7\n4 × 2\n7\n= −1\n2\nCalculate the product of the two gradients:\nmRT × mPQ = −1\n2 × 2\n= −1\nTherefore PQ is perpendicular to RT.\nQuadrilaterals\n• A quadrilateral is a closed shape consisting of four straight line segments.\n• A parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n– Both pairs of opposite sides are equal in length.\n– Both pairs of opposite angles are equal.\n– The diagonals bisect each other.\n• A rectangle is a parallelogram that has all four angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other.\n– The diagonals are equal in length.\n108\n4.1.\nRevision\n\n• A rhombus is a parallelogram that has all four sides equal in length.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n×\n×\n××\n•\n•\n•\n•\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals of a rhombus bisect both pairs of opposite angles.\n• A square is a rhombus that has all four interior angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n•\n•\n••\n••\n••\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals are equal in length.\n– The diagonals bisect both pairs of interior opposite angles (that is, all angles\nare 45◦).\n• A trapezium is a quadrilateral with one pair of opposite sides parallel.\n• A kite is a quadrilateral with two pairs of adjacent sides equal.\nA\nB\nC\nD\nb\nb\n××\n– One pair of opposite angles are equal (the angles are between unequal\nsides).\n– The diagonal between equal sides bisects the other diagonal.\n– The diagonal between equal sides bisects the interior angles.\n– The diagonals intersect at 90◦.\n109\nChapter 4.\nAnalytical geometry\n\nWorked example 2: Quadrilaterals\nQUESTION\nPoints A (−1; 0), B (0; 3), C (8; 11) and D (x; y) are points on the Cartesian plane.\nDetermine D (x; y) if ABCD is a parallelogram.\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n−1\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\nb\nb\nx\ny\nC(8; 11)\n0\nD(x; y)\nA(−1; 0)\nb\nB(0; 3)\nM\nThe mid-point of AC will be the same as the mid-point of BD. We first find the\nmid-point of AC and then use it to determine the coordinates of point D.\nStep 2: Assign values to (x1; y1) and (x2; y2)\nLet the mid-point of AC be M(x; y)\nx1 = −1;\ny1 = 0;\nx2 = 8;\ny2 = 11\nStep 3: Write down the mid-point formula\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nStep 4: Substitute the values and calculate the coordinates of M\nM(x; y) =\n\u0012−1 + 8\n2\n; 0 + 11\n2\n\u0013\n=\n\u00127\n2; 11\n2\n\u0013\n110\n4.1.\nRevision\n\nStep 5: Use the coordinates of M to determine D\nM is also the mid-point of BD so we use M\n\u0000 7\n2; 11\n2\n\u0001\nand B (0; 3) to find D (x; y)\nStep 6: Substitute values and determine x and y\nM =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n∴\n\u00127\n2; 11\n2\n\u0013\n=\n\u00120 + x\n2\n; 3 + y\n2\n\u0013\n7\n2 = 0 + x\n2\n7 = 0 + x\n∴x = 7\n11\n2 = 3 + y\n2\n11 = 3 + y\n∴y = 8\nStep 7: Alternative method: inspection\nSince we are given that ABCD is a parallelogram, we can use the properties of a\nparallelogram and the given points to determine the coordinates of D.\nFrom the sketch we expect that point D will lie below C.\nConsider the given points A, B and C:\n• Opposite sides of a parallelogram are parallel, therefore BC must be parallel to\nAD and their gradients must be equal.\n• The vertical change from B to C is 8 units up.\n• Therefore the vertical change from A to D is also 8 units up (y = 0 + 8 = 8).\n• The horizontal change from B to C is 8 units to the right.\n• Therefore the horizontal change from A to D is also 8 units to the right (x =\n−1 + 8 = 7).\nor\n• Opposite sides of a parallelogram are parallel, therefore AB must be parallel to\nDC and their gradients must be equal.\n• The vertical change from A to B is 3 units up.\n111\nChapter 4.\nAnalytical geometry\n\n• Therefore the vertical change from C to D is 3 units down (y = 11 −3 = 8).\n• The horizontal change from A to B is 1 unit to the right.\n• Therefore the horizontal change from C to D is 1 unit to the left (x = 8 −1 = 7).\nStep 8: Write the final answer\nThe coordinates of D are (7; 8).\nExercise 4 – 1: Revision\n1. Determine the length of the line segment between the following points:\na) P(−3; 5) and Q(−1; −5)\nb) R(0,75; 3) and S(0,75; −4)\nc) T(2x; y −2) and U(3x + 1; y −2)\n2. Given Q(4; 1), T(p; 3) and length QT =\n√\n8 units, determine the value of p.\n3. Determine the gradient of the line AB if:\na) A(−5; 3) and B(−7; 4)\nb) A(3; −2) and B(1; −8)\n4. Prove that the line PQ, with P(0; 3) and Q(5; 5), is parallel to the line 5y + 5 =\n2x.\n5. Given the points A(−1; −1), B(2; 5), C(−1; −5\n2) and D(x; −4) and AB ⊥CD,\ndetermine the value of x.\n6. Calculate the coordinates of the mid-point P(x; y) of the line segment between\nthe points:\na) M(3; 5) and N(−1; −1)\nb) A(−3; −4) and B(2; 3)\n7. The line joining A(−2; 4) and B(x; y) has the mid-point C(1; 3). Determine the\nvalues of x and y.\n8. Given\nquadrilateral\nABCD\nwith\nvertices\nA(0; 3), B(4; 3), C(5; −1)\nand\nD(1; −1).\na) Determine the equation of the line AD and the line BC.\nb) Show that AD ∥BC.\nc) Calculate the lengths of AD and BC.\nd) Determine the equation of the diagonal BD.\ne) What type of quadrilateral is ABCD?\n112\n4.1.\nRevision\n\n9. MPQN is a parallelogram with points M(−5; 3), P(−1; 5) and Q(4; 5). Draw a\nsketch and determine the coordinates of N(x; y).\n10. PQRS is a quadrilateral with points P(−3; 1), Q(1; 3), R(6; 1) and S(2; −1) in\nthe Cartesian plane.\na) Determine the lengths of PQ and SR.\nb) Determine the mid-point of PR.\nc) Show that PQ ∥SR.\nd) Determine the equations of the line PS and the line SR.\ne) Is PS ⊥SR? Explain your answer.\nf) What type of quadrilateral is PQRS?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22JG\n1b. 22JH\n1c. 22JJ\n2. 22JK\n3a. 22JM\n3b. 22JN\n4. 22JP\n5. 22JQ\n6a. 22JR\n6b. 22JS\n7. 22JT\n8. 22JV\n9. 22JW\n10. 22JX\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.2\nEquation of a line\nEMBG8\nWe can derive different forms of the straight line equation. The different forms are\nused depending on the information provided in the problem:\n• The two-point form of the straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• The gradient–point form of the straight line equation: y −y1 = m(x −x1)\n• The gradient–intercept form of the straight line equation: y = mx + c\nThe two-point form of the straight line equation\nEMBG9\nb\nb\n(x1; y1)\n(x2; y2)\nx\ny\n0\n113\nChapter 4.\nAnalytical geometry\n\nGiven any two points (x1; y1) and (x2; y2), we can determine the equation of the line\npassing through the two points using the equation:\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 3: The two-point form of the straight line equation\nQUESTION\nFind the equation of the straight line passing through P (−1; −5) and Q (5; 4).\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nP(−1; −5)\nQ(5; 4)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q(x2; y2)\nx1 = −1;\ny1 = −5;\nx2 = 5;\ny2 = 4\nStep 3: Write down the two-point form of the straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\n114\n4.2.\nEquation of a line\n\nStep 4: Substitute the values and make y the subject of the equation\ny −(−5)\nx −(−1) = 4 −(−5)\n5 −(−1)\ny + 5\nx + 1 = 9\n6\ny + 5 = 3\n2(x + 1)\ny + 5 = 3\n2x + 3\n2\ny = 3\n2x −7\n2\nStep 5: Write the final answer\ny = 3\n2x −31\n2\nExercise 4 – 2: The two-point form of the straight line equation\nDetermine the equation of the straight line passing through the points:\n1. (3; 7) and (−6; 1)\n2. (1; −11\n4 ) and (2\n3; −7\n4)\n3. (−2; 1) and (3; 6)\n4. (2; 3) and (3; 5)\n5. (1; −5) and (−7; −5)\n6. (−4; 0) and (1; 15\n4 )\n7. (s; t) and (t; s)\n8. (−2; −8) and (1; 7)\n9. (2p; q) and (0; −q)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22JY\n2. 22JZ\n3. 22K2\n4. 22K3\n5. 22K4\n6. 22K5\n7. 22K6\n8. 22K7\n9. 22K8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n115\nChapter 4.\nAnalytical geometry\n\nThe gradient–point form of the straight line equation\nEMBGB\nWe derive the gradient–point form of the straight line equation using the definition of\ngradient and the two-point form of a straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nSubstitute m = y2 −y1\nx2 −x1\non the right-hand side of the equation\ny −y1\nx −x1\n= m\nMultiply both sides of the equation by (x −x1)\ny −y1 = m(x −x1)\nTo use this equation, we need to know the gradient of the line and the coordinates of\none point on the line.\nSee video: 22K9 at www.everythingmaths.co.za\nWorked example 4: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −1\n3 and passing through\nthe point (−1; 1).\nSOLUTION\nStep 1: Draw a sketch\nWe notice that m < 0, therefore the graph decreases as x increases.\n1\n2\n3\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\n(−1; 1)\nx\ny\n0\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\n116\n4.2.\nEquation of a line\n\nSubstitute the value of the gradient\ny −y1 = −1\n3(x −x1)\nSubstitute the coordinates of the given point\ny −1 = −1\n3(x −(−1))\ny −1 = −1\n3(x + 1)\ny = −1\n3x −1\n3 + 1\n= −1\n3x + 2\n3\nStep 3: Write the final answer\nThe equation of the straight line is y = −1\n3x + 2\n3.\nIf we are given two points on a straight line, we can also use the gradient–point form\nto determine the equation of a straight line. We first calculate the gradient using the\ntwo given points and then substitute either of the two points into the gradient–point\nform of the equation.\nWorked example 5: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line passing through (−3; 2) and (5; 8).\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n8\n2\n4\n6\n−2\n−4\nb\nb\n(−3; 2)\n(5; 8)\nx\ny\n0\n117\nChapter 4.\nAnalytical geometry\n\nStep 2: Assign variables to the coordinates of the given points\nx1 = −3;\ny1 = 2;\nx2 = 5;\ny2 = 8\nStep 3: Calculate the gradient using the two given points\nm = y2 −y1\nx2 −x1\n=\n8 −2\n5 −(−3)\n= 6\n8\n= 3\n4\nStep 4: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the value of the gradient\ny −y1 = 3\n4(x −x1)\nSubstitute the coordinates of a given point\ny −y1 = 3\n4(x −x1)\ny −2 = 3\n4(x −(−3))\ny −2 = 3\n4(x + 3)\ny = 3\n4x + 9\n4 + 2\n= 3\n4x + 17\n4\nStep 5: Write the final answer\nThe equation of the straight line is y = 3\n4x + 41\n4.\nSee video: 22KB at www.everythingmaths.co.za\n118\n4.2.\nEquation of a line\n\nExercise 4 – 3: Gradient–point form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (−1; 10\n3 ) and with m = 2\n3.\n2. with m = −1 and passing through the point (−2; 0).\n3. passing through the point (3; −1) and with m = −1\n3.\n4. parallel to the x-axis and passing through the point (0; 11).\n5. passing through the point (1; 5) and with m = −2.\n6. perpendicular to the x-axis and passing through the point (−3\n2; 0).\n7. with m = −0,8 and passing through the point (10; −7).\n8. with undefined gradient and passing through the point (4; 0).\n9. with m = 3a and passing through the point (−2; −6a + b).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KC\n2. 22KD\n3. 22KF\n4. 22KG\n5. 22KH\n6. 22KJ\n7. 22KK\n8. 22KM\n9. 22KN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe gradient–intercept form of a straight line equation EMBGC\nUsing the gradient–point form, we can also derive the gradient–intercept form of the\nstraight line equation.\nStarting with the equation\ny −y1 = m(x −x1)\nExpand the brackets and make y the subject of the formula\ny −y1 = mx −mx1\ny = mx −mx1 + y1\ny = mx + (y1 −mx1)\nWe define constant c such that c = y1 −mx1 so that we get the equation\ny = mx + c\nThis is also called the standard form of the straight line equation.\n119\nChapter 4.\nAnalytical geometry\n\nNotice that when x = 0, we have\ny = m(0) + c\n= c\nTherefore c is the y-intercept of the straight line.\nWorked example 6: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −2 and passing through\nthe point (−1; 7).\nSOLUTION\nStep 1: Slope of the line\nWe notice that m < 0, therefore the graph decreases as x increases.\n2\n4\n6\n8\n−2\n2\n4\n−2\n−4\nb\n(−1; 7)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\nSubstitute the value of the gradient\ny = −2x + c\n120\n4.2.\nEquation of a line\n\nSubstitute the coordinates of the given point and find c\ny = −2x + c\n7 = −2(−1) + c\n7 −2 = c\n∴c = 5\nThis gives the y-intercept (0; 5).\nStep 3: Write the final answer\nThe equation of the straight line is y = −2x + 5.\nIf we are given two points on a straight line, we can also use the gradient–intercept\nform to determine the equation of a straight line. We solve for the two unknowns m\nand c using simultaneous equations — using the methods of substitution or elimina-\ntion.\nWorked example 7: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line passing through the points (−2; −7) and\n(3; 8).\nSOLUTION\nStep 1: Draw a sketch\n4\n8\n−4\n−8\n2\n4\n−2\n−4\nb\nb\n(−2; −7)\n(3; 8)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\n121\nChapter 4.\nAnalytical geometry\n\nStep 3: Substitute the coordinates of the given points\n−7 = m(−2) + c\n−7 = −2m + c\n. . . (1)\n8 = m(3) + c\n8 = 3m + c\n. . . (2)\nWe have two equations with two unknowns; we can therefore solve using simultane-\nous equations.\nStep 4: Make the coefficient of one of the variables the same in both equations\nWe notice that the coefficient of c in both equations is 1, therefore we can subtract\none equation from the other to eliminate c:\n−7 = −2m + c\n−(8 = 3m + c)\n−15 = −5m\n∴3 = m\nSubstitute m = 3 into either of the two equations and determine c:\n−7 = −2m + c\n−7 = −2(3) + c\n∴c = −1\nor\n8 = 3m + c\n8 = 3(3) + c\n∴c = −1\nStep 5: Write the final answer\nThe equation of the straight line is y = 3x −1.\n122\n4.2.\nEquation of a line\n\nExercise 4 – 4: The gradient–intercept form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (1\n2; 4) and\nwith m = 2.\n2. passing through the points (1\n2; −2)\nand (2; 4).\n3. passing through the points (2; −3)\nand (−1; 0).\n4. passing through the point (2; −6\n7)\nand with m = −3\n7.\n5. which cuts the y-axis at y = −1\n5 and\nwith m = 1\n2.\n6.\nb\nb\n(−1; −4)\n(2; 2)\nx\ny\n0\n7.\nb −3\n2\nx\ny\n0\n8.\nb\n(−2; −2)\n4\nx\ny\n0\n9.\nb (−2; 10)\nx\ny\n0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KP\n2. 22KQ\n3. 22KR\n4. 22KS\n5. 22KT\n6. 22KV\n7. 22KW\n8. 22KX\n9. 22KY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n123\nChapter 4.\nAnalytical geometry\n\n4.3\nInclination of a line\nEMBGD\n1\n2\n3\n1\n2\n3\nθ\n∆y\n∆x\nx\ny\nThe diagram shows that a straight line makes an angle θ with the positive x-axis. This\nis called the angle of inclination of a straight line.\nWe notice that if the gradient changes, then the value of θ also changes, therefore the\nangle of inclination of a line is related to its gradient. We know that gradient is the\nratio of a change in the y-direction to a change in the x-direction:\nm = ∆y\n∆x\nFrom trigonometry we know that the tangent function is defined as the ratio:\ntan θ = opposite side\nadjacent side\nAnd from the diagram we see that\ntan θ = ∆y\n∆x\n∴m = tan θ\nfor 0◦≤θ < 180◦\nTherefore the gradient of a straight line is equal to the tangent of the angle formed\nbetween the line and the positive direction of the x-axis.\nVertical lines\n• θ = 90◦\n• Gradient is undefined since there is no change in the x-values (∆x = 0).\n• Therefore tan θ is also undefined (the graph of tan θ has an asymptote at θ =\n90◦).\n124\n4.3.\nInclination of a line\n\nHorizontal lines\n• θ = 0◦\n• Gradient is equal to 0 since there is no change in the y-values (∆y = 0).\n• Therefore tan θ is also equal to 0 (the graph of tan θ passes through the origin\n(0◦; 0).\nLines with negative gradients\nIf a straight line has a negative gradient (m < 0, tan θ < 0), then the angle formed\nbetween the line and the positive direction of the x-axis is obtuse.\nθ\nx\ny\n0\nFrom the CAST diagram in trigonometry, we know that the tangent function is negative\nin the second and fourth quadrant. If we are calculating the angle of inclination for a\nline with a negative gradient, we must add 180◦to change the negative angle in the\nfourth quadrant to an obtuse angle in the second quadrant:\nIf we are given a straight line with gradient m = −0,7, then we can determine the\nangle of inclination using a calculator:\ntan θ = m\n= −0,7\n∴θ = tan−1(−0,7)\n= −35,0◦\nThis negative angle lies in the fourth quadrant. We must add 180◦to get an obtuse\nangle in the second quadrant:\nθ = −35,0◦+ 180◦\n= 145◦\n125\nChapter 4.\nAnalytical geometry\n\nAnd we can always use our calculator to check that the obtuse angle θ = 145◦gives a\ngradient of m = −0,7.\n35◦\n180◦−35◦= 145◦\nx\ny\n0\nExercise 4 – 5: Angle of inclination\n1. Determine the gradient (correct to 1 decimal place) of each of the following\nstraight lines, given that the angle of inclination is equal to:\na) 60◦\nb) 135◦\nc) 0◦\nd) 54◦\ne) 90◦\nf) 45◦\ng) 140◦\nh) 180◦\ni) 75◦\n2. Determine the angle of inclination (correct to 1 decimal place) for each of the\nfollowing:\na) a line with m = 3\n4\nb) 2y −x = 6\nc) the line passes through the points (−4; −1) and (2; 5)\nd) y = 4\ne) x = 3y + 1\n2\nf) x = −0,25\ng) the line passes through the points (2; 5) and (2\n3; 1)\nh) a line with gradient equal to 0,577\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22KZ\n1b. 22M2\n1c. 22M3\n1d. 22M4\n1e. 22M5\n1f. 22M6\n1g. 22M7\n1h. 22M8\n1i. 22M9\n2a. 22MB\n2b. 22MC\n2c. 22MD\n2d. 22MF\n2e. 22MG\n2f. 22MH\n2g. 22MJ\n2h. 22MK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n126\n4.3.\nInclination of a line\n\nWorked example 8: Inclination of a straight line\nQUESTION\nDetermine the angle of inclination (correct to 1 decimal place) of the straight line\npassing through the points (2; 1) and (−3; −9).\nSOLUTION\nStep 1: Draw a sketch\n2\n−2\n−4\n−6\n−8\n1\n2\n−1\n−2\n−3\nb\nb\n(2; 1)\n(−3; −9)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nx1 = 2;\ny1 = 1;\nx2 = −3;\ny2 = −9\nStep 3: Determine the gradient of the line\nm = y2 −y1\nx2 −x1\n= −9 −1\n−3 −2\n= −10\n−5\n∴m = 2\nStep 4: Use the gradient to determine the angle of inclination of the line\ntan θ = m\n= 2\n∴θ = tan−1 2\n= 63,4◦\nImportant: make sure your calculator is in DEG (degrees) mode.\nStep 5: Write the final answer\nThe angle of inclination of the straight line is 63,4◦.\n127\nChapter 4.\nAnalytical geometry\n\nWorked example 9: Inclination of a straight line\nQUESTION\nDetermine the equation of the straight line passing through the point (3; 1) and with\nan angle of inclination of 135◦.\nSOLUTION\nStep 1: Use the angle of inclination to determine the gradient of the line\nm = tan θ\n= tan 135◦\n∴m = −1\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = −1\ny −y1 = −(x −x1)\nSubstitute the given point (3; 1)\ny −1 = −(x −3)\ny = −x + 3 + 1\n= −x + 4\nStep 3: Write the final answer\nThe equation of the straight line is y = −x + 4.\nWorked example 10: Inclination of a straight line\nQUESTION\nDetermine the acute angle (correct to 1 decimal place) between the line passing\nthrough the points M(−1; 13\n4) and N(4; 3) and the straight line y = −3\n2x + 4.\nSOLUTION\nStep 1: Draw a sketch\nDraw the line through points M(−1; 13\n4) and N(4; 3) and the line y = −3\n2x + 4 on a\nsuitable system of axes. Label α and β, the angles of inclination of the two lines. Label\nθ, the acute angle between the two straight lines.\n128\n4.3.\nInclination of a line\n\n2\n4\n6\n−2\n2\n4\n−2\n−4\n−6\n−8\nb\nb\nβ\nˆB1\nα\nθ\nx\ny\n0\nM(−1; 7\n4)\nN(4; 3)\nNotice that α and θ are acute angles and β is an obtuse angle.\nˆB1 = 180◦−β\n(∠on str. line)\nand θ = α + ˆB1\n(ext. ∠of △= sum int. opp)\n∴θ = α + (180◦−β)\n= 180◦+ α −β\nStep 2: Use the gradient to determine the angle of inclination β\nFrom the equation y = −3\n2x + 4 we see that m < 0, therefore β is an obtuse angle\nsuch that 90◦< β < 180◦.\ntan β = m\n= −3\n2\ntan−1\n\u0012\n−3\n2\n\u0013\n= −56,3◦\nThis negative angle lies in the fourth quadrant. We know that the angle of inclination\nβ is an obtuse angle that lies in the second quadrant, therefore\nβ = −56,3◦+ 180◦\n= 123,7◦\nStep 3: Determine the gradient and angle of inclination of the line through M and\nN\n129\nChapter 4.\nAnalytical geometry\n\nDetermine the gradient\nm = y2 −y1\nx2 −x1\n=\n3 −7\n4\n4 −(−1)\n=\n5\n4\n5\n= 1\n4\nDetermine the angle of inclination\ntan α = m\n= 1\n4\n∴α = tan−1\n\u00121\n4\n\u0013\n= 14,0◦\nStep 4: Write the final answer\nθ = 180◦+ α −β\n= 180◦+ 14,0◦−123,7◦\n= 70,3◦\nThe acute angle between the two straight lines is 70,3◦.\nExercise 4 – 6: Inclination of a straight line\n1. Determine the angle of inclination for each of the following:\na) a line with m = 4\n5\nb) x + y + 1 = 0\nc) a line with m = 5,69\nd) the line that passes through (1; 1) and (−2; 7)\ne) 3 −2y = 9x\nf) the line that passes through (−1; −6) and (−1\n2; −11\n2 )\ng) 5 = 10y −15x\n130\n4.3.\nInclination of a line\n\nh)\nb\nx\ny\n(2; 3)\n−1\n0\ni)\nb\nx\ny\n(6; 0)\n2\n0\nj)\nb\nx\ny\n(−3; 3)\n−3\n0\n2. Determine the acute angle between the line passing through the points A(−2; 1\n5)\nand B(0; 1) and the line passing through the points C(1; 0) and D(−2; 6).\n3. Determine the angle between the line y + x = 3 and the line x = y + 1\n2.\n4. Find the angle between the line y = 2x and the line passing through the points\n(−1; 7\n3) and (0; 2).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22MM\n1b. 22MN\n1c. 22MP\n1d. 22MQ\n1e. 22MR\n1f. 22MS\n1g. 22MT\n1h. 22MV\n1i. 22MW\n1j. 22MX\n2. 22MY\n3. 22MZ\n4. 22N2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n131\nChapter 4.\nAnalytical geometry\n\n4.4\nParallel lines\nEMBGF\nInvestigation: Parallel lines\n1. Draw a sketch of the line passing through the points P(−1; 0) and Q(1; 4) and\nthe line passing through the points R(1; 2) and S(2; 4).\n2. Label and measure α and β, the angles of inclination of straight lines PQ and\nRS respectively.\n3. Describe the relationship between α and β.\n4. “α and β are alternate angles, therefore PQ ∥RS.” Is this a true statement? If\nnot, provide a correct statement.\n5. Use your calculator to determine tan α and tan β.\n6. Complete the sentence: . . . . . . lines have . . . . . . angles of inclination.\n7. Determine the equations of the straight lines PQ and RS.\n8. What do you notice about mPQ and mRS?\n9. Complete the sentence: . . . . . . lines have . . . . . . gradients.\nAnother method of determining the equation of a straight line is to be given a point on\nthe unknown line, (x1; y1), and the equation of a line which is parallel to the unknown\nline.\nLet the equation of the unknown line be y = m1x + c1 and the equation of the given\nline be y = m2x + c2.\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are parallel then\nm1 = m2\n132\n4.4.\nParallel lines\n\nImportant: when determining the gradient of a line using the coefficient of x, make\nsure the given equation is written in the gradient–intercept (standard) form. y = mx+c\nSubstitute the value of m2 and the given point (x1; y1), into the gradient–intercept form\nof a straight line equation\ny −y1 = m(x −x1)\nand determine the equation of the unknown line.\nWorked example 11: Parallel lines\nQUESTION\nDetermine the equation of the line that passes through the point (−1; 1) and is parallel\nto the line y −2x + 1 = 0.\nSOLUTION\nStep 1: Write the equation in gradient–intercept form\nWe write the given equation in gradient–intercept form and determine the value of m.\ny = 2x −1\nWe know that the two lines are parallel, therefore m1 = m2 = 2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = 2\ny −y1 = 2(x −x1)\nSubstitute the given point (−1; 1)\ny −1 = 2(x −(−1))\ny −1 = 2x + 2\ny = 2x + 2 + 1\n= 2x + 3\n133\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\n(−1; 1)\ny = 2x −1\ny = 2x + 3\nx\ny\n0\nA sketch was not required, but it is always helpful and can be used to check answers.\nStep 3: Write the final answer\nThe equation of the straight line is y = 2x + 3.\nWorked example 12: Parallel lines\nQUESTION\nLine AB passes through the point A(0; 3) and has an angle of inclination of 153,4◦.\nDetermine the equation of the line CD which passes through the point C(2; −3) and\nis parallel to AB.\nSOLUTION\nStep 1: Use the given angle of inclination to determine the gradient\nmAB = tan θ\n= tan 153,4◦\n= −0,5\nStep 2: Parallel lines have equal gradients\nSince we are given AB ∥CD,\nmCD = mAB = −0,5\n134\n4.4.\nParallel lines\n\nStep 3: Write down the gradient–point form of a straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient mCD = −0,5.\ny −y1 = −1\n2(x −x1)\nSubstitute the given point (2; −3).\ny −(−3) = −1\n2(x −2)\ny + 3 = −1\n2x + 1\ny = −1\n2x −2\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\nb C(2; −3)\ny = −1\n2x −2\ny = −1\n2x + 3\nx\ny\n0\nA sketch was not required, but it is always useful.\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n2x −2.\nSee video: 22N3 at www.everythingmaths.co.za\n135\nChapter 4.\nAnalytical geometry\n\nExercise 4 – 7: Parallel lines\n1. Determine whether or not the following two lines are parallel:\na) y + 2x = 1 and −2x + 3 = y\nb)\ny\n3 + x + 5 = 0 and 2y + 6x = 1\nc) y = 2x −7 and the line passing through (1; −2) and (1\n2; −1)\nd) y + 1 = x and x + y = 3\ne) The line passing through points (−2; −1) and (−4; −3) and the line −y +\nx −4 = 0\nf) y −1 = 1\n3x and the line passing through points (−2; 4) and (1; 5)\n2. Determine the equation of the straight line that passes through the point (1; −5)\nand is parallel to the line y + 2x −1 = 0.\n3. Determine the equation of the straight line that passes through the point (−2; −6)\nand is parallel to the line 2y + 1 = 6x.\n4. Determine the equation of the straight line that passes through the point (−2; −2)\nand is parallel to the line with angle of inclination θ = 56,31◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is parallel to the line with angle of inclination θ = 145◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22N4\n1b. 22N5\n1c. 22N6\n1d. 22N7\n1e. 22N8\n1f. 22N9\n2. 22NB\n3. 22NC\n4. 22ND\n5. 22NF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.5\nPerpendicular lines\nEMBGG\nInvestigation: Perpendicular lines\n1. Draw a sketch of the line passing through the points A(−2; −3) and B(2; 5) and\nthe line passing through the points C(−1; 1\n2) and D(4; −2).\n2. Label and measure α and β, the angles of inclination of straight lines AB and\nCD respectively.\n3. Label and measure θ, the angle between the lines AB and CD.\n4. Describe the relationship between the lines AB and CD.\n5. “θ is a reflex angle, therefore AB ⊥CD.” Is this a true statement? If not, provide\na correct statement.\n136\n4.5.\nPerpendicular lines\n\n6. Determine the equation of the straight line AB and the line CD.\n7. Use your calculator to determine tan α × tan β.\n8. Determine mAB × mCD.\n9. What do you notice about these products?\n10. Complete the sentence: if two lines are . . . . . . to each other, then the product of\ntheir . . . . . . is equal . . . . . .\n11. Complete the sentence: if the gradient of a straight line is equal to the negative\n. . . . . . of the gradient of another straight line, then the two lines are . . . . . .\nDeriving the formula: m1 × m2 = −1\nb\nb\nA(4; 3)\nB(−3; 4)\nθ\n90◦+ θ\nO\ny\nx\nConsider the point A(4; 3) on the Cartesian plane with an angle of inclination A ˆOX =\nθ. Rotate through an angle of 90◦and place point B at (−3; 4) so that we have the\nangle of inclination B ˆOX = 90◦+ θ.\nWe determine the gradient of OA:\nmOA = y2 −y1\nx2 −x1\n= 3 −0\n4 −0\n= 3\n4\nAnd determine the gradient of OB:\nmOB = y2 −y1\nx2 −x1\n= 4 −0\n−3 −0\n= 4\n−3\n137\nChapter 4.\nAnalytical geometry\n\nBy rotating through an angle of 90◦we know that OB ⊥OA:\nmOA × mOB = 3\n4 × 4\n−3\n= −1\nWe can also write that\nmOA = −\n1\nmOB\nb\nb\nA(x; y)\nB(−y; x)\nθ\n90◦+ θ\nO\ny\nx\nIf we have the general point A(x; y) with an angle of inclination A ˆOX = θ and point\nB(−y; x) such that B ˆOX = 90◦+ θ, then we know that\nmOA = y\nx\nmOB = −x\ny\n∴mOA × mOB = y\nx × −x\ny\n= −1\nAnother method of determining the equation of a straight line is to be given a point on\nthe line, (x1; y1), and the equation of a line which is perpendicular to the unknown\nline. Let the equation of the unknown line be y = m1x + c1 and the equation of the\ngiven line be y = m2x + c2.\nθ2\nθ1\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are perpendicular then\nm1 × m2 = −1\nNote: this rule does not apply to vertical or horizontal lines.\n138\n4.5.\nPerpendicular lines\n\nWhen determining the gradient of a line using the coefficient of x, make sure the\ngiven equation is written in the gradient–intercept (standard) form y = mx + c. Then\nwe know that\nm1 = −1\nm2\nSubstitute the value of m1 and the given point (x1; y1), into the gradient–intercept form\nof the straight line equation y −y1 = m(x −x1) and determine the equation of the\nunknown line.\nWorked example 13: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point T(2; 2) and per-\npendicular to the line 3y + 2x −6 = 0.\nSOLUTION\nStep 1: Write the equation in standard form\nLet the gradient of the unknown line be m1 and the given gradient be m2. We write\nthe given equation in gradient–intercept form and determine the value of m2.\n3y + 2x −6 = 0\n3y = −2x + 6\ny = −2\n3x + 2\n∴m2 = −2\n3\nWe know that the two lines are perpendicular, therefore m1 × m2 = −1. Therefore\nm1 = 3\n2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m1 = 3\n2.\ny −y1 = 3\n2(x −x1)\nSubstitute the given point T(2; 2).\ny −2 = 3\n2(x −2)\ny −2 = 3\n2x −3\ny = 3\n2x −1\n139\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\nb T(2; 2)\ny = −2\n3x + 2\ny = 3\n2x −1\nx\ny\n0\nA sketch was not required, but it is useful for checking the answer.\nStep 3: Write the final answer\nThe equation of the straight line is y = 3\n2x −1.\nWorked example 14: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point (2; 1\n3) and per-\npendicular to the line with an angle of inclination of 71,57◦.\nSOLUTION\nStep 1: Use the given angle of inclination to determine gradient\nLet the gradient of the unknown line be m1 and let the given gradient be m2.\nm2 = tan θ\n= tan 71,57◦\n= 3,0\nStep 2: Determine the unknown gradient\nSince we are given that the two lines are perpendicular,\nm1 × m2 = −1\n∴m1 = −1\n3\n140\n4.5.\nPerpendicular lines\n\nStep 3: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient m1 = −1\n3.\ny −y1 = −1\n3(x −x1)\nSubstitute the given point (2; 1\n3).\ny −\n\u00121\n3\n\u0013\n= −1\n3(x −2)\ny −1\n3 = −1\n3x + 2\n3\ny = −1\n3x + 1\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n3x + 1.\nSee video: 22NG at www.everythingmaths.co.za\nExercise 4 – 8: Perpendicular lines\n1. Calculate whether or not the following two lines are perpendicular:\na) y −1 = 4x and 4y + x + 2 = 0\nb) 10x = 5y −1 and 5y −x −10 = 0\nc) x = y −5 and the line passing through (−1; 5\n4) and (3; −11\n4 )\nd) y = 2 and x = 1\ne)\ny\n3 = x and 3y + x = 9\nf) 1 −2x = y and the line passing through (2; −1) and (−1; 5)\ng) y = x + 2 and 2y + 1 = 2x\n2. Determine the equation of the straight line that passes through the point (−2; −4)\nand is perpendicular to the line y + 2x = 1.\n3. Determine the equation of the straight line that passes through the point (2; −7)\nand is perpendicular to the line 5y −x = 0.\n141\nChapter 4.\nAnalytical geometry\n\n4. Determine the equation of the straight line that passes through the point (3; −1)\nand is perpendicular to the line with angle of inclination θ = 135◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is perpendicular to the line y = 4\n3.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NH\n1b. 22NJ\n1c. 22NK\n1d. 22NM\n1e. 22NN\n1f. 22NP\n1g. 22NQ\n2. 22NR\n3. 22NS\n4. 22NT\n5. 22NV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.6\nSummary\nEMBGH\nSee presentation: 22NW at www.everythingmaths.co.za\n• Distance between two points: d =\np\n(x2 −x1)2 + (y2 −y1)2\n• Gradient of a line between two points: m = y2 −y1\nx2 −x1\n• Mid-point of a line: M(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n• Parallel lines: m1 = m2\n• Perpendicular lines: m1 × m2 = −1\n• General form of a straight line equation: ax + by + c = 0\n• Two-point form of a straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• Gradient–point form of a straight line equation: y −y1 = m(x −x1)\n• Gradient–intercept form of a straight line equation (standard form): y = mx + c\n• Angle of inclination of a straight line: θ, the angle formed between the line and\nthe positive x-axis; m = tan θ\n142\n4.6.\nSummary\n\nExercise 4 – 9: End of chapter exercises\n1. Determine the equation of the line:\na) through points (−1; 3) and (1; 4)\nb) through points (7; −3) and (0; 4)\nc) parallel to y = 1\n2x + 3 and passing through (−2; 3)\nd) perpendicular to y = −1\n2x + 3 and passing through (−1; 2)\ne) perpendicular to 3y + x = 6 and passing through the origin\n2. Determine the angle of inclination of the following lines:\na) y = 2x −3\nb) y = 1\n3x −7\nc) 4y = 3x + 8\nd) y = −2\n3x + 3\ne) 3y + x −3 = 0\n3. P(2; 3), Q(−4; 0) and R(5; −3) are the vertices of △PQR in the Cartesian plane.\nPR intersects the x-axis at S. Determine the following:\na) the equation of the line PR\nb) the coordinates of point S\nc) the angle of inclination of PR (correct to two decimal places)\nd) the gradient of line PQ\ne) Q ˆPR\nf) the equation of the line perpendicular to PQ and passing through the origin\ng) the mid-point M of QR\nh) the equation of the line parallel to PR and passing through point M\n4. Points A(−3; 5), B(−7; −4) and C(2; 0) are given.\na) Plot the points on the Cartesian plane.\nb) Determine the coordinates of D if ABCD is a parallelogram.\nc) Prove that ABCD is a rhombus.\n5.\nb\nb\nb\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\nx\ny\n0\nM\nN\nP\n143\nChapter 4.\nAnalytical geometry\n\nConsider the sketch above, with the following lines shown:\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\na) Determine the coordinates of the point N.\nb) Determine the coordinates of the point P.\nc) Determine the equation of the vertical line MN.\nd) Determine the length of the vertical line MN.\ne) Find M ˆNP.\nf) Determine the equation of the line parallel to NP and passing through the\npoint M.\n6. The following points are given: A(−2; 3), B(2; 4), C(3; 0).\na) Plot the points on the Cartesian plane.\nb) Prove that △ABC is a right-angled isosceles triangle.\nc) Determine the equation of the line AB.\nd) Determine the coordinates of D if ABCD is a square.\ne) Determine the coordinates of E, the mid-point of BC.\n7. Given points S(2; 5), T(−3; −4) and V (4; −2).\na) Determine the equation of the line ST.\nb) Determine the size of T ˆSV .\n8. Consider triangle FGH with vertices F(−1; 3), G(2; 1) and H(4; 4).\na) Sketch △FGH on the Cartesian plane.\nb) Show that △FGH is an isosceles triangle.\nc) Determine the equation of the line PQ, perpendicular bisector of FH.\nd) Does G lie on the line PQ?\ne) Determine the equation of the line parallel to GH and passing through\npoint F.\n9. Given the points A(−1; 5), B(5; −3) and C(0; −6). M is the mid-point of AB\nand N is the mid-point of AC.\na) Draw a sketch on the Cartesian plane.\nb) Show that the coordinates of M and N are (2; 1) and (−1\n2; −1\n2) respectively.\nc) Use analytical geometry methods to prove the mid-point theorem. (Prove\nthat NM ∥CB and NM = 1\n2CB.)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NX\n1b. 22NY\n1c. 22NZ\n1d. 22P2\n1e. 22P3\n2a. 22P4\n2b. 22P5\n2c. 22P6\n2d. 22P7\n2e. 22P8\n3. 22P9\n4. 22PB\n5. 22PC\n6. 22PD\n7. 22PF\n8. 22PG\n9. 22PH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n144\n4.6.\nSummary\n\nCHAPTER\n5\nFunctions\n5.1\nQuadratic functions\n146\n5.2\nAverage gradient\n164\n5.3\nHyperbolic functions\n170\n5.4\nExponential functions\n184\n5.5\nThe sine function\n197\n5.6\nThe cosine function\n209\n5.7\nThe tangent function\n222\n5.8\nSummary\n235\n\n5\nFunctions\nA function describes a specific relationship between two variables; where an indepen-\ndent (input) variable has exactly one dependent (output) variable. Every element in the\ndomain maps to only one element in the range. Functions can be one-to-one relations\nor many-to-one relations. A many-to-one relation associates two or more values of the\nindependent variable with a single value of the dependent variable. Functions allow\nus to visualise relationships in the form of graphs, which are much easier to read and\ninterpret than lists of numbers.\n5.1\nQuadratic functions\nEMBGJ\nRevision\nEMBGK\nFunctions of the form y = ax2 + q\nFunctions of the general form y = ax2 + q are called parabolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = ax2 + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\nThe turning point of f(x) is\nabove the x-axis.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\nThe turning point of f(x) is be-\nlow the x-axis.\n– q is also the y-intercept of the\nparabola.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\n• The effect of a on shape\n– For a > 0; the graph of f(x) is a “smile” and has a minimum turning point\n(0; q). As the value of a becomes larger, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n– For a < 0; the graph of f(x) is a “frown” and has a maximum turning point\n(0; q). As the value of a becomes smaller, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n146\n5.1.\nQuadratic functions\n\nExercise 5 – 1: Revision\n1. On separate axes, accurately draw each of the following functions.\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = x2\nb) y2 = 1\n2x2\nc) y3 = −x2 −1\nd) y4 = −2x2 + 4\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nyint = 0\nvalue of a\na = 1\neffect of a\nstandard\nparabola\nturning point\n(0; 0)\naxis of symmetry\nx = 0\n(y-axis)\ndomain\n{x : x ∈R}\nrange\n{y : y ≥0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22PJ\n1b. 22PK\n1c. 22PM\n1d. 22PN\n2. 22PP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22PQ at www.everythingmaths.co.za\n147\nChapter 5.\nFunctions\n\nFunctions of the form y = a(x + p)2 + q\nEMBGM\nWe now consider parabolic functions of the form y = a(x + p)2 + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a parabolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = x2\nb) y2 = (x −2)2\nc) y3 = (x −1)2\nd) y4 = (x + 1)2\ne) y5 = (x + 2)2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = x2 + 2\nb) y2 = (x −2)2 −1\nc) y3 = (x −1)2 + 1\nd) y4 = (x + 1)2 + 1\ne) y5 = (x + 2)2 −1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of q\n3. Consider the three functions given below and answer the questions that follow:\n• y1 = (x −2)2 + 1\n• y2 = 2(x −2)2 + 1\n• y3 = −1\n2(x −2)2 + 1\na) What is the value of a for y2?\nb) Does y1 have a minimum or maximum turning point?\n148\n5.1.\nQuadratic functions\n\nc) What are the coordinates of the turning point of y2?\nd) Compare the graphs of y1 and y2. Discuss the similarities and differences.\ne) What is the value of a for y3?\nf) Will the graph of y3 be narrower or wider than the graph of y1?\ng) Determine the coordinates of the turning point of y3.\nh) Compare the graphs of y1 and y3. Describe any differences.\nSee video: 22PR at www.everythingmaths.co.za\nThe effect of the parameters on y = a(x + p)2 + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects whether the turning point is to the left of the y-axis (p > 0)\nor to the right of the y-axis (p < 0). The axis of symmetry is the line x = −p.\nThe effect of q is a vertical shift. The value of q affects whether the turning point of the\ngraph is above the x-axis (q > 0) or below the x-axis (q < 0).\nThe value of a affects the shape of the graph. If a < 0, the graph is a “frown” and has\na maximum turning point. If a > 0 then the graph is a “smile” and has a minimum\nturning point. When a = 0, the graph is a horizontal line y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\nSee simulation: 22PS at www.everythingmaths.co.za\n149\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form f(x) = y = a(x + p)2 + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative. If a > 0\nwe have:\n(x + p)2 ≥0\n(perfect square is always positive)\n∴a(x + p)2 ≥0\n(a is positive)\n∴a(x + p)2 + q ≥q\n∴f(x) ≥q\nThe range is therefore {y : y ≥q, y ∈R} if a > 0. Similarly, if a < 0, the range is\n{y : y ≤q, y ∈R}.\nWorked example 1: Domain and range\nQUESTION\nState the domain and range for g(x) = −2(x −1)2 + 3.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n(x −1)2 ≥0\n−2(x −1)2 ≤0\n−2(x −1)2 + 3 ≤3\ng(x) ≤3\nTherefore the range is {g(x) : g(x) ≤3} or in interval notation (−∞; 3].\nNotice in the example above that it helps to have the function in the form y = a(x +\np)2 + q.\nWe use the method of completing the square to write a quadratic function of the\ngeneral form y = ax2 + bx + c in the form y = a(x + p)2 + q (see Chapter 2).\n150\n5.1.\nQuadratic functions\n\nExercise 5 – 2: Domain and range\nGive the domain and range for each of the following functions:\n1. f(x) = (x −4)2 −1\n2. g(x) = −(x −5)2 + 4\n3. h(x) = x2 −6x + 9\n4. j(x) = −2(x + 1)2\n5. k(x) = −x2 + 2x −3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PT\n2. 22PV\n3. 22PW\n4. 22PX\n5. 22PY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nEvery point on the y-axis has an x-coordinate of 0, therefore to calculate the y-intercept\nwe let x = 0.\nFor example, the y-intercept of g(x) = (x −1)2 + 5 is determined by setting x = 0:\ng(x) = (x −1)2 + 5\ng(0) = (0 −1)2 + 5\n= 6\nThis gives the point (0; 6).\nThe x-intercept:\nEvery point on the x-axis has a y-coordinate of 0, therefore to calculate the x-intercept\nwe let y = 0.\nFor example, the x-intercept of g(x) = (x −1)2 + 5 is determined by setting y = 0:\ng(x) = (x −1)2 + 5\n0 = (x −1)2 + 5\n−5 = (x −1)2\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n151\nChapter 5.\nFunctions\n\nExercise 5 – 3: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = (x + 4)2 −1\n2. g(x) = 16 −8x + x2\n3. h(x) = −x2 + 4x −3\n4. j(x) = 4(x −3)2 −1\n5. k(x) = 4(x −3)2 + 1\n6. l(x) = 2x2 −3x −4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PZ\n2. 22Q2\n3. 22Q3\n4. 22Q4\n5. 22Q5\n6. 22Q6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTurning point\nThe turning point of the function f(x) = a(x+p)2 +q is determined by examining the\nrange of the function:\n• If a > 0, f(x) has a minimum turning point and the range is [q; ∞):\nThe minimum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\n• If a < 0, f(x) has a maximum turning point and the range is (−∞; q]:\nThe maximum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\nTherefore the turning point of the quadratic function f(x) = a(x + p)2 + q is (−p; q).\nAlternative form for quadratic equations:\nWe can also write the quadratic equation in the form\ny = a(x −p)2 + q\nThe effect of p is still a horizontal shift, however notice that:\n• For p > 0, the graph is shifted to the right by p units.\n• For p < 0, the graph is shifted to the left by p units.\nThe turning point is (p; q) and the axis of symmetry is the line x = p.\n152\n5.1.\nQuadratic functions\n\nWorked example 2: Turning point\nQUESTION\nDetermine the turning point of g(x) = 3x2 −6x −1.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q\nWe use the method of completing the square:\ng(x) = 3x2 −6x −1\n= 3(x2 −2x) −1\n= 3\n\u0000(x −1)2 −1\n\u0001\n−1\n= 3(x −1)2 −3 −1\n= 3(x −1)2 −4\nStep 2: Determine turning point (−p; q)\nFrom the equation g(x) = 3(x −1)2 −4 we know that the turning point for g(x) is\n(1; −4).\nWorked example 3: Turning point\nQUESTION\n1. Show that the x-value for the turning point of h(x) = ax2 + bx + c is given by\nx = −b\n2a.\n2. Hence, determine the turning point of k(x) = 2 −10x + 5x2.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q and show that p =\nb\n2a\nWe use the method of completing the square:\nh(x) = ax2 + bx + c\n= a\n\u0012\nx2 + b\nax + c\na\n\u0013\nTake half the coefficient of the x term and square it; then add and subtract it from the\n153\nChapter 5.\nFunctions\n\nexpression.\nh(x) = a\n \nx2 + b\nax +\n\u0012 b\n2a\n\u00132\n−\n\u0012 b\n2a\n\u00132\n+ c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2\n4a2 + c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a2\n!\n= a\n\u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a\nFrom the above we have that the turning point is at x = −p = −b\n2a and y = q =\n−b2−4ac\n4a\n.\nStep 2: Determine the turning point of k(x)\nWrite the equation in the general form y = ax2 + bx + c.\nk(x) = 5x2 −10x + 2\nTherefore a = 5; b = −10; c = 2.\nUse the results obtained above to determine x = −b\n2a:\nx = −\n\u0012−10\n2(5)\n\u0013\n= 1\nSubstitute x = 1 to obtain the corresponding y-value :\ny = 5x2 −10x + 2\n= 5(1)2 −10(1) + 2\n= 5 −10 + 2\n= −3\nThe turning point of k(x) is (1; −3).\nExercise 5 – 4: Turning points\nDetermine the turning point of each of the following:\n1. y = x2 −6x + 8\n2. y = −x2 + 4x −3\n3. y = 1\n2(x + 2)2 −1\n4. y = 2x2 + 2x + 1\n154\n5.1.\nQuadratic functions\n\n5. y = 18 + 6x −3x2\n6. y = −2[(x + 1)2 + 3]\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22Q7\n2. 22Q8\n3. 22Q9\n4. 22QB\n5. 22QC\n6. 22QD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxis of symmetry\nThe axis of symmetry for f(x) = a(x + p)2 + q is the vertical line x = −p. The axis of\nsymmetry passes through the turning point (−p; q) and is parallel to the y-axis.\ny\nx\n0\nb\nx = −p\nf(x) = a(x + p)2 + q\nExercise 5 – 5: Axis of symmetry\n1. Determine the axis of symmetry of each of the following:\na) y = 2x2 −5x −18\nb) y = 3(x −2)2 + 1\nc) y = 4x −x2\n2. Write down the equation of a parabola where the y-axis is the axis of symmetry.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QF\n1b. 22QG\n1c. 22QH\n2. 22QJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n155\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) = a(x + p)2 + q\nIn order to sketch graphs of the form f(x) = a(x + p)2 + q, we need to determine five\ncharacteristics:\n• sign of a\n• turning point\n• y-intercept\n• x-intercept(s) (if they exist)\n• domain and range\nSee video: 22QK at www.everythingmaths.co.za\nWorked example 4: Sketching a parabola\nQUESTION\nSketch the graph of y = −1\n2(x + 1)2 −3.\nMark the intercepts, turning point and the axis of symmetry. State the domain and\nrange of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = a(x + p)2 + q\nWe notice that a < 0, therefore the graph is a “frown” and has a maximum turning\npoint.\nStep 2: Determine the turning point (−p; q)\nFrom the equation we know that the turning point is (−1; −3).\nStep 3: Determine the axis of symmetry x = −p\nFrom the equation we know that the axis of symmetry is x = −1.\nStep 4: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = −1\n2 ((0) + 1)2 −3\n= −1\n2 −3\n= −31\n2\nThis gives the point (0; −31\n2).\n156\n5.1.\nQuadratic functions\n\nStep 5: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = −1\n2 (x + 1)2 −3\n3 = −1\n2 (x + 1)2\n−6 = (x + 1)2\nwhich has no real solutions. Therefore, there are no x-intercepts and the graph lies\nbelow the x-axis.\nStep 6: Plot the points and sketch the graph\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\nb\ny\nx\n0\n(0; −3 1\n2)\n(−1; −3)\nStep 7: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≤−3, y ∈R}\nSee video: 22QM at www.everythingmaths.co.za\nWorked example 5: Sketching a parabola\nQUESTION\nSketch the graph of y = 1\n2x2 −4x + 7\n2.\nDetermine the intercepts, turning point and the axis of symmetry. Give the domain\nand range of the function.\nSOLUTION\n157\nChapter 5.\nFunctions\n\nStep 1: Examine the equation of the form y = ax2 + bx + c\nWe notice that a > 0, therefore the graph is a “smile” and has a minimum turning\npoint.\nStep 2: Determine the turning point and the axis of symmetry\nCheck that the equation is in standard form and identify the coefficients.\na = 1\n2;\nb = −4;\nc = 7\n2\nCalculate the x-value of the turning point using\nx = −b\n2a\n= −\n \n−4\n2\n\u0000 1\n2\n\u0001\n!\n= 4\nTherefore the axis of symmetry is x = 4.\nSubstitute x = 4 into the original equation to obtain the corresponding y-value.\ny = 1\n2x2 −4x + 7\n2\n= 1\n2(4)2 −4(4) + 7\n2\n= 8 −16 + 7\n2\n= −41\n2\nThis gives the point\n\u00004; −41\n2\n\u0001\n.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = 1\n2(0)2 −4(0) + 7\n2\n= 7\n2\nThis gives the point\n\u00000; 7\n2\n\u0001\n.\nStep 4: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = 1\n2x2 −4x + 7\n2\n= x2 −8x + 7\n= (x −1)(x −7)\nTherefore x = 1 or x = 7. This gives the points (1; 0) and (7; 0).\nStep 5: Plot the points and sketch the graph\n158\n5.1.\nQuadratic functions\n\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\nb\nb\ny\nx\n0\n(4; −41\n2)\n(0; 3 1\n2)\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≥−41\n2, y ∈R}\nSee video: 22QN at www.everythingmaths.co.za\nInvestigation: Shifting the equation of a parabola\nCarl and Eric are doing their Mathematics homework and decide to check each others\nanswers.\nHomework question:\nIf the parabola y = 3x2 + 1 is shifted 2 units to the right, determine the equation of the\nnew parabola.\n• Carl’s answer:\nA shift to the right means moving in the positive x direction, therefore x is re-\nplaced with x + 2 and the new equation is y = 3(x + 2)2 + 1.\n• Eric’s answer:\nWe replace x with x −2, therefore the new equation is y = 3(x −2)2 + 1.\nWork together in pairs. Discuss the two different answers and decide which one is\ncorrect. Use calculations and sketches to help explain your reasoning.\n159\nChapter 5.\nFunctions\n\nWriting an equation of a shifted parabola\nThe parabola is shifted horizontally:\n• If the parabola is shifted m units to the right, x is replaced by (x −m).\n• If the parabola is shifted m units to the left, x is replaced by (x + m).\nThe parabola is shifted vertically:\n• If the parabola is shifted n units down, y is replaced by (y + n).\n• If the parabola is shifted n units up, y is replaced by (y −n).\nWorked example 6: Shifting a parabola\nQUESTION\nGiven y = x2 −2x −3.\n1. If the parabola is shifted 1 unit to the right, determine the new equation of the\nparabola.\n2. If the parabola is shifted 3 units down, determine the new equation of the\nparabola.\nSOLUTION\nStep 1: Determine the new equation of the shifted parabola\n1. The parabola is shifted 1 unit to the right, so x must be replaced by (x −1).\ny = x2 −2x −3\n= (x −1)2 −2(x −1) −3\n= x2 −2x + 1 −2x + 2 −3\n= x2 −4x\nBe careful not to make a common error: replacing x with x + 1 for a shift to the\nright.\n2. The parabola is shifted 3 units down, so y must be replaced by (y + 3).\ny + 3 = x2 −2x −3\ny = x2 −2x −3 −3\n= x2 −2x −6\n160\n5.1.\nQuadratic functions\n\nExercise 5 – 6: Sketching parabolas\n1. Sketch graphs of the following functions and determine:\n• intercepts\n• turning point\n• axes of symmetry\n• domain and range\na) y = −x2 + 4x + 5\nb) y = 2(x + 1)2\nc) y = 3x2 −2(x + 2)\nd) y = 3(x −2)2 + 1\n2. Draw the following graphs on the same system of axes:\nf(x) = −x2 + 7\ng(x) = −(x −2)2 + 7\nh(x) = (x −2)2 −7\n3. Draw a sketch of each of the following graphs:\na) y = ax2 + bx + c if a > 0, b > 0, c < 0.\nb) y = ax2 + bx + c if a < 0, b = 0, c > 0.\nc) y = ax2 + bx + c if a < 0, b < 0, b2 −4ac < 0.\nd) y = (x + p)2 + q if p < 0, q < 0 and the x-intercepts have different signs.\ne) y = a(x + p)2 + q if a < 0, p < 0, q > 0 and one root is zero.\nf) y = a(x + p)2 + q if a > 0, p = 0, b2 −4ac > 0.\n4. Determine the new equation (in the form y = ax2 + bx + c) if:\na) y = 2x2 + 4x + 2 is shifted 3 units to the left.\nb) y = −(x + 1)2 is shifted 1 unit up.\nc) y = 3(x −1)2 + 2\n\u0000x −1\n2\n\u0001\nis shifted 2 units to the right.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QP\n1b. 22QQ\n1c. 22QR\n1d. 22QS\n2. 22QT\n3a. 22QV\n3b. 22QW\n3c. 22QX\n3d. 22QY\n3e. 22QZ\n3f. 22R2\n4a. 22R3\n4b. 22R4\n4c. 22R5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n161\nChapter 5.\nFunctions\n\nFinding the equation of a parabola from the graph\nIf the intercepts are given, use y = a(x −x1)(x −x2).\nExample:\ny\nx\n0\n(0; 2)\n(4; 0)\n(−1; 0)\nx-intercepts: (−1; 0) and (4; 0)\ny = a(x −x1)(x −x2)\n= a(x + 1)(x −4)\n= ax2 −3ax −4a\ny-intercept: (0; 2)\n−4a = 2\na = −1\n2\nEquation of the parabola:\ny = ax2 −3ax −4a\n= −1\n2x2 −3\n\u0012\n−1\n2\n\u0013\nx −4\n\u0012\n−1\n2\n\u0013\n= −1\n2x2 + 3\n2x + 2\nIf the x-intercepts and another point are given, use y = a(x −x1)(x −x2).\nExample:\nb\ny\nx\n0\n(−1; 12)\n(1; 0)\n(5; 0)\nx-intercepts: (1; 0) and (5; 0)\ny = a(x −x1)(x −x2)\n= a(x −1)(x −5)\n= ax2 −6ax + 5a\nSubstitute the point: (−1; 12)\n12 = a(−1)2 −6a(−1) + 5a\n12 = a + 6a + 5a\n12 = 12a\n1 = a\nEquation of the parabola:\ny = ax2 −6ax + 5a\n= x2 −6x + 5\nIf the turning point and another point are given, use y = a(x + p)2 + q.\nExample:\nb\nb\ny\nx\n0\n(1; 5)\n(−3; 1)\nTurning point: (−3; 1)\ny = a(x + p)2 + q\n= a(x + 3)2 + 1\n= ax2 + 6ax + 9a + 1\nSubstitute the point: (1; 5)\n5 = a(1)2 + 6a(1) + 9a + 1\n4 = 16a\n1\n4 = a\nEquation of the parabola:\ny = 1\n4(x + 3)2 + 1\n162\n5.1.\nQuadratic functions\n\nExercise 5 – 7: Finding the equation\nDetermine the equations of the following graphs. Write your answers in the form\ny = a(x + p)2 + q.\n1.\nb\nb\ny\nx\n0\n3\n(−1; 6)\n2.\nb\ny\nx\n0\n(−1; 3)\n5\n3.\nb\nb\ny\nx\n0\n(1; 6)\n−2\n4.\nb\nb\nb\ny\nx\n0\n(1; 6)\n(3; 4)\n4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22R6\n2. 22R7\n3. 22R8\n4. 22R9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n163\nChapter 5.\nFunctions\n\n5.2\nAverage gradient\nEMBGN\nWe notice that the gradient of a curve changes at every point on the curve, therefore\nwe need to work with the average gradient. The average gradient between any two\npoints on a curve is the gradient of the straight line passing through the two points.\ny\nx\n0\nb\nb\nA(−3; 7)\nC(−1; −1)\nFor the diagram above, the gradient of the line AC is\nGradient = yA −yC\nxA −xC\n= 7 −(−1)\n−3 −(−1)\n= 8\n−2\n= −4\nThis is the average gradient of the curve between the points A and C.\nWhat happens to the gradient if we fix the position of one point and move the second\npoint closer to the fixed point?\nSee video: 22RB at www.everythingmaths.co.za\nInvestigation: Gradient at a single point on a curve\nThe curve shown here is defined by y = −2x2 −5.\nPoint B is fixed at (0; −5) and the position of point\nA varies.\nComplete the table below by calculating the y-\ncoordinates of point A for the given x-coordinates\nand then calculating the average gradient between\npoints A and B.\ny\nx\n0\nb\nb\nA\nB(0; −5)\n164\n5.2.\nAverage gradient\n\nxA\nyA\nAverage gradient\n−2\n−1,5\n−1\n−0,5\n0\n0,5\n1\n1,5\n2\n1. What happens to the average gradient as A moves towards B?\n2. What happens to the average gradient as A moves away from B?\n3. What is the average gradient when A overlaps with B?\nIn the example above, the gradient of the straight line that passes through points A and\nC changes as A moves closer to C. At the point where A and C overlap, the straight\nline only passes through one point on the curve. This line is known as a tangent to the\ncurve.\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb A\nC\ny\nx\n0\nb\nb\nA\nC\nWe therefore introduce the idea of the gradient at a single point on a curve. The\ngradient at a point on a curve is the gradient of the tangent to the curve at the given\npoint.\n165\nChapter 5.\nFunctions\n\nWorked example 7: Average gradient\nQUESTION\ny\nx\n0\nb\nb\nP(a; g(a))\nQ(a + h; g(a + h))\ng(x) = x2\n1. Find the average gradient between two points P (a; g(a)) and Q (a + h; g(a + h))\non a curve g(x) = x2.\n2. Determine the average gradient between P (2; g(2)) and Q (5; g(5)).\n3. Explain what happens to the average gradient if Q moves closer to P.\nSOLUTION\nStep 1: Assign labels to the x-values for the given points\nx1 = a\nx2 = a + h\nStep 2: Determine the corresponding y-coordinates\nUsing the function g(x) = x2, we can determine:\ny1 = g(a)\n= a2\ny2 = g(a + h)\n= (a + h)2\n= a2 + 2ah + h2\n166\n5.2.\nAverage gradient\n\nStep 3: Calculate the average gradient\ny2 −y1\nx2 −x1\n=\n\u0000a2 + 2ah + h2\u0001\n−\n\u0000a2\u0001\n(a + h) −(a)\n= a2 + 2ah + h2 −a2\na + h −a\n= 2ah + h2\nh\n= h(2a + h)\nh\n= 2a + h\nThe average gradient between P (a; g(a)) and Q (a + h; g(a + h)) on the curve g(x) =\nx2 is 2a + h.\nStep 4: Calculate the average gradient between P (2; g(2)) and Q (5; g(5))\nThe x-coordinate of P is a and the x-coordinate of Q is a+h therefore if we know that\na = 2 and a + h = 5, then h = 3.\nThe average gradient is therefore 2a + h = 2 (2) + (3) = 7\nStep 5: When Q moves closer to P\nWhen point Q moves closer to point P, h gets smaller.\nWhen the point Q overlaps with the point P, h = 0 and the gradient is given by 2a.\nWe can write the equation for average gradient in another form. Given a curve f(x)\nwith two points P and Q with P (a; f(a)) and Q (a + h; f(a + h)). The average gradi-\nent between P and Q is:\nAverage gradient = yQ −yP\nxQ −xP\n= f(a + h) −f(a)\n(a + h) −(a)\n= f(a + h) −f(a)\nh\nThis result is important for calculating the gradient at a point on a curve and will be\nexplored in greater detail in Grade 12.\n167\nChapter 5.\nFunctions\n\nWorked example 8: Average gradient\nQUESTION\nGiven f(x) = −2x2.\n1. Draw a sketch of the function and determine the average gradient between the\npoints A, where x = 1, and B, where x = 3.\n2. Determine the gradient of the curve at point A.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that a < 0, therefore the graph is a “frown” and has a\nmaximum turning point. We also see that when x = 0, y = 0, therefore the graph\npasses through the origin.\nStep 2: Draw a rough sketch\ny = −2x2\nx\ny\n0\nb\nb\nA\nB\nStep 3: Calculate the average gradient between A and B\nAverage gradient = f(3) −f(1)\n3 −1\n= −2(3)2 −(−2(1)2)\n2\n= −18 + 2\n2\n= −16\n2\n= −8\n168\n5.2.\nAverage gradient\n\nStep 4: Calculate the average gradient for f(x)\nAverage gradient = f(a + h) −f(a)\n(a + h) −a\n= −2(a + h)2 −(−2a2)\nh\n= −2a2 −4ah −2h2 + 2a2\nh\n= −4ah −2h2\nh\n= h(−4a −2h)\nh\n= −4a −2h\nAt point A, h = 0 and a = 1. Therefore\nAverage gradient = −4a −2h\n= −4(1) −2(0)\n= −4\nExercise 5 – 8:\n1.\na) Determine the average gradient of the curve f(x) = x (x + 3) between\nx = 5 and x = 3.\nb) Hence, state what you can deduce about the function f between x = 5 and\nx = 3.\n2. A (1; 3) is a point on f(x) = 3x2.\na) Draw a sketch of f(x) and label point A.\nb) Determine the gradient of the curve at point A.\nc) Determine the equation of the tangent line at A.\n3. Given: g(x) = −x2 + 1.\na) Draw a sketch of g(x).\nb) Determine the average gradient of the curve between x = −2 and x = 1.\nc) Determine the gradient of g at x = 2.\nd) Determine the gradient of g at x = 0.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RC\n2. 22RD\n3. 22RF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n169\nChapter 5.\nFunctions\n\n5.3\nHyperbolic functions\nEMBGP\nRevision\nEMBGQ\nFunctions of the form y = a\nx + q\nFunctions of the general form y = a\nx + q are called hyperbolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = a\nx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted vertically upwards by q units.\n– For q < 0, f(x) is shifted vertically downwards by q units.\n– The horizontal asymptote is the line y = q.\n– The vertical asymptote is the y-axis, the line x = 0.\n• The effect of a on shape and quad-\nrants\n– For a > 0, f(x) lies in the first\nand third quadrants.\n– For a > 1, f(x) will be further\naway from both axes than y =\n1\nx.\n– For 0 < a < 1, as a tends to\n0, f(x) moves closer to the axes\nthan y = 1\nx.\n– For a < 0, f(x) lies in the sec-\nond and fourth quadrants.\n– For a < −1, f(x) will be further\naway from both axes than y =\n−1\nx.\n– For −1 < a < 0, as a tends to\n0, f(x) moves closer to the axes\nthan y = −1\nx.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\nExercise 5 – 9: Revision\n1. Consider the following hyperbolic functions:\n• y1 = 1\nx\n• y2 = −4\nx\n• y3 = 4\nx −2\n• y4 = −4\nx + 1\n170\n5.3.\nHyperbolic functions\n\nComplete the table to summarise the properties of the hyperbolic function:\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nlies in I and III quad\nasymptotes\ny-axis, x = 0\nx-axis, y = 0\naxes of symmetry\ny = x\ny = −x\ndomain\n{x : x ∈R, x ̸= 0}\nrange\n{y : y ∈R, y ̸= 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22RH at www.everythingmaths.co.za\nFunctions of the form y =\na\nx+p + q\nEMBGR\nWe now consider hyperbolic functions of the form y =\na\nx+p + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a hyperbolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 1\nx\nb) y2 =\n1\nx−2\nc) y3 =\n1\nx−1\nd) y4 =\n1\nx+1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nintercept(s)\nasymptotes\naxes of symmetry\ndomain\nrange\neffect of p\n171\nChapter 5.\nFunctions\n\n2. Complete the following sentences for functions of the form y =\na\nx+p + q:\na) A change in p causes a . . . . . . shift.\nb) If the value of p increases, the graph and the vertical asymptote . . . . . .\nc) If the value of q changes, then the . . . . . . asymptote of the hyperbola will\nshift.\nd) If the value of p decreases, the graph and the vertical asymptote . . . . . .\nThe effect of the parameters on y =\na\nx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects the vertical asymptote, the line x = −p.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position on the Cartesian plane.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n172\n5.3.\nHyperbolic functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y =\na\nx+p + q:\nDomain and range\nThe domain is {x : x ∈R, x ̸= −p}. If x = −p, the dominator is equal to zero and the\nfunction is undefined.\nWe see that\ny =\na\nx + p + q\ncan be re-written as:\ny −q =\na\nx + p\nIf x ̸= −p then:\n(y −q) (x + p) = a\nx + p =\na\ny −q\nThe range is therefore {y : y ∈R, y ̸= q}.\nThese restrictions on the domain and range determine the vertical asymptote x = −p\nand the horizontal asymptote y = q.\nWorked example 9: Domain and range\nQUESTION\nDetermine the domain and range for g(x) =\n2\nx+1 + 2.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R, x ̸= −1} since g(x) is undefined for x = −1.\nStep 2: Determine the range\nLet g(x) = y:\ny =\n2\nx + 1 + 2\ny −2 =\n2\nx + 1\n(y −2)(x + 1) = 2\nx + 1 =\n2\ny −2\nTherefore the range is {g(x) : g(x) ∈R, g(x) ̸= 2}.\n173\nChapter 5.\nFunctions\n\nExercise 5 – 10: Domain and range\nDetermine the domain and range for each of the following functions:\n1. y = 1\nx + 1\n2. g(x) =\n8\nx−8 + 4\n3. y = −\n4\nx+1 −3\n4. x =\n2\n3−y + 5\n5. (y −2)(x + 2) = 3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RJ\n2. 22RK\n3. 22RM\n4. 22RN\n5. 22RP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n2\nx+1+2\nis determined by setting x = 0:\ng(x) =\n2\nx + 1 + 2\ng(0) =\n2\n0 + 1 + 2\n= 2 + 2\n= 4\nThis gives the point (0; 4).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n2\nx+1+2\nis determined by setting y = 0:\ng(x) =\n2\nx + 1 + 2\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\n174\n5.3.\nHyperbolic functions\n\nExercise 5 – 11: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) =\n1\nx+4 −2\n2. g(x) = −5\nx + 2\n3. j(x) =\n2\nx−1 + 3\n4. h(x) =\n3\n6−x + 1\n5. k(x) =\n5\nx+2 −1\n2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RQ\n2. 22RR\n3. 22RS\n4. 22RT\n5. 22RV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptotes\nThere are two asymptotes for functions of the form y =\na\nx+p + q. The asymptotes\nindicate the values of x for which the function does not exist. In other words, the\nvalues that are excluded from the domain and the range. The horizontal asymptote is\nthe line y = q and the vertical asymptote is the line x = −p.\nExercise 5 – 12: Asymptotes\nDetermine the asymptotes for each of the following functions:\n1. y =\n1\nx+4 −2\n2. y = −5\nx\n3. y =\n3\n2−x + 1\n4. y = 1\nx −8\n5. y = −\n2\nx−2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RW\n2. 22RX\n3. 22RY\n4. 22RZ\n5. 22S2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxes of symmetry\nThere are two lines about which a hyperbola is symmetrical.\nFor the standard hyperbola y = 1\nx, we see that if we replace x ⇒y and y ⇒x, we get\ny = 1\nx. Similarly, if we replace x ⇒−y and y ⇒−x, the function remains the same.\nTherefore the function is symmetrical about the lines y = x and y = −x.\nFor the shifted hyperbola y =\na\nx+p + q, the axes of symmetry intersect at the point\n(−p; q).\n175\nChapter 5.\nFunctions\n\nTo determine the axes of symmetry we define the two straight lines y1 = m1x + c1 and\ny2 = m2x + c2. For the standard and shifted hyperbolic function, the gradient of one\nof the lines of symmetry is 1 and the gradient of the other line of symmetry is −1. The\naxes of symmetry are perpendicular to each other and the product of their gradients\nequals −1. Therefore we let y1 = x+c1 and y2 = −x+c2. We then substitute (−p; q),\nthe point of intersection of the axes of symmetry, into both equations to determine the\nvalues of c1 and c2.\nWorked example 10: Axes of symmetry\nQUESTION\nDetermine the axes of symmetry for y =\n2\nx+1 −2.\nSOLUTION\nStep 1: Determine the point of intersection (−p; q)\nFrom the equation we see that p = 1 and q = −2. So the axes of symmetry will\nintersect at (−1; −2).\nStep 2: Define two straight line equations\ny1 = x + c1\ny2 = −x + c2\nStep 3: Solve for c1 and c2\nUse (−1; −2) to solve for c1:\ny1 = x + c1\n−2 = −1 + c1\n−1 = c1\nUse (−1; −2) to solve for c2:\ny2 = −x + c2\n−2 = −(−1) + c2\n−3 = c2\nStep 4: Write the final answer\nThe axes of symmetry for y =\n2\nx+1 −2 are the lines\ny1 = x −1\ny2 = −x −3\n176\n5.3.\nHyperbolic functions\n\n1\n2\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny1 = x −1\ny2 = −x −3\nExercise 5 – 13: Axes of symmetry\n1. Complete the following for f(x) and g(x):\n• Sketch the graph.\n• Determine (−p; q).\n• Find the axes of symmetry.\nCompare f(x) and g(x) and also their axes of symmetry. What do you notice?\na) f(x) = 2\nx\ng(x) = 2\nx + 1\nb) f(x) = −3\nx\ng(x) = −\n3\nx+1\nc) f(x) = 5\nx\ng(x) =\n5\nx−1 −1\n2. A hyperbola of the form k(x) =\na\nx+p + q passes through the point (4; 3). If the\naxes of symmetry intersect at (−1; 2), determine the equation of k(x).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S3\n1b. 22S4\n1c. 22S5\n2. 22S6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n177\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) =\na\nx+p + q\nIn order to sketch graphs of functions of the form, f(x) =\na\nx+p +q, we need to calculate\nfive characteristics:\n• quadrants\n• asymptotes\n• y-intercept\n• x-intercept\n• domain and range\nWorked example 11: Sketching a hyperbola\nQUESTION\nSketch the graph of y =\n2\nx+1 + 2. Determine the intercepts, asymptotes and axes of\nsymmetry. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Determine the asymptotes\nFrom the equation we know that p = 1 and q = 2.\nTherefore the horizontal asymptote is the line y = 2 and the vertical asymptote is the\nline x = −1.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n2\n0 + 1 + 2\n= 4\nThis gives the point (0; 4).\nStep 4: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n178\n5.3.\nHyperbolic functions\n\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\nStep 5: Determine the axes of symmetry\nUsing (−1; 2) to solve for c1:\ny1 = x + c1\n2 = −1 + c1\n3 = c1\ny2 = −x + c2\n2 = −(−1) + c2\n1 = c2\nTherefore the axes of symmetry are y = x + 3 and y = −x + 1.\nStep 6: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny = x + 3\ny = −x + 1\nStep 7: State the domain and range\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 2}\n179\nChapter 5.\nFunctions\n\nWorked example 12: Sketching a hyperbola\nQUESTION\nUse horizontal and vertical shifts to sketch the graph of f(x) =\n1\nx−2 + 3.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Sketch the standard hyperbola y = 1\nx\nStart with a sketch of the standard hyperbola g(x) = 1\nx. The vertical asymptote is x = 0\nand the horizontal asymptote is y = 0.\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 3: Determine the vertical shift\nFrom the equation we see that q = 3, which means g(x) must shifted 3 units up. The\nhorizontal asymptote is also shifted 3 units up to y = 3 .\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 4: Determine the horizontal shift\nFrom the equation we see that p = −2, which means g(x) must shifted 2 units to the\nright. The vertical asymptote is also shifted 2 units to the right.\n180\n5.3.\nHyperbolic functions\n\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n5\n6\n−1\n−2\ny\nx\n0\nStep 5: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n1\n0 −2 + 3\n= 21\n2\nThis gives the point (0; 21\n2).\nStep 6: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n1\nx −2 + 3\n−3 =\n1\nx −2\n−3(x −2) = 1\n−3x + 6 = 1\n−3x = −5\nx = 5\n3\nThis gives the point (5\n3; 0).\nStep 7: Determine the domain and range\nDomain: {x : x ∈R, x ̸= 2}\nRange: {y : y ∈R, y ̸= 3}\n181\nChapter 5.\nFunctions\n\nWorked example 13: Finding the equation of a hyperbola from the graph\nQUESTION\nUse the graph below to determine the values of a, p and q for y =\na\nx+p + q.\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\nSOLUTION\nStep 1: Examine the graph and deduce the sign of a\nWe notice that the graph lies in the second and fourth quadrants, therefore a < 0.\nStep 2: Determine the asymptotes\nFrom the graph we see that the vertical asymptote is x = −1, therefore p = 1. The\nhorizontal asymptote is y = 3, and therefore q = 3.\ny =\na\nx + 1 + 3\nStep 3: Determine the value of a\nTo determine the value of a we substitute a point on the graph, namely (0; 0):\ny =\na\nx + 1 + 3\n0 =\na\n0 + 1 + 3\n∴−3 = a\nStep 4: Write the final answer\ny = −\n3\nx + 1 + 3\n182\n5.3.\nHyperbolic functions\n\nExercise 5 – 14: Sketching graphs\n1. Draw the graphs of the following functions and indicate:\n• asymptotes\n• intercepts, where applicable\n• axes of symmetry\n• domain and range\na) y = 1\nx + 2\nb) y =\n1\nx+4 −2\nc) y = −\n1\nx+1 + 3\nd) y = −\n5\nx−2 1\n2 −2\ne) y =\n8\nx−8 + 4\n2. Given the graph of the hyperbola of the form y =\n1\nx+p + q, determine the values\nof p and q.\ny\nx\n−2\n−1\n3. Given a sketch of the function of the form y =\na\nx+p + q, determine the values of\na, p and q.\ny\nx\n2\n2\n4.\na) Draw the graph of f(x) = −3\nx, x > 0.\nb) Determine the average gradient of the graph between x = 1 and x = 3.\nc) Is the gradient at (1\n2; −6) less than or greater than the average gradient be-\ntween x = 1 and x = 3? Illustrate this on your graph.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S7\n1b. 22S8\n1c. 22S9\n1d. 22SB\n1e. 22SC\n2. 22SD\n3. 22SF\n4. 22SG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n183\nChapter 5.\nFunctions\n\n5.4\nExponential functions\nEMBGS\nRevision\nEMBGT\nFunctions of the form y = abx + q\nFunctions of the general form y = abx + q, for b > 0, are called exponential functions,\nwhere a, b and q are constants.\nThe effects of a, b and q on f(x) = abx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\n– The horizontal asymptote is the\nline y = q.\n• The effect of a on shape\n– For a > 0, f(x) is increasing.\n– For a < 0, f(x) is decreasing.\nThe graph is reflected about the\nhorizontal asymptote.\n• The effect of b on direction\nAssuming a > 0:\n– If b > 1, f(x) is an increasing\nfunction.\n– If 0 < b < 1, f(x) is a decreas-\ning function.\n– If b ≤0, f(x) is not defined.\nb > 1\na < 0\na > 0\nq > 0\nq < 0\n0 < b < 1\na < 0\na > 0\nq > 0\nq < 0\nExercise 5 – 15: Revision\n1. On separate axes, accurately draw each of the following functions:\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = 3x\nb) y2 = −2 × 3x\nc) y3 = 2 × 3x + 1\nd) y4 = 3x −2\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\n184\n5.4.\nExponential functions\n\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nincreasing\nasymptote\nx-axis, y = 0\ndomain\n{x : x ∈R}\nrange\n{y : y ∈R, y > 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22SH\n1b. 22SJ\n1c. 22SK\n1d. 22SM\n2. 22SN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = ab(x+p) + q\nEMBGV\nWe now consider exponential functions of the form y = ab(x+p) + q and the effects of\nparameter p.\nSee video: 22SP at www.everythingmaths.co.za\nInvestigation: The effects of a, p and q on an exponential graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 2x\nb) y2 = 2(x−2)\nc) y3 = 2(x−1)\nd) y4 = 2(x+1)\ne) y5 = 2(x+2)\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptote\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = 2(x−1) + 2\n185\nChapter 5.\nFunctions\n\nb) y2 = 3 × 2(x−1) + 2\nc) y3 = 1\n2 × 2(x−1) + 2\nd) y4 = 0 × 2(x−1) + 2\ne) y5 = −3 × 2(x−1) + 2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptotes\ndomain\nrange\neffect of a\nThe effect of the parameters on y = abx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position relative to the horizontal\nasymptote.\n• For a > 0, the graph lies above the horizontal asymptote, y = q.\n• For a < 0, the graph lies below the horizontal asymptote, y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n186\n5.4.\nExponential functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y = ab(x+p) + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative.\nIf a > 0 we have:\nb(x+p) > 0\nab(x+p) > 0\nab(x+p) + q > q\nf(x) > q\nThe range is therefore {y : y > q, y ∈R}.\nSimilarly, if a < 0, the range is {y : y < q, y ∈R}.\nWorked example 14: Domain and range\nQUESTION\nState the domain and range for g(x) = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n3(x+1) > 0\n5 × 3(x+1) > 0\n5 × 3(x+1) −1 > −1\n∴g(x) > −1\nTherefore the range is {g(x) : g(x) > −1} or in interval notation (−1; ∞).\n187\nChapter 5.\nFunctions\n\nExercise 5 – 16: Domain and range\nGive the domain and range for each of the following functions:\n1. y =\n\u0000 3\n2\n\u0001(x+3)\n2. f(x) = −5(x−2) + 1\n3. y + 3 = 2(x+1)\n4. y = n + 3(x−m)\n5.\ny\n2 = 3(x−1) −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SQ\n2. 22SR\n3. 22SS\n4. 22ST\n5. 22SV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting x = 0:\ng(0) = 3 × 2(0+1) + 2\n= 3 × 2 + 2\n= 8\nThis gives the point (0; 8).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting y = 0:\n0 = 3 × 2(x+1) + 2\n−2 = 3 × 2(x+1)\n−2\n3 = 2(x+1)\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n188\n5.4.\nExponential functions\n\nExercise 5 – 17: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = 2(x+1) −8\n2. y = 2 × 3(x−1) −18\n3. y + 5(x+2) = 5\n4. y = 1\n2\n\u0000 3\n2\n\u0001(x+3) −0,75\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SW\n2. 22SX\n3. 22SY\n4. 22SZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptote\nExponential functions of the form y = ab(x+p) + q have a horizontal asymptote, the\nline y = q.\nWorked example 15: Asymptote\nQUESTION\nDetermine the asymptote for y = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the asymptote\nThe asymptote of g(x) can be calculated as:\n3(x+1) ̸= 0\n5 × 3(x+1) ̸= 0\n5 × 3(x+1) −1 ̸= −1\n∴y ̸= −1\nTherefore the asymptote is the line y = −1.\n189\nChapter 5.\nFunctions\n\nExercise 5 – 18: Asymptote\nGive the asymptote for each of the following functions:\n1. y = −5(x+1)\n2. y = 3(x−2) + 1\n3.\n\u0010\n3y\n2\n\u0011\n= 5(x+3) −1\n4. y = 7(x+1) −2\n5.\ny\n2 + 1 = 3(x+2)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T2\n2. 22T3\n3. 22T4\n4. 22T5\n5. 22T6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching graphs of the form f(x) = ab(x+p) + q\nIn order to sketch graphs of functions of the form, f(x) = ab(x+p) + q, we need to\ndetermine five characteristics:\n• shape\n• y-intercept\n• x-intercept\n• asymptote\n• domain and range\nWorked example 16: Sketching an exponential graph\nQUESTION\nSketch the graph of 2y = 10 × 2(x+1) −5.\nMark the intercept(s) and asymptote. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nWe notice that a > 0 and b > 1, therefore the function is increasing.\n190\n5.4.\nExponential functions\n\nStep 2: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\n2y = 10 × 2(0+1) −5\n= 10 × 2 −5\n= 15\n∴y = 71\n2\nThis gives the point (0; 71\n2).\nStep 3: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 = 10 × 2(x+1) −5\n5 = 10 × 2(x+1)\n1\n2 = 2(x+1)\n2−1 = 2(x+1)\n∴−1 = x + 1\n(same base)\n−2 = x\nThis gives the point (−2; 0).\nStep 4: Determine the asymptote\nThe horizontal asymptote is the line y = −5\n2.\nStep 5: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\n−3\n1\n2\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y > −5\n2, y ∈R}\n191\nChapter 5.\nFunctions\n\nWorked example 17: Finding the equation of an exponential function from a\ngraph\nQUESTION\nUse the given graph of y = −2 × 3(x+p) + q to determine the values of p and q.\n1\n2\n3\n4\n5\n6\n7\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nFrom the graph we see that the function is decreasing. We also note that a = −2 and\nb = 3. We need to solve for p and q.\nStep 2: Use the asymptote to determine q\nThe horizontal asymptote y = 6 is given, therefore we know that q = 6.\ny = −2 × 3(x+p) + 6\nStep 3: Use the x-intercept to determine p\nSubstitute (2; 0) into the equation and solve for p:\ny = −2 × 3(x+p) + 6\n0 = −2 × 3(2+p) + 6\n−6 = −2 × 3(2+p)\n3 = 3(2+p)\n∴1 = 2 + p\n(same base)\n∴p = −1\nStep 4: Write the final answer\ny = −2 × 3(x−1) + 6\n192\n5.4.\nExponential functions\n\nExercise 5 – 19: Mixed exercises\n1. Given the graph of the hyperbola of the form h(x) = k\nx, x < 0, which passes\nthough the point A(−1\n2; −6).\nb\ny\nx\n0\nA(−1\n2; −6)\na) Show that k = 3.\nb) Write down the equation for the new function formed if h(x):\ni. is shifted 3 units vertically upwards\nii. is shifted to the right by 3 units\niii. is reflected about the y-axis\niv. is shifted so that the asymptotes are x = 0 and y = −1\n4\nv. is shifted upwards to pass through the point (−1; 1)\nvi. is shifted to the left by 2 units and 1 unit vertically downwards (for\nx < 0)\n2. Given the graphs of f(x) = a(x + p)2 and g(x) = a\nx.\nThe axis of symmetry for f(x) is x = −1 and f(x) and g(x) intersect at point M.\nThe line y = 2 also passes through M.\nb\ny\nx\n0\nM\n−1\n2\nf\ng\n193\nChapter 5.\nFunctions\n\nDetermine:\na) the coordinates of M\nb) the equation of g(x)\nc) the equation of f(x)\nd) the values for which f(x) < g(x)\ne) the range of f(x)\n3. On the same system of axes, sketch:\na) the graphs of k(x) = 2(x + 1\n2)2 −41\n2 and h(x) = 2(x+ 1\n2 ). Determine all\nintercepts, turning point(s) and asymptotes.\nb) the reflection of h(x) about the x-axis. Label this function as j(x).\n4. Sketch the graphs of y = ax2 + bx + c for:\na) a < 0, b > 0, b2 < 4ac\nb) a > 0, b > 0, one root = 0\n5. On separate systems of axes, sketch the graphs:\ny =\n2\nx−2\ny = 2\nx −2\ny = −2(x−2)\n6. For the diagrams shown below, determine:\n• the equations of the functions; f(x) = a(x + p)2 + q, g(x) = ax2 + q,\nh(x) = a\nx, x < 0 and k(x) = bx + q\n• the axes of symmetry of each function\n• the domain and range of each function\na)\nb\ny\nx\n0\n(2; 3)\nf\n194\n5.4.\nExponential functions\n\nb)\nb\ny\nx\n0\n(−2; −1)\ng\nh\n−2\nc)\ny\nx\n0\nk\ny = 2x + 1\n2\n7. Given the graph of the function Q(x) = ax.\nb\nb\nb\ny\nx\n0\nQ = ax\n(−2; p)\n1\n(1; 1\n3)\na) Show that a = 1\n3.\nb) Find the value of p if the point (−2; p) is on Q.\nc) Calculate the average gradient of the curve between x = −2 and x = 1.\nd) Determine the equation of the new function formed if Q is shifted 2 units\nvertically downwards and 2 units to the left.\n8. Find the equation for each of the functions shown below:\na) f(x) = 2x + q\ng(x) = mx + c\n195\nChapter 5.\nFunctions\n\nb\ny\nx\n0\nf\ng\n−1\n2\n−2\nb) h(x) =\nk\nx+p + q\nb\nb\ny\nx\n0\n1\n−2\n−1\n2\nh\n9. Given: the graph of k(x) = −x2 + 3x + 10 with turning point at D. The graph of\nthe straight line h(x) = mx + c passing through points B and C is also shown.\nb\nb\nb\ny\nx\n0\nB\nA\nE\nF\nD\nC\nk\nh\nDetermine:\na) the lengths AO, OB, OC and DE\nb) the equation of DE\nc) the equation of h(x)\nd) the x-values for which k(x) < 0\n196\n5.4.\nExponential functions\n\ne) the x-values for which k(x) ≥h(x)\nf) the length of DF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T7\n2. 22T8\n3. 22T9\n4. 22TB\n5. 22TC\n6a. 22TD\n6b. 22TF\n6c. 22TG\n7. 22TH\n8a. 22TJ\n8b. 22TK\n9. 22TM\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIMPORTANT: Trigonometric functions are examined in PAPER 2.\n5.5\nThe sine function\nEMBGW\nRevision\nEMBGX\nFunctions of the form y = sin θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• Period of one complete wave is 360◦.\n• Amplitude is the maximum height of the wave above and below the x-axis and\nis always positive. Amplitude = 1.\n• Domain: [0◦; 360◦]\nFor y = sin θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Maximum turning point: (90◦; 1)\n• Minimum turning point: (270◦; −1)\n197\nChapter 5.\nFunctions\n\nFunctions of the form y = a sin θ + q\nThe effects of a and q on f(θ) = a sin θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 20: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦.\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function also determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n198\n5.5.\nThe sine function\n\n1. y1 = sin θ\n2. y2 = −2 sin θ\n3. y3 = sin θ + 1\n4. y4 = 1\n2 sin θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TN\n2. 22TP\n3. 22TQ\n4. 22TR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin kθ\nEMBGY\nWe now consider cosine functions of the form y = sin kθ and the effects of k.\nInvestigation: The effects of k on a sine graph\n1. Complete the following table for y1 = sin θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−270◦\n−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n2. Use the table of values to plot the graph of y1 = sin θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = sin(−θ)\nb) y3 = sin 2θ\nc) y4 = sin θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n199\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = sin θ and y2 = sin(−θ)?\n6. Is sin(−θ) = −sin θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = sin kθ?\nThe effect of the parameter on y = sin kθ\nThe value of k affects the period of the sine function. If k is negative, then the graph is\nreflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the sine function decreases.\nFor 0 < k < 1, the period of the sine function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\nsin(−θ) = −sin θ\nCalculating the period:\nTo determine the period of y = sin kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k (this means that k is always considered to be\npositive).\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n200\n5.5.\nThe sine function\n\nWorked example 18: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = sin θ\nb) y2 = sin 3θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin kθ\nNotice that k > 1 for y2 = sin 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\nsin θ\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\nsin 3θ\n2\n1\n0,38\n−0,71\n−0,92\n0\n0,92\n0,71\n−0,38\n−1\nStep 3: Sketch the sine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin 3\n2θ\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin 3θ\n2\nperiod\n360◦\n240◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(90◦; 1)\n(−180◦; 1) and (60◦; 1)\nminimum turning points\n(−90◦; −1)\n(−60◦; −1) and (180◦; 1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\n201\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = sin kθ\n= sin 0◦\n= 0\nThis gives the point (0◦; 0).\nExercise 5 – 21: Sine functions of the form y = sin kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦and for each graph deter-\nmine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = sin 3θ\nb) g(θ) = sin θ\n3\nc) h(θ) = sin(−2θ)\nd) k(θ) = sin 3θ\n4\n2. For each graph of the form f(θ) = sin kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n202\n5.5.\nThe sine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22TS\n1b. 22TT\n1c. 22TV\n1d. 22TW\n2a. 22TX\n2b. 22TY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin(θ + p)\nEMBGZ\nInvestigation: The effects of p on a sine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = sin θ\nb) y2 = sin(θ −90◦)\nc) y3 = sin(θ −60◦)\nd) y4 = sin(θ + 90◦)\ne) y5 = sin(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of p\n203\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = sin(θ + p)\nThe effect of p on the sine function is a horizontal shift, also called a phase shift; the\nentire graph slides to the left or to the right.\n• For p > 0, the graph of the sine function shifts to the left by p.\n• For p < 0, the graph of the sine function shifts to the right by p.\np > 0\np < 0\nWorked example 19: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = sin θ\nb) y2 = sin(θ −30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin(θ + p)\nNotice that for y1 = sin θ we have p = 0 (no phase shift) and for y2 = sin(θ −30◦),\np < 0 therefore the graph shifts to the right by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n0\n1\n0\n−1\n0\n1\n0\n−1\n0\nsin(θ −30◦)\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n204\n5.5.\nThe sine function\n\nStep 3: Sketch the sine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = sin θ\ny2 = sin(θ −30◦)\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin(θ −30◦)\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−270◦; 1) and (90◦; 1)\n(−240◦; 1) and\n(120◦; 1)\nminimum turning points\n(−90◦; −1) and\n(270◦; −1)\n(−60◦; −1) and\n(300◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; −1\n2)\nx-intercept(s)\n(−360◦; 0), (−180◦; 0),\n(0◦; 0), (180◦; 0) and\n(360◦; 0)\n(−330◦; 0), (−150◦; 0),\n(30◦; 0) and (210◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\n205\nChapter 5.\nFunctions\n\nExercise 5 – 22: Sine functions of the form y = sin(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = sin(θ + 30◦)\n2. g(θ) = sin(θ −45◦)\n3. h(θ) = sin(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TZ\n2. 22V2\n3. 22V3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching sine graphs\nEMBH2\nWorked example 20: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(45◦−θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin(θ + p).\nf(θ) = sin(45◦−θ)\n= sin(−θ + 45◦)\n= sin (−(θ −45◦))\n= −sin(θ −45◦)\nTo draw a graph of the above function, we know that the standard sine graph, y = sin θ,\n206\n5.5.\nThe sine function\n\nmust:\n• be reflected about the x-axis\n• be shifted to the right by 45◦\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n0,71\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = −sin(θ −45◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 360◦\nAmplitude: 1\nDomain: [−360◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (315◦; 1)\nMinimum turning point: (135◦; −1)\ny-intercepts: (0◦; 0,71)\nx-intercept: (45◦; 0) and (225◦; 0)\nWorked example 21: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(3θ + 60◦) for 0◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin k(θ + p).\nf(θ) = sin(3θ + 60◦)\n= sin 3(θ + 20◦)\n207\nChapter 5.\nFunctions\n\nTo draw a graph of the above equation, the standard sine graph, y = sin θ, must be\nchanged in the following ways:\n• decrease the period by a factor of 3;\n• shift to the left by 20◦.\nStep 2: Complete a table of values\nθ\n0◦\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nf(θ)\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nθ\nf(θ)\nf(θ) = sin 3(θ + 20◦)\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 180◦]\nRange: [−1; 1]\nMaximum turning point: (10◦; 1) and (130◦; 1)\nMinimum turning point: (70◦; −1)\ny-intercept: (0◦; 0,87)\nx-intercepts: (40◦; 0), (100◦; 0) and (160◦; 0)\nExercise 5 – 23: The sine function\n1. Sketch the following graphs on separate axes:\na) y = 2 sin θ\n2 for −360◦≤θ ≤360◦\nb) f(θ) = 1\n2 sin(θ −45◦) for −90◦≤θ ≤90◦\nc) y = sin(θ + 90◦) + 1 for 0◦≤θ ≤360◦\nd) y = sin(−3θ\n2 ) for −180◦≤θ ≤180◦\ne) y = sin(30◦−θ) for −360◦≤θ ≤360◦\n2. Given the graph of the function y = a sin(θ + p), determine the values of a and\np.\n208\n5.5.\nThe sine function\n\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nθ\nf(θ)\nCan you describe this graph in terms of cos θ?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22V4\n1b. 22V5\n1c. 22V6\n1d. 22V7\n1e. 22V8\n2. 22V9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n5.6\nThe cosine function\nEMBH3\nRevision\nEMBH4\nFunctions of the form y = cos θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• The period is 360◦and the amplitude is 1.\n• Domain: [0◦; 360◦]\nFor y = cos θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (90◦; 0), (270◦; 0)\n• y-intercept: (0◦; 1)\n• Maximum turning points: (0◦; 1), (360◦; 1)\n• Minimum turning point: (180◦; −1)\n209\nChapter 5.\nFunctions\n\nFunctions of the form y = a cos θ + q\nCosine functions of the general form y = a cos θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a cos θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 24: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function in the previous problem determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n210\n5.6.\nThe cosine function\n\n1. y1 = cos θ\n2. y2 = −3 cos θ\n3. y3 = cos θ + 2\n4. y4 = 1\n2 cos θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VB\n2. 22VC\n3. 22VD\n4. 22VF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos(kθ)\nEMBH5\nWe now consider cosine functions of the form y = cos kθ and the effects of k.\nInvestigation: The effects of k on a cosine graph\n1. Complete the following table for y1 = cos θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ncos θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ncos θ\n2. Use the table of values to plot the graph of y1 = cos θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = cos(−θ)\nb) y3 = cos 3θ\nc) y4 = cos 3θ\n4\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n211\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = cos θ and y2 = cos(−θ)?\n6. Is cos(−θ) = −cos θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = cos kθ?\nThe effect of the parameter k on y = cos kθ\nThe value of k affects the period of the cosine function.\n• For k > 0:\nFor k > 1, the period of the cosine function decreases.\nFor 0 < k < 1, the period of the cosine function increases.\n• For k < 0:\nFor −1 < k < 0, the period increases.\nFor k < −1, the period decreases.\nNegative angles:\ncos(−θ) = cos θ\nNotice that for negative values of θ, the graph is not reflected about the x-axis.\nCalculating the period:\nTo determine the period of y = cos kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k.\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n212\n5.6.\nThe cosine function\n\nWorked example 22: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = cos θ\nb) y2 = cos θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos kθ\nNotice that for y2 = cos θ\n2, k < 1 therefore the period of the graph increases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ncos θ\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\n−0,71\n−1\ncos θ\n2\n0\n0,38\n0,71\n0,92\n1\n0,92\n0,71\n0,38\n0\nStep 3: Sketch the cosine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = cos θ\n2\ny2 = cos θ\nStep 4: Complete the table\ny1 = cos θ\ny2 = cos θ\n2\nperiod\n360◦\n720◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[0; 1]\nmaximum turning points\n(0◦; 1)\n(0◦; 1)\nminimum turning points\n(−180◦; −1) and (180◦; −1)\nnone\ny-intercept(s)\n(0◦; 1)\n(0◦; 1)\nx-intercept(s)\n(−90◦; 0) and (90◦; 0)\n(−180◦; 0) and (180◦; 0)\n213\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos kθ\n= cos 0◦\n= 1\nThis gives the point (0◦; 1).\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = cos 2θ\nb) g(θ) = cos θ\n3\nc) h(θ) = cos(−2θ)\nd) k(θ) = cos 3θ\n4\n2. For each graph of the form f(θ) = cos kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n214\n5.6.\nThe cosine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nbA(135◦; 0)\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VG\n1b. 22VH\n1c. 22VJ\n1d. 22VK\n2a. 22VM\n2b. 22VN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos (θ + p)\nEMBH6\nWe now consider cosine functions of the form y = cos(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a cosine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = cos θ\nb) y2 = cos(θ −90◦)\nc) y3 = cos(θ −60◦)\nd) y4 = cos(θ + 90◦)\ne) y5 = cos(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning points\nminimum turning points\ny-intercept(s)\nx-intercept(s)\neffect of p\n215\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = cos(θ + p)\nThe effect of p on the cosine function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the cosine function shifts to the left by p degrees.\n• For p < 0, the graph of the cosine function shifts to the right by p degrees.\np > 0\np < 0\nWorked example 23: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = cos θ\nb) y2 = cos(θ + 30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos(θ + p)\nNotice that for y1 = cos θ we have p = 0 (no phase shift) and for y2 = cos(θ + 30◦),\np < 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\ncos θ\n1\n0\n−1\n0\n1\n0\n−1\n0\n1\ncos(θ +30◦)\n0,87\n−0,5\n−0,87\n0,5\n0,87\n−0,5\n−0,87\n0,5\n0,87\n216\n5.6.\nThe cosine function\n\nStep 3: Sketch the cosine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = cos θ\ny2 = cos(θ + 30◦)\nStep 4: Complete the table\ny1\ny2\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−360◦; 1), (0◦; 1) and\n(360◦; 1)\n(−30◦; 1) and (330◦; 1)\nminimum turning points\n(−180◦; −1) and\n(180◦; −1)\n(−210◦; −1) and\n(150◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,87)\nx-intercept(s)\n(−270◦; 0), (−90◦; 0),\n(90◦; 0) and (270◦; 0)\n(−300◦; 0), (−120◦; 0),\n(60◦; 0) and (240◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos(θ + p)\n= cos(0◦+ p)\n= cos p\nThis gives the point (0◦; cos p).\n217\nChapter 5.\nFunctions\n\nExercise 5 – 26: Cosine functions of the form y = cos(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = cos(θ + 45◦)\n2. g(θ) = cos(θ −30◦)\n3. h(θ) = cos(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VP\n2. 22VQ\n3. 22VR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching cosine graphs\nEMBH7\nWorked example 24: Sketching a cosine graph\nQUESTION\nSketch the graph of f(θ) = cos(180◦−3θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = cos k(θ + p).\nf(θ) = cos(180◦−3θ)\n= cos(−3θ + 180◦)\n= cos (−3(θ −60◦))\n= cos 3(θ −60◦)\nTo draw a graph of the above function, the standard cosine graph, y = cos θ, must be\nchanged in the following ways:\n218\n5.6.\nThe cosine function\n\n• decrease the period by a factor of 3\n• shift to the right by 60◦.\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n−1\n0,71\n0\n−0,71\n1\n−0,71\n0\n0,71\n−1\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = cos 3(θ −60◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (60◦; 1), (180◦; 1) and (300◦; 1)\nMinimum turning point: (0◦; −1), (120◦; −1), (240◦; −1) and (360◦; −1)\ny-intercepts: (0◦; −1)\nx-intercept: (30◦; 0), (90◦; 0), (150◦; 0), (210◦; 0), (270◦; 0) and (330◦; 0)\nWorked example 25: Finding the equation of a cosine graph\nQUESTION\nGiven the graph of y = a cos(kθ+p), determine the values of a, k, p and the minimum\nturning point.\nθ\ny\ny = a cos(θ + p)\nb\n(45◦; 2)\n−45◦\n315◦\n219\nChapter 5.\nFunctions\n\nSOLUTION\nStep 1: Determine the value of k\nFrom the sketch we see that the period of the graph is 360◦, therefore k = 1.\ny = a cos(θ + p)\nStep 2: Determine the value of a\nFrom the sketch we see that the maximum turning point is (45◦; 2), so we know that\nthe amplitude of the graph is 2 and therefore a = 2.\ny = 2 cos(θ + p)\nStep 3: Determine the value of p\nCompare the given graph with the standard cosine function y = cos θ and notice the\ndifference in the maximum turning points. We see that the given function has been\nshifted to the right by 45◦, therefore p = 45◦.\ny = 2 cos(θ −45◦)\nStep 4: Determine the minimum turning point\nAt the minimum turning point, y = −2:\ny = 2 cos(θ −45◦)\n−2 = 2 cos(θ −45◦)\n−1 = cos(θ −45◦)\ncos−1(−1) = θ −45◦\n180◦= θ −45◦\n225◦= θ\nThis gives the point (225◦; −2).\n220\n5.6.\nThe cosine function\n\nExercise 5 – 27: The cosine function\n1. Sketch the following graphs on separate axes:\na) y = cos(θ + 15◦) for −180◦≤θ ≤180◦\nb) f(θ) = 1\n3 cos(θ −60◦) for −90◦≤θ ≤90◦\nc) y = −2 cos θ for 0◦≤θ ≤360◦\nd) y = cos(30◦−θ) for −360◦≤θ ≤360◦\ne) g(θ) = 1 + cos(θ −90◦) for 0◦≤θ ≤360◦\nf) y = cos(2θ + 60◦) for −360◦≤θ ≤360◦\n2. Two girls are given the following graph:\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\ny\nb\na) Audrey decides that the equation for the graph is a cosine function of the\nform y = a cos θ. Determine the value of a.\nb) Megan thinks that the equation for the graph is a cosine function of the\nform y = cos(θ + p). Determine the value of p.\nc) What can they conclude?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VS\n1b. 22VT\n1c. 22VV\n1d. 22VW\n1e. 22VX\n1f. 22VY\n2. 22VZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n221\nChapter 5.\nFunctions\n\n5.7\nThe tangent function\nEMBH8\nRevision\nEMBH9\nFunctions of the form y = tan θ for 0◦≤θ ≤360◦\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nf(θ)\nθ\nThe dashed vertical lines are called the asymptotes. The asymptotes are at the values\nof θ where tan θ is not defined.\n• Period: 180◦\n• Domain: {θ : 0◦≤θ ≤360◦, θ ̸= 90◦; 270◦}\n• Range: {f(θ) : f(θ) ∈R}\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Asymptotes: the lines θ = 90◦and θ = 270◦\nFunctions of the form y = a tan θ + q\nTangent functions of the general form y = a tan θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a tan θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted vertically upwards by q units.\n– For q < 0, f(θ) is shifted vertically downwards by q units.\n• The effect of a on shape\n– For a > 1, branches of f(θ) are steeper.\n– For 0 < a < 1, branches of f(θ) are less steep and curve more.\n222\n5.7.\nThe tangent function\n\n– For a < 0, there is a reflection about the x-axis.\n– For −1 < a < 0, there is a reflection about the x-axis and the branches of\nthe graph are less steep.\n– For a < −1, there is a reflection about the x-axis and the branches of the\ngraph are steeper.\na < 0\na > 0\nq > 0\nb\n0\nb\n0\nq = 0\nb\n0\nb\n0\nq < 0\nb\n0\nb\n0\nExercise 5 – 28: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function determine the following:\n• Period\n• Domain and range\n223\nChapter 5.\nFunctions\n\n• x- and y-intercepts\n• Asymptotes\n1. y1 = tan θ −1\n2\n2. y2 = −3 tan θ\n3. y3 = tan θ + 2\n4. y4 = 2 tan θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W2\n2. 22W3\n3. 22W4\n4. 22W5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = tan(kθ)\nEMBHB\nInvestigation: The effects of k on a tangent graph\n1. Complete the following table for y1 = tan θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ntan θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ntan θ\n2. Use the table of values to plot the graph of y1 = tan θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = tan(−θ)\nb) y3 = tan 3θ\nc) y4 = tan θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of k\n224\n5.7.\nThe tangent function\n\n5. What do you notice about y1 = tan θ and y2 = tan(−θ)?\n6. Is tan(−θ) = −tan θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = tan kθ?\nThe effect of the parameter on y = tan kθ\nThe value of k affects the period of the tangent function. If k is negative, then the\ngraph is reflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the tangent function decreases.\nFor 0 < k < 1, the period of the tangent function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\ntan(−θ) = −tan θ\nCalculating the period:\nTo determine the period of y = tan kθ we use,\nPeriod = 180◦\n|k|\nwhere |k| is the absolute value of k.\nk > 0\nk < 0\n225\nChapter 5.\nFunctions\n\nWorked example 26: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan 3θ\n2\n2. For each function determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan kθ\nNotice that k > 1 for y2 = tan 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan 3θ\n2\nUNDEF\n−0,41\n1\n−2,41\n0\n2,41\n−1\n0,41\nUNDEF\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan 3θ\n2\nperiod\n180◦\n120◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦< θ < 180◦, θ ̸=\n−60◦; 60◦}\nrange\n{f(θ) : f(θ) ∈R}\n{f(θ) : f(θ) ∈R}\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −180◦; −60◦and 180◦\n226\n5.7.\nThe tangent function\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan kθ:\nDomain and range\nThe domain of one branch is {θ : −90◦\nk\n< θ < 90◦\nk , θ ∈R} because f(θ) is undefined\nfor θ = −90◦\nk and θ = 90◦\nk .\nThe range is {f(θ) : f(θ) ∈R} or (−∞; ∞).\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0 and solving for f(θ).\ny = tan kθ\n= tan 0◦\n= 0\nThis gives the point (0◦; 0).\nAsymptotes\nThese are the values of kθ for which tan kθ is undefined.\nExercise 5 – 29: Tangent functions of the form y = tan kθ\nSketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan 2θ\n2. g(θ) = tan 3θ\n4\n3. h(θ) = tan(−2θ)\n4. k(θ) = tan 2θ\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W6\n2. 22W7\n3. 22W8\n4. 22W9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n227\nChapter 5.\nFunctions\n\nFunctions of the form y = tan (θ + p)\nEMBHC\nWe now consider tangent functions of the form y = tan(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a tangent graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = tan θ\nb) y2 = tan(θ −60◦)\nc) y3 = tan(θ −90◦)\nd) y4 = tan(θ + 60◦)\ne) y5 = tan(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of p\nThe effect of the parameter on y = tan(θ + p)\nThe effect of p on the tangent function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the tangent function shifts to the left by p.\n• For p < 0, the graph of the tangent function shifts to the right by p.\np > 0\np < 0\n228\n5.7.\nThe tangent function\n\nWorked example 27: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan(θ + 30◦)\nFor each function determine the following:\n2.\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan(θ + p)\nNotice that for y1 = tan θ we have p = 0◦(no phase shift) and for y2 = tan(θ + 30◦),\np > 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−180◦−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan(θ+30◦)\n0,58\n3,73\n−1,73\n−0,27\n0,58\n3,73\n−1,73\n−0,27\n0,58\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan(θ + 30◦)\nperiod\n180◦\n180◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦≤θ ≤\n180◦, θ ̸= −120◦; 60◦}\nrange\n(−∞; ∞)\n(−∞; ∞)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,58)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and (180◦; 0)\n(−30◦; 0) and (150◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −120◦and θ = 60◦\n229\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan(θ + p):\nDomain and range\nThe domain of one branch is {θ : θ ∈(−90◦−p; 90◦−p)} because the function is\nundefined for θ = −90◦−p and θ = 90◦−p.\nThe range is {f(θ) : f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = tan(θ + p)\n= tan(0◦+ p)\n= tan p\nThis gives the point (0◦; tan p).\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan(θ + 45◦)\n2. g(θ) = tan(θ −30◦)\n3. h(θ) = tan(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WB\n2. 22WC\n3. 22WD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n230\n5.7.\nThe tangent function\n\nSketching tangent graphs\nEMBHD\nWorked example 28: Sketching a tangent graph\nQUESTION\nSketch the graph of f(θ) = tan 1\n2(θ −30◦) for −180◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that 0 < k < 1, therefore the branches of the graph will be\nless steep than the standard tangent graph y = tan θ. We also notice that p < 0 so the\ngraph will be shifted to the right on the x-axis.\nStep 2: Determine the period\nThe period for f(θ) = tan 1\n2(θ −30◦) is:\nPeriod = 180◦\n|k|\n= 180◦\n1\n2\n= 360◦\nStep 3: Determine the asymptotes\nThe standard tangent graph, y = tan θ, for −180◦≤θ ≤180◦is undefined at θ = −90◦\nand θ = 90◦. Therefore we can determine the asymptotes of f(θ) = tan 1\n2(θ −30◦):\n•\n−90◦\n0,5 + 30◦= −150◦\n•\n90◦\n0,5 + 30◦= 210◦\nThe asymptote at θ = 210◦lies outside the required interval.\n231\nChapter 5.\nFunctions\n\nStep 4: Plot the points and join with a smooth curve\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 360◦\nDomain: {θ : −180◦≤θ ≤180◦, θ ̸= −150◦}\nRange: (−∞; ∞)\ny-intercepts: (0◦; −0,27)\nx-intercept: (30◦; 0)\nAsymptotes: θ = −150◦\nExercise 5 – 31: The tangent function\n1. Sketch the following graphs on separate axes:\na) y = tan θ −1 for −90◦≤θ ≤90◦\nb) f(θ) = −tan 2θ for 0◦≤θ ≤90◦\nc) y = 1\n2 tan(θ + 45◦) for 0◦≤θ ≤360◦\nd) y = tan(30◦−θ) for −180◦≤θ ≤180◦\n2. Given the graph of y = a tan kθ, determine the values of a and k.\nθ\nf(θ)\nb\nb\n(90◦; −1)\n360◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WF\n1b. 22WG\n1c. 22WH\n1d. 22WJ\n2. 22WK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n232\n5.7.\nThe tangent function\n\nExercise 5 – 32: Mixed exercises\n1. Determine the equation for each of the following:\na) f(θ) = a sin kθ and g(θ) = a tan θ\nθ\ny\nb\nb\nf\n(45◦; −3\n2 )\ng\n(180◦; 0)\n(135◦; −1 1\n2 )\nb) f(θ) = a sin kθ and g(θ) = a cos(θ + p)\nθ\n0\ny\nb\n(−90◦; 2)\n−180◦\n180◦\nf and g\nc) y = a tan kθ\nθ\n0\ny\nb\n(90◦; 3)\n360◦\n180◦\nd) y = a cos θ + q\nθ\n0\ny\n4\n360◦\n180◦\n233\nChapter 5.\nFunctions\n\n2. Given the functions f(θ) = 2 sin θ and g(θ) = cos θ + 1:\na) Sketch the graphs of both functions on the same system of axes, for 0◦≤\nθ ≤360◦. Indicate the turning points and intercepts on the diagram.\nb) What is the period of f?\nc) What is the amplitude of g?\nd) Use your sketch to determine how many solutions there are for the equation\n2 sin θ −cos θ = 1. Give one of the solutions.\ne) Indicate on your sketch where on the graph the solution to 2 sin θ = −1 is\nfound.\n3. The sketch shows the two functions f(θ) = a cos θ and g(θ) = tan θ for 0◦≤θ ≤\n360◦. Points P(135◦; b) and Q(c; −1) lie on g(θ) and f(θ) respectively.\nθ\n0\ny\nb\nb\n360◦\n180◦\nP\nQ\ng\nf\n−1\n2\n−2\na) Determine the values of a, b and c.\nb) What is the period of g?\nc) Solve the equation cos θ = 1\n2 graphically and show your answer(s) on the\ndiagram.\nd) Determine the equation of the new graph if g is reflected about the x-axis\nand shifted to the right by 45◦.\n4. Sketch the graphs of y1 = −1\n2 sin(θ + 30◦) and y2 = cos(θ −60◦), on the same\nsystem of axes for 0◦≤θ ≤360◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WM\n1b. 22WN\n1c. 22WP\n1d. 22WQ\n2. 22WR\n3. 22WS\n4. 22WT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n234\n5.7.\nThe tangent function\n\n5.8\nSummary\nEMBHF\nSee presentation: 22WV at www.everythingmaths.co.za\n1. Parabolic functions:\nStandard form: y = ax2 + bx + c\n• y-intercept: (0; c)\n• x-intercept: x = −b±\n√\nb2−4ac\n2a\n• Turning point:\n\u0010\n−b\n2a; −b2\n4a + c\n\u0011\n• Axis of symmetry: x = −b\n2a\nCompleted square form: y = a(x + p)2 + q\n• Turning point: (−p; q)\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n2. Average gradient:\n• Average gradient = y2−y1\nx2−x1\n3. Hyperbolic functions:\nStandard form: y = k\nx\n• k > 0: first and third quadrant\n• k < 0: second and fourth quadrant\nShifted form: y =\nk\nx+p + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: x = −p and y = q\n4. Exponential functions:\nStandard form: y = abx\n• a > 0: above x-axis\n• a < 0: below x-axis\n• b > 1: increasing function if a > 0; decreasing function if a < 0\n• 0 < b < 1: decreasing function if a > 0; increasing function if a < 0\nShifted form: y = ab(x+p) + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n235\nChapter 5.\nFunctions\n\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: y = q\n5. Sine functions:\nShifted form: y = a sin(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• sin(−θ) = −sin θ\n6. Cosine functions:\nShifted form: y = a cos(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• cos(−θ) = cos θ\n7. Tangent functions:\nShifted form: y = a tan(kθ + p) + q\n• Period = 180◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• tan(−θ) = −tan θ\n• Asymptotes: 90◦−p\nk\n± 180◦n\nk\n, n ∈Z\n236\n5.8.\nSummary\n\nExercise 5 – 33: End of chapter exercises\n1. Show that if a\n<\n0,\nthen the range of f(x)\n=\na(x + p)2 + q is\n{f(x) : f(x) ∈(−∞, q]}.\n2. If (2; 7) is the turning point of f(x) = −2x2 −4ax + k, find the values of the\nconstants a and k.\n3. The following graph is represented by the equation f(x) = ax2 + bx. The coor-\ndinates of the turning point are (3; 9). Show that a = −1 and b = 6.\nb (3; 9)\nx\n0\ny\n4. Given: f(x) = x2 −2x + 3. Give the equation of the new graph originating if:\na) the graph of f is moved three units to the left.\nb) the x-axis is moved down three units.\n5. A parabola with turning point (−1; −4) is shifted vertically by 4 units upwards.\nWhat are the coordinates of the turning point of the shifted parabola?\n6. Plot the graph of the hyperbola defined by y = 2\nx for −4 ≤x ≤4. Suppose\nthe hyperbola is shifted 3 units to the right and 1 unit down. What is the new\nequation then?\n7. Based on the graph of y =\nk\n(x+p) + q, determine the equation of the graph with\nasymptotes y = 2 and x = 1 and passing through the point (2; 3).\ny\nx\n0\n2\n1\nb (2; 3)\n237\nChapter 5.\nFunctions\n\n8. The columns in the table below give the y-values for the following functions:\ny = ax, y = ax+1 and y = ax + 1. Match each function to the correct column.\nx\nA\nB\nC\n−2\n7,25\n6,25\n2,5\n−1\n3,5\n2,5\n1\n0\n2\n1\n0,4\n1\n1,4\n0,4\n0,16\n2\n1,16\n0,16\n0,064\n9. The graph of f(x) = 1 + a . 2x (a is a constant) passes through the origin.\na) Determine the value of a.\nb) Determine the value of f(−15) correct to five decimal places.\nc) Determine the value of x, if P (x; 0,5) lies on the graph of f.\nd) If the graph of f is shifted 2 units to the right to give the function h, write\ndown the equation of h.\n10. The graph of f(x) = a . bx (a ̸= 0) has the point P (2; 144) on f.\na) If b = 0,75, calculate the value of a.\nb) Hence write down the equation of f.\nc) Determine, correct to two decimal places, the value of f(13).\nd) Describe the transformation of the curve of f to h if h(x) = f(−x).\n11. Using your knowledge of the effects of p and k draw a rough sketch of the fol-\nlowing graphs without a table of values.\na) y = sin 3θ for −180◦≤θ ≤180◦\nb) y = −cos 2θ for 0◦≤θ ≤180◦\nc) y = tan 1\n2θ for 0◦≤θ ≤360◦\nd) y = sin(θ −45◦) for −360◦≤θ ≤360◦\ne) y = cos(θ + 45◦) for 0◦≤θ ≤360◦\nf) y = tan(θ −45◦) for 0◦≤θ ≤360◦\ng) y = 2 sin 2θ for −180◦≤θ ≤180◦\nh) y = sin(θ + 30◦) + 1 for −360◦≤θ ≤0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WW\n2. 22WX\n3. 22WY\n4. 22WZ\n5. 22X2\n6. 22X3\n7. 22X4\n8. 22X5\n9. 22X6\n10. 22X7\n11a. 22X8\n11b. 22X9\n11c. 22XB\n11d. 22XC\n11e. 22XD\n11f. 22XF\n11g. 22XG\n11h. 22XH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n238\n5.8.\nSummary\n\nCHAPTER\n6\nTrigonometry\n6.1\nRevision\n240\n6.2\nTrigonometric identities\n247\n6.3\nReduction formula\n253\n6.4\nTrigonometric equations\n266\n6.5\nArea, sine, and cosine rules\n280\n6.6\nSummary\n301\n\n6\nTrigonometry\n6.1\nRevision\nEMBHG\nTrigonometric ratios\nb\nb\nb\ny\nx\nP(x; y)\nQ(−x; y)\nO\nα\nβ\nr\nr\nWe plot the points P(x; y) and Q(−x; y) in the Cartesian plane and measure the angles\nfrom the positive x-axis to the terminal arms (OP and OQ).\nP(x; y) lies in the first quadrant with P ˆOX = α and Q(−x; y) lies in the second\nquadrant with Q ˆOX = β.\nUsing the theorem of Pythagoras we have that\nOP 2 = x2 + y2\nAnd OQ2 = (−x)2 + y2\n= x2 + y2\n∴OP = OQ\nLet OP = OQ = r.\nTrigonometric ratios\nsin α = y\nr\ncos α = x\nr\ntan α = y\nx\nIn the second quadrant we notice that −x < 0\nsin β = y\nr\ncos β = −x\nr\ntan β = −y\nx\n240\n6.1.\nRevision\n\nSimilarily, in the third and fourth quadrants the sign of the trigonometric ratios depends\non the signs of x and y:\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nSpecial angles\n30◦\n60◦\n1\n√\n3\n2\n45◦\n45◦\n1\n1\n√\n2\nθ\n0◦\n30◦\n45◦\n60◦\n90◦\ncos θ\n1\n√\n3\n2\n1\n√\n2\n1\n2\n0\nsin θ\n0\n1\n2\n1\n√\n2\n√\n3\n2\n1\ntan θ\n0\n1\n√\n3\n1\n√\n3\nundef\nSee video: 22XJ at www.everythingmaths.co.za\n241\nChapter 6.\nTrigonometry\n\nSolving equations\nWorked example 1: Solving equations\nQUESTION\nDetermine the values of a and b in the right-angled triangle TUW (correct to one\ndecimal place):\nT\nW\nU\n47◦\nb\n30\na\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of a\nsin θ = opposite side\nhypotenuse\nsin 47◦= 30\na\na =\n30\nsin 47◦\n∴a = 41,0\nStep 3: Determine the value of b\nAlways try to use the information that is given for calculations and not answers that you\nhave worked out in case you have made an error. For example, avoid using a = 41,0\nto determine the value of b.\ntan θ = opposite side\nadjacent side\ntan 47◦= 30\nb\nb =\n30\ntan 47◦\n∴b = 28,0\nStep 4: Write the final answer\na = 41,0 units and b = 28,0 units.\n242\n6.1.\nRevision\n\nFinding an angle\nWorked example 2: Finding an angle\nQUESTION\nCalculate the value of θ in the right-angled triangle MNP (correct to one decimal\nplace):\nP\nN\nM\n41\nθ\n24\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of θ\ntan θ = opposite side\nadjacent side\ntan θ = 41\n24\n∴θ = tan−1\n\u001241\n24\n\u0013\nθ = 59,7◦\n243\nChapter 6.\nTrigonometry\n\nWorked example 3: Finding an angle\nQUESTION\nGiven 2 sin θ\n2 = cos 43◦, for θ ∈[0◦; 90◦], determine the value of θ (correct to one\ndecimal place).\nSOLUTION\nStep 1: Simplify the equation\nAvoiding rounding off in calculations until you have determined the final answer. In\nthe calculation below, the dots indicate that the number has not been rounded so that\nthe answer is as accurate as possible.\n2 sin θ\n2 = cos 43◦\nsin θ\n2 = cos 43◦\n2\nθ\n2 = sin−1(0,365 . . .)\nθ = 2(21,449 . . .)\n∴θ = 42,9◦\nTwo-dimensional problems\nWorked example 4: Flying a kite\nQUESTION\nThelma flies a kite on a 22 m piece of string and the height of the kite above the\nground is 20,4 m. Determine the angle of inclination of the string (correct to one\ndecimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the opposite and adjacent sides and the hy-\npotenuse\nLet the angle of inclination of the string be θ.\n244\n6.1.\nRevision\n\nKite\nThelma\n20,4\nθ\n22\nStep 2: Use an appropriate trigonometric ratio to find θ\nsin θ = opposite side\nhypotenuse\n= 20,4\n22\nθ = sin−1(0,927 . . .)\n∴θ = 68,0◦\nExercise 6 – 1: Revision\n1. If p = 49◦and q = 32◦, use a calculator to determine whether the following\nstatements are true of false:\na) sin p + 3 sin p = 4 sin p\nb) sin q\ncos q = tan q\nc) cos(p −q) = cos p −cos q\nd) sin(2p) = 2 sin p cos p\n2. Determine the following angles (correct to one decimal place):\na) cos α = 0,64\nb) sin θ + 2 = 2,65\nc) 1\n2 cos 2β = 0,3\nd) tan θ\n3 = sin 48◦\ne) cos 3p = 1,03\nf) 2 sin 3β + 1 = 2,6\ng) sin θ\ncos θ = 42\n3\n3. In △ABC, A ˆCB = 30◦, AC = 20 cm and BC = 22 cm. The perpendicular\nline from A intersects BC at T.\n245\nChapter 6.\nTrigonometry\n\nDetermine:\nA\nC\nB\nT\n20 cm\n22 cm\n30◦\na) the length TC\nb) the length AT\nc) the angle B ˆAT\n4. A rhombus has a perimeter of 40 cm and one of the internal angles is 30◦.\na) Determine the length of the sides.\nb) Determine the lengths of the diagonals.\nc) Calculate the area of the rhombus.\n5. Simplify the following without using a calculator:\na) 2 sin 45◦× 2 cos 45◦\nb) cos2 30◦−sin2 60◦\nc) sin 60◦cos 30◦−cos 60◦sin 30◦−tan 45◦\nd) 4 sin 60◦cos 30◦−2 tan 45◦+ tan 60◦−2 sin 60◦\ne) sin 60◦×\n√\n2 tan 45◦+ 1 −sin 30◦\n6. Given the diagram below.\nb\nx\ny\n0\nB(2; 2\n√\n3)\nβ\nDetermine the following without using a calculator:\na) β\nb) cos β\nc) cos2 β + sin2 β\n7. The 10 m ladder of a fire truck leans against the wall of a burning building at an\nangle of 60◦. The height of an open window is 9 m from the ground. Will the\nladder reach the window?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22XK\n2a. 22XM\n2b. 22XN\n2c. 22XP\n2d. 22XQ\n2e. 22XR\n2f. 22XS\n2g. 22XT\n3. 22XV\n4. 22XW\n5a. 22XX\n5b. 22XY\n5c. 22XZ\n5d. 22Y2\n5e. 22Y3\n6. 22Y4\n7. 22Y5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n246\n6.1.\nRevision\n\n6.2\nTrigonometric identities\nEMBHH\nAn identity is a mathematical statement that equates one quantity with another. Trigono-\nmetric identities allow us to simplify a given expression so that it contains sine and co-\nsine ratios only. This enables us to solve equations and also to prove other identities.\nQuotient identity\nInvestigation: Quotient identity\n1. Complete the table without using a calculator, leaving your answer in surd form\nwhere applicable:\nθ = 45◦\nθ\n3\n5\nx\ny\n(3; 2)\nθ\nb\nsin θ\ncos θ\nsin θ\ncos θ\ntan θ\n2. Examine the last two rows of the table and make a conjecture.\n3. Are there any values of θ for which your conjecture would not be true? Explain\nyour answer.\nWe know that tan θ is defined as:\ntan θ = opposite side\nadjacent side\nUsing the diagram below and the theorem of Pythagoras, we can write the tangent\nfunction in terms of x, y and r:\nx\ny\n(x; y)\nθ\nb\nO\n247\nChapter 6.\nTrigonometry\n\ntan θ = y\nx\n= y\nx × r\nr\n= y\nr × r\nx\n= y\nr ÷ x\nr\n= sin θ ÷ cos θ\n= sin θ\ncos θ\nThis is the quotient identity:\ntan θ = sin θ\ncos θ\nNotice that tan θ is undefined if cos θ = 0, therefore θ ̸= k × 90◦, where k is an odd\ninteger.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\ntan θ\nSquare identity\nInvestigation: Square identity\n1. Use a calculator to complete the following table:\nsin2 80◦+ cos2 80◦=\ncos2 23◦+ sin2 23◦=\nsin 50◦+ cos 50◦=\nsin2 67◦−cos2 67◦=\nsin2 67◦+ cos2 67◦=\n2. What do you notice? Make a conjecture.\n3. Draw a sketch and prove your conjecture in general terms, using x, y and r.\n248\n6.2.\nTrigonometric identities\n\nx\ny\n(x; y)\nα\nb\nO\nr\nUsing the theorem of Pythagoras, we can write the sine and cosine functions in terms\nof x, y and r:\nsin2 θ + cos2 θ =\n\u0010y\nr\n\u00112\n+\n\u0010x\nr\n\u00112\n= y2\nr2 + x2\nr2\n= y2 + x2\nr2\n= r2\nr2\n= 1\nThis is the square identity:\nsin2 θ + cos2 θ = 1\nOther forms of the square identity\nComplete the following:\n1. sin2 θ = 1 −. . . . . .\n2. cos θ = ±√. . . . . .\n3. sin2 θ = (1 + . . . . . .)(1 −. . . . . .)\n4. cos2 θ −1 = . . . . . .\nHere are some useful tips for proving identities:\n• Change all trigonometric ratios to sine and cosine.\n• Choose one side of the equation to simplify and show that it is equal to the other\nside.\n• Usually it is better to choose the more complicated side to simplify.\n• Sometimes we need to simplify both sides of the equation to show that they are\nequal.\n• A square root sign often indicates that we need to use the square identity.\n• We can also add to the expression to make simplifying easier:\n– replace 1 with sin2 θ + cos2 θ.\n– multiply by 1 in the form of a suitable fraction, for example 1 + sin θ\n1 + sin θ.\n249\nChapter 6.\nTrigonometry\n\nSee video: 22Y6 at www.everythingmaths.co.za\nWorked example 5: Trigonometric identities\nQUESTION\nSimplify the following:\n1. tan2 θ × cos2 θ\n2.\n1\ncos2 θ −tan2 θ\nSOLUTION\nStep 1: Write the expression in terms of sine and cosine only\nWe use the square and quotient identities to write the given expression in terms of sine\nand cosine and then simplify as far as possible.\n1.\ntan2 θ × cos2 θ =\n\u0012 sin θ\ncos θ\n\u00132\n× cos2 θ\n= sin2 θ\ncos2 θ × cos2 θ\n= sin2 θ\n2.\n1\ncos2 θ −tan2 θ =\n1\ncos2 θ −\n\u0012 sin θ\ncos θ\n\u00132\n=\n1\ncos2 θ −sin2 θ\ncos2 θ\n= 1 −sin2 θ\ncos2 θ\n= cos2 θ\ncos2 θ\n= 1\n250\n6.2.\nTrigonometric identities\n\nWorked example 6: Trigonometric identities\nQUESTION\nProve: 1 −sin α\ncos α\n=\ncos α\n1 + sin α\nSOLUTION\nStep 1: Note restrictions\nWhen working with fractions, we must be careful that the denominator does not equal\n0. Therefore cos θ ̸= 0 for the fraction on the left-hand side and sin θ + 1 ̸= 0 for the\nfraction on the right-hand side.\nStep 2: Simplify the left-hand side\nThis is not an equation that needs to be solved. We are required to show that one side\nof the equation is equal to the other. We can choose either of the two sides to simplify.\nLHS = 1 −sin α\ncos α\n= 1 −sin α\ncos α\n× 1 + sin α\n1 + sin α\nNotice that we have not changed the equation — this is the same as multiplying by 1\nsince the numerator and the denominator are the same.\nStep 3: Determine the lowest common denominator and simplify\nLHS =\n1 −sin2 α\ncos α(1 + sin α)\n=\ncos2 α\ncos α(1 + sin α)\n=\ncos α\n1 + sin α\n= RHS\n251\nChapter 6.\nTrigonometry\n\nExercise 6 – 2: Trigonometric identities\n1. Reduce the following to one trigonometric ratio:\na) sin α\ntan α\nb) cos2 θ tan2 θ + tan2 θ sin2 θ\nc) 1 −sin θ cos θ tan θ\nd)\n\u00121 −cos2 β\ncos2 β\n\u0013\n−tan2 β\n2. Prove the following identities and state restrictions where appropriate:\na) 1 + sin θ\ncos θ\n=\ncos θ\n1 −sin θ\nb) sin2α + (cos α −tan α) (cos α + tan α) = 1 −tan2α\nc)\n1\ncos θ −cos θtan2θ\n1\n= cos θ\nd)\n2 sin θ cos θ\nsin θ + cos θ = sin θ + cos θ −\n1\nsin θ + cos θ\ne)\n\u0012cos β\nsin β + tan β\n\u0013\ncos β =\n1\nsin β\nf)\n1\n1 + sin θ +\n1\n1 −sin θ = d\n2 tan θ\nsin θ cos θ\ng) (1 + tan2 α) cos α\n(1 −tan α)\n=\n1\ncos α −sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Y7\n1b. 22Y8\n1c. 22Y9\n1d. 22YB\n2a. 22YC\n2b. 22YD\n2c. 22YF\n2d. 22YG\n2e. 22YH\n2f. 22YJ\n2g. 22YK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n252\n6.2.\nTrigonometric identities\n\n6.3\nReduction formula\nEMBHJ\nAny trigonometric function whose argument is 90◦± θ; 180◦± θ and 360◦± θ can be\nwritten simply in terms of θ.\nDeriving reduction formulae\nEMBHK\nInvestigation: Reduction formulae for function values of 180◦± θ\n1. Function values of 180◦−θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the y-axis, determine the coordi-\nnates of P ′.\nb) Write down values for sin θ, cos θ and tan θ.\nc) Use the coordinates for P ′ to determine sin(180◦−θ), cos(180◦−θ),\ntan(180◦−θ).\nd) From your results determine a relationship between the trigonometric func-\ntion values of (180◦−θ) and θ.\n253\nChapter 6.\nTrigonometry\n\n2. Function values of 180◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦+ θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the origin (the two points are sym-\nmetrical about both the x-axis and the y-axis), determine the coordinates of\nP ′.\nb) Use the coordinates for P ′ to determine sin(180◦+ θ), cos(180◦+ θ) and\ntan(180◦+ θ).\nc) From your results determine a relationship between the trigonometric func-\ntion values of (180◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(180◦−θ) = . . . . . .\nb) cos(180◦−θ) = . . . . . .\nc) tan(180◦−θ) = . . . . . .\nd) sin(180◦+ θ) = . . . . . .\ne) cos(180◦+ θ) = . . . . . .\nf) tan(180◦+ θ) = . . . . . .\n254\n6.3.\nReduction formula\n\nWorked example 7: Reduction formulae for function values of 180◦± θ\nQUESTION\nWrite the following as a single trigonometric ratio:\nsin 163◦\ncos 197◦+ tan 17◦+ cos(180◦−θ) × tan(180◦+ θ)\nSOLUTION\nStep 1: Use reduction formulae to write the trigonometric function values in terms\nof acute angles and θ\n= sin(180◦−17◦)\ncos(180◦+ 17◦) + tan 17◦+ (−cos θ) × tan θ\nStep 2: Simplify\n=\nsin 17◦\n−cos 17◦+ tan 17◦−cos θ × sin θ\ncos θ\n= −tan 17◦+ tan 17◦−sin θ\n= −sin θ\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1. Determine the value of the following expressions without using a calculator:\na) tan 150◦sin 30◦−cos 210◦\nb) (1 + cos 120◦)(1 −sin2 240◦)\nc) cos2 140◦+ sin2 220◦\n2. Write the following in terms of a single trigonometric ratio:\na) tan(180◦−θ) × sin(180◦+ θ)\nb) tan(180◦+ θ) cos(180◦−θ)\nsin(180◦−θ)\n255\nChapter 6.\nTrigonometry\n\n3. If t = tan 40◦, express the following in terms of t:\na) tan 140◦+ 3 tan 220◦\nb) cos 220◦\nsin 140◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YM\n1b. 22YN\n1c. 22YP\n2a. 22YQ\n2b. 22YR\n3. 22YS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of (360◦± θ) and (−θ)\n1. Function values of (360◦−θ) and (−θ)\nIn the Cartesian plane we measure angles from the positive x-axis to the terminal\narm, which means that an anti-clockwise rotation gives a positive angle. We can\ntherefore measure negative angles by rotating in a clockwise direction.\nFor an acute angle θ, we know that −θ will lie in the fourth quadrant.\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n360◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the x-axis (y = 0), determine the\ncoordinates of P ′.\nb) Use the coordinates of P ′ to determine sin(360◦−θ), cos(360◦−θ) and\ntan(360◦−θ).\nc) Use the coordinates of P ′ to determine sin(−θ), cos(−θ) and tan(−θ).\nd) From your results determine a relationship between the function values of\n(360◦−θ) and −θ.\n256\n6.3.\nReduction formula\n\ne) Complete the following reduction formulae:\ni. sin(360◦−θ) = . . . . . .\nii. cos(360◦−θ) = . . . . . .\niii. tan(360◦−θ) = . . . . . .\niv. sin(−θ) = . . . . . .\nv. cos(−θ) = . . . . . .\nvi. tan(−θ) = . . . . . .\n2. Function values of 360◦+ θ\nWe can also have an angle that is larger than 360◦. The angle completes a\nrevolution of 360◦and then continues to give an angle of θ.\nComplete the following reduction formulae:\na) sin(360◦+ θ) = . . . . . .\nb) cos(360◦+ θ) = . . . . . .\nc) tan(360◦+ θ) = . . . . . .\nFrom working with functions, we know that the graph of y = sin θ has a period of\n360◦. Therefore, one complete wave of a sine graph is the same as one complete\nrevolution for sin θ in the Cartesian plane.\n0\n1\n−1\n90◦\n180◦\n270◦\n360◦\n1st\n2nd\n3rd\n4th\npositive\npositive\nnegative\nnegative\n0◦/360◦\n90◦\n180◦\n270◦\n2nd\npos.\nneg.\n3rd\nneg.\n4th\n1st\npos.\nWe can also have multiple revolutions. The periodicity of the trigonometric graphs\nshows this clearly. A complete sine or cosine curve is completed in 360◦.\ny = cos θ\ny = sin θ\nθ\ny\n257\nChapter 6.\nTrigonometry\n\nIf k is any integer, then\nsin(k . 360◦+ θ) = sin θ\ncos(k . 360◦+ θ) = cos θ\ntan(k . 360◦+ θ) = tan θ\nWorked example 8: Reduction formulae for function values of 360◦± θ\nQUESTION\nIf f = tan 67◦, express the following in terms of f\nsin 293◦\ncos 427◦+ tan(−67◦) + tan 1147◦\nSOLUTION\nStep 1: Using reduction formula\n= sin(360◦−67◦)\ncos(360◦+ 67◦) −tan(67◦) + tan (3(360◦) + 67◦)\n= −sin 67◦\ncos 67◦−tan 67◦+ tan 67◦\n= −tan 67◦\n= −f\nWorked example 9: Using reduction formula\nQUESTION\nEvaluate without using a calculator:\ntan2 210◦−(1 + cos 120◦) sin2 405◦\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and special angles\n258\n6.3.\nReduction formula\n\n= tan2(180◦+ 30◦) −(1 + cos(180◦−60◦)) sin2(360◦+ 45◦)\n= tan2 30◦−(1 + (−cos 60◦)) sin2 45◦\n=\n\u0012 1\n√\n3\n\u00132\n−\n\u0012\n1 −1\n2\n\u0013 \u0012 1\n√\n2\n\u00132\n= 1\n3 −\n\u00121\n2\n\u0013 \u00121\n2\n\u0013\n= 1\n3 −1\n4\n= 1\n12\nExercise 6 – 4: Using reduction formula\n1. Simplify the following:\na) tan(180◦−θ) sin(360◦+ θ)\ncos(180◦+ θ) tan(360◦−θ)\nb) cos2(360◦+ θ) + cos(180◦+ θ) tan(360◦−θ) sin(360◦+ θ)\nc)\nsin(360◦+ α) tan(180◦+ α)\ncos(360◦−α) tan2(360◦+ α)\n2. Write the following in terms of cos β:\ncos(360◦−β) cos(−β) −1\nsin(360◦+ β) tan(360◦−β)\n3. Simplify the following without using a calculator:\na)\ncos 300◦tan 150◦\nsin 225◦cos(−45◦)\nb) 3 tan 405◦+ 2 tan 330◦cos 750◦\nc) cos 315◦cos 405◦+ sin 45◦sin 135◦\nsin 750◦\nd) tan 150◦cos 390◦−2 sin 510◦\ne) 2 sin 120◦+ 3 cos 765◦−2 sin 240◦−3 cos 45◦\n5 sin 300◦+ 3 tan 225◦−6 cos 60◦\n4. Given 90◦< α < 180◦, use a sketch to help explain why:\na) sin(−α) = −sin α\nb) cos(−α) = −cos α\n259\nChapter 6.\nTrigonometry\n\n5. If t = sin 43◦, express the following in terms of t:\na) sin 317◦\nb) cos2 403◦\nc) tan(−43◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YT\n1b. 22YV\n1c. 22YW\n2. 22YX\n3a. 22YY\n3b. 22YZ\n3c. 22Z2\n3d. 22Z3\n3e. 22Z4\n4. 22Z5\n5. 22Z6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of 90◦± θ\nIn any right-angled triangle, the two acute angles are complements of each other, ˆA +\nˆC = 90◦\nA\nB\nC\nb\na\nc\nComplete the following:\nIn △ABC\nsin ˆC = c\nb = cos . . .\ncos ˆC = a\nb = sin . . .\nComplementary angles are positive acute angles that add up to 90◦. For example 20◦\nand 70◦are complementary angles.\n260\n6.3.\nReduction formula\n\nIn the figure P(\n√\n3; 1) and P ′ lie on a circle with radius 2. OP makes an angle of\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\nP ′\n2\n2\n90◦−θ\n1. Function values of 90◦−θ\na) If points P and P ′ are symmetrical about the line y = x, determine the\ncoordinates of P ′.\nb) Use the coordinates for P ′ to determine sin(90◦−θ) and cos(90◦−θ).\nc) From your results determine a relationship between the function values of\n(90◦−θ) and θ.\n2. Function values of 90◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nθ\nb\nP\nO\nx\ny\nθ\nb\nP ′\n90◦+ θ\n2\n2\n261\nChapter 6.\nTrigonometry\n\na) If point P is rotated through 90◦to get point P ′, determine the coordinates\nof P ′.\nb) Use the coordinates for P ′ to determine sin(90◦+ θ) and cos(90◦+ θ).\nc) From your results determine a relationship between the function values of\n(90◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(90◦−θ) = . . . . . .\nb) cos(90◦−θ) = . . . . . .\nc) sin(90◦+ θ) = . . . . . .\nd) cos(90◦+ θ) = . . . . . .\nSine and cosine are known as co-functions. Two functions are called co-functions if\nf (A) = g (B) whenever A + B = 90◦(that is, A and B are complementary angles).\nThe function value of an angle is equal to the co-function of its complement.\nThus for sine and cosine we have\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nThe sine and cosine graphs illustrate this clearly: the two graphs are identical except\nthat they have a 90◦phase difference.\nθ\ny\ny = cos θ\ny = sin θ\n262\n6.3.\nReduction formula\n\nWorked example 10: Using the co-function rule\nQUESTION\nWrite each of the following in terms of sin 40◦:\n1. cos 50◦\n2. sin 320◦\n3. cos 230◦\n4. cos 130◦\nSOLUTION\n1. cos 50◦= sin(90◦−50◦) = sin 40◦\n2. sin 320◦= sin(360◦−40◦) = −sin 40◦\n3. cos 230◦= cos(180◦+ 50◦) = −cos 50◦= −cos(90◦−40◦) = −sin 40◦\n4. cos 130◦= cos(90◦+ 40◦) = −sin 40◦\nFunction values of θ −90◦\nWe can write sin(θ −90◦) as\nsin(θ −90◦) = sin [−(90◦−θ)]\n= −sin(90◦−θ)\n= −cos θ\nsimilarly, we can show that cos (θ −90◦) = sin θ\nTherefore, sin (θ −90◦) = −cos θ and cos (θ −90◦) = sin θ.\nWorked example 11: Co-functions\nQUESTION\nExpress the following in terms of t if t = sin θ:\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and co-functions\n263\nChapter 6.\nTrigonometry\n\nUse the CAST diagram to check in which quadrants the trigonometric ratios are positive\nand negative.\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\n=cos[−(90◦−θ)] cos[2(360◦) + θ] tan[−(360◦−θ)]\nsin2(360◦+ θ) cos(90◦+ θ)\n=sin θ cos θ tan θ\nsin2 θ(−sin θ)\n= −cos θ\n\u0000 sin θ\ncos θ\n\u0001\nsin2 θ\n= −\n1\nsin θ\n= −1\nt\nExercise 6 – 5: Co-functions\n1. Simplify the following:\na) cos(90◦+ θ) sin(θ + 90◦)\nsin(−θ)\nb) 2 sin(90◦−x) + sin(90◦+ x)\nsin(90◦−x) + cos(180◦+ x)\n2. Given cos 36◦= p, express the following in terms on p:\na) sin 54◦\nb) sin 36◦\nc) tan 126◦\nd) cos 324◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Z7\n1b. 22Z8\n2. 22Z9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n264\n6.3.\nReduction formula\n\nReduction formulae and co-functions:\n1. The reduction formulae hold for any angle θ. For convenience, we assume θ is\nan acute angle (0◦< θ < 90◦).\n2. When determining function values of (180◦±θ), (360◦±θ) and (−θ) the function\ndoes not change.\n3. When determining function values of (90◦±θ) and (θ±90◦) the function changes\nto its co-function.\nsecond quadrant (180◦−θ) or (90◦+ θ)\nfirst quadrant (θ) or (90◦−θ)\nsin(180◦−θ) = + sin θ\nall trig functions are positive\ncos(180◦−θ) = −cos θ\nsin(360◦+ θ) = sin θ\ntan(180◦−θ) = −tan θ\ncos(360◦+ θ) = cos θ\nsin(90◦+ θ) = + cos θ\ntan(360◦+ θ) = tan θ\ncos(90◦+ θ) = −sin θ\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nthird quadrant (180◦+ θ)\nfourth quadrant (360◦−θ)\nsin(180◦+ θ) = −sin θ\nsin(360◦−θ) = −sin θ\ncos(180◦+ θ) = −cos θ\ncos(360◦−θ) = + cos θ\ntan(180◦+ θ) = + tan θ\ntan(360◦−θ) = −tan θ\nExercise 6 – 6: Reduction formulae\n1. Write A and B as a single trigonometric ratio:\na) A = sin(360◦−θ) cos(180◦−θ) tan(360◦+ θ)\nb) B = cos(360◦+ θ) cos(−θ) sin(−θ)\ncos(90◦+ θ)\nc) Hence, determine:\ni. A + B = . . .\nii.\nA\nB = . . .\n2. Write the following as a function of an acute angle:\na) sin 163◦\nb) cos 327◦\nc) tan 248◦\nd) cos(−213◦)\n3. Determine the value of the following, without using a calculator:\na) sin(−30◦)\ntan(150◦) + cos 330◦\nb) tan 300◦cos 120◦\nc) (1 −cos 30◦)(1 −cos 210◦)\nd) cos 780◦−(sin 315◦)(cos 405◦)\n4. Prove that the following identity is true and state any restrictions:\nsin(180◦+ α) tan(360◦+ α) cos α\ncos(90◦−α)\n= sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22ZB\n2a. 22ZC\n2b. 22ZD\n2c. 22ZF\n2d. 22ZG\n3a. 22ZH\n3b. 22ZJ\n3c. 22ZK\n3d. 22ZM\n4. 22ZN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n265\nChapter 6.\nTrigonometry\n\n6.4\nTrigonometric equations\nEMBHM\nSolving trigonometric equations requires that we find the value of the angles that satisfy\nthe equation. If a specific interval for the solution is given, then we need only find the\nvalue of the angles within the given interval that satisfy the equation. If no interval is\ngiven, then we need to find the general solution. The periodic nature of trigonometric\nfunctions means that there are many values that satisfy a given equation, as shown in\nthe diagram below.\n1\n−1\n90◦180◦270◦360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nθ\n0\ny\ny = 0,5\ny = sin θ\nWorked example 12: Solving trigonometric equations\nQUESTION\nSolve for θ (correct to one decimal place), given tan θ = 5 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to solve for θ\ntan θ = 5\n∴θ = tan−1 5\n= 78,7◦\nThis value of θ is an acute angle which lies in the first quadrant and is called the\nreference angle.\nStep 2: Use the CAST diagram to determine in which quadrants tan θ is positive\nThe CAST diagram indicates that tan θ is positive in the first and third quadrants, there-\nfore we must determine the value of θ such that 180◦< θ < 270◦.\nUsing reduction formulae, we know that tan(180◦+ θ) = tan θ\nθ = 180◦+ 78,7◦\n∴θ = 258,7◦\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 78,7◦or θ = 258,7◦.\n266\n6.4.\nTrigonometric equations\n\nWorked example 13: Solving trigonometric equations\nQUESTION\nSolve for α (correct to one decimal place), given cos α = −0,7 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we do not include the negative sign. The reference\nangle must be an acute angle in the first quadrant, where all the trigonometric functions\nare positive.\nref ∠= cos−1 0,7\n= 45,6◦\nStep 2: Use the CAST diagram to determine in which quadrants cos α is negative\nThe CAST diagram indicates that cos α is negative in the second and third quadrants,\ntherefore we must determine the value of α such that 90◦< α < 270◦.\nUsing reduction formulae, we know that cos(180◦−α) = −cos α and cos(180◦+α) =\n−cos α\nIn the second quadrant:\nα = 180◦−45,6◦\n= 134,4◦\nIn the third quadrant:\nα = 180◦+ 45,6◦\n= 225,6◦\nNote: the reference angle (45,6◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nα = 134,4◦or α = 225,6◦.\n267\nChapter 6.\nTrigonometry\n\nWorked example 14: Solving trigonometric equations\nQUESTION\nSolve for β (correct to one decimal place), given sin β = −0,5 and β ∈[−360◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we use a positive value.\nref ∠= sin−1 0,5\n= 30◦\nStep 2: Use the CAST diagram to determine in which quadrants sin β is negative\nThe CAST diagram indicates that sin β is negative in the third and fourth quadrants.\nWe also need to find the values of β such that −360◦≤β ≤360◦.\nUsing reduction formulae, we know that sin(180◦+β) = −sin β and sin(360◦−β) =\n−sin β\nIn the third quadrant:\nβ = 180◦+ 30◦\n= 210◦\nor β = −180◦+ 30◦\n= −150◦\nIn the fourth quadrant:\nβ = 360◦−30◦\n= 330◦\nor β = 0◦−30◦\n= −30◦\nNotice: the reference angle (30◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nβ = −150◦, −30◦, 210◦or 330◦.\n268\n6.4.\nTrigonometric equations\n\nExercise 6 – 7: Solving trigonometric equations\n1. Determine the values of α for α ∈[0◦; 360◦] if:\na) 4 cos α = 2\nb) sin α + 3,65 = 3\nc) tan α = 51\n4\nd) cos α + 0,939 = 0\ne) 5 sin α = 3\nf)\n1\n2 tan α = −1,4\n2. Determine the values of θ for θ ∈[−360◦; 360◦] if:\na) sin θ = 0,6\nb) cos θ + 3\n4 = 0\nc) 3 tan θ = 20\nd) sin θ = cos 180◦\ne) 2 cos θ = 4\n5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22ZP\n1b. 22ZQ\n1c. 22ZR\n1d. 22ZS\n1e. 22ZT\n1f. 22ZV\n2a. 22ZW\n2b. 22ZX\n2c. 22ZY\n2d. 22ZZ\n2e. 2322\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe general solution\nEMBHN\nIn the previous worked example, the solution was restricted to a certain interval. How-\never, the periodicity of the trigonometric functions means that there are an infinite\nnumber of positive and negative angles that satisfy an equation. If we do not restrict\nthe solution, then we need to determine the general solution to the equation. We know\nthat the sine and cosine functions have a period of 360◦and the tangent function has\na period of 180◦.\nMethod for finding the general solution:\n1. Determine the reference angle (use a positive value).\n2. Use the CAST diagram to determine where the function is positive or negative\n(depending on the given equation).\n3. Find the angles in the interval [0◦; 360◦] that satisfy the equation and add multi-\nples of the period to each answer.\n4. Check answers using a calculator.\n269\nChapter 6.\nTrigonometry\n\nWorked example 15: Finding the general solution\nQUESTION\nDetermine the general solution for sin θ = 0,3 (correct to one decimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nsin θ = 0,3\n∴ref ∠= sin−1 0,3\n= 17,5◦\nStep 2: Use CAST diagram to determine in which quadrants sin θ is positive\nThe CAST diagram indicates that sin θ is positive in the first and second quadrants.\nUsing reduction formulae, we know that sin(180◦−θ) = sin θ.\nIn the first quadrant:\nθ = 17,5◦\n∴θ = 17,5◦+ k . 360◦\nIn the second quadrant:\nθ = 180◦−17,5◦\n∴θ = 162,5◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 4:\nθ = 17,5◦+ 4(360)◦\n∴θ = 1457,5◦\nAnd sin 1457,5◦= 0,3007 . . .\nThis solution is correct.\nSimilarly, if we let k = −2:\nθ = 162,5◦−2(360)◦\n∴θ = −557,5◦\nAnd sin(−557,5◦) = 0,3007 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 17,5◦+ k . 360◦or θ = 162,5◦+ k . 360◦.\n270\n6.4.\nTrigonometric equations\n\nWorked example 16: Finding the general solution\nQUESTION\nDetermine the general solution for cos 2θ = −0,6427 (give answers correct to one\ndecimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nref ∠= sin−1 0,6427\n= 50,0◦\nStep 2: Use CAST diagram to determine in which quadrants cos θ is negative\nThe CAST diagram shows that cos θ is negative in the second and third quadrants.\nTherefore we use the reduction formulae cos(180◦−θ) = −cos θ and cos(180◦+θ) =\n−cos θ.\nIn the second quadrant:\n2θ = 180◦−50◦+ k . 360◦\n= 130◦+ k . 360◦\n∴θ = 65◦+ k . 180◦\nIn the third quadrant:\n2θ = 180◦+ 50◦+ k . 360◦\n= 230◦+ k . 360◦\n∴θ = 115◦+ k . 180◦\nwhere k ∈Z.\nRemember: also divide the period (360◦) by the coefficient of θ.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 2:\nθ = 65◦+ 2(180◦)\n∴θ = 425◦\nAnd cos 2(425)◦= −0,6427 . . .\nThis solution is correct.\n271\nChapter 6.\nTrigonometry\n\nSimilarly, if we let k = −5:\nθ = 115◦−5(180◦)\n∴θ = −785◦\nAnd cos 2(−785◦) = −0,6427 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 65◦+ k . 180◦or θ = 115◦+ k . 180◦.\nWorked example 17: Finding the general solution\nQUESTION\nDetermine the general solution for tan(2α −10◦) = 2,5 such that −180◦≤α ≤180◦\n(give answers correct to one decimal place).\nSOLUTION\nStep 1: Make a substitution\nTo solve this equation, it can be useful to make a substitution: let x = 2α −10◦.\ntan(x) = 2,5\nStep 2: Use a calculator to find the reference angle\ntan x = 2,5\n∴ref ∠= tan−1 2,5\n= 68,2◦\nStep 3: Use CAST diagram to determine in which quadrants the tangent function is\npositive\nWe see that tan x is positive in the first and third quadrants, so we use the reduction\nformula tan(180◦+ x) = tan x. It is also important to remember that the period of the\ntangent function is 180◦.\n272\n6.4.\nTrigonometric equations\n\nIn the first quadrant:\nx = 68,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 68,2◦+ k . 180◦\n2α = 78,2◦+ k . 180◦\n∴α = 39,1◦+ k . 90◦\nIn the third quadrant:\nx = 180◦+ 68,2◦+ k . 180◦\n= 248,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 248,2◦+ k . 180◦\n2α = 258,2◦+ k . 180◦\n∴α = 129,1◦+ k . 90◦\nwhere k ∈Z.\nRemember: to divide the period (180◦) by the coefficient of α.\nStep 4: Find the answers within the given interval\nSubstitute suitable values of k to determine the values of α that lie within the interval\n(−180◦≤α ≤180◦).\nI: α = 39,1◦+ k . 90◦\nIII: α = 129,1◦+ k . 90◦\nk = 0\n39,1◦\n129,1◦\nk = 1\n129,1◦\n219,1◦\n(outside)\nk = 2\n219,1◦\n(outside)\nk = −1\n−50,9◦\n39,1◦\nk = −2\n−140,9◦\n−50,9◦\nk = −3\n−230,9◦\n(outside)\n−140,9◦\nk = −4\n−230,9◦\n(outside)\nNotice how some of the values repeat. This is because of the periodic nature of the\ntangent function. Therefore we need only determine the solution:\nα = 39,1◦+ k . 90◦\nfor k ∈Z.\nStep 5: Write the final answer\nα = −140,9◦; −50,9◦; 39,1◦or 129,1◦.\n273\nChapter 6.\nTrigonometry\n\nWorked example 18: Finding the general solution using co-functions\nQUESTION\nDetermine the general solution for sin(θ −20◦) = cos 2θ.\nSOLUTION\nStep 1: Use co-functions to simplify the equation\nsin(θ −20◦) = cos 2θ\n= sin(90◦−2θ)\n∴θ −20◦= 90◦−2θ + k . 360◦,\nk ∈Z\n3θ = 110◦+ k . 360◦\n∴θ = 36,7◦+ k . 120◦\nStep 2: Use the CAST diagram to determine the correct quadrants\nSince the original equation equates a sine and cosine function, we need to work in the\nquadrant where both functions are positive or in the quadrant where both functions\nare negative so that the equation holds true. We therefore determine the solution using\nthe first and third quadrants.\nIn the first quadrant: θ = 36,7◦+ k . 120◦.\nIn the third quadrant:\n3θ = 180◦+ 110◦+ k . 360◦\n= 290◦+ k . 360◦\n∴θ = 96,6◦+ k . 120◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 36,7◦+ k . 120◦or θ = 96,6◦+ k . 120◦\n274\n6.4.\nTrigonometric equations\n\nExercise 6 – 8: General solution\n1.\n• Find the general solution for each equation.\n• Hence, find all the solutions in the interval [−180◦; 180◦].\na) cos(θ + 25◦) = 0,231\nb) sin 2α = −0,327\nc) 2 tan β = −2,68\nd) cos α = 1\ne) 4 sin θ = 0\nf) cos θ = −1\ng) tan θ\n2 = 0,9\nh) 4 cos θ + 3 = 1\ni) sin 2θ = −\n√\n3\n2\n2. Find the general solution for each equation.\na) cos(θ + 20◦) = 0\nb) sin 3α = −1\nc) tan 4β = 0,866\nd) cos(α −25◦) = 0,707\ne) 2 sin 3θ\n2 = −1\nf) 5 tan(β + 15◦) =\n5\n√\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2323\n1b. 2324\n1c. 2325\n1d. 2326\n1e. 2327\n1f. 2328\n1g. 2329\n1h. 232B\n1i. 232C\n2a. 232D\n2b. 232F\n2c. 232G\n2d. 232H\n2e. 232J\n2f. 232K\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSolving quadratic trigonometric equations\nWe can use our knowledge of algebraic equations to solve quadratic trigonometric\nequations.\nWorked example 19: Quadratic trigonometric equations\nQUESTION\nFind the general solution of 4 sin2 θ = 3.\nSOLUTION\nStep 1: Simplify the equation and determine the reference angle\n4 sin2 θ = 3\nsin2 θ = 3\n4\n∴sin θ = ±\nr\n3\n4\n= ±\n√\n3\n2\n∴ref ∠= 60◦\n275\nChapter 6.\nTrigonometry\n\nStep 2: Determine in which quadrants the sine function is positive and negative\nThe CAST diagram shows that sin θ is positive in the first and second quadrants and\nnegative in the third and fourth quadrants.\nPositive in the first and second quadrants:\nθ = 60◦+ k . 360◦\nor θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nNegative in the third and fourth quadrants:\nθ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor θ = 360◦−60◦+ k . 360◦\n= 300◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 60◦+ k . 360◦or 120◦+ k . 360◦or 240◦+ k . 360◦or 300◦+ k . 360◦\nWorked example 20: Quadratic trigonometric equations\nQUESTION\nFind θ if 2 cos2 θ −cos θ −1 = 0 for θ ∈[−180◦; 180◦].\nSOLUTION\nStep 1: Factorise the equation\n2 cos2 θ −cos θ −1 = 0\n(2 cos θ + 1)(cos θ −1) = 0\n∴2 cos θ + 1 = 0 or cos θ −1 = 0\n276\n6.4.\nTrigonometric equations\n\nStep 2: Simplify the equations and solve for θ\n2 cos θ + 1 = 0\n2 cos θ = −1\ncos θ = −1\n2\n∴ref ∠= 60◦\nII quadrant: θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nIII quadrant: θ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor\ncos θ −1 = 0\ncos θ = 1\n∴ref ∠= 0◦\nII and IV quadrants: θ = k . 360◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of θ that lie within the the given interval θ ∈[−180◦; 180◦] by\nsubstituting suitable values of k.\nIf k = −1,\nθ = 240◦+ k . 360◦\n= 240◦−(360◦)\n= −120◦\nIf k = 0,\nθ = 120◦+ k . 360◦\n= 120◦+ 0(360◦)\n= 120◦\nIf k = 1,\nθ = k . 360◦\n= 0(360◦)\n= 0◦\n277\nChapter 6.\nTrigonometry\n\nStep 4: Alternative method: substitution\nWe can simplify the given equation by letting y = cos θ and then factorising as:\n2y2 −y −1 = 0\n(2y + 1)(y −1) = 0\n∴y = −1\n2 or y = 1\nWe substitute y = cos θ back into these two equations and solve for θ.\nStep 5: Write the final answer\nθ = −120◦; 0◦; 120◦\nWorked example 21: Quadratic trigonometric equations\nQUESTION\nFind α if 2 sin2 α −sin α cos α = 0 for α ∈[0◦; 360◦].\nSOLUTION\nStep 1: Factorise the equation by taking out a common factor\n2 sin2 α −sin α cos α = 0\nsin α(2 sin α −cos α) = 0\n∴sin α = 0 or 2 sin α −cos α = 0\nStep 2: Simplify the equations and solve for α\nsin α = 0\n∴ref ∠= 0◦\n∴α = 0◦+ k . 360◦\nor α = 180◦+ k . 360◦\nand since 360◦= 2 × 180◦\nwe therefore have α = k . 180◦\n278\n6.4.\nTrigonometric equations\n\nor\n2 sin α −cos α = 0\n2 sin α = cos α\nTo simplify further, we divide both sides of the equation by cos α.\n2 sin α\ncos α = cos α\ncos α\n(cos α ̸= 0)\n2 tan α = 1\ntan α = 1\n2\n∴ref ∠= 26,6◦\n∴α = 26,6◦+ k . 180◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of α that lie within the the given interval α ∈[0◦; 360◦] by\nsubstituting suitable values of k.\nIf k = 0:\nα = 0◦\nor α = 26,6◦\nIf k = 1:\nα = 180◦\nor α = 26,6◦+ 180◦\n= 206,6◦\nIf k = 2:\nα = 360◦\nStep 4: Write the final answer\nα = 0◦; 26,6◦; 180◦; 206,6◦; 360◦\n279\nChapter 6.\nTrigonometry\n\nExercise 6 – 9: Solving trigonometric equations\n1. Find the general solution for each of the following equations:\na) cos 2θ = 0\nb) sin(α + 10◦) =\n√\n3\n2\nc) 2 cos θ\n2 −\n√\n3 = 0\nd)\n1\n2 tan(β −30◦) = −1\ne) 5 cos θ = tan 300◦\nf) 3 sin α = −1,5\ng) sin 2β = cos(β + 20◦)\nh) 0,5 tan θ + 2,5 = 1,7\ni) sin(3α −10◦) = sin(α + 32◦)\nj) sin 2β = cos 2β\n2. Find θ if sin2 θ + 1\n2 sin θ = 0 for θ ∈[0◦; 360◦].\n3. Determine the general solution for each of the following:\na) 2 cos2 θ −3 cos θ = 2\nb) 3 tan2 θ + 2 tan θ = 0\nc) cos2 α = 0,64\nd) sin(4β + 35◦) = cos(10◦−β)\ne) sin(α + 15◦) = 2 cos(α + 15◦)\nf) sin2 θ −4 cos2 θ = 0\ng) cos(2θ + 30◦)\n2\n+ 0,38 = 0\n4. Find β if 1\n3 tan β = cos 200◦for β ∈[−180◦; 180◦].\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 232M\n1b. 232N\n1c. 232P\n1d. 232Q\n1e. 232R\n1f. 232S\n1g. 232T\n1h. 232V\n1i. 232W\n1j. 232X\n2. 232Y\n3a. 232Z\n3b. 2332\n3c. 2333\n3d. 2334\n3e. 2335\n3f. 2336\n3g. 2337\n4. 2338\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n6.5\nArea, sine, and cosine rules\nEMBHP\nThere are three identities relating to the trigonometric functions that make working\nwith triangles easier:\n1. the area rule\n2. the sine rule\n3. the cosine rule\n280\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nEMBHQ\nInvestigation: The area rule\n1. Consider △ABC:\nB\nA\nC\n10\n54◦\n7\nComplete the following:\na) Area △ABC = 1\n2 × . . . × AC\nb) sin ˆB = . . . and AC = . . . × . . .\nc) Therefore area △ABC = . . . × . . . × . . . × . . .\n2. Consider △A′B′C′:\nB′\nA′\nC′\n10\n54◦\n7\nComplete the following:\na) How is △A′B′C′ different from △ABC?\nb) Calculate area △A′B′C′.\n3. Use your results to write a general formula for determining the area of △PQR:\nQ\nP\nR\nr\np\nq\n281\nChapter 6.\nTrigonometry\n\nFor any △ABC with AB = c, BC = a and AC = b, we can construct a perpendicular\nheight (h) from vertex A to the line BC:\nB\nA\nC\nc\na\nb\nh\nIn △ABC:\nsin ˆB = h\nc\n∴h = c sin ˆB\nAnd we know that\nArea △ABC = 1\n2 × a × h\n= 1\n2 × a × c sin ˆB\n∴Area △ABC = 1\n2ac sin ˆB\nAlternatively, we could write that\nsin ˆC = h\nb\n∴h = b sin ˆC\nAnd then we would have that\nArea △ABC = 1\n2 × a × h\n= 1\n2ab sin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nThe area rule\nIn any △ABC:\nArea △ABC = 1\n2bc sin ˆA\n= 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n282\n6.5.\nArea, sine, and cosine rules\n\nWorked example 22: The area rule\nQUESTION\nFind the area of △ABC (correct to two decimal places):\nA\n7\nB\nC\n50◦\nSOLUTION\nStep 1: Use the given information to determine unknown angles and sides\nAB = AC = 7\n(given)\n∴ˆB = ˆC = 50◦\n(∠s opp. equal sides)\nAnd ˆA = 180◦−50◦−50◦\n(∠s sum of △ABC)\n∴ˆA = 80◦\nStep 2: Use the area rule to calculate the area of △ABC\nNotice that we do not know the length of side a and must therefore choose the form\nof the area rule that does not include this side of the triangle.\nIn △ABC:\nArea = 1\n2bc sin ˆA\n= 1\n2(7)(7) sin 80◦\n= 24,13\nStep 3: Write the final answer\nArea of △ABC = 24,13 square units.\n283\nChapter 6.\nTrigonometry\n\nWorked example 23: The area rule\nQUESTION\nShow that the area of △DEF = 1\n2df sin ˆE.\nD\nF\nE\nH\ne\nd\nf\nh\n1\n2\nSOLUTION\nStep 1: Construct a perpendicular height h\nDraw DH such that DH ⊥EF and let DH = h, D ˆEF = ˆE1 and D ˆEH = ˆE2.\nIn △DHE:\nsin ˆE2 = h\nf\nh = f sin(180◦−ˆE1)\n(∠s on str. line)\n= f sin ˆE1\nStep 2: Use the area rule to calculate the area of △DEF\nIn △DEF:\nArea = 1\n2d × h\n= 1\n2df sin ˆE1\n284\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nIn any △PQR:\nP\nP\nQ\nQ\nR\nR\nq\nq\nr\nr\np\np\nArea △PQR = 1\n2qr sin ˆP\n= 1\n2pr sin ˆQ\n= 1\n2pq sin ˆR\nThe area rule states that the area of any triangle is equal to half the product of the\nlengths of the two sides of the triangle multiplied by the sine of the angle included by\nthe two sides.\nExercise 6 – 10: The area rule\n1. Draw a sketch and calculate the area of △PQR given:\na) ˆQ = 30◦; r = 10 and p = 7\nb) ˆR = 110◦; p = 8 and q = 9\n2. Find the area of △XY Z given XZ = 52 cm, XY = 29 cm and ˆX = 58,9◦.\n3. Determine the area of a parallelogram in which two adjacent sides are 10 cm\nand 13 cm and the angle between them is 55◦.\n4. If the area of △ABC is 5000 m2 with a = 150 m and b = 70 m, what are the two\npossible sizes of ˆC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2339\n1b. 233B\n2. 233C\n3. 233D\n4. 233F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n285\nChapter 6.\nTrigonometry\n\nThe sine rule\nEMBHR\nSo far we have only applied the trigonometric ratios to right-angled triangles. We now\nexpand the application of the trigonometric ratios to triangles that do not have a right\nangle:\nInvestigation: The sine rule\nIn △ABC, AC = 15, BC = 11 and ˆA = 48◦. Find ˆB.\nA\nC\nB\nb = 15\nF\na = 11\n48◦\n1. Method 1: using the sine ratio\na) Draw a sketch of △ABC.\nb) Construct CF ⊥AB.\nc) In △CBF:\nCF\n. . . = sin ˆB\n∴CF = . . . × sin ˆB\nd) In △CAF:\nCF\n15 = . . .\n∴CF = 15 × . . .\ne) Therefore we have that:\nCF = 15 × . . .\nand CF = . . . × sin ˆB\n∴15 × . . . = . . . × sin ˆB\n∴sin ˆB = . . . . . . . . .\n∴ˆB = . . .\n2. Method 2: using the area rule\n286\n6.5.\nArea, sine, and cosine rules\n\na) In △ABC:\nArea △ABC = 1\n2AB × AC × . . .\n= 1\n2AB × . . . × . . .\nb) And we also know that\nArea △ABC = 1\n2AB × . . . × sin ˆB\nc) We can equate these two equations and solve for ˆB:\n1\n2AB × . . . × sin ˆB = 1\n2AB × . . . × . . .\n∴. . . × sin ˆB = . . . × . . .\n∴sin ˆB = . . . × . . .\n∴ˆB = . . .\n3. Use your results to write a general formula for the sine rule given △PQR:\nP\nQ\nR\nq\nr\np\nFor any triangle ABC with AB = c, BC = a and AC = b, we can construct a perpen-\ndicular height (h) at F:\nA\nC\nB\nb\nF\na\nh\nc\nMethod 1: using the sine ratio\nIn △ABF:\nsin ˆB = h\nc\n∴h = c sin ˆB\n287\nChapter 6.\nTrigonometry\n\nIn △ACF:\nsin ˆC = h\nb\n∴h = b sin ˆC\nWe can equate the two equations\nc sin ˆB = b sin ˆC\n∴sin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that:\nsin ˆA\na\n= sin ˆC\nc\nor\na\nsin ˆA\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nMethod 2: using the area rule\nIn △ABC:\nArea △ABC = 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n∴1\n2ac sin ˆB = 1\n2ab sin ˆC\nc sin ˆB = b sin ˆC\nsin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\n288\n6.5.\nArea, sine, and cosine rules\n\nThe sine rule\nIn any △ABC:\nA\nC\nB\nb\na\nc\nsin ˆA\na\n= sin ˆB\nb\n= sin ˆC\nc\na\nsin ˆA\n=\nb\nsin ˆB\n=\nc\nsin ˆC\nSee video: 233G at www.everythingmaths.co.za\nWorked example 24: The sine rule\nQUESTION\nGiven △TRS with S ˆTR = 55◦, TR = 30 and R ˆST = 40◦, determine RS, ST and\nT ˆRS.\nSOLUTION\nStep 1: Draw a sketch\nLet RS = t, ST = r and TR = s.\nR\nS\nT\n55◦\n30\n40◦\nStep 2: Find T ˆRS using angles in a triangle\nT ˆRS + R ˆST + S ˆTR = 180◦\n(∠s sum of △TRS)\n∴T ˆRS = 180◦−40◦−55◦\n= 85◦\n289\nChapter 6.\nTrigonometry\n\nStep 3: Determine t and r using the sine rule\nt\nsin ˆT\n=\ns\nsin ˆS\nt\nsin 55◦=\n30\nsin 40◦\n∴t =\n30\nsin 40◦× sin 55◦\n= 38,2\nr\nsin ˆR\n=\ns\nsin ˆS\nr\nsin 85◦=\n30\nsin 40◦\n∴r =\n30\nsin 40◦× sin 85◦\n= 46,5\nWorked example 25: The sine rule\nQUESTION\nProve the sine rule for △MNP with MS ⊥NP.\nM\nP\nN\nS\nn\nm\np\nh\n1\n2\nSOLUTION\nStep 1: Use the sine ratio to express the angles in the triangle in terms of the length\nof the sides\nIn △MSN:\nsin ˆN2 = h\np\n∴h = p sin ˆN2\nand ˆN2 = 180◦−ˆN1\n∠s on str. line\n∴h = p sin(180◦−ˆN1)\n= p sin ˆN1\n290\n6.5.\nArea, sine, and cosine rules\n\nIn △MSP:\nsin ˆP = h\nn\n∴h = n sin ˆP\nStep 2: Equate the two equations to derive the sine rule\np sin ˆN1 = n sin ˆP\n∴sin ˆN1\nn\n= sin ˆP\np\nor\nn\nsin ˆN1\n=\np\nsin ˆP\nThe ambiguous case\nIf two sides and an interior angle of a triangle are given, and the side opposite the given\nangle is the shorter of the two sides, then we can draw two different triangles (△NMP\nand △NMP ′), both having the given dimensions. We call this the ambiguous case\nbecause there are two ways of interpreting the given information and it is not certain\nwhich is the required solution.\nM\nP ′\np\nN\nP\nn\nn\nWorked example 26: The ambiguous case\nQUESTION\nIn △ABC, AB = 82, BC = 65 and ˆA = 50◦. Draw △ABC and find ˆC (correct to\none decimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the ambiguous case\nWe notice that for the given dimensions of △ABC, the side BC opposite ˆA is shorter\nthan AB. This means that we can draw two different triangles with the given dimen-\nsions.\nA\nB\nC\n50◦\n65\n82\nA′\nC′\nB′\n65\n82\n50◦\n291\nChapter 6.\nTrigonometry\n\nStep 2: Solve for unknown angle using the sine rule\nIn △ABC:\nsin ˆA\nBC = sin ˆC\nAB\nsin 50◦\n65\n= sin ˆC\n82\n∴sin 50◦\n65\n× 82 = sin ˆC\n∴ˆC = 75,1◦\nIn △A′B′C′:\nWe know that sin(180 −ˆC) = sin ˆC, which means we can also have the solution\nˆC′ = 180◦−75,1◦\n= 104,9◦\nBoth solutions are correct.\nWorked example 27: Lighthouses\nQUESTION\nThere is a coastline with two lighthouses, one on either side of a beach. The two\nlighthouses are 0,67 km apart and one is exactly due east of the other. The lighthouses\ntell how close a boat is by taking bearings to the boat (a bearing is an angle measured\nclockwise from north). These bearings are shown on the diagram below.\nCalculate how far the boat is from each lighthouse.\nˆA = 127◦\nˆB = 255◦\nC\nSOLUTION\nWe see that the two lighthouses and the boat form a triangle. Since we know the\ndistance between the lighthouses and we have two angles we can use trigonometry\n292\n6.5.\nArea, sine, and cosine rules\n\nto find the remaining two sides of the triangle, the distance of the boat from the two\nlighthouses.\nb\nA\nb B\nb\nC\n15◦\n37◦\n128◦\n0,67 km\nWe need to determine the lengths of the two sides AC and BC. We can use the sine\nrule to find the missing lengths.\nBC\nsin ˆA\n= AB\nsin ˆC\nBC = AB . sin ˆA\nsin ˆC\n= (0,67 km) sin 37◦\nsin 128◦\n= 0,51 km\nAC\nsin ˆB\n= AB\nsin ˆC\nAC = AB . sin ˆB\nsin ˆC\n= (0,67 km) sin 15◦\nsin 128◦\n= 0,22 km\nExercise 6 – 11: Sine rule\n1. Find all the unknown sides and angles of the following triangles:\na) △PQR in which ˆQ = 64◦; ˆR = 24◦and r = 3\nb) △KLM in which ˆK = 43◦; ˆ\nM = 50◦and m = 1\nc) △ABC in which ˆA = 32,7◦; ˆC = 70,5◦and a = 52,3\nd) △XY Z in which ˆX = 56◦; ˆZ = 40◦and x = 50\n2. In △ABC, ˆA = 116◦; ˆC = 32◦and AC = 23 m. Find the lengths of the sides\nAB and BC.\n3. In △RST, ˆR = 19◦; ˆS = 30◦and RT = 120 km. Find the length of the side\nST.\n4. In △KMS, ˆK = 20◦; ˆ\nM = 100◦and s = 23 cm. Find the length of the side m.\n293\nChapter 6.\nTrigonometry\n\n5. In △ABD, ˆB = 90◦, AB = 10 cm and A ˆDB = 40◦. In △BCD, ˆC = 106◦and\nC ˆDB = 15◦. Determine BC.\nA\nB\nD\nC\n10\n106◦\n15◦\n40◦\n6. In △ABC, ˆA = 33◦, AC = 21 mm and AB = 17 mm. Can you determine BC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233H\n1b. 233J\n1c. 233K\n1d. 233M\n2. 233N\n3. 233P\n4. 233Q\n5. 233R\n6. 233S\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe cosine rule\nEMBHS\nInvestigation: The cosine rule\nIf a triangle is given with two sides and the included angle known, then we can not\nsolve for the remaining unknown sides and angles using the sine rule. We therefore\ninvestigate the cosine rule:\nIn △ABC, AB = 21, AC = 17 and ˆA = 33◦. Find ˆB.\nA\nC\nB\nH\n21\nc\n17\n33◦\n1. Determine CB:\na) Construct CH ⊥AB.\nb) Let AH = c and therefore HB = . . .\n294\n6.5.\nArea, sine, and cosine rules\n\nc) Applying the theorem of Pythagoras in the right-angled triangles:\nIn△CHB:\nCB2 = BH2 + CH2\n= (. . .)2 + CH2\n= 212 −(2)(21)c + c2 + CH2 . . . . . . (1)\nIn △CHA:\nCA2 = c2 + CH2\n172 = c2 + CH2 . . . . . . (2)\nSubstitute equation (2) into equation (1):\nCB2 = 212 −(2)(21)c + 172\nNow c is the only remaining unknown. In △CHA:\nc\n17 = cos 33◦\n∴c = 17 cos 33◦\nTherefore we have that\nCB2 = 212 −(2)(21)c + 172\n= 212 −(2)(21)(17 cos 33◦) + 172\n= 212 + 172 −(2)(21)(17) cos 33◦\n= 131,189 . . .\n∴CB = 11,5\n2. Use your results to write a general formula for the cosine rule given △PQR:\nP\nQ\nR\nq\nr\np\nThe cosine rule relates the length of a side of a triangle to the angle opposite it and the\nlengths of the other two sides.\n295\nChapter 6.\nTrigonometry\n\nConsider △ABC with CD ⊥AB:\nb\nD\nb\nA\nb\nB\nbC\nh\na\nb\nc\nc −d\nd\nIn △DCB: a2 = (c −d)2 + h2 from the theorem of Pythagoras.\nIn △ACD: b2 = d2 + h2 from the theorem of Pythagoras.\nSince h2 is common to both equations we can write:\na2 = (c −d)2 + h2\n∴h2 = a2 −(c −d)2\nAnd b2 = d2 + h2\n∴h2 = b2 −d2\n∴b2 −d2 = a2 −(c −d)2\na2 = b2 + (c2 −2cd + d2) −d2\n= b2 + c2 −2cd\nIn order to eliminate d we look at △ACD, where we have: cos ˆA = d\nb. So, d = b cos ˆA.\nSubstituting back we get: a2 = b2 + c2 −2bc cos ˆA.\nThe cosine rule\nIn any △ABC:\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\na2 = b2 + c2 −2bc cos ˆA\nb2 = a2 + c2 −2ac cos ˆB\nc2 = a2 + b2 −2ab cos ˆC\nSee video: 233T at www.everythingmaths.co.za\n296\n6.5.\nArea, sine, and cosine rules\n\nWorked example 28: The cosine rule\nQUESTION\nDetermine the length of QR.\nP\nR\nQ\n13 cm\n4 cm\n70◦\nSOLUTION\nStep 1: Use the cosine rule to solve for the unknown side\nQR2 = PR2 + QP 2 −2(PR)(QP) cos ˆP\n= 42 + 132 −2(4)(13) cos 70◦\n= 149,42 . . .\n∴QR = 12,2\nStep 2: Write the final answer\nQR = 12,2 cm\nWorked example 29: The cosine rule\nQUESTION\nDetermine ˆA.\n5\n7\n8\nA\nB\nC\nSOLUTION\nApplying the cosine rule:\na2 = b2 + c2 −2bc cos ˆA\n∴cos ˆA = b2 + c2 −a2\n2bc\n= 82 + 52 −72\n2 . 8 . 5\n= 0,5\n∴ˆA = 60◦\n297\nChapter 6.\nTrigonometry\n\nIt is very important:\n• not to round off before the final answer as this will affect accuracy;\n• to take the square root;\n• to remember to give units where applicable.\nHow to determine which rule to use:\n1. Area rule:\n• if no perpendicular height is given\n2. Sine rule:\n• if no right angle is given\n• if two sides and an angle are given (not the included angle)\n• if two angles and a side are given\n3. Cosine rule:\n• if no right angle is given\n• if two sides and the included angle are given\n• if three sides are given\nExercise 6 – 12: The cosine rule\n1. Solve the following triangles (that is, find all unknown sides and angles):\na) △ABC in which ˆA = 70◦; b = 4 and c = 9\nb) △RST in which RS = 14; ST = 26 and RT = 16\nc) △KLM in which KL = 5; LM = 10 and KM = 7\nd) △JHK in which ˆH = 130◦; JH = 13 and HK = 8\ne) △DEF in which d = 4; e = 5 and f = 7\n2. Find the length of the third side of the △XY Z where:\na) ˆX = 71,4◦; y = 3,42 km and z = 4,03 km\nb) x = 103,2 cm; ˆY = 20,8◦and z = 44,59 cm\n3. Determine the largest angle in:\na) △JHK in which JH = 6; HK = 4 and JK = 3\nb) △PQR where p = 50; q = 70 and r = 60\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233V\n1b. 233W\n1c. 233X\n1d. 233Y\n1e. 233Z\n2a. 2342\n2b. 2343\n3a. 2344\n3b. 2345\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n298\n6.5.\nArea, sine, and cosine rules\n\nSee video: 2346 at www.everythingmaths.co.za\nExercise 6 – 13: Area, sine and cosine rule\n1. Q is a ship at a point 10 km due south of another ship P. R is a lighthouse on\nthe coast such that ˆP = ˆQ = 50◦.\n10 km\nP\nQ\nR\n50◦\n50◦\nDetermine:\na) the distance QR\nb) the shortest distance from the lighthouse to the line joining the two ships\n(PQ).\n2. WXY Z is a trapezium, WX ∥Y Z with WX = 3 m; Y Z = 1,5 m; ˆZ = 120◦\nand ˆW = 30◦.\nDetermine the distances XZ and XY .\n1,5 m\n3 m\n30◦\n120◦\nW\nX\nY\nZ\n3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has\nbeen flying 3◦off course. At this point the plane is 500 km from Johannesburg.\nThe direct distance between Cape Town and Johannesburg airports is 1552 km.\nDetermine, to the nearest km:\na) The distance the plane has to travel to get to Cape Town and hence the\nextra distance that the plane has had to travel due to the pilot’s error.\nb) The correction, to one hundredth of a degree, to the plane’s heading (or\ndirection).\n4. ABCD is a trapezium (meaning that AB ∥CD). AB = x; B ˆAD = a; B ˆCD = b\nand B ˆDC = c.\nFind an expression for the length of CD in terms of x, a, b and c.\nA\nB\nC\nD\na\nb\nc\nx\n299\nChapter 6.\nTrigonometry\n\n5. A surveyor is trying to determine the distance between points X and Z. However\nthe distance cannot be determined directly as a ridge lies between the two points.\nFrom a point Y which is equidistant from X and Z, he measures the angle X ˆY Z.\nY\nX\nZ\nx\nθ\na) If XY = x and X ˆY Z = θ, show that XZ = x\np\n2(1 −cos θ).\nb) Calculate XZ (to the nearest kilometre) if x = 240 km and θ = 132◦.\n6. Find the area of WXY Z (to two decimal places):\nW\nX\nY\nZ\n120◦\n3\n4\n3,5\n7. Find the area of the shaded triangle in terms of x, α, β, θ and φ:\nA\nB\nC\nD\nE\nx\nα\nβ\nθ\nφ\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2347\n2. 2348\n3. 2349\n4. 234B\n5. 234C\n6. 234D\n7. 234F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n300\n6.5.\nArea, sine, and cosine rules\n\n6.6\nSummary\nEMBHT\nSee presentation: 234G at www.everythingmaths.co.za\nsquare identity\nquotient identity\ncos2 θ + sin2 θ = 1\ntan θ = sin θ\ncos θ\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\nnegative angles\nperiodicity identities\nco-function identities\nsin(−θ) = −sin θ\nsin(θ ± 360◦) = sin θ\nsin(90◦−θ) = cos θ\ncos(−θ) = cos θ\ncos(θ ± 360◦) = cos θ\ncos(90◦−θ) = sin θ\nsine rule\narea rule\ncosine rule\nsin A\na\n= sin B\nb\n= sin C\nc\narea △ABC = 1\n2bc sin A\na2 = b2 + c2 −2bc cos A\na\nsin A =\nb\nsin B =\nc\nsin C\narea △ABC = 1\n2ac sin B\nb2 = a2 + c2 −2ac cos B\narea △ABC = 1\n2ab sin C\nc2 = a2 + b2 −2ab cos C\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nGeneral solution:\n301\nChapter 6.\nTrigonometry\n\n1.\nIf sin θ = x\nθ = sin−1 x + k . 360◦\nor θ =\n\u0000180◦−sin−1 x\n\u0001\n+ k . 360◦\n2.\nIf cos θ = x\nθ = cos−1 x + k . 360◦\nor θ =\n\u0000360◦−cos−1 x\n\u0001\n+ k . 360◦\n3.\nIf tan θ = x\nθ = tan−1 x + k . 180◦\nfor k ∈Z.\nHow to determine which rule to use:\n1. Area rule:\n• no perpendicular height is given\n2. Sine rule:\n• no right angle is given\n• two sides and an angle are given (not the included angle)\n• two angles and a side are given\n3. Cosine rule:\n• no right angle is given\n• two sides and the included angle angle are given\n• three sides are given\nExercise 6 – 14: End of chapter exercises\n1. Write the following as a single trigonometric ratio:\ncos(90◦−A) sin 20◦\nsin(180◦−A) cos 70◦+ cos(180◦+ A) sin(90◦+ A)\n2. Determine the value of the following expression without using a calculator:\nsin 240◦cos 210◦−tan2 225◦cos 300◦cos 180◦\n302\n6.6.\nSummary\n\n3. Simplify:\nsin(180◦+ θ) sin(θ + 360◦)\nsin(−θ) tan(θ −360◦)\n4. Without the use of a calculator, evaluate:\n3 sin 55◦sin2 325◦\ncos(−145◦)\n−3 cos 395◦sin 125◦\n5. Prove the following identities:\na)\n1\n(cos x −1)(cos x + 1) =\n−1\ntan2 x cos2 x\nb) (1 −tan α) cos α = sin(90 + α) + cos(90 + α)\n6.\na) Prove: tan y +\n1\ntan y =\n1\ncos2 y tan y\nb) For which values of y ∈[0◦; 360◦] is the identity above undefined?\n7.\na) Simplify: sin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\nb) Hence, solve the equation\nsin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\n= tan θ\nfor θ ∈[0◦; 360◦].\n8. Given 12 tan θ = 5 and θ > 90◦.\na) Draw a sketch.\nb) Determine without using a calculator sin θ and cos(180◦+ θ).\nc) Use a calculator to find θ (correct to two decimal places).\n9.\nθ\nP(a; b)\n2\nx\ny\nO\nb\nIn the figure, P is a point on the Cartesian plane such that OP = 2 units and\nθ = 300◦. Without the use of a calculator, determine:\na) the values of a and b\nb) the value of sin(180◦−θ)\n10. Solve for x with x ∈[−180◦; 180◦] (correct to one decimal place):\na) 2 sin x\n2 = 0,86\n303\nChapter 6.\nTrigonometry\n\nb) tan(x + 10◦) = cos 202,6◦\nc) cos2 x −4 sin2 x = 0\n11. Find the general solution for the following equations:\na)\n1\n2 sin(x −25◦) = 0,25\nb) sin2 x + 2 cos x = −2\n12. Given the equation: sin 2α = 0,84\na) Find the general solution of the equation.\nb) Illustrate how this equation could be solved graphically for α ∈[0◦; 360◦].\nc) Write down the solutions for sin 2α = 0,84 for α ∈[0◦; 360◦].\n13.\nA\nT\nG\nN\nH\nn\nα\nβ\nA is the highest point of a vertical tower AT. At point N on the tower, n metres\nfrom the top of the tower, a bird has made its nest. The angle of inclination from\nG to point A is α and the angle of inclination from G to point N is β.\na) Express A ˆGN in terms of α and β.\nb) Express ˆA in terms of α and/or β.\nc) Show that the height of the nest from the ground (H) can be determined by\nthe formula\nH = n cos α sin β\nsin(α −β)\nd) Calculate the height of the nest H if n = 10 m, α = 68◦and β = 40◦(give\nyour answer correct to the nearest metre).\n304\n6.6.\nSummary\n\n14.\nA\nD\nB\nC\n11\n8\n5\nMr. Collins wants to pave his trapezium-shaped backyard, ABCD. AB ∥DC\nand ˆB = 90◦. DC = 11 m, AB = 8 m and BC = 5 m.\na) Calculate the length of the diagonal AC.\nb) Calculate the length of the side AD.\nc) Calculate the area of the patio using geometry.\nd) Calculate the area of the patio using trigonometry.\n15.\nA\nC\nB\n2t\nF\nt\nn\nn\n2n\nα\nIn △ABC, AC = 2A, AF = BF, A ˆFB = α and FC = 2AF. Prove that\ncos α = 1\n4.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 234H\n2. 234J\n3. 234K\n4. 234M\n5a. 234N\n5b. 234P\n6. 234Q\n7. 234R\n8. 234S\n9. 234T\n10a. 234V\n10b. 234W\n10c. 234X\n11a. 234Y\n11b. 234Z\n12. 2352\n13. 2353\n14. 2354\n15. 2355\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n305\nChapter 6.\nTrigonometry\n\n\nCHAPTER\n7\nMeasurement\n7.1\nArea of a polygon\n308\n7.2\nRight prisms and cylinders\n311\n7.3\nRight pyramids, right cones and spheres\n318\n7.4\nMultiplying a dimension by a constant factor\n322\n7.5\nSummary\n326\n\n7\nMeasurement\nThis chapter is a revision of perimeters and areas of two dimensional objects and\nvolumes of three dimensional objects. We also examine different combinations of\ngeometric objects and calculate areas and volumes in a variety of real-life contexts.\nSee video: 2356 at www.everythingmaths.co.za\n7.1\nArea of a polygon\nEMBHV\nSquare\ns\ns\nArea = s2\nRectangle\nh\nb\nArea = b × h\nTriangle\nh\nb\nArea = 1\n2b × h\nSee video: 2357 at www.everythingmaths.co.za\nTrapezium\nh\nb\na\nArea = 1\n2 (a + b) × h\nParallelogram\nh\nb\nArea = b × h\nCircle\nb r\nArea = πr2\n(Circumference = 2πr)\nSee video: 2358 at www.everythingmaths.co.za\n308\n7.1.\nArea of a polygon\n\nWorked example 1: Finding the area of a polygon\nQUESTION\nABCD is a parallelogram with DC = 15 cm, h = 8 cm and BF = 9 cm.\nA\nB\nC\nD\nH\n9 cm\n15 cm\nh\nF\nCalculate:\n1. the area of ABCD\n2. the perimeter of ABCD\nSOLUTION\nStep 1: Determine the area\nThe area of a parallelogram ABCD = base × height:\nArea = 15 × 8\n= 120 cm2\nStep 2: Determine the perimeter\nThe perimeter of a parallelogram ABCD = 2DC + 2BC.\nTo find the length of BC, we use AF ⊥BC and the theorem of Pythagoras.\nIn △ABF:\nAF 2 = AB2 −BF 2\n= 152 −92\n= 144\n∴AF = 12 cm\nAreaABCD = BC × AF\n120 = BC × 12\n∴BC = 10 cm\n∴PerimeterABCD = 2(15) + 2(10)\n= 50 cm\n309\nChapter 7.\nMeasurement\n\nExercise 7 – 1: Area of a polygon\n1. Vuyo and Banele are having a competition to see who can build the best kite\nusing balsa wood (a lightweight wood) and paper. Vuyo decides to make his kite\nwith one diagonal 1 m long and the other diagonal 60 cm long. The intersection\nof the two diagonals cuts the longer diagonal in the ratio 1 : 3.\nBanele also uses diagonals of length 60 cm and 1 m, but he designs his kite to\nbe rhombus-shaped.\na) Draw a sketch of Vuyo’s kite and write down all the known measurements.\nb) Determine how much balsa wood Vuyo will need to build the outside frame\nof the kite (give answer correct to the nearest cm).\nc) Calculate how much paper he will need to cover the frame of the kite.\nd) Draw a sketch of Banele’s kite and write down all the known measure-\nments.\ne) Determine how much wood and paper Banele will need for his kite.\nf) Compare the two designs and comment on the similarities and differences.\nWhich do you think is the better design? Motivate your answer.\n2. O is the centre of the bigger semi-circle with a radius of 10 units. Two smaller\nsemi-circles are inscribed into the bigger one, as shown on the diagram. Calcu-\nlate the following (in terms of π):\nO\nb\nb\na) The area of the shaded figure.\nb) The perimeter enclosing the shaded area.\n3. Karen’s engineering textbook is 30 cm long and 20 cm wide. She notices that\nthe dimensions of her desk are in the same proportion as the dimensions of her\ntextbook.\na) If the desk is 90 cm wide, calculate the area of the top of the desk.\nb) Karen uses some cardboard to cover each corner of her desk with an isosce-\nles triangle, as shown in the diagram:\n150 mm\n150 mm\ndesk\nCalculate the new perimeter and area of the visible part of the top of her\ndesk.\n310\n7.1.\nArea of a polygon\n\nc) Use this new area to calculate the dimensions of a square desk with the\nsame desk top area.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2359\n2. 235B\n3. 235C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.2\nRight prisms and cylinders\nEMBHW\nA right prism is a geometric solid that has a polygon as its base and vertical sides\nperpendicular to the base. The base and top surface are the same shape and size. It is\ncalled a “right” prism because the angles between the base and sides are right angles.\nA triangular prism has a triangle as its base, a rectangular prism has a rectangle as its\nbase, and a cube is a rectangular prism with all its sides of equal length. A cylinder is\nanother type of right prism which has a circle as its base. Examples of right prisms are\ngiven below: a rectangular prism, a cube, a triangular prism and a cylinder.\nSurface area of prisms and cylinders\nEMBHX\nSurface area is the total area of the exposed or outer surfaces of a prism. This is easier\nto understand if we imagine the prism to be a cardboard box that we can unfold. A\nsolid that is unfolded like this is called a net. When a prism is unfolded into a net, we\ncan clearly see each of its faces. In order to calculate the surface area of the prism, we\ncan then simply calculate the area of each face, and add them all together.\nFor example, when a triangular prism is unfolded into a net, we can see that it has\ntwo faces that are triangles and three faces that are rectangles. To calculate the surface\narea of the prism, we find the area of each triangle and each rectangle, and add them\ntogether.\nIn the case of a cylinder the top and bottom faces are circles and the curved surface\nflattens into a rectangle with a length that is equal to the circumference of the circular\nbase. To calculate the surface area we therefore find the area of the two circles and the\nrectangle and add them together.\n311\nChapter 7.\nMeasurement\n\nBelow are examples of right prisms that have been unfolded into nets. A rectangular\nprism unfolded into a net is made up of six rectangles.\nA cube unfolded into a net is made up of six identical squares.\nA triangular prism unfolded into a net is made up of two triangles and three rectangles.\nThe sum of the lengths of the rectangles is equal to the perimeter of the triangles.\nA cylinder unfolded into a net is made up of two identical circles and a rectangle with\nlength equal to the circumference of the circles.\n312\n7.2.\nRight prisms and cylinders\n\nWorked example 2: Calculating surface area\nQUESTION\nA box of chocolates has the following dimensions:\nlength = 25 cm\nwidth = 20 cm\nheight = 4 cm\n25 cm\n20 cm\n4 cm\nAnd a cylindrical tin of biscuits has the following dimensions:\ndiameter = 20 cm\nheight = 20 cm\nb\n20 cm\n20 cm\n1. Calculate the area of the wrapping paper needed to cover the entire box (assume\nno overlapping at the corners).\n2. Determine if this same sheet of wrapping paper would be enough to cover the\ntin of biscuits.\nSOLUTION\nStep 1: Determine the area of the rectangular box\nSurface area = 2 × (25 × 20) + 2 × (20 × 4) + 2 × (25 × 4)\n= 1360 cm2\n313\nChapter 7.\nMeasurement\n\nStep 2: Determine the area of the cylindrical tin\nThe radius of the cylinder = 20\n2 = 10 cm.\nSurface area = 2 × π(10)2 + 2π(10)(20)\n= 1885 cm2\nStep 3: Write the final answer\nNo, the area of the sheet of wrapping paper used to cover the box is not big enough to\ncover the tin.\nExercise 7 – 2: Calculating surface area\n1. A popular chocolate container is an equilateral right triangular prism with sides\nof 34 mm. The box is 170 mm long. Calculate the surface area of the box (to the\nnearest square centimetre).\n34 mm\n34 mm\n34 mm\n170 mm\n2. Gordon buys a cylindrical water tank to catch rain water off his roof. He discov-\ners a full 2 ℓtin of green paint in his garage and decides to paint the tank (not the\nbase). If he uses 250 ml to cover 1 m2, will he have enough green paint to cover\nthe tank with one layer of paint?\nDimensions of the tank:\ndiameter = 1,1 m\nheight = 1,4 m\nb\n1,1 m\n1,4 m\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235D\n2. 235F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n314\n7.2.\nRight prisms and cylinders\n\nVolume of prisms and cylinders\nEMBHY\nVolume, sometimes also called capacity, is the three dimensional space occupied by\nan object, or the contents of an object. It is measured in cubic units.\nThe volume of a right prism is simply calculated by multiplying the area of the base of\na solid by the height of the solid.\nRectangular\nprism\nl\nb\nh\nVolume = area of base × height\n= area of rectangle × height\n= l × b × h\nTriangular\nprism\nH\nb\nh\nVolume = area of base × height\n= area of triangle × height\n=\n\u00121\n2b × h\n\u0013\n× H\nCylinder\nh\nr\nVolume = area of base × height\n= area of circle × height\n= πr2 × h\nSee video: 235G at www.everythingmaths.co.za\n315\nChapter 7.\nMeasurement\n\nWorked example 3: Calculating volume\nQUESTION\nA rectangular glass vase with dimensions 28 cm × 18 cm × 8 cm is used for flower\narrangements. A florist uses a platic cylindrical jug to pour water into the glass vase.\nThe jug has a diameter of 142 mm and a height of 28 cm.\n28 cm\n18 cm\n8 cm\n142 mm\n28 cm\njug\nvase\n1. Will the plastic jug hold 5 ℓof water?\n2. Will a full jug of water be enough to fill the glass vase?\nSOLUTION\nStep 1: Determine the volume of the plastic jug\nThe diameter of the jug is 142 mm, therefore the radius =\n142\n2×10 = 7,1 cm.\nVolume of a cylinder = area of the base × height\nVolume of the jug = πr2 × h\n= π × (7,1)2 × 28\n= 4434 cm3\nAnd 1000 cm3 = 1 ℓ\n∴Volume of the jug = 4434\n1000\n= 4,434 ℓ\nNo, the capacity of the jug is not enough to hold 5 ℓof water.\n316\n7.2.\nRight prisms and cylinders\n\nStep 2: Determine the volume of the glass vase\nVolume of a rectangular prism = area of the base × height\nVolume of the vase = l × b × h\n= 28 × 18 × 8\n= 4032 cm3\n∴Volume of the vase = 4032\n1000\n= 4,032 ℓ\nYes, the volume of the jug is greater than the volume of the vase.\nExercise 7 – 3: Calculating volume\n1. The roof of Phumza’s house is the shape of a right-angled trapezium. A cylindri-\ncal water tank is positioned next to the house so that the rain on the roof runs\ninto the tank. The diameter of the tank is 140 cm and the height is 2,2 m.\n10 m\n8 m\n7,5 m\n2,2 m\n140 cm\na) Determine the area of the roof.\nb) Determine how many litres of water the tank can hold.\n2. The length of a side of a hexagonal sweet tin is 8 cm and its height is equal to\nhalf of the side length.\nA\nB\nC\nD\nE\nF\n8 cm\nh\na) Show that the interior angles are equal to 120◦.\n317\nChapter 7.\nMeasurement\n\nb) Determine the length of the line AE.\nc) Calculate the volume of the tin.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235H\n2. 235J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.3\nRight pyramids, right cones and spheres\nEMBHZ\nA pyramid is a geometric solid that has a polygon as its base and sides that converge\nat a point called the apex. In other words the sides are not perpendicular to the base.\nb\nThe triangular pyramid and square pyramid take their names from the shape of their\nbase. We call a pyramid a “right pyramid” if the line between the apex and the centre\nof the base is perpendicular to the base. Cones are similar to pyramids except that\ntheir bases are circles instead of polygons. Spheres are solids that are perfectly round\nand look the same from any direction.\nSurface area of pyramids, cones and spheres\nEMBJ2\nSquare\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n= b2 + 4\n\u0000 1\n2bhs\n\u0001\n= b (b + 2hs)\n318\n7.3.\nRight pyramids, right cones and spheres\n\nTriangular\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n=\n\u0000 1\n2b × hb\n\u0001\n+ 3\n\u0000 1\n2b × hs\n\u0001\n= 1\n2b (hb + 3hs)\nRight cone\nh\nr\nH\nSurface area = area of base +\narea of walls\n= πr2 + 1\n2 × 2πrh\n= πr (r + h)\nSphere\nb\nr\nSurface area = 4πr2\nVolume of pyramids, cones and spheres\nEMBJ3\nSquare\npyramid\nb\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × b2 × H\nTriangular\npyramid\nb\nh\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × 1\n2bh × H\n319\nChapter 7.\nMeasurement\n\nRight cone\nr\nH\nVolume = 1\n3 × area of base ×\nheight of cone\n= 1\n3 × πr2 × H\nSphere\nb\nr\nVolume = 4\n3πr3\nSee video: 235K at www.everythingmaths.co.za\nWorked example 4: Finding surface area and volume\nQUESTION\nThe Southern African Large Telescope (SALT) is housed in a cylindrical building with\na domed roof in the shape of a hemisphere. The height of the building wall is 17 m\nand the diameter is 26 m.\n17 m\n26 m\n1. Calculate the total surface area of the building.\n2. Calculate the total volume of the building.\n320\n7.3.\nRight pyramids, right cones and spheres\n\nSOLUTION\nStep 1: Calculate the total surface area\nTotal surface area = area of the dome + area of the cylinder\nSurface area =\n\u00141\n2(4πr2)\n\u0015\n+ [2πr × h]\n= 1\n2(4π)(13)2 + 2π(13)(17)\n= 2450 m2\nStep 2: Calculate the total volume\nTotal volume = volume of the dome + volume of the cylinder\nVolume =\n\u00141\n2 ×\n\u00124\n3πr3\n\u0013\u0015\n+\n\u0002\nπr2h\n\u0003\n= 2\n3π(13)3 + π(11)2(13)\n= 9543 m3\nExercise 7 – 4: Finding surface area and volume\n1. An ice-cream cone has a diameter of 52,4 mm and a total height of 146 mm.\n52,4 mm\n146 mm\na) Calculate the surface area of the ice-cream and the cone.\nb) Calculate the total volume of the ice-cream and the cone.\nc) How many ice-cream cones can be made from a 5 ℓtub of ice-cream (as-\nsume the cone is completely filled with ice-cream)?\n321\nChapter 7.\nMeasurement\n\nd) Consider the net of the cone given below. R is the length from the tip of\nthe cone to its perimeter, P.\nP\nR\nb\nM\ni. Determine the value of R.\nii. Calculate the length of arc P.\niii. Determine the length of arc M.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.4\nMultiplying a dimension by a constant factor\nEMBJ4\nWhen one or more of the dimensions of a prism or cylinder is multiplied by a constant,\nthe surface area and volume will change. The new surface area and volume can be\ncalculated by using the formulae from the preceding section.\nIt is important to see a relationship between the change in dimensions and the resulting\nchange in surface area and volume. These relationships make it simpler to calculate\nthe new volume or surface area of an object when its dimensions are scaled up or\ndown.\nConsider a rectangular prism of dimensions l, b and h. Below we multiply one, two\nand three of its dimensions by a constant factor of 5 and calculate the new volume and\nsurface area.\n322\n7.4.\nMultiplying a dimension by a constant factor\n\nDimensions\nVolume\nSurface\nOriginal dimensions\nl\nb\nh\nV = l × b × h\n= lbh\nA\n= 2 [(l × h) + (l × b) + (b × h)]\n= 2 (lh + lb + bh)\nMultiply one\ndimension by 5\nl\nb\n5h\nV1 = l × b × 5h\n= 5 (lbh)\n= 5V\nA1\n= 2 [(l × 5h) + (l × b) + (b × 5h)]\n= 2 (5lh + lb + 5bh)\nMultiply two\ndimensions by 5\n5l\nb\n5h\nV = 5l × b × 5h\n= 5 . 5(lbh)\n= 52V\nA2\n= 2 [(5l × 5h) + (5l × b) + (b × 5h)]\n= 2 × 5(5lh + lb + bh)\nMultiply all three\ndimensions by 5\n5l\n5b\n5h\nV = 5l × 5b × 5h\n= 53(lbh)\n= 53V\nA3\n= 2 [(5l × 5h) + (5l × 5b) + (5b × 5h)]\n= 2 × (52lh + 52lb + 52bh)\n= 52 × 2(lh + lb + bh)\n= 52A\nMultiply all three\ndimensions by k\nkl\nkb\nkh\nV = kl × kb × kh\n= k3(lbh)\n= k3V\nAk\n= 2 [(kl × kh) + (kl × kb) + (kb × kh)]\n= 2 × (k2lh + k2lb + k2bh)\n= k2 × 2(lh + lb + bh)\n= k2A\n323\nChapter 7.\nMeasurement\n\nWorked example 5: The effects of k\nQUESTION\nThe Nash family wants to build a television room onto their house. The dad draws up\nthe plans for the new square room of length k metres. The mum looks at the plans and\ndecides that the area of the room needs to be doubled. To achieve this:\n• the mum suggests doubling the length of the sides of the room\n• the dad recommends adding 2 m to the length of the sides\n• the daughter suggests multiplying the length of the sides by a factor of\n√\n2\n• the son suggests doubling only the width of the room\nWho’s suggestion will double the area of the square room? Show all calculations.\nSOLUTION\nStep 1: Draw a sketch\nk\nk\n2k\n2k\nk + 2\nk + 2\n√\n2k\n√\n2k\n2k\nk\nArea O\nArea M\nArea D\nArea d\nArea s\nStep 2: Calculate and compare\nFirst calculate the area of the square room in the original plan:\nArea O = length × length\n= k2\nTherefore, double the area of the room would be 2k2.\n324\n7.4.\nMultiplying a dimension by a constant factor\n\nConsider the mum’s suggestion of doubling the length of the sides of the room:\nArea M = length × length\n= 2k × 2k\n= 4k2\nThis area would be 4 times the original area.\nThe dad suggests adding 2 m to the length of the sides of the room:\nArea D = length × length\n= (k + 2) × (k + 2)\n= k2 + 4k + 2\n̸= 2k2\nThis is not double the original area.\nThe daughter suggests multiplying the length of the sides by a factor of\n√\n2:\nArea d = length × length\n=\n√\n2k ×\n√\n2k\n= 2k2\nThe daughter’s suggestion would double the area of the room. Practically, the length\nof the room could be multiplied by\n√\n2 ≈1,41 which would given an area of 1,96 m2.\nThe son suggests doubling only the width of the room:\nArea s = length × length\n= 2k × k\n= 2k2\nThe son’s suggestion would double the area of the room, however the room would no\nlonger be a square.\nStep 3: Write the final answer\nThe daughter’s suggestion of multiplying the length of the sides of the room by a factor\nof\n√\n2 would keep the shape of the room a square and would double the area of the\nroom.\nExercise 7 – 5: The effects of k\n1. Complete the following sentences:\na) If one dimension of a cube is multiplied by a factor 1\n2, the volume of the\ncube . . .\nb) If two dimensions of a cube are multiplied by a factor 7, the volume of the\ncube . . .\n325\nChapter 7.\nMeasurement\n\nc) If three dimensions of a cube are multiplied by a factor 3, then:\ni. each side of the cube will . . .\nii. the outer surface area of the cube will . . .\niii. the volume of the cube will . . .\nd) If each side of a cube is halved, then:\ni. the outer surface area of the cube will . . .\nii. the volume of the cube will . . .\n2. The municipality intends building a swimming pool of volume W 3 cubic metres.\nHowever, they realise that it will be very expensive to fill the pool with water, so\nthey decide to make the pool smaller.\na) The length and breadth of the pool are reduced by a factor of\n7\n10. Express\nthe new volume in terms of W.\nb) The dimensions of the pool are reduced so that the volume of the pool\ndecreases by a factor of 0,8. Determine the new dimensions of the pool in\nterms of W (remember that the pool must be a cube).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235N\n2. 235P\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.5\nSummary\nEMBJ5\nSee presentation: 235Q at www.everythingmaths.co.za\n1. Area is the two dimensional space inside the boundary of a flat object.\n2. Area formulae:\n• square: s2\n• rectangle: b × h\n• triangle: 1\n2b × h\n• trapezium: 1\n2 (a + b) × h\n• parallelogram: b × h\n• circle: πr2\n3. Surface area is the total area of the exposed or outer surfaces of a prism.\n4. A net is the unfolded “plan” of a solid.\n5. Volume is the three dimensional space occupied by an object, or the contents\nof an object.\n• Volume of a rectangular prism: l × b × h\n326\n7.5.\nSummary\n\n• Volume of a triangular prism:\n\u0000 1\n2b × h\n\u0001\n× H\n• Volume of a square prism or cube: s3\n• Volume of a cylinder: πr2 × h\n6. A pyramid is a geometric solid that has a polygon as its base and sides that\nconverge at a point called the apex. The sides are not perpendicular to the base.\n7. Surface area formulae:\n• square pyramid: b (b + 2h)\n• triangular pyramid: 1\n2b (hb + 3hs)\n• right cone: πr (r + hs)\n• sphere: 4πr2\n8. Volume formulae:\n• square pyramid: 1\n3 × b2 × H\n• triangular pyramid: 1\n3 × 1\n2bh × H\n• right cone: 1\n3 × πr2 × H\n• sphere: 4\n3πr3\nExercise 7 – 6: End of chapter exercises\n1.\na) Describe this figure in terms of a prism.\nb) Draw a net of this figure.\n2. Which of the following is a net of a cube?\na)\nb)\nc)\nd)\ne)\n327\nChapter 7.\nMeasurement\n\n3. Name and draw the following figures:\na) A prism with the least number of sides.\nb) A pyramid with the least number of vertices.\nc) A right prism with a kite base.\n4.\na)\ni. Determine how much paper is needed to make a box of width 16 cm,\nheight 3 cm and length 20 cm (assume no overlapping at corners).\nii. Give a mathematical name for the shape of the box.\niii. Calculate the volume of the box.\nb) Determine how much paper is needed to make a cube with a capacity of\n1 ℓ.\nc) Compare the box and the cube. Which has the greater volume and which\nrequires the most paper to make?\n5. ABCD is a rhombus with sides of length 3\n2x millimetres. The diagonals intersect\nat O and length DO = x millimetres. Express the area of ABCD in terms of x.\nO\nB\nD\nx\nC\nA\n3\n2x\n6. The diagram shows a rectangular pyramid with a base of length 80 cm and\nbreadth 60 cm. The vertical height of the pyramid is 45 cm.\n60 cm\n80 cm\n45 cm\nb\nh\nH\na) Calculate the volume of the pyramid.\nb) Calculate H and h.\nc) Calculate the surface area of the pyramid.\n7. A group of children are playing soccer in a field. The soccer ball has a capacity\nof 5000 cc (cubic centimetres). A drain pipe in the corner of the field has a\ndiameter of 20 cm. Is it possible for the children to lose their ball down the pipe?\nShow your calculations.\n328\n7.5.\nSummary\n\n8. A litre of washing powder goes into a standard cubic container at the factory.\na) Determine the length of the sides of the container.\nb) Determine the dimensions of the cubic container required to hold double\nthe volume of washing powder.\n9. A cube has sides of length k units.\na) Describe the effect on the volume of the cube if the height is tripled.\nb) If all three dimensions of the cube are tripled, determine the effect on the\nouter surface area.\nc) If all three dimensions of the cube are tripled, determine the effect on the\nvolume.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235R\n2. 235S\n3a. 235T\n3b. 235V\n3c. 235W\n4. 235X\n5. 235Y\n6. 235Z\n7. 2362\n8. 2363\n9. 2364\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n329\nChapter 7.\nMeasurement\n\n\nCHAPTER\n8\nEuclidean geometry\n8.1\nRevision\n332\n8.2\nCircle geometry\n333\n8.3\nSummary\n363\n\n8\nEuclidean geometry\n8.1\nRevision\nEMBJ6\nParallelogram\nEMBJ7\nA parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nSummary of the properties of a parallelogram:\n• Both pairs of opposite sides are parallel.\n• Both pairs of opposite sides are equal in length.\n• Both pairs of opposite angles are equal.\n• Both diagonals bisect each other.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\nThe mid-point theorem\nEMBJ8\nThe line joining the mid-points of two sides of a triangle is parallel to the third side\nand equal to half the length of the third side.\nA\nB\nC\nD\nE\nGiven: AD = DB and AE = EC, we can conclude that DE ∥BC and DE = 1\n2BC.\n332\n8.1.\nRevision\n\n8.2\nCircle geometry\nEMBJ9\nTerminology\nThe following terms are regularly used when referring to circles:\n• Arc — a portion of the circumference of a circle.\n• Chord — a straight line joining the ends of an arc.\n• Circumference — the perimeter or boundary line of a circle.\n• Radius (r) — any straight line from the centre of the circle to a point on the\ncircumference.\n• Diameter — a special chord that passes through the centre of the circle. A di-\nameter is a straight line segment from one point on the circumference to another\npoint on the circumference that passes through the centre of the circle.\n• Segment — part of the circle that is cut off by a chord. A chord divides a circle\ninto two segments.\n• Tangent — a straight line that makes contact with a circle at only one point on\nthe circumference.\nb\nb\nA\nB\nO\nP\na\nr\nc\nchord\ntangent\ndiameter\nradius\nsegment\nSee video: 2365 at www.everythingmaths.co.za\nAxioms\nAn axiom is an established or accepted principle. For this section, the following are\naccepted as axioms.\n333\nChapter 8.\nEuclidean geometry\n\n1. The theorem of Pythagoras states that the square of the hypotenuse of a right-\nangled triangle is equal to the sum of the squares of the other two sides.\n(AC)2 = (AB)2 + (BC)2\nC\nB\nA\n(AC)2\n(AB)2\n(BC)2\n2. A tangent is perpendicular to the radius (OT ⊥ST), drawn at the point of contact\nwith the circle.\nT\nS\nb\nO\nTheorems\nEMBJB\nA theorem is a hypothesis (proposition) that can be shown to be true by accepted\nmathematical operations and arguments. A proof is the process of showing a theorem\nto be correct.\nThe converse of a theorem is the reverse of the hypothesis and the conclusion. For\nexample, given the theorem “if A, then B”, the converse is “if B, then A”.\n334\n8.2.\nCircle geometry\n\nTheorem: Perpendicular line from circle centre bisects chord\nSTATEMENT\nIf a line is drawn from the centre of a circle perpendicular to a chord, then it bisects\nthe chord.\n(Reason: ⊥from centre bisects chord)\nGiven:\nCircle with centre O and line OP perpendicular to chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = PB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA2 = OP 2 + AP 2\n(Pythagoras)\nOB2 = OP 2 + BP 2\n(Pythagoras)\nand\nOA = OB\n(equal radii)\n∴AP 2 = BP 2\n∴AP = BP\nTherefore OP bisects AB.\nAlternative proof:\nIn △OPA and in △OPB,\nO ˆPA = O ˆPB\n(given OP ⊥AB)\nOA = OB\n(equal radii)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(RHS)\n∴AP = PB\nTherefore OP bisects AB.\n335\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Line from circle centre to mid-point of\nchord is perpendicular\nSTATEMENT\nIf a line is drawn from the centre of a circle to the mid-point of a chord, then the line\nis perpendicular to the chord.\n(Reason: line from centre to mid-point ⊥)\nGiven:\nCircle with centre O and line OP to mid-point P on chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nOP ⊥AB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA = OB\n(equal radii)\nAP = PB\n(given)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(SSS)\n∴O ˆPA = O ˆPB\nand O ˆPA + O ˆPB = 180◦\n(∠on str. line)\n∴O ˆPA = O ˆPB = 90◦\nTherefore OP ⊥AB.\nSee video: 2366 at www.everythingmaths.co.za\n336\n8.2.\nCircle geometry\n\nTheorem: Perpendicular bisector of chord passes through circle centre\nSTATEMENT\nIf the perpendicular bisector of a chord is drawn, then the line will pass through the\ncentre of the circle.\n(Reason: ⊥bisector through centre)\nGiven:\nCircle with mid-point P on chord AB.\nLine QP is drawn such that Q ˆPA = Q ˆPB = 90◦.\nLine RP is drawn such that R ˆPA = R ˆPB = 90◦.\nb\nb\nA\nB\nQ\nP\nR\nRequired to prove:\nCircle centre O lies on the line PR\nPROOF\nDraw lines QA and QB.\nDraw lines RA and RB.\nIn △QPA and in △QPB,\nAP = PB\n(given)\nQP = QP\n(common side)\nQ ˆPA = Q ˆPB = 90◦\n(given)\n∴△QPA ≡△QPB\n(SAS)\n∴QA = QB\nSimilarly it can be shown that in △RPA and in △RPB, RA = RB.\nWe conclude that all the points that are equidistant from A and B will lie on the\nline PR extended. Therefore the centre O, which is equidistant to all points on the\ncircumference, must also lie on the line PR.\n337\nChapter 8.\nEuclidean geometry\n\nWorked example 1: Perpendicular line from circle centre bisects chord\nQUESTION\nGiven OQ ⊥PR and PR = 8 units, determine the value of x.\nO\nx\n5\nP\nQ\nR\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nPQ = QR = 4\n(⊥from centre bisects chord)\nStep 2: Solve for x\nIn △OQP:\nPQ = 4\n(⊥from centre bisects chord)\nOP 2 = OQ2 + QP 2\n(Pythagoras)\n52 = x2 + 42\n∴x2 = 25 −16\nx2 = 9\nx = 3\nStep 3: Write the final answer\nx = 3 units.\n338\n8.2.\nCircle geometry\n\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. In the circle with centre O, OQ ⊥PR,\nOQ = 4 units and PR = 10. Determine\nx.\nO\n4\nP\nQ\nR\nx\n2. In the circle with centre O and radius\n= 10 units, OQ ⊥PR and PR = 8. De-\ntermine x.\nO\nx\n10\nP\nQ\nR\n3. In the circle with centre O, OQ ⊥PR,\nPR = 12 units and SQ = 2 units. Deter-\nmine x.\nO\nx\nP\nQ\nR\nS\n4. In the circle with centre O, OT ⊥SQ,\nOT ⊥PR, OP = 10 units, ST = 5 units\nand PU = 8 units. Determine TU.\nO\nV\n8\nP\nR\nU\n10\n5\nT\nQ\nS\n5. In the circle with centre O, OT ⊥QP,\nOS ⊥PR, OT = 5 units, PQ = 24 units\nand PR = 25 units. Determine OS = x.\nO\nx\nP\nS\n5\nT\nQ\nR\n25\n24\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2367\n2. 2368\n3. 2369\n4. 236B\n5. 236C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n339\nChapter 8.\nEuclidean geometry\n\nInvestigation: Angles subtended by an arc at the centre and the circumference of\na circle\n1. Measure angles x and y in each of the following graphs:\nb\nx1\ny1\nb\nx2\ny2\nb\nx3\ny3\n2. Complete the table:\nx\ny\n3. Use your results to make a conjecture about the relationship between angles\nsubtended by an arc at the centre of a circle and angles at the circumference of\na circle.\n4. Now draw three of your own similar diagrams and measure the angles to check\nyour conjecture.\n340\n8.2.\nCircle geometry\n\nTheorem: Angle at the centre of a circle is twice the size of the angle at the cir-\ncumference\nSTATEMENT\nIf an arc subtends an angle at the centre of a circle and at the circumference, then the\nangle at the centre is twice the size of the angle at the circumference.\n(Reason: ∠at centre = 2∠at circum.)\nGiven:\nCircle with centre O, arc AB subtending A ˆOB at the centre of the circle, and A ˆPB at\nthe circumference.\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nRequired to prove:\nA ˆOB = 2A ˆPB\nPROOF\nDraw PO extended to Q and let A ˆOQ = ˆO1 and B ˆOQ = ˆO2.\nˆO1 = A ˆPO + P ˆAO\n(ext. ∠△= sum int. opp. ∠s)\nand A ˆPO = P ˆAO\n(equal radii, isosceles △APO)\n∴ˆO1 = A ˆPO + A ˆPO\nˆO1 = 2A ˆPO\nSimilarly, we can also show that ˆO2 = 2B ˆPO.\nFor the first two diagrams shown above we have that:\nA ˆOB = ˆO1 + ˆO2\n= 2A ˆPO + 2B ˆPO\n= 2(A ˆPO + B ˆPO)\n∴A ˆOB = 2(A ˆPB)\nAnd for the last diagram:\nA ˆOB = ˆO2 −ˆO1\n= 2B ˆPO −2A ˆPO\n= 2(B ˆPO −A ˆPO)\n∴A ˆOB = 2(A ˆPB)\n341\nChapter 8.\nEuclidean geometry\n\nWorked example 2: Angle at the centre of circle is twice angle at circumference\nQUESTION\nGiven HK, the diameter of the circle passing through centre O.\nb\nJ\nH\nK\nO\na\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nStep 2: Solve for a\nIn △HJK:\nH ˆOK = 180◦\n(∠on str. line)\n= 2a\n(∠at centre = 2∠at circum.)\n∴2a = 180◦\na = 180◦\n2\n= 90◦\nStep 3: Conclusion\nThe diameter of a circle subtends a right angle at the circumference (angles in a semi-\ncircle).\n342\n8.2.\nCircle geometry\n\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\nGiven O is the centre of the circle, determine the unknown angle in each of the fol-\nlowing diagrams:\n1.\nb\nJ\nH\nK\nO\nb\n45◦\n2.\nbO\nJ\nK\nH\n45◦\nc\n3.\nb\nO\nK\nJ\n100◦\nH\nd\n4.\nb\nO\nH\nJ\ne\nK\n35◦\n5.\nb\nO\nJ\nK\nH\n120◦\nf\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236D\n2. 236F\n3. 236G\n4. 236H\n5. 236J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n343\nChapter 8.\nEuclidean geometry\n\nInvestigation: Subtended angles in the same segment of a circle\n1. Measure angles a, b, c, d and e in the diagram below:\na\ne\nd\nb\nc\nP\nQ\n2. Choose any two points on the circumference of the circle and label them A and\nB.\n3. Draw AP and BP, and measure A ˆPB.\n4. Draw AQ and BQ, and measure A ˆQB.\n5. What do you observe? Make a conjecture about these types of angles.\nTheorem: Subtended angles in the same segment of a circle are equal\nSTATEMENT\nIf the angles subtended by a chord of the circle are on the same side of the chord, then\nthe angles are equal.\n(Reason: ∠s in same seg.)\nGiven:\nCircle with centre O, and points P and Q on the circumference of the circle. Arc AB\nsubtends A ˆPB and A ˆQB in the same segment of the circle.\n344\n8.2.\nCircle geometry\n\nbO\nA\nB\nP\nQ\nRequired to prove:\nA ˆPB = A ˆQB\nPROOF\nA ˆOB = 2A ˆPB\n(∠at centre = 2∠at circum.)\nA ˆOB = 2A ˆQB\n(∠at centre = 2∠at circum.)\n∴2A ˆPB = 2A ˆQB\nA ˆPB = A ˆQB\nEqual arcs subtend equal angles\nFrom the theorem above we can deduce that if angles at the circumference of a circle\nare subtended by arcs of equal length, then the angles are equal. In the figure below,\nnotice that if we were to move the two chords with equal length closer to each other,\nuntil they overlap, we would have the same situation as with the theorem above. This\nshows that the angles subtended by arcs of equal length are also equal.\nb\nb\n345\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Concyclic points\nSTATEMENT\nIf a line segment subtends equal angles at two other points on the same side of the line\nsegment, then these four points are concyclic (lie on a circle).\nGiven:\nLine segment AB subtending equal angles at points P and Q on the same side of the\nline segment AB.\nA\nB\nR\nQ\nP\nRequired to prove:\nA, B, P and Q lie on a circle.\nPROOF\nProof by contradiction:\nPoints on the circumference of a circle: we know that there are only two possible\noptions regarding a given point — it either lies on circumference or it does not.\nWe will assume that point P does not lie on the circumference.\nWe draw a circle that cuts AP at R and passes through A, B and Q.\nA ˆQB = A ˆRB\n(∠s in same seg.)\nbut A ˆQB = A ˆPB\n(given)\n∴A ˆRB = A ˆPB\nbut A ˆRB = A ˆPB + R ˆBP\n(ext. ∠△= sum int. opp.)\n∴R ˆBP = 0◦\nTherefore the assumption that the circle does not pass through P must be false.\nWe can conclude that A, B, Q and P lie on a circle (A, B, Q and P are concyclic).\n346\n8.2.\nCircle geometry\n\nWorked example 3: Concyclic points\nQUESTION\nGiven FH ∥EI and E ˆIF = 15◦, determine the value of b.\nE\nF\nG\nH\nI\n15◦\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nH ˆFI = 15◦\n(alt. ∠, FH ∥EI)\nand b = H ˆFI\n(∠s in same seg.)\n∴b = 15◦\nExercise 8 – 3: Subtended angles in the same segment\n1. Find the values of the unknown angles.\na)\nA\nB\nC\nD\n21◦\na\nb)\nJ\nK\nL\nM\n24◦\nc\n102◦\nd\nc)\nN\nO\nP\nQ\n17◦\nd\n347\nChapter 8.\nEuclidean geometry\n\n2.\nR\nS\nT\nU\nV\n45◦\n35◦\n15◦\ne\na) Given T ˆV S = S ˆV R, deter-\nmine the value of e.\nb) Is TV a diameter of the cir-\ncle? Explain your answer.\n3.\nb\nW\nX\nY\nZ\nO\n35◦\nf\nT\n1\n2\nGiven circle with centre O, WT =\nTY and X ˆWT = 35◦. Determine\nf.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236K\n1b. 236M\n1c. 236N\n2. 236P\n3. 236Q\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nCyclic quadrilaterals\nCyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference\nof a circle (concyclic).\nInvestigation: Cyclic quadrilaterals\nConsider the diagrams given below:\nCircle 1\nCircle 2\nCircle 3\nA\nB\nC\nD\nA\nB\nC\nD\nA\nB\nC\nD\n348\n8.2.\nCircle geometry\n\n1. Complete the following:\nABCD is a cyclic quadrilateral because . . . . . .\n2. Complete the table:\nCircle 1\nCircle 2\nCircle 3\nˆA =\nˆB =\nˆC =\nˆD =\nˆA + ˆC =\nˆB + ˆD =\n3. Use your results to make a conjecture about the relationship between angles of\ncyclic quadrilaterals.\nTheorem: Opposite angles of a cyclic quadrilateral\nSTATEMENT\nThe opposite angles of a cyclic quadrilateral are supplementary.\n(Reason: opp. ∠s cyclic quad.)\nGiven:\nCircle with centre O with points A, B, P and Q on the circumference such that ABPQ\nis a cyclic quadrilateral.\nbO\nA\nB\nP\nQ\n1\n2\nRequired to prove:\nA ˆBP + A ˆQP = 180◦and Q ˆAB + Q ˆPB = 180◦\n349\nChapter 8.\nEuclidean geometry\n\nPROOF\nDraw AO and OP. Label ˆO1 and ˆO2.\nˆO1 = 2A ˆBP\n(∠at centre = 2∠at circum.)\nˆO2 = 2A ˆQP\n(∠at centre = 2∠at circum.)\nand ˆO1 + ˆO2 = 360◦\n(∠s around a point)\n∴2A ˆBP + 2A ˆQP = 360◦\nA ˆBP + A ˆQP = 180◦\nSimilarly, we can show that Q ˆAB + Q ˆPB = 180◦.\nConverse: interior opposite angles of a quadrilateral\nIf the interior opposite angles of a quadrilateral are supplementary, then the quadrilat-\neral is cyclic.\nExterior angle of a cyclic quadrilateral\nIf a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle.\nb\nb\nWorked example 4: Opposite angles of a cyclic quadrilateral\nQUESTION\nGiven the circle with centre O and cyclic quadrilateral PQRS. SQ is drawn and\nS ˆPQ = 34◦. Determine the values of a, b and c.\nbO\nP\nQ\nR\nS\na\nb\nc\n34◦\n350\n8.2.\nCircle geometry\n\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nS ˆPQ + c = 180◦\n(opp. ∠s cyclic quad supp.)\n∴c = 180◦−34◦\n= 146◦\na = 90◦\n(∠in semi circle)\nIn △PSQ:\na + b + 34◦= 180◦\n(∠sum of △)\n∴b = 180◦−90◦−34◦\n= 56◦\nMethods for proving a quadrilateral is cyclic\nThere are three ways to prove that a quadrilateral is a cyclic quadrilateral:\nMethod of proof\nReason\nR\nQ\nS\nP\nIf ˆP + ˆR = 180◦or ˆS +\nˆQ = 180◦, then PQRS is\na cyclic quad.\nopp.\nint.\nangles\nsuppl.\nR\nQ\nS\nP\nIf ˆP = ˆQ or ˆS = ˆR, then\nPQRS is a cyclic quad.\nangles in the same\nseg.\nR\nQ\nS\nP\nT\nIf T ˆQR = ˆS, then PQRS\nis a cyclic quad.\next.\nangle equal to\nint. opp. angle\n351\nChapter 8.\nEuclidean geometry\n\nWorked example 5: Proving a quadrilateral is a cyclic quadrilateral\nQUESTION\nProve that ABDE is a cyclic quadrilateral.\nbO\nE\nC\nD\nA\nB\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Prove that ABDE is a cyclic quadrilateral\nD ˆBC = 90◦\n(∠in semi circle)\nand ˆE = 90◦\n(given)\n∴D ˆBC = ˆE\n∴ABDE is a cyclic quadrilateral\n(ext. ∠equals int. opp. ∠)\nExercise 8 – 4: Cyclic quadrilaterals\n1. Find the values of the unknown angles.\na)\nX\nY\nZ\nW\na\nb\n106◦\n87◦\nb)\nH\nI\nJ\nK\nL\n114◦\na\nc)\nU\nV\nW\nX\n57◦\na\n86◦\n352\n8.2.\nCircle geometry\n\n2. Prove that ABCD is a cyclic quadrilateral:\na) D\nC\n72◦\nB\nA\n32◦\nM\n40◦\nb) D\nC\n70◦\nB\nA\n35◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236R\n1b. 236S\n1c. 236T\n2a. 236V\n2b. 236W\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTangent line to a circle\nA tangent is a line that touches the circumference of a circle at only one place. The\nradius of a circle is perpendicular to the tangent at the point of contact.\nb\nO\n353\nChapter 8.\nEuclidean geometry\n\nTheorem: Two tangents drawn from the same point outside a circle\nSTATEMENT\nIf two tangents are drawn from the same point outside a circle, then they are equal in\nlength.\n(Reason: tangents from same point equal)\nGiven:\nCircle with centre O and tangents PA and PB, where A and B are the respective\npoints of contact for the two lines.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = BP\nPROOF\nIn △AOP and △BOP,\nO ˆAP = O ˆBP = 90◦\n(tangent ⊥radius)\nAO = BO\n(equal radii)\nOP = OP\n(common side)\n∴△AOP ≡△BOP\n(RHS)\n∴AP = BP\n354\n8.2.\nCircle geometry\n\nWorked example 6: Tangents from the same point outside a circle\nQUESTION\nIn the diagram below AE = 5 cm, AC = 8 cm and CE = 9 cm. Determine the values\nof a, b and c.\nA\nB\nC\nD\nE\nF\nAE = 5 cm\nAC = 8 cm\nCE = 9 cm\na\nb\nc\nb\nb\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for a, b and c\nAB = AF = a\n(tangents from A)\nEF = ED = c\n(tangents from E)\nCB = CD = b\n(tangents from C)\n∴AE = a + c = 5\nand AC = a + b = 8\nand CE = b + c = 9\nStep 3: Solve for the unknown variables using simultaneous equations\na + c = 5\n. . . (1)\na + b = 8\n. . . (2)\nb + c = 9\n. . . (3)\nSubtract equation (1) from equation (2) and then substitute into equation (3):\n(2) −(1)\nb −c = 8 −5\n= 3\n∴b = c + 3\nSubstitute into (3)\nc + 3 + c = 9\n2c = 6\nc = 3\n∴a = 2\nand b = 6\n355\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 5: Tangents to a circle\nFind the values of the unknown lengths.\n1.\nb\nG\nH\nI\nJ\nd\n5 cm\n8 cm\n2.\nb\nK\nL\nM\nN\nO\nP\ne\nLN = 7,5 cm\n2 cm\n6 cm\n3.\nb\nb\nR\nQ\nS\nf\n3 cm\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236X\n2. 236Y\n3. 236Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Tangent-chord theorem\nConsider the diagrams given below:\nDiagram 1\nDiagram 2\nDiagram 3\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\n1. Measure the following angles with a protractor and complete the table:\nDiagram 1\nDiagram 2\nDiagram 3\nA ˆBC =\nˆD =\nˆE =\n2. Use your results to complete the following: the angle between a tangent to a\ncircle and a chord is . . . . . . to the angle in the alternate segment.\n356\n8.2.\nCircle geometry\n\nTheorem: Tangent-chord theorem\nSTATEMENT\nThe angle between a tangent to a circle and a chord drawn at the point of contact, is\nequal to the angle which the chord subtends in the alternate segment.\n(Reason: tan. chord theorem)\nGiven:\nCircle with centre O and tangent SR touching the circle at B. Chord AB subtends ˆP1\nand ˆQ1.\nb\nO\nA\nB\nP\n1\n1\nQ\nT\n1\nS\nR\nRequired to prove:\n1. A ˆBR = A ˆPB\n2. A ˆBS = A ˆQB\nPROOF\nDraw diameter BT and join T to A.\nLet A ˆTB = T1.\nA ˆBS + A ˆBT = 90◦\n(tangent ⊥radius)\nB ˆAT = 90◦\n(∠in semi circle)\n∴A ˆBT + T1 = 90◦\n(∠sum of △BAT)\n∴A ˆBS = T1\nbut Q1 = T1\n(∠s in same segment)\n∴Q1 = A ˆBS\nA ˆBS + A ˆBR = 180◦\n(∠s on str. line)\nˆQ1 + ˆP1 = 180◦\n(opp. ∠s cyclic quad. supp.)\n∴A ˆBS + A ˆBR = Q1 + P1\nand A ˆBS = Q1\n∴A ˆBR = P1\n357\nChapter 8.\nEuclidean geometry\n\nWorked example 7: Tangent-chord theorem\nQUESTION\nDetermine the values of h and s.\nP\nO\nQ\nS\nR\nh + 20◦s\n4h\n4h −70◦\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for h\nO ˆQS = S ˆRQ\n(tangent chord theorem)\nh + 20◦= 4h −70◦\n90◦= 3h\n∴h = 30◦\nStep 3: Solve for s\nP ˆQR = Q ˆSR\n(tangent chord theorem)\ns = 4h\n= 4(30◦)\n= 120◦\n358\n8.2.\nCircle geometry\n\nExercise 8 – 6: Tangent-chord theorem\n1. Find the values of the unknown letters, stating reasons.\nQ\nR\nS\nO\nP\na\nb\n33◦\na)\nO\nP\nQ\nR\nS\nc\nd\n72◦\nb)\nO\nP\nQ\nR\nS\ng\nf\n38◦\n47◦\nc)\nR\nP\nO\nQ\nl\n1\n1\n66◦\nd)\nO\nP\nQ\nR\nS\ni\nj\nk\n39◦\n101◦\ne)\nO\nR\nQ\nS\nT\nm\nn\no\n34◦\nf)\nO\n•\nP\nR\nQ\nS\nT\np\nq\nr\n52◦\ng)\n359\nChapter 8.\nEuclidean geometry\n\n2. O is the centre of the circle and SPT is a tangent, with OP ⊥ST. Determine\na, b and c, giving reasons.\nO•\nS\nT\nP\nM\nN\na\nb\nc\n64◦\n3.\nP\nL\nA\nB\nC\n1 2\n3\n1\n2\nD\nGiven AB = AC, AP ∥BC and ˆA2 = ˆB2. Prove:\na) PAL is a tangent to the circle ABC.\nb) AB is a tangent to the circle ADP.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2372\n1b. 2373\n1c. 2374\n1d. 2375\n1e. 2376\n1f. 2377\n1g. 2378\n2. 2379\n3. 237B\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nConverse: tangent-chord theorem\nIf a line drawn through the end point of a chord forms an angle equal to the angle\nsubtended by the chord in the alternate segment, then the line is a tangent to the\ncircle.\n(Reason: ∠between line and chord = ∠in alt. seg. )\n360\n8.2.\nCircle geometry\n\nWorked example 8: Applying the theorems\nQUESTION\nA\nD\nB\nC\nO\nE\nF\nBD is a tangent to the circle with centre O, with BO ⊥AD.\nProve that:\n1. CFOE is a cyclic quadrilateral\n2. FB = BC\n3. ∠A ˆOC = 2B ˆFC\n4. Will DC be a tangent to the circle passing through C, F, O and E? Motivate your\nanswer.\nSOLUTION\nStep 1: Prove CFOE is a cyclic quadrilateral by showing opposite angles are supple-\nmentary\nBO ⊥OD\n(given)\n∴F ˆOE = 90◦\nF ˆCE = 90◦\n(∠in semi circle)\n∴CFOE is a cyclic quad.\n(opp. ∠s suppl.)\nStep 2: Prove BFC is an isosceles triangle\nTo show that FB = BC we first prove △BFC is an isosceles triangle by showing that\nB ˆFC = B ˆCF.\nB ˆCF = C ˆEO\n(tangent-chord)\nC ˆEO = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴B ˆFC = B ˆCF\n∴FB = BC\n(△BFC isosceles)\n361\nChapter 8.\nEuclidean geometry\n\nStep 3: Prove A ˆOC = 2B ˆFC\nA ˆOC = 2A ˆEC\n(∠at centre = 2∠at circum.)\nand A ˆEC = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴A ˆOC = 2B ˆFC\nStep 4: Determine if DC is a tangent to the circle through C, F, O and E\nProof by contradiction.\nLet us assume that DC is a tangent to the circle passing through the points C, F, O\nand E:\n∴D ˆCE = C ˆOE\n(tangent-chord)\nAnd using the circle with centre O and tangent BD we have that:\nD ˆCE = C ˆAE\n(tangent-chord)\nbut C ˆAE = 1\n2C ˆOE\n(∠at centre = 2∠at circum.)\n∴D ˆCE ̸= C ˆOE\nTherefore our assumption is not correct and we can conclude that DC is not a tangent\nto the circle passing through the points C, F, O and E.\nWorked example 9: Applying the theorems\nQUESTION\nA\nB\nC\nD\nE\nF\nG\nH\nFD is drawn parallel to the tangent CB\n362\n8.2.\nCircle geometry\n\nProve that:\n1. FADE is a cyclic quadrilateral\n2. F ˆEA = ˆB\nSOLUTION\nStep 1: Prove FADE is a cyclic quadrilateral using angles in the same segment\nF ˆDC = D ˆCB\n(alt. ∠s FD ∥CB)\nand D ˆCB = C ˆAE\n(tangent-chord)\n∴F ˆDC = C ˆAE\n∴FADE is a cyclic quad.\n(∠s in same seg.)\nStep 2: Prove F ˆEA = ˆB\nF ˆDA = ˆB\n(corresp. ∠s FD ∥CB)\nand F ˆEA = F ˆDA\n(∠s same seg. cyclic quad. FADE)\n∴F ˆEA = ˆB\n8.3\nSummary\nEMBJC\nSee presentation: 237C at www.everythingmaths.co.za\n• Arc An arc is a portion of the circumference of a circle.\n• Chord - a straight line joining the ends of an arc.\n• Circumference - perimeter or boundary line of a circle.\n• Radius (r) - any straight line from the centre of the circle to a point on the cir-\ncumference.\n• Diameter - a special chord that passes through the centre of the circle. A diame-\nter is the length of a straight line segment from one point on the circumference to\nanother point on the circumference, that passes through the centre of the circle.\n• Segment A segment is a part of the circle that is cut off by a chord. A chord\ndivides a circle into two segments.\n• Tangent - a straight line that makes contact with a circle at only one point on the\ncircumference.\n• A tangent line is perpendicular to the radius, drawn at the point of contact with\nthe circle.\n363\nChapter 8.\nEuclidean geometry\n\nb O\nM\nA\nB\n• If O is the centre and OM ⊥AB, then AM =\nMB.\n• If O is the centre and AM\n= MB, then\nA ˆ\nMO = B ˆ\nMO = 90◦.\n• If AM = MB and OM ⊥AB, then ⇒MO\npasses through centre O.\nb\n2x\nx\n2y\ny\nx\nIf an arc subtends an angle at the centre of a cir-\ncle and at the circumference, then the angle at the\ncentre is twice the size of the angle at the circum-\nference.\nb\nb\nAngles at the circumference subtended by the same\narc (or arcs of equal length) are equal.\nA\nB\nC\nD\n1\n2\nE\nThe four sides of a cyclic quadrilateral ABCD are\nchords of the circle with centre O.\n• ˆA + ˆC = 180◦(opp. ∠s supp.)\n• ˆB + ˆD = 180◦(opp. ∠s supp.)\n• E ˆBC = ˆD (ext. ∠cyclic quad.)\n• ˆA1 = ˆA2 = ˆC (vert. opp. ∠, ext. ∠cyclic\nquad.)\nA\nB\nC\nD\nProving a quadrilateral is cyclic: If ˆA + ˆC = 180◦or\nˆB+ ˆD = 180◦, then ABCD is a cyclic quadrilateral.\n364\n8.3.\nSummary\n\nA\nB\nC\nD\n1\n1\nIf ˆA1 = ˆC or ˆD1 = ˆB, then ABCD is a cyclic\nquadrilateral.\nA\nB\nC\nD\nIf ˆA = ˆB or ˆC = ˆD, then ABCD is a cyclic quadri-\nlateral.\nb\nA\nB\nO\nT\nIf AT and BT are tangents to circle O, then\n• OA ⊥AT (tangent ⊥radius)\n• OB ⊥BT (tangent ⊥radius)\n• TA = TB (tangents from same point equal)\nA\nB\nT\nD\nC\nx\ny\nx\ny\n• If DC is a tangent, then D ˆTA = T ˆBA and\nC ˆTB = T ˆAB\n• If D ˆTA = T ˆBA or C ˆTB = T ˆAB, then DC is\na tangent touching at T\n365\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 7: End of chapter exercises\n1.\nO\n•\nA\nB\nC\nD\nE\nF\n×\n×\nx\nAOC is a diameter of the circle with centre O. F is the mid-point of chord EC.\nB ˆOC = C ˆOD and ˆB = x. Express the following angles in terms of x, stating\nreasons:\na) ˆA\nb) C ˆOD\nc) ˆD\n2.\nM•\nD\nE\nF\nG\n1 2\n1\n2\n1\n2\n1 2\nD, E, F and G are points on circle with centre M.\nˆF1 = 7◦and ˆD2 = 51◦.\nDetermine the sizes of the following angles, stating reasons:\na)\nˆ\nM1\nb) ˆD1\nc) ˆF2\nd) ˆG\ne) ˆE1\n366\n8.3.\nSummary\n\n3.\nM\n•\nO•\nD\nA\nB\nC\n1 2\nO is a point on the circle with centre M. O is also the centre of a second circle.\nDA cuts the smaller circle at C and ˆD1 = x. Express the following angles in\nterms of x, stating reasons:\na) ˆD2\nb) O ˆAB\nc) O ˆBA\nd) A ˆOB\ne) ˆC\n4.\nO•\nA\nB\nC\nM\nO is the centre of the circle with radius 5 cm and chord BC = 8 cm. Calculate\nthe lengths of:\na) OM\nb) AM\nc) AB\n5.\nO•\nA\nB\nC\n70◦\nx\nAO ∥CB in circle with centre O. A ˆOB = 70◦and O ˆAC = x. Calculate the\nvalue of x, giving reasons.\n367\nChapter 8.\nEuclidean geometry\n\n6.\nO\n•\nP\nQ\nR\nS\nT\nx\nPQ is a diameter of the circle with centre O. SQ bisects P ˆQR and P ˆQS = x.\na) Write down two other angles that are also equal to x.\nb) Calculate P ˆOS in terms of x, giving reasons.\nc) Prove that OS is a perpendicular bisector of PR.\n7.\nO•\nA\nB\nC\nD\n35◦\nB ˆOD is a diameter of the circle with centre O. AB = AD and O ˆCD = 35◦.\nCalculate the value of the following angles, giving reasons:\na) O ˆDC\nb) C ˆOD\nc) C ˆBD\nd) B ˆAD\ne) A ˆDB\n8.\nO\n•\nR\nP\nT\nQ\nx\ny\nQP in the circle with centre O is protracted to T so that PR = PT. Express y in\nterms of x.\n368\n8.3.\nSummary\n\n9.\nO•\nA\nB\nC\nD\nE\nP\nF\nO is the centre of the circle with diameter AB. CD ⊥AB at P and chord DE\ncuts AB at F. Prove that:\na) C ˆBP = D ˆPB\nb) C ˆED = 2C ˆBA\nc) A ˆBD = 1\n2C ˆOA\n10.\nO\n•\nP\nQ\nR\nx\nS\nIn the circle with centre O, OR ⊥QP, PQ = 30 mm and RS = 9 mm. Deter-\nmine the length of OQ.\n11.\nM •\nP\nQ\nR\nS\nT\nP, Q, R and S are points on the circle with centre M. PS and QR are extended\nand meet at T. PQ = PR and P ˆQR = 70◦.\na) Determine, stating reasons, three more angles equal to 70◦.\nb) If Q ˆPS = 80◦, calculate S ˆRT, S ˆTR and P ˆQS.\nc) Explain why PQ is a tangent to the circle QST at point Q.\nd) Determine P ˆ\nMQ.\n369\nChapter 8.\nEuclidean geometry\n\n12.\nO\n•\nA\nP\nQ\nC\nB\nPOQ is a diameter of the circle with centre O. QP is protruded to A and AC is\na tangent to the circle. BA ⊥AQ and BCQ is a straight line. Prove:\na) P ˆCQ = B ˆAP\nb) BAPC is a cyclic quadrilateral\nc) AB = AC\n13.\nO•\nT\nC\nA\nB\nx\nTA and TB are tangents to the circle with centre O. C is a point on the circum-\nference and A ˆTB = x. Express the following in terms of x, giving reasons:\na) A ˆBT\nb) O ˆBA\nc) ˆC\n14.\nO•\nA\nB\nC\nE\nD\nAOB is a diameter of the circle\nAECB with centre O. OE ∥BC\nand cuts AC at D.\na) Prove AD = DC\nb) Show that A ˆBC is bisected\nby EB\nc) If O ˆEB = x, express B ˆAC\nin terms of x\nd) Calculate the radius of the\ncircle if AC = 10 cm and\nDE = 1 cm\n370\n8.3.\nSummary\n\n15.\nV\nQ\nS\nR\nP\nT\nW\nx\ny\nPQ and RS are chords of the circle and PQ ∥RS. The tangent to the circle at\nQ meets RS protruded at T. The tangent at S meets QT at V . QS and PR are\ndrawn.\nLet T ˆQS = x and Q ˆRP = y. Prove that:\na) T ˆV S = 2Q ˆRS\nb) QV SW is a cyclic quadrilateral\nc) Q ˆPS + ˆT = P ˆRT\nd) W is the centre of the circle\n16.\nF\nD\nB\nC\nE\nA\nK\nT\n1\n2\n1\n2\n1\n2\n3\n4\nThe two circles shown intersect at points F and D. BFT is a tangent to the\nsmaller circle at F. Straight line AFE is drawn such that DF = EF. CDE is a\nstraight line and chord AC and BF cut at K. Prove that:\na) BT ∥CE\nb) BCEF is a parallelogram\nc) AC = BF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237D\n2. 237F\n3. 237G\n4. 237H\n5. 237J\n6. 237K\n7. 237M\n8. 237N\n9. 237P\n10. 237Q\n11. 237R\n12. 237S\n13. 237T\n14. 237V\n15. 237W\n16. 237X\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n371\nChapter 8.\nEuclidean geometry\n\n\nCHAPTER\n9\nFinance, growth and decay\n9.1\nRevision\n374\n9.2\nSimple and compound depreciation\n377\n9.3\nTimelines\n388\n9.4\nNominal and effective interest rates\n394\n9.5\nSummary\n398\n\n9\nFinance, growth and decay\n9.1\nRevision\nEMBJD\nSimple interest is the interest calculated only on the initial amount invested, the prin-\ncipal amount. Compound interest is the interest earned on the principal amount and\non its accumulated interest. This means that interest is being earned on interest. The\naccumulated amount is the final amount; the sum of the principal amount and the\namount of interest earned.\nFormula for simple interest:\nA = P(1 + in)\nFormula for compound interest:\nA = P(1 + i)n\nwhere\nA = accumulated amount\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nWorked example 1: Simple and compound interest\nQUESTION\nSam wants to invest R 3450 for 5 years. Wise Bank offers a savings account which pays\nsimple interest at a rate of 12,5% per annum, and Grand Bank offers a savings account\npaying compound interest at a rate of 10,4% per annum. Which bank account would\ngive Sam the greatest accumulated balance at the end of the 5 year period?\nSOLUTION\nStep 1: Calculation using the simple interest formula\nWrite down the known variables and the simple interest formula\nP = 3450\ni = 0,125\nn = 5\nA = P(1 + in)\nSubstitute the values to determine the accumulated amount for the Wise Bank savings\n374\n9.1.\nRevision\n\naccount.\nA = 3450(1 + 0,125 × 5)\n= R 5606,25\nStep 2: Calculation using the compound interest formula\nWrite down the known variables and the compound interest formula.\nP = 3450\ni = 0,104\nn = 5\nA = P(1 + i)n\nSubstitute the values to determine the accumulated amount for the Grand Bank savings\naccount.\nA = 3450(1 + 0,104)5\n= R 5658,02\nStep 3: Write the final answer\nThe Grand Bank savings account would give Sam the highest accumulated balance at\nthe end of the 5 year period.\nWorked example 2: Finding i\nQUESTION\nBongani decides to put R 30 000 in an investment account. What compound interest\nrate must the investment account achieve for Bongani to double his money in 6 years?\nGive your answer correct to one decimal place.\nSOLUTION\nStep 1: Write down the known variables and the compound interest formula\nA = 60 000\nP = 30 000\nn = 6\nA = P(1 + i)n\n375\nChapter 9.\nFinance, growth and decay\n\nStep 2: Substitute the values and solve for i\n60 000 = 30 000(1 + i)6\n60 000\n30 000 = (1 + i)6\n2 = (1 + i)6\n6√\n2 = 1 + i\n6√\n2 −1 = i\n∴i = 0,122 . . .\nStep 3: Write the final answer and comment\nWe round up to a rate of 12,3% p.a. to make sure that Bongani doubles his invest-\nment.\nExercise 9 – 1: Revision\n1. Determine the value of an investment of R 10 000 at 12,1% p.a. simple interest\nfor 3 years.\n2. Calculate the value of R 8000 invested at 8,6% p.a. compound interest for 4\nyears.\n3. Calculate how much interest John will earn if he invests R 2000 for 4 years at:\na) 6,7% p.a. simple interest\nb) 5,4% p.a. compound interest\n4. The value of an investment grows from R 2200 to R 3850 in 8 years. Determine\nthe simple interest rate at which it was invested.\n5. James had R 12 000 and invested it for 5 years. If the value of his investment is\nR 15 600, what compound interest rate did it earn?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237Y\n2. 237Z\n3. 2382\n4. 2383\n5. 2384\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n376\n9.1.\nRevision\n\n9.2\nSimple and compound depreciation\nEMBJF\nAs soon as a new car leaves the dealership, its value decreases and it is considered\n“second-hand”.\nVehicles, equipment, machinery and other similar assets, all lose\nvalue over time as a result of usage and age. This loss in value is called deprecia-\ntion. Assets that have a relatively long useful lifetime, such as machines, trucks, farm-\ning equipment etc., depreciate slower than assets like office equipment, computers,\nfurniture etc. which need to be replaced more often and therefore depreciate more\nquickly.\nDepreciation is used to calculate the value of a company’s assets, which determines\nhow much tax a company must pay. Companies can take depreciation into account as\nan expense, and thereby reduce their taxable income. A lower taxable income means\nthat the company will pay less income tax to SARS (South African Revenue Service).\nWe can calculate two different kinds of depreciation: simple decay and compound\ndecay. Decay is also a term used to describe a reduction or decline in value. Simple\ndecay is also called straight-line depreciation and compound decay can also be re-\nferred to as reducing-balance depreciation. In the straight-line method the value of the\nasset is reduced by a constant amount each year, which is calculated on the principal\namount. In reducing-balance depreciation we calculate the depreciation on the re-\nduced value of the asset. This means that the value of an asset decreases by a different\namount each year.\nInvestigation: Simple and compound depreciation\n1. Mr. Sontange buys an Opel Fiesta for R 72 000. He expects that the value of the\ncar will depreciate by R 6000 every year. He draws up a table to calculate the\ndepreciated value of his Opel Fiesta.\nComplete Mr. Sontange’s table of values for the 7 year period:\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 6000\nR 66 000\n2\nR 66 000\nR 6000\n3\n4\n5\n6\n7\n2. His son, David, does not agree that the value of the car will reduce by the same\namount each year. David thinks that the car will depreciate by 10% every year.\nComplete David’s table of values:\n377\nChapter 9.\nFinance, growth and decay\n\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 7200\nR 64 800\n2\nR 64 800\nR 6480\n3\n4\n5\n6\n7\n3. Compare and discuss the results of the two different tables.\n4. Consider the graph below, which represents Mr. Sontange’s table of values:\n10 000\n20 000\n30 000\n40 000\n50 000\n60 000\n70 000\n80 000\n1\n2\n3\n4\n5\n6\n7\n8\n0\nTime (years)\nValue (Rands)\na) Draw a similar graph using David’s table of values.\nb) Interpret the two graphs and discuss the differences between them.\nc) Explain how the graphs can be used to determine the total depreciation in\neach case.\nd)\ni. Draw two new graphs by plotting the maximum value of each bar.\nii. Join the points with a line to show the general trend.\niii. Is it mathematically correct to join these points? Explain your answer.\n378\n9.2.\nSimple and compound depreciation\n\nSimple depreciation\nEMBJG\nWorked example 3: Straight-line depreciation\nQUESTION\nA new smartphone costs R 6000 and depreciates at 22% p.a. on a straight-line basis.\nDetermine the value of the smartphone at the end of each year over a 4 year period.\nSOLUTION\nStep 1: Calculate depreciation amount\nDepreciation = 6000 × 22\n100\n= 1320\nTherefore the smartphone depreciates by R 1320 every year.\nStep 2: Complete a table of values\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 6000\nR 1320\nR 4680\n2\nR 4680\nR 1320\nR 3360\n3\nR 3360\nR 1320\nR 2040\n4\nR 2040\nR 1320\nR 720\nWe notice that\nTotal depreciation = P × i × n\nwhere\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nTherefore the depreciated value of the asset (also called the book value) can be calcu-\nlated as:\nA = P(1 −in)\nNote the similarity to the simple interest formula A = P(1 + in). Interest increases the\nvalue of the principal amount, whereas with simple decay, depreciation reduces the\nvalue of the principal amount.\nImportant: to get an accurate answer do all calculations in one step on your calculator.\nDo not round off answers in your calculations until the final answer. In the worked\nexamples in this chapter, we use dots to show that the answer has not been rounded\noff. We always round the final answer to two decimal places (cents).\n379\nChapter 9.\nFinance, growth and decay\n\nWorked example 4: Straight-line depreciation method\nQUESTION\nA car is valued at R 240 000. If it depreciates at 15% p.a. using straight-line deprecia-\ntion, calculate the value of the car after 5 years.\nSOLUTION\nStep 1: Write down the known variables and the simple decay formula\nP = 240 000\ni = 0,15\nn = 5\nA = P(1 −in)\nStep 2: Substitute the values and solve for A\nA = 240 000(1 −0,15 × 5)\n= 240 000(0,25)\n= 60 000\nStep 3: Write the final answer\nAt the end of 5 years, the car is worth R 60 000.\nWorked example 5: Simple decay\nQUESTION\nA small business buys a photocopier for R 12 000. For the tax return the owner depre-\nciates this asset over 3 years using a straight-line depreciation method. What amount\nwill he fill in on his tax form at the end of each year?\nSOLUTION\nStep 1: Write down the known variables\nThe owner of the business wants the photocopier to have a book value of R 0 after 3\nyears.\nA = 0\nP = 12 000\nn = 3\n380\n9.2.\nSimple and compound depreciation\n\nTherefore we can calculate the annual depreciation as\nDepreciation = P\nn\n= 12 000\n3\n= R 4000\nStep 2: Determine the book value at the end of each year\nBook value end of first year = 12 000 −4000\n= R 8000\nBook value end of second year = 8000 −4000\n= R 4000\nBook value end of third year = 4000 −4000\n= R 0\nExercise 9 – 2: Simple decay\n1. A business buys a truck for R 560 000. Over a period of 10 years the value of\nthe truck depreciates to R 0 using the straight-line method. What is the value of\nthe truck after 8 years?\n2. Harry wants to buy his grandpa’s donkey for R 800. His grandpa is quite pleased\nwith the offer, seeing that it only depreciated at a rate of 3% per year using the\nstraight-line method. Grandpa bought the donkey 5 years ago. What did grandpa\npay for the donkey then?\n3. Seven years ago, Rocco’s drum kit cost him R 12 500. It has now been valued at\nR 2300. What rate of simple depreciation does this represent?\n4. Fiona buys a DStv satellite dish for R 3000. Due to weathering, its value depre-\nciates simply at 15% per annum. After how long will the satellite dish have a\nbook value of zero?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2385\n2. 2386\n3. 2387\n4. 2388\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n381\nChapter 9.\nFinance, growth and decay\n\nCompound depreciation\nEMBJH\nWorked example 6: Reducing-balance depreciation\nQUESTION\nA second-hand farm tractor worth R 60 000 has a limited useful life of 5 years and\ndepreciates at 20% p.a. on a reducing-balance basis. Determine the value of the\ntractor at the end of each year over the 5 year period.\nSOLUTION\nStep 1: Write down the known variables\nP = 60 000\ni = 0,2\nn = 5\nWhen we calculate depreciation using the reducing-balance method:\n1. the depreciation amount changes for each year.\n2. the depreciation amount gets smaller each year.\n3. the book value at the end of a year becomes the principal amount for the next\nyear.\n4. the asset will always have some value (the book value will never equal zero).\nStep 2: Complete a table of values\nYear\nBook value\nDepreciation\nValue at end of\nyear\n1\nR 60 000\n60 000 × 0,2 = 12 000\nR 48 000\n2\nR 48 000\n48 000 × 0,2 = 9600\nR 38 400\n3\nR 38 400\n38 400 × 0,2 = 7680\nR 30 720\n4\nR 30 720\n30 720 × 0,2 = 6144\nR 24 576\n5\nR 24 576\n24 576 × 0,2 = 4915,20\nR 19 660,80\n382\n9.2.\nSimple and compound depreciation\n\nNotice in the example above that we could also write the book value at the end of\neach year as:\nBook value end of first year\n= 60 000(1 −0,2)\nBook value end of second year = 48 000(1 −0,2) = 60 000(1 −0,2)2\nBook value end of third year\n= 38 400(1 −0,2) = 60 000(1 −0,2)3\nBook value end of fourth year = 30 720(1 −0,2) = 60 000(1 −0,2)4\nBook value end of fifth year\n= 24 576(1 −0,2) = 60 000(1 −0,2)5\nUsing the formula for simple decay and the observed pattern in the calculation above,\nwe obtain the following formula for compound decay:\nA = P(1 −i)n\nwhere\nA = book value or depreciated value\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nAgain, notice the similarity to the compound interest formula A = P(1 + i)n.\nWorked example 7: Reducing-balance depreciation\nQUESTION\nThe number of pelicans at the Berg river mouth is decreasing at a compound rate of\n12% p.a. If there are currently 3200 pelicans in the wetlands of the Berg river mouth,\nwhat will the population be in 5 years?\nSOLUTION\nStep 1: Write down the known variables and the compound decay formula\nP = 3200\ni = 0,12\nn = 5\nA = P(1 −i)n\nStep 2: Substitute the values and solve for A\nA = 3200(1 −0,12)5\n= 3200(0,88)5\n= 1688,7421 . . .\nStep 3: Write the final answer\nIn 5 years, the pelican population will be approximately 1689.\n383\nChapter 9.\nFinance, growth and decay\n\nWorked example 8: Compound decay\nQUESTION\n1. A school buys a minibus for R 950 000, which depreciates at 13,5% per annum.\nDetermine the value of the minibus after 3 years if the depreciation is calculated:\na) on a straight-line basis.\nb) on a reducing-balance basis.\n2. Which is the better option?\nSOLUTION\nStep 1: Write down known variables\nP = 950 000\ni = 0,135\nn = 3\nStep 2: Use the simple decay formula and solve for A\nA = 950 000(1 −3 × 0,135)\n= 950 000(0,865)\n= 565 250\n∴A = R 565 250\nStep 3: Use the compound decay formula and solve for A\nA = 950 000(1 −0,135)3\n= 950 000(0,865)3\n= 614 853,89\n∴A = R 614 853,89\nStep 4: Interpret the answers\nAfter a period of 3 years, the value of the minibus calculated on the straight-line\nmethod is less than the value of the minibus calculated on the reducing-balance\nmethod. The value of the minibus depreciated less on the reducing-balance basis\nbecause the amount of depreciation is calculated on a smaller amount every year,\nwhereas the straight-line method is based on the full value of the minibus every year.\n384\n9.2.\nSimple and compound depreciation\n\nWorked example 9: Compound depreciation\nQUESTION\nFarmer Jack bought a tractor and it has depreciated by 20% p.a. on a reducing-balance\nbasis. If the current value of the tractor is R 52 429, calculate how much Farmer Jack\npaid for his tractor if he bought it 7 years ago.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 52 429\ni = 0,2\nn = 7\nA = P(1 −i)n\nStep 2: Substitute the values and solve for P\n52 429 = P(1 −0,2)7\n= P(0,8)7\n∴P = 52 429\n(0,8)7\n= 250 000,95 . . .\nStep 3: Write the final answer\n7 years ago, Farmer Jack paid R 250 000 for his tractor.\nExercise 9 – 3: Compound depreciation\n1. Jwayelani buys a truck for R 89 000 and depreciates it by 9% p.a. using the\ncompound depreciation method. What is the value of the truck after 14 years?\n2. The number of cormorants at the Amanzimtoti river mouth is decreasing at a\ncompound rate of 8% p.a. If there are now 10 000 cormorants, how many will\nthere be in 18 years’ time?\n3. On January 1, 2008 the value of my Kia Sorento is R 320 000. Each year after\nthat, the car’s value will decrease 20% of the previous year’s value. What is the\nvalue of the car on January 1, 2012?\n385\nChapter 9.\nFinance, growth and decay\n\n4. The population of Bonduel decreases at a reducing-balance rate of 9,5% per\nannum as people migrate to the cities. Calculate the decrease in population over\na period of 5 years if the initial population was 2 178 000.\n5. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses\n3% of its water each day. How much does it weigh after a month of 31 days?\n6. Richard bought a car 15 years ago and it depreciated by 17% p.a. on a com-\npound depreciation basis. How much did he pay for the car if it is now worth\nR 5256?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2389\n2. 238B\n3. 238C\n4. 238D\n5. 238F\n6. 238G\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFinding i\nEMBJJ\nWorked example 10: Finding i for simple decay\nQUESTION\nAfter 4 years, the value of a computer is halved. Assuming simple decay, at what\nannual rate did it depreciate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and simple decay formula\nLet the value of the computer be x, therefore:\nA = x\n2\nP = x\nn = 4\nA = P(1 −in)\nStep 2: Substitute the values and solve for i\n386\n9.2.\nSimple and compound depreciation\n\nx\n2 = x(1 −3i)\n1\n2 = 1 −3i\n∴3i = 1 −1\n2\n∴i = 0,1667\nStep 3: Write the final answer\nThe computer depreciated at a rate of 16,67% p.a.\nWorked example 11: Finding i for compound decay\nQUESTION\nCristina bought a fridge at the beginning of 2009 for R 8999 and sold it at the end\nof 2011 for R 4500. At what rate did the value of her fridge depreciate assuming a\nreducing-balance method? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 4500\nP = 8999\nn = 3\nA = P(1 −i)n\nStep 2: Substitute the values and solve for i\n4500 = 8999(1 −i)3\n4500\n8999 = (1 −i)3\n3\nr\n4500\n8999 = 1 −i\n∴i = 1 −\n3\nr\n4500\n8999\n= 0,206\nStep 3: Write the final answer\nCristina’s fridge depreciated at a rate of 20,6% p.a.\n387\nChapter 9.\nFinance, growth and decay\n\nExercise 9 – 4: Finding i\n1. A machine costs R 45 000 and has a scrap value of R 9000 after 10 years. Deter-\nmine the annual rate of depreciation if it is calculated on the reducing balance\nmethod.\n2. After 15 years, an aeroplane is worth 1\n6 of its original value. At what annual rate\nwas depreciation compounded?\n3. Mr. Mabula buys furniture for R 20 000. After 6 years he sells the furniture for\nR 9300. Calculate the annual compound rate of depreciation of the furniture.\n4. Ayanda bought a new car 7 years ago for double what it is worth today. At what\nyearly compound rate did her car depreciate?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238H\n2. 238J\n3. 238K\n4. 238M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.3\nTimelines\nEMBJK\nInterest can be compounded more than once a year. For example, an investment can\nbe compounded monthly or quarterly. Below is a table of compounding terms and\ntheir corresponding numeric value (p). When amounts are compounded more than\nonce per annum, we multiply the number of years by p and we also divide the interest\nrate by p.\nTerm\np\nyearly / annually\n1\nhalf-yearly / bi-annually\n2\nquarterly\n4\nmonthly\n12\nweekly\n52\ndaily\n365\nWorked example 12: Timelines\nQUESTION\nR 5500 is invested for a period of 4 years in a savings account. For the first year, the\ninvestment grows at a simple interest rate of 11% p.a. and then at a rate of 12,5%\np.a. compounded quarterly for the rest of the period. Determine the value of the\ninvestment at the end of the 4 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\n388\n9.3.\nTimelines\n\nT0\nT1\nT2\nT3\nT4\n11% p.a. simple interest\n12,5% p.a. compounded quarterly\nR 5500\nIn the timeline above, the intervals are given in years. For example, T0 is the start of\nthe investment, T1 is the end of the first year and T4 is the end of the fourth year.\nStep 2: Use the simple interest formula to calculate A at T1\nA = P(1 + in)\n= 5500(1 + 0,11)\n= R 6105\nStep 3: Use the compound interest formula to calculate A at T4\nThe investment is compounded quarterly, therefore:\nn = 3 × 4\n= 12\nand i = 0,125\n4\nAlso notice that the accumulated amount at the end of the first year becomes the\nprincipal amount at the beginning of the second year.\nA = P(1 + i)n\n= 6105\n\u0012\n1 + 0,125\n4\n\u001312\n= R 8831,88\nStep 4: Write the final answer\nThe value of the investment at the end of the 4 years is R 8831,88.\n389\nChapter 9.\nFinance, growth and decay\n\nWorked example 13: Timelines\nQUESTION\nR 150 000 is deposited in an investment account for a period of 6 years at an interest\nrate of 12% p.a. compounded half-yearly for the first 4 years and then 8,5% p.a.\ncompounded yearly for the rest of the period. A deposit of R 8000 is made into the\naccount after the first year and then another deposit of R 2000 is made 5 years after\nthe initial investment. Calculate the value of the investment at the end of the 6 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n8,5% p.a. compounded yearly\nR 15 000\nT5\nT6\n12% p.a. compounded half-yearly\n+R 8000\n+R 2000\nRemember to show when the additional deposits of R 8000 and R 2000 where made\ninto the account. It is very important to note that the interest rate changes at T4.\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nBetween T0 and T4:\nWe notice that interest for the first 4 years is compounded half-yearly, therefore:\nn1 = 4 × 2\n= 8\nand i1 = 0,12\n2\nBetween T4 and T6:\nn2 = 2\nand i2 = 0,085\nTherefore the total growth of the initial deposit over the 6 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\nStep 3: The deposit at T1\nBetween T1 and T4:\n390\n9.3.\nTimelines\n\nInterest on this deposit is compounded half-yearly for 3 years, therefore:\nn3 = 3 × 2\n= 6\nand i3 = 0,12\n2\nBetween T4 and T6:\nn4 = 2\nand i4 = 0,085\nTherefore the total growth of the deposit over the 5 years is:\nA = P(1 + i3)n3(1 + i4)n4\n= 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2\nStep 4: The deposit at T5\nAccumulate interest for only 1 year:\nA = P(1 + i)n\n= 2000(1 + 0,085)1\nStep 5: Determine the total calculation\nTo get as accurate an answer as possible, we do the the calculation on the calculator\nin one step. Using the memory and answer recall function on the calculator, we avoid\nrounding off until we get the final answer.\nA = 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\n+ 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2 + 2000(1 + 0,085)1\n= R 296 977,00\nStep 6: Write the final answer\nThe value of the investment at the end of the 6 years is R 296 977,00.\n391\nChapter 9.\nFinance, growth and decay\n\nWorked example 14: Timelines\nQUESTION\nR 60 000 is invested in an account which offers interest at 7% p.a.\ncompounded\nquarterly for the first 18 months. Thereafter the interest rate changes to 5% p.a. com-\npounded monthly. Three years after the initial investment, R 5000 is withdrawn from\nthe account. How much will be in the account at the end of 5 years?\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n5% p.a. compounded monthly\nR 60 000\nT5\n7% p.a. compounded quarterly\n−R 5000\nRemember to show when the withdrawal of R 5000 was taken out of the account. It is\nalso important to note that the interest rate changes after 18 months (T1 1\n2 ).\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nInterest for the first 1,5 years is compounded quarterly, therefore:\nn1 = 1,5 × 4\n= 6\nand i1 = 0,07\n4\nInterest for the remaining 3,5 years is compounded monthly, therefore:\nn2 = 3,5 × 12\n= 42\nand i2 = 0,05\n12\nTherefore the total growth of the initial deposit over the 5 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n392\n9.3.\nTimelines\n\nStep 3: The withdrawal at T3\nWe calculate the interest that the R 5000 would have earned if it had remained in the\naccount:\nn = 2 × 12\n= 24\nand i = 0,05\n12\nTherefore we have that:\nA = P(1 + i)n\n= 5000\n\u0012\n1 + 0,05\n12\n\u001324\nStep 4: Determine the total calculation\nWe subtract the withdrawal and the interest it would have earned from the accumu-\nlated amount at the end of the 5 years:\nA = 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n−5000\n\u0012\n1 + 0,05\n12\n\u001324\n= R 73 762,19\nStep 5: Write the final answer\nThe value of the investment at the end of the 5 years is R 73 762,19.\nExercise 9 – 5: Timelines\n1. After a 20-year period Josh’s lump sum investment matures to an amount of\nR 313 550. How much did he invest if his money earned interest at a rate of\n13,65% p.a. compounded half yearly for the first 10 years, 8,4% p.a. com-\npounded quarterly for the next five years and 7,2% p.a. compounded monthly\nfor the remaining period?\n2. Sindisiwe wants to buy a motorcycle. The cost of the motorcycle is R 55 000.\nIn 1998 Sindisiwe opened an account at Sutherland Bank with R 16 000. Then\nin 2003 she added R 2000 more into the account. In 2007 Sindisiwe made\nanother change: she took R 3500 from the account. If the account pays 6% p.a.\ncompounded half-yearly, will Sindisiwe have enough money in the account at\nthe end of 2012 to buy the motorcycle?\n3. A loan has to be returned in two equal semi-annual instalments. If the rate of\ninterest is 16% per annum, compounded semi-annually and each instalment is\nR 1458, find the sum borrowed.\n393\nChapter 9.\nFinance, growth and decay\n\n4. A man named Phillip invests R 10 000 into an account at North Bank at an\ninterest rate of 7,5% p.a. compounded monthly. After 5 years the bank changes\nthe interest rate to 8% p.a. compounded quarterly. How much money will\nPhillip have in his account 9 years after the original deposit?\n5. R 75 000 is invested in an account which offers interest at 11% p.a.\ncom-\npounded monthly for the first 24 months.\nThen the interest rate changes to\n7,7% p.a. compounded half-yearly. If R 9000 is withdrawn from the account\nafter one year and then a deposit of R 3000 is made three years after the initial\ninvestment, how much will be in the account at the end of 6 years?\n6. Christopher wants to buy a computer, but right now he doesn’t have enough\nmoney. A friend told Christopher that in 5 years the computer will cost R 9150.\nHe decides to start saving money today at Durban United Bank. Christopher\ndeposits R 5000 into a savings account with an interest rate of 7,95% p.a. com-\npounded monthly.\nThen after 18 months the bank changes the interest rate\nto 6,95% p.a. compounded weekly. After another 6 months, the interest rate\nchanges again to 7,92% p.a.\ncompounded two times per year.\nHow much\nmoney will Christopher have in the account after 5 years, and will he then have\nenough money to buy the computer?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238N\n2. 238P\n3. 238Q\n4. 238R\n5. 238S\n6. 238T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.4\nNominal and effective interest rates\nEMBJM\nWe have seen that although interest is quoted as a percentage per annum it can be\ncompounded more than once a year. We therefore need a way of comparing interest\nrates. For example, is an annual interest rate of 8% compounded quarterly higher or\nlower than an interest rate of 8% p.a. compounded yearly?\nInvestigation: Nominal and effective interest rates\n1. Calculate the accumulated amount at the end of one year if R 1000 is invested\nat 8% p.a. compound interest:\nA = P(1 + i)n\n= . . . . . .\n2. Calculate the value of R 1000 if it is invested for one year at 8% p.a. com-\npounded:\n394\n9.4.\nNominal and effective interest rates\n\nFrequency\nCalculation\nAccumulated\namount\nInterest\namount\nhalf-yearly\nA = 1000\n\u0010\n1 + 0,08\n2\n\u00111×2\nR 1081,60\nR 81,60\nquarterly\nmonthly\nweekly\ndaily\n3. Use your results from the table above to calculate the effective rate that the\ninvestment of R 1000 earns in one year:\nFrequency\nAccumulated\namount\nCalculation\nEffective\ninterest\nrate\nhalf-yearly\nR 1081,60\n1081,60 = 1000(1 + i)\n1081,60\n1000\n= 1 + i\n1081,60\n1000\n−1 = i\n∴i = 0,0816\ni = 8,16%\nquarterly\nmonthly\nweekly\ndaily\n4. If you wanted to borrow R 10 000 from the bank, would it be better to pay it\nback at an interest rate of 22% p.a. compounded quarterly or 22% compounded\nmonthly? Show your calculations.\nAn interest rate compounded more than once a year is called the nominal interest rate.\nIn the investigation above, we determined that the nominal interest rate of 8% p.a.\ncompounded half-yearly is actually an effective rate of 8,16% p.a.\nGiven a nominal interest rate i(m) compounded at a frequency of m times per year\nand the effective interest rate i, the accumulated amount calculated using both interest\nrates will be equal so we can write:\nP(1 + i) = P\n \n1 + i(m)\nm\n!m\n∴1 + i =\n \n1 + i(m)\nm\n!m\n395\nChapter 9.\nFinance, growth and decay\n\nWorked example 15: Nominal and effective interest rates\nQUESTION\nInterest on a credit card is quoted as 23% p.a. compounded monthly. What is the\neffective annual interest rate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down the known variables\nInterest is being added monthly, therefore:\nm = 12\ni(12) = 0,23\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i\n1 + i =\n\u0012\n1 + 0,23\n12\n\u001312\n∴i = 1 −\n\u0012\n1 + 0,23\n12\n\u001312\n= 25,59%\nStep 3: Write the final answer\nThe effective interest rate is 25,59% per annum.\nWorked example 16: Nominal and effective interest rates\nQUESTION\nDetermine the nominal interest rate compounded quarterly if the effective interest rate\nis 9% per annum (correct to two decimal places).\nSOLUTION\nStep 1: Write down the known variables\n396\n9.4.\nNominal and effective interest rates\n\nInterest is being added quarterly, therefore:\nm = 4\ni = 0,09\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i(m)\n1 + 0,09 =\n \n1 + i(4)\n4\n!4\n4p\n1,09 = 1 + i(4)\n4\n4p\n1,09 −1 = i(4)\n4\n4\n\u0010\n4p\n1,09 −1\n\u0011\n= i(4)\n∴i(4) = 8,71%\nStep 3: Write the final answer\nThe nominal interest rate is 8,71% p.a. compounded quarterly.\nExercise 9 – 6: Nominal and effect interest rates\n1. Determine the effective annual interest rate if the nominal interest rate is:\na) 12% p.a. compounded quarterly.\nb) 14,5% p.a. compounded weekly.\nc) 20% p.a. compounded daily.\n2. Consider the following:\n• 16,8% p.a. compounded annually.\n• 16,4% p.a. compounded monthly.\n• 16,5% p.a. compounded quarterly.\na) Determine the effective annual interest rate of each of the nominal rates\nlisted above.\nb) Which is the best interest rate for an investment?\nc) Which is the best interest rate for a loan?\n397\nChapter 9.\nFinance, growth and decay\n\n3. Calculate the effective annual interest rate equivalent to a nominal interest rate\nof 8,75% p.a. compounded monthly.\n4. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every\nfour months on her investment of R 85 000.\nCalculate the effective rate per\nannum.\n5. Determine which of the following would be the better agreement for paying back\na student loan:\na) 9,1% p.a. compounded quarterly.\nb) 9% p.a. compounded monthly.\nc) 9,3% p.a. compounded half-yearly.\n6. Miranda invests R 8000 for 5 years for her son’s study fund. Determine how\nmuch money she will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 6% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 238V\n1b. 238W\n1c. 238X\n2. 238Y\n3. 238Z\n4. 2392\n5. 2393\n6. 2394\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.5\nSummary\nEMBJN\nSee presentation: 2395 at www.everythingmaths.co.za\n• Simple interest: A = P(1 + in)\n• Compound interest: A = P(1 + i)n\n• Simple depreciation: A = P(1 −in)\n• Compound depreciation: A = P(1 −i)n\n• Nominal and effective annual interest rates: 1 + i =\n\u0010\n1 + i(m)\nm\n\u0011m\n398\n9.5.\nSummary\n\nExercise 9 – 7: End of chapter exercises\n1. Thabang buys a Mercedes worth R 385 000 in 2007. What will the value of the\nMercedes be at the end of 2013 if:\na) the car depreciates at 6% p.a. straight-line depreciation.\nb) the car depreciates at 6% p.a. reducing-balance depreciation.\n2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R 8900.\nThe interest rate is quoted as 11% per annum based on simple interest. Calculate\nthe required monthly payment for this contract.\n3. A computer is purchased for R 16 000. It depreciates at 15% per annum.\na) Determine the book value of the computer after 3 years if depreciation is\ncalculated according to the straight-line method.\nb) Find the rate according to the reducing-balance method that would yield,\nafter 3 years, the same book value as calculated in the previous question.\n4. Maggie invests R 12 500 for 5 years at 12% per annum compounded monthly\nfor the first 2 years and 14% per annum compounded semi-annually for the next\n3 years. How much will Maggie receive in total after 5 years?\n5. Tintin invests R 120 000. He is quoted a nominal interest rate of 7,2% per an-\nnum compounded monthly.\na) Calculate the effective rate per annum (correct to two decimal places).\nb) Use the effective rate to calculate the value of Tintin’s investment if he\ninvested the money for 3 years.\nc) Suppose Tintin invests his money for a total period of 4 years, but after 18\nmonths makes a withdrawal of R 20 000, how much will he receive at the\nend of the 4 years?\n6. Ntombi opens accounts at a number of clothing stores and spends freely. She\ngets herself into terrible debt and she cannot pay off her accounts. She owes\nFashion World R 5000 and the shop agrees to let her pay the bill at a nominal\ninterest rate of 24% compounded monthly.\na) How much money will she owe Fashion World after two years?\nb) What is the effective rate of interest that Fashion World is charging her?\n7. John invests R 30 000 in the bank for a period of 18 months. Calculate how\nmuch money he will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 8% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\ndaily\n399\nChapter 9.\nFinance, growth and decay\n\n8. Convert an effective annual interest rate of 11,6% p.a. to a nominal interest rate\ncompounded:\na) half-yearly\nb) quarterly\nc) monthly\n9. Joseph must sell his plot on the West Coast and he needs to get R 300 000 on the\nsale of the land. If the estate agent charges him 7% commission on the selling\nprice, what must the buyer pay for the plot?\n10. Mrs. Brown retired and received a lump sum of R 200 000. She deposited the\nmoney in a fixed deposit savings account for 6 years. At the end of the 6 years\nthe value of the investment was R 265 000. If the interest on her investment was\ncompounded monthly, determine:\na) the nominal interest rate per annum\nb) the effective annual interest rate\n11. R 145 000 is invested in an account which offers interest at 9% p.a.\ncom-\npounded half-yearly for the first 2 years. Then the interest rate changes to 4%\np.a. compounded quarterly. Four years after the initial investment, R 20 000 is\nwithdrawn. 6 years after the initial investment, a deposit of R 15 000 is made.\nDetermine the balance of the account at the end of 8 years.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2396\n2. 2397\n3. 2398\n4. 2399\n5. 239B\n6. 239C\n7. 239D\n8. 239F\n9. 239G\n10. 239H\n11. 239J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n400\n9.5.\nSummary\n\nCHAPTER\n10\nProbability\n10.1\nRevision\n402\n10.2\nDependent and independent events\n411\n10.3\nMore Venn diagrams\n419\n10.4\nTree diagrams\n426\n10.5\nContingency tables\n431\n10.6\nSummary\n435\n\n10\nProbability\n10.1\nRevision\nEMBJP\nTerminology\nEMBJQ\nOutcome: a single observation of an uncertain or random process (called an experi-\nment). For example, when you accidentally drop a book, it might fall on its cover, on\nits back or on its side. Each of these options is a possible outcome.\nSample space of an experiment: the set of all possible outcomes of the experiment. For\nexample, the sample space when you roll a single 6-sided die is the set {1; 2; 3; 4; 5; 6}.\nFor a given experiment, there is exactly one sample space. The sample space is de-\nnoted by the letter S.\nEvent: a set of outcomes of an experiment. For example, during radioactive decay of\n1 gramme of uranium-234, one possible event is that the number of alpha-particles\nemitted during 1 microsecond is between 225 and 235.\nProbability of an event: a real number between 0 and 1 that describes how likely it\nis that the event will occur. A probability of 0 means the outcome of the experiment\nwill never be in the event set. A probability of 1 means the outcome of the experiment\nwill always be in the event set. When all possible outcomes of an experiment have\nequal chance of occurring, the probability of an event is the number of outcomes in\nthe event set as a fraction of the number of outcomes in the sample space.\nRelative frequency of an event: the number of times that the event occurs during\nexperimental trials, divided by the total number of trials conducted. For example, if\nwe flip a coin 10 times and it landed on heads 3 times, then the relative frequency of\nthe heads event is 3\n10 = 0,3.\nUnion of events: the set of all outcomes that occur in at least one of the events. For\n2 events called A and B, we write the union as “A or B”. Another way of writing the\nunion is using set notation: A ∪B.\nIntersection of events: the set of all outcomes that occur in all of the events. For 2\nevents called A and B, we write the intersection as “A and B”. Another way of writing\nthe intersection is using set notation: A ∩B.\nMutually exclusive events: events with no outcomes in common, that is (A and B) =\n∅. Mutually exclusive events can never occur simultaneously. For example the event\nthat a number is even and the event that the same number is odd are mutually exclu-\nsive, since a number can never be both even and odd.\nComplementary events: two mutually exclusive events that together contain all the\noutcomes in the sample space. For an event called A, we write the complement as\n“not A”. Another way of writing the complement is as A′.\nSee video: 239K at www.everythingmaths.co.za\n402\n10.1.\nRevision\n\nIdentities\nEMBJR\nThe addition rule (also called the sum rule) for any 2 events, A and B is\nP(A or B) = P(A) + P(B) −P(A and B)\nThis rule relates the probabilities of 2 events with the probabilities of their union and\nintersection.\nThe addition rule for 2 mutually exclusive events is\nP(A or B) = P(A) + P(B)\nThis rule is a special case of the previous rule. Because the events are mutually exclu-\nsive, P(A and B) = 0.\nThe complementary rule is\nP(not A) = 1 −P(A)\nThis rule is a special case of the previous rule. Since A and (not A) are mutually\nexclusive, P(A or (not A)) = 1.\nSee video: 239M at www.everythingmaths.co.za\nWorked example 1: Events\nQUESTION\nYou take all the hearts from a deck of cards. You then select a random card from the set\nof hearts. What is the sample space? What is the probability of each of the following\nevents?\n1. The card is the ace of hearts.\n2. The card has a prime number on it.\n3. The card has a letter of the alphabet on it.\nSOLUTION\nStep 1: Write down the sample space\nSince we are considering only one suit from the deck of cards (the hearts), we need to\nwrite down only the letters and numbers on the cards. Therefore the sample space is\nS = {A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K}\nStep 2: Write down the event sets\n• ace of hearts: {A}\n• prime number: {2; 3; 5; 7}\n• letter of alphabet: {A; J; Q; K}\n403\nChapter 10.\nProbability\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are 13 elements in the\nsample space. So the probability of each event is\n• ace of hearts:\n1\n13\n• prime number:\n4\n13\n• letter of alphabet:\n4\n13\nWorked example 2: Events\nQUESTION\nYou roll two 6-sided dice. Let E be the event that the total number of dots on the dice\nis 10. Let F be the event that at least one die is a 3.\n1. Write down the event sets for E and F.\n2. Determine the probabilities for E and F.\n3. Are E and F mutually exclusive? Why or why not?\nSOLUTION\nStep 1: Write down the sample space\nThe sample space of a single 6-sided die is just {1; 2; 3; 4; 5; 6}. To get the sample\nspace of two 6-sided dice, we have to take every possible pair of numbers from 1 to 6.\nS =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(1; 1)\n(1; 2)\n(1; 3)\n(1; 4)\n(1; 5)\n(1; 6)\n(2; 1)\n(2; 2)\n(2; 3)\n(2; 4)\n(2; 5)\n(2; 6)\n(3; 1)\n(3; 2)\n(3; 3)\n(3; 4)\n(3; 5)\n(3; 6)\n(4; 1)\n(4; 2)\n(4; 3)\n(4; 4)\n(4; 5)\n(4; 6)\n(5; 1)\n(5; 2)\n(5; 3)\n(5; 4)\n(5; 5)\n(5; 6)\n(6; 1)\n(6; 2)\n(6; 3)\n(6; 4)\n(6; 5)\n(6; 6)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nStep 2: Write down the events\nFor E the dice have to add to 10.\nE = {(4; 6); (5; 5); (6; 4)}\nFor F at least one die has to be 3.\nF = {(1; 3); (3; 1); (2; 3); (3; 2); (3; 3); (4; 3); (3; 4); (5; 3); (3; 5); (6; 3); (3; 6)}\n404\n10.1.\nRevision\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are\n• 6 × 6 = 36 outcomes in the sample space, S;\n• 3 outcomes in event E; and\n• 11 outcomes in event F.\nTherefore\nP(E) = 3\n36 = 1\n12\nand\nP(F) = 11\n36\nStep 4: Are they mutually exclusive\nTo test whether two events are mutually exclusive, we have to test whether their in-\ntersection is empty. Since E has no outcomes that contain a 3 on one of the dice,\nthe intersection of E and F is empty: (E and F) = ∅. This means that the events are\nmutually exclusive.\nSee video: 239N at www.everythingmaths.co.za\nExercise 10 – 1: Revision\n1. A bag contains r red balls, b blue balls and y yellow balls. What is the probability\nthat a ball drawn from the bag at random is yellow?\n2. A packet has yellow and pink sweets. The probability of taking out a pink sweet\nis 7\n12. What is the probability of taking out a yellow sweet?\n3. You flip a coin 4 times. What is the probability that you get 2 heads and 2 tails?\nWrite down the sample space and the event set to determine the probability of\nthis event.\n4. In a class of 37 children, 15 children walk to school, 20 children have pets at\nhome and 12 children who have a pet at home also walk to school. How many\nchildren walk to school and do not have a pet at home?\n5. You roll two 6-sided dice and are interested in the following two events:\n• A: the sum of the dice equals 8\n• B: at least one of the dice shows a 1\nShow that these events are mutually exclusive.\n405\nChapter 10.\nProbability\n\n6. You ask a friend to think of a number from 1 to 100. You then ask her the\nfollowing questions:\n• Is the number even?\n• Is the number divisible by 7?\nHow many possible numbers are less than 80 if she answered “yes” to both\nquestions?\n7. In a group of 42 pupils, all but 3 had a packet of chips or a Fanta or both. If 23\nhad a packet of chips and 7 of these also had a Fanta, what is the probability that\none pupil chosen at random has:\na) both chips and Fanta\nb) only Fanta\n8. Tamara has 18 loose socks in a drawer. Eight of these are orange and two are\npink. Calculate the probability that the first sock taken out at random is:\na) orange\nb) not orange\nc) pink\nd) not pink\ne) orange or pink\nf) neither orange nor pink\n9. A box contains coloured blocks. The number of blocks of each colour is given\nin the following table.\nColour\nPurple\nOrange\nWhite\nPink\nNumber of blocks\n24\n32\n41\n19\nA block is selected randomly. What is the probability that the block will be:\na) purple\nb) purple or white\nc) pink and orange\nd) not orange?\n10. The surface of a soccer ball is made up of 32 faces. 12 faces are regular pen-\ntagons, each with a surface area of about 37 cm2. The other 20 faces are regular\nhexagons, each with a surface area of about 56 cm2.\nYou roll the soccer ball. What is the probability that it stops with a pentagon\ntouching the ground?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 239P\n2. 239Q\n3. 239R\n4. 239S\n5. 239T\n6. 239V\n7. 239W\n8. 239X\n9. 239Y\n10. 239Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n406\n10.1.\nRevision\n\nVenn diagrams\nEMBJS\nA Venn diagram is used to show how events are related to one another.\nA Venn\ndiagram can be very helpful when doing calculations with probabilities. In a Venn\ndiagram each event is represented by a shape, often a circle or a rectangle. The region\ninside the shape represents the outcomes included in the event and the region outside\nthe shape represents the outcomes that are not in the event.\nS\nA\nB\nA and B\nA Venn diagram representing a sample space, S, as a square; and two events, A and\nB, as circles. The intersection of the two circles contains outcomes that are in both A\nand B.\nVenn diagrams can be used in slightly different ways and it is important to notice the\ndifferences between them. The following 3 examples show how a Venn diagram is\nused to represent\n• the outcomes included in each event;\n• the number of outcomes in each event; and\n• the probability of each event.\nWorked example 3: Venn diagrams with outcomes\nQUESTION\nChoose a number between 1 and 20. Draw a Venn diagram to answer the following\nquestions.\n1. What is the probability that the number is a multiple of 3?\n2. What is the probability that the number is a multiple of 5?\n3. What is the probability that the number is a multiple of 3 or 5?\n4. What is the probability that the number is a multiple of 3 and 5?\nSOLUTION\nStep 1: Draw a Venn diagram\nThe Venn diagram should show the sample space of all numbers from 1 to 20. It should\nalso show an event set that contains all the multiples of 3, let A = {3; 6; 9; 12; 15; 18},\n407\nChapter 10.\nProbability\n\nand another event set that contains all the multiples of 5, let B = {5; 10; 15; 20}. Note\nthat there is one shared outcome between these two events, namely 15.\n3\n18\n15\n12\n9\n6\n10\n20\n5\n1\n2\n4\n7\n8\n11\n13\n14\n16\n17\n19\nStep 2: Compute probabilities\nThe probability of an event is the number of outcomes in the event set divided by the\nnumber of outcomes in the sample space. There are 20 outcomes in the sample space.\n1. Since there are 6 outcomes in the multiples of 3 event set, the probability of a\nmultiple of 3 is P(A) = 6\n20 = 3\n10.\n2. Since there are 4 outcomes in the multiples of 5 event set, the probability of a\nmultiple of 5 is P(B) = 4\n20 = 1\n5.\n3. The event that the number is a multiple of 3 or 5 is the union of the above two\nevent sets. There are 9 elements in the union of the event sets, so the probability\nis 9\n20.\n4. The event that the number is a multiple of 3 and 5 is the intersection of the\ntwo event sets. There is 1 element in the intersection of the event sets, so the\nprobability is 1\n20.\nWorked example 4: Venn diagrams with counts\nQUESTION\nIn a group of 50 learners, 35 take Mathematics and 30 take History, while 12 take\nneither of the two subjects. Draw a Venn diagram representing this information. If a\nlearner is chosen at random from this group, what is the probability that he takes both\nMathematics and History?\nSOLUTION\nStep 1: Draw outline of Venn diagram\nThere are 2 events in this question, namely\n• M: that a learner takes Mathematics; and\n• H: that a learner takes History.\n408\n10.1.\nRevision\n\nWe need to do some calculations before drawing the full Venn diagram, but with the\ninformation above we can already draw the outline.\nS\nM\nH\nStep 2: Write down sizes of the event sets, their union and intersection\nWe are told that 12 learners take neither of the two subjects. Graphically we can\nrepresent this as:\nS\nM\nH\n12\nSince there are 50 elements in the sample space, we can see from this figure that there\nare 50 −12 = 38 elements in (M or H). So far we know\n• n(M) = 35\n• n(H) = 30\n• n(M or H) = 38\nFrom the addition rule,\nn(M or H) = n(M) + n(H) −n(M and H)\n∴n(M and H) = 35 + 30 −38\n= 27\nStep 3: Draw the final Venn diagram\nS\nM\nH\n12\n27\n8\n3\n409\nChapter 10.\nProbability\n\nWorked example 5: Venn diagrams with probabilities\nQUESTION\nDraw a Venn diagram to represent the same information as in the previous example,\nexcept showing the probabilities of the different events, rather than the counts.\nIf a learner is chosen at random from this group, what is the probability that she takes\nboth Mathematics and History?\nSOLUTION\nStep 1: Use counts to compute probabilities\nSince there are 50 elements (learners) in the sample space, we can compute the prob-\nability of any event by dividing the size of the event set by 50. This gives the following\nprobabilities:\n• P(M) = 35\n50 = 7\n10\n• P(H) = 30\n50 = 3\n5\n• P(M or H) = 38\n50 = 19\n25\n• P(M and H) = 27\n50\nStep 2: Draw the Venn diagram\nNext we replace each count from the Venn diagram in the previous example with a\nprobability.\nS\nM\nH\n6\n25\n27\n50\n4\n25\n3\n50\nStep 3: Find the answer\nThe probability that a random learner will take both Mathematics and History is\nP(M and H) = 27\n50.\nSee video: 23B2 at www.everythingmaths.co.za\n410\n10.1.\nRevision\n\nExercise 10 – 2: Venn diagram revision\n1. Given the following information:\n• P(A) = 0,3\n• P(B and A) = 0,2\n• P(B) = 0,7\nFirst draw a Venn diagram to represent this information. Then compute the value\nof P(B and (not A)).\n2. You are given the following information:\n• P(A) = 0,5\n• P(A and B) = 0,2\n• P(not B) = 0,6\nDraw a Venn diagram to represent this information and determine P(A or B).\n3. A study was undertaken to see how many people in Port Elizabeth owned either\na Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60%\nowned a Volkswagen. What percentage of people owned neither car?\n4. Let S denote the set of whole numbers from 1 to 15, X denote the set of even\nnumbers from 1 to 15 and Y denote the set of prime numbers from 1 to 15.\nDraw a Venn diagram depicting S, X and Y .\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B3\n2. 23B4\n3. 23B5\n4. 23B6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.2\nDependent and independent events\nEMBJT\nSometimes the presence or absence of one event tells us something about other events.\nWe call events dependent if knowing whether one of them happened tells us some-\nthing about whether the others happened. Independent events give us no information\nabout one another; the probability of one event occurring does not affect the probabil-\nity of the other events occurring.\nDEFINITION: Independent events\nTwo events, A and B are independent if and only if\nP(A and B) = P(A) × P(B)\nAt first it might not be clear why we should call events that satisfy the equation above\nindependent. We will explore this further using a number of examples.\n411\nChapter 10.\nProbability\n\nInvestigation: Independence\nRoll a single 6-sided die and consider the following two events:\n• E: you get an even number\n• T: you get a number that is divisible by three\nNow answer the following questions:\n• What is the probability of E?\n• What is the probability of getting an even number if you are told that the number\nwas also divisible by three?\n• Does knowing that the number was divisible by three change the probability that\nthe number was even?\nAre the events E and T dependent or independent according to the definition (hint:\ncompute the probabilities in the definition of independence)?\nSee video: 23B7 at www.everythingmaths.co.za\nSo, why do we call it independence when P(A and B) = P(A) × P(B)? For two\nevents, A and B, independence means that knowing the outcome of B does not affect\nthe probability of A.\nConsider the following Venn diagram.\nS\nA\nB\nA and B\nThe probability of A is the ratio between the number of outcomes in A and the number\nof outcomes in the sample space, S.\nP(A) = n(A)\nn(S)\n412\n10.2.\nDependent and independent events\n\nNow, let’s say that we know that event B happened. How does this affect the proba-\nbility of A? Here is how the Venn diagram changes:\nS\nA\nB\nA and B\nA lot of the possible outcomes (all of the outcomes outside B) are now out of the pic-\nture, because we know that they did not happen. Now the probability of A happening,\ngiven that we know that B happened, is the ratio between the size of the region where\nA is present (A and B) and the size of all possible events (B).\nP(A if we know B) = n(A and B)\nn(B)\nIf P(A) = P(A if we know B) we call them independent, because knowing B does\nnot change the probability of A.\nWith some algebra, we can prove that this statement of independence is the same\nas the definition of independence that we saw at the beginning of this section. For\nindependent events\nP(A and B) = P(A) × P(B)\nThis is equivalent to\nP(A) = P(A and B) ÷ P(B)\n= n(A and B)\nn(S)\n÷ n(B)\nn(S)\n= n(A and B)\nn(B)\n= P(A if we know B)\nThat is why we call events independent!\n(For enrichment only):\nThe ratio\nP(A and B)\nP(B)\nis called a conditional probability and written using the notation P(A | B). This\nnotation is read as “the probability of A given B.”\nIf (and only if) A and B are independent: P(A | B) = P(A) and P(B | A) = P(B).\nTry to prove this using the definition of independence.\n413\nChapter 10.\nProbability\n\nWorked example 6: Independent and dependent events\nQUESTION\nA bag contains 5 red and 5 blue balls. We remove a random ball from the bag, record\nits colour and put it back into the bag. We then remove another random ball from the\nbag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Probability of a red ball first\nSince there are a total of 10 balls, of which 5 are red, the probability of getting a red\nball is\nP(first ball red) = 5\n10 = 1\n2\nStep 2: Probability of a blue ball second\nThe problem states that the first ball is placed back into the bag before we take the\nsecond ball. This means that when we draw the second ball, there are again a total of\n10 balls in the bag, of which 5 are blue. Therefore the probability of drawing a blue\nball is\nP(second ball blue) = 5\n10 = 1\n2\nStep 3: Probability of red first and blue second\nWhen drawing two balls from the bag, there are 4 possibilities. We can get\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nWe want to know the probability of the second outcome, where we have to get a red\nball first. Since there are 5 red balls and 10 balls in total, there are\n5\n10 ways to get a\nred ball first. Now we put the first ball back, so there are again 5 red balls and 5 blue\nballs in the bag. Therefore there are\n5\n10 ways to get a blue ball second if the first ball\nwas red. This means that there are\n5\n10 × 5\n10 = 25\n100\n414\n10.2.\nDependent and independent events\n\nways to get a red ball first and a blue ball second. So, the probability of getting a red\nball first and a blue ball second is 1\n4.\nStep 4: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 1\n4\nSince 1\n4 = 1\n2 × 1\n2, the events are independent.\nSee video: 23B8 at www.everythingmaths.co.za\nWorked example 7: Independent and dependent events\nQUESTION\nIn the previous example, we picked a random ball and put it back into the bag before\ncontinuing. This is called sampling with replacement. In this example, we will follow\nthe same process, except that we will not put the first ball back into the bag. This is\ncalled sampling without replacement.\nSo, from a bag with 5 red and 5 blue balls, we remove a random ball and record its\ncolour. Then, without putting back the first ball, we remove another random ball from\nthe bag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Count the number of outcomes\nWe will look directly at the number of possible ways in which we can get the 4 possible\noutcomes when removing 2 balls. In the previous example, we saw that the 4 possible\noutcomes are\n415\nChapter 10.\nProbability\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nFor the first outcome, we have to get a red ball first. Since there are 5 red balls and\n10 balls in total, there are\n5\n10 ways to get a red ball first. After we have taken out a red\nball, there are now 4 red balls and 5 blue balls left. Therefore there are 4\n9 ways to get\na red ball second if the first ball was also red. This means that there are\n5\n10 × 4\n9 = 20\n90\nways to get a red ball first and a red ball second. The probability of the first outcome\nis 2\n9.\nFor the second outcome, we have to get a red ball first. As in the first outcome, there\nare\n5\n10 ways to get a red ball first; and there are now 4 red balls and 5 blue balls left.\nTherefore there are 5\n9 ways to get a blue ball second if the first ball was red. This means\nthat there are\n5\n10 × 5\n9 = 25\n90\nways to get a red ball first and a blue ball second. The probability of the second\noutcome is 5\n18.\nWe can compute the probabilities of the third and fourth outcomes in the same way as\nthe first two, but there is an easier way. Notice that there are only 2 types of ball and\nthat there are exactly equal numbers of them at the start. This means that the problem\nis completely symmetric in red and blue. We can use this symmetry to compute the\nprobabilities of the other two outcomes.\nIn the third outcome, the first ball is blue and the second ball is red. Because of\nsymmetry this outcome must have the same probability as the second outcome (when\nthe first ball is red and the second ball is blue). Therefore the probability of the third\noutcome is 5\n18.\nIn the fourth outcome, the first and second balls are both blue. From symmetry, this\noutcome must have the same probability as the first outcome (when both balls are red).\nTherefore the probability of the fourth outcome is 2\n9.\nTo summarise, these are the possible outcomes and their probabilities:\n• first ball red and second ball red: 2\n9;\n• first ball red and second ball blue:\n5\n18;\n• first ball blue and second ball red:\n5\n18;\n• first ball blue and second ball blue: 2\n9.\nStep 2: Probability of a red ball first\nTo determine the probability of getting a red ball on the first draw, we look at all of the\noutcomes that contain a red ball first. These are\n416\n10.2.\nDependent and independent events\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball.\nThe probability of the first outcome is 2\n9 and the probability of the second outcome is\n5\n18. By adding these two probabilities, we see that the probability of getting a red ball\nfirst is\nP(first ball red) = 2\n9 + 5\n18 = 1\n2\nThis is the same as in the previous exercise, which should not be too surprising since\nthe probability of the first ball being red is not affected by whether or not we put it\nback into the bag before drawing the second ball.\nStep 3: Probability of a blue ball second\nTo determine the probability of getting a blue ball on the second draw, we look at all\nof the outcomes that contain a blue ball second. These are\n• a red ball and then a blue ball;\n• a blue ball and then another blue ball.\nThe probability of the first outcome is 5\n18 and the probability of the second outcome is\n2\n9. By adding these two probabilities, we see that the probability of getting a blue ball\nsecond is\nP(second ball blue) = 5\n18 + 2\n9 = 1\n2\nThis is also the same as in the previous exercise! You might find it surprising that the\nprobability of the second ball is not affected by whether or not we replace the first ball.\nThe reason why this probability is still 1\n2 is that we are computing the probability that\nthe second ball is blue without knowing the colour of the first ball. Because there are\nonly two equal possibilities for the second ball (red and blue) and because we don’t\nknow whether the first ball is red or blue, there is an equal chance that the second ball\nwill be one colour or the other.\nStep 4: Probability of red first and blue second\nWe have already calculated the probability that the first ball is red and the second ball\nis blue. It is 5\n18.\nStep 5: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 5\n18\nSince 5\n18 ̸= 1\n2 × 1\n2, the events are dependent.\n417\nChapter 10.\nProbability\n\nWARNING!\nJust because two events are mutually exclusive does not necessarily mean that they\nare independent. To test whether events are mutually exclusive, always check that\nP(A and B) = 0. To test whether events are independent, always check that P(A and B) =\nP(A) × P(B). See the exercises below for examples of events that are mutually ex-\nclusive and independent in different combinations.\nExercise 10 – 3: Dependent and independent events\n1. Use the following Venn diagram to determine whether events X and Y are\na) mutually exclusive or not mutually exclusive;\nb) dependent or independent.\nS\nX\nY\n11\n7\n3\n14\n2. Of the 30 learners in a class 17 have black hair, 11 have brown hair and 2 have\nred hair. A learner is selected from the class at random.\na) What is the probability that the learner has black hair?\nb) What is the probability that the learner has brown hair?\nc) Are these two events mutually exclusive?\nd) Are these two events independent?\n3. P(M) = 0,45; P(N) = 0,3 and P(M or N) = 0,615. Are the events M and N\nmutually exclusive, independent or neither mutually exclusive nor independent?\n4. (For enrichment)\nProve that if event A and event B are mutually exclusive with P(A) ̸= 0 and\nP(B) ̸= 0, then A and B are always dependent.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B9\n2. 23BB\n3. 23BC\n4. 23BD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n418\n10.2.\nDependent and independent events\n\n10.3\nMore Venn diagrams\nEMBJV\nIn the rest of this chapter we will look at tools and techniques for working with proba-\nbility problems.\nWhen working with more complex problems, we can have three or more events that\nintersect in various ways.\nTo solve these problems, we usually want to count the\nnumber (or percentage) of outcomes in an event, or a combination of events. Venn\ndiagrams are a useful tool for recording and visualising the counts.\nInvestigation: Venn diagram for 3 events\nThe diagram below shows a general Venn diagram for 3 events.\nS\nA\nB\nC\nWrite down the sets corresponding to each of the three coloured regions and also\nto the shaded region. Remember that the intersections between circles represent the\nintersections between the different events.\nWhat is the event for\n• the red region;\n• the green region;\n• the blue region; and\n• the shaded region?\n419\nChapter 10.\nProbability\n\nWorked example 8: Venn diagram for 3 events\nQUESTION\nDraw a Venn diagram that shows the following sample space and events:\n• S: all the integers from 1 to 30\n• P: prime numbers\n• M: multiples of 3\n• F: factors of 30\nSOLUTION\nStep 1: Write down the sample space and event sets\nThe sample space contains all the positive integers up to 30.\nS = {1; 2; 3; . . . ; 30}\nThe prime numbers between 1 and 30 are\nP = {2; 3; 5; 7; 11; 13; 17; 19; 23; 29}\nThe multiples of 3 between 1 and 30 are\nM = {3; 6; 9; 12; 15; 18; 21; 24; 27; 30}\nThe factors of 30 are\nF = {1; 2; 3; 5; 6; 10; 15; 30}\nStep 2: Draw the outline of the Venn diagram\nThere are 3 events, namely P, M and F, and the sample space, S. Put this information\non a Venn diagram:\nS\nP\nM\nF\n420\n10.3.\nMore Venn diagrams\n\nStep 3: Place the outcomes in the appropriate event sets\nS\nP\nM\nF\n3\n2\n5\n6\n15\n30\n1\n10\n7\n11\n13\n17\n19\n23\n29\n9\n12\n21\n24\n18\n27\n4\n8\n14\n16\n20\n22\n25\n26\n28\nWorked example 9: Venn diagram for 3 events\nQUESTION\nAt Dawnview High there are 400 Grade 11 learners. 270 do Computer Science, 300\ndo English and 50 do Business studies. All those doing Computer Science do English,\n20 take Computer Science and Business studies and 35 take English and Business\nstudies. Using a Venn diagram, calculate the probability that a pupil drawn at random\nwill take:\n1. English, but not Business studies or Computer Science\n2. English but not Business studies\n3. English or Business studies but not Computer Science\n4. English or Business studies\nSOLUTION\nStep 1: Draw the outline of the Venn diagram\nWe need to be careful with this problem. In the question statement we are told that all\nthe learners who do Computer Science also do English. This means that the circle for\nComputer Science on the Venn diagram needs to be inside the circle for English.\n421\nChapter 10.\nProbability\n\nS\nE\nC\nB\nStep 2: Fill in the counts on the Venn diagram\nS\nE\nC\nB\n20\n250\n15\n15\n15\n85\nStep 3: Compute probabilities\nTo find the number of learners taking English, but not Business studies or Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 15 and there are a total of 400 learners in the grade. There-\nfore the probability that a learner will take English but not Business studies or Computer\nScience is\n15\n400 = 3\n80.\nTo find the number of learners taking English but not Business studies, we need to look\nat this region of the Venn diagram:\n422\n10.3.\nMore Venn diagrams\n\nThe count in this region is 265. Therefore the probability that a learner will take\nEnglish but not Business studies is 265\n400 = 53\n80.\nTo find the number of learners taking English or Business studies but not Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 45. Therefore the probability that a learner will take English\nor Business studies but not Computer Science is\n45\n400 = 9\n80.\nTo find the number of learners taking English or Business studies, we need to look at\nthis region of the Venn diagram:\nThe count in this region is 315. Therefore the probability that a learner will take\nEnglish or Business studies is 315\n400 = 63\n80.\n423\nChapter 10.\nProbability\n\nThere are some words that tell you which part of the Venn diagram should be filled in.\nThe following table summarises the most important ones:\nWords\nSymbols\nVenn diagram\n“all”\nA and B and C / A ∩B ∩C\n“none”\n“at least one”\nA or B or C / A ∪B ∪C\n“both A and B”\nA and B / A ∩B\n“A or B”\nA or B / A ∪B\nExercise 10 – 4: Venn diagrams\n1. Use the Venn diagram below to answer the following questions. Also given:\nn(S) = 120.\nS\nF\n8\n10\nG\n24\n15\nH\n14\n7\n2\na) Compute P(F).\nb) Compute P(G or H).\nc) Compute P(F and G).\nd) Are F and G dependent or independent?\n424\n10.3.\nMore Venn diagrams\n\n2. The Venn diagram below shows the probabilities of 3 events. Complete the Venn\ndiagram using the additional information provided.\nS\nZ\n1\n25\nY\n17\n100\nX\n17\n100\n3\n20\n• P(Z and (not Y )) =\n31\n100\n• P(Y and X) =\n23\n100\n• P(Y ) =\n39\n100\nAfter completing the Venn diagram, compute the following:\nP (Z and not (X or Y ))\n3. There are 79 Grade 10 learners at school. All of these take some combination of\nMaths, Geography and History. The number who take Geography is 41; those\nwho take History is 36; and 30 take Maths. The number who take Maths and\nHistory is 16; the number who take Geography and History is 6, and there are 8\nwho take Maths only and 16 who take History only.\na) Draw a Venn diagram to illustrate all this information.\nb) How many learners take Maths and Geography but not History?\nc) How many learners take Geography only?\nd) How many learners take all three subjects?\n4. Draw a Venn diagram with 3 mutually exclusive events. Use the diagram to\nshow that for 3 mutually exclusive events, A, B and C, the following is true:\nP(A or B or C) = P(A) + P(B) + P(C)\nThis is the addition rule for 3 mutually exclusive events.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BF\n2. 23BG\n3. 23BH\n4. 23BJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n425\nChapter 10.\nProbability\n\n10.4\nTree diagrams\nEMBJW\nTree diagrams are useful for organising and visualising the different possible outcomes\nof a sequence of events. For each possible outcome of the first event, we draw a line\nwhere we write down the probability of that outcome and the state of the world if that\noutcome happened. Then, for each possible outcome of the second event we do the\nsame thing.\nBelow is an example of a simple tree diagram, showing the possible outcomes of\nrolling a 6-sided die.\n1\n1\n6\n2\n1\n6\n3\n1\n6\n4\n1\n6\n5\n1\n6\n6\n1\n6\noutcomes\nprobabilities\nNote that each outcome (the numbers 1 to 6) is shown at the end of a line; and that\nthe probability of each outcome (all 1\n6 in this case) is shown shown on a line. The\nprobabilities have to add up to 1 in order to cover all of the possible outcomes. In the\nexamples below, we will see how to draw tree diagrams with multiple events and how\nto compute probabilities using the diagrams.\nEarlier in this chapter you learned about dependent and independent events. Tree\ndiagrams are very helpful for analysing dependent events. A tree diagram allows you\nto show how each possible outcome of one event affects the probabilities of the other\nevents.\nTree diagrams are not so useful for independent events since we can just multiply the\nprobabilities of separate events to get the probability of the combined event. Remem-\nber that for independent events:\nP(A and B) = P(A) × P(B)\nSo if you already know that events are independent, it is usually easier to solve a\nproblem without using tree diagrams. But if you are uncertain about whether events\nare independent or if you know that they are not, you should use a tree diagram.\nWorked example 10: Drawing a tree diagram\nQUESTION\nIf it rains on a given day, the probability that it rains the next day is 1\n3. If it does not rain\non a given day, the probability that it rains the next day is 1\n6. The probability that it will\nrain tomorrow is 1\n5. What is the probability that it will rain the day after tomorrow?\nDraw a tree diagram of all the possibilities to determine the answer.\nSOLUTION\nStep 1: Draw the first level of the tree diagram\nBefore we can determine what happens on the day after tomorrow, we first have to\ndetermine what might happen tomorrow. We are told that there is a 1\n5 probability that\n426\n10.4.\nTree diagrams\n\nit will rain tomorrow. Here is how to represent this information using a tree diagram:\n1\n5\nrain\n4\n5\nno rain\ntoday:\ntomorrow:\nStep 2: Draw the second level of the tree diagram\nWe are also told that if it does rain on one day, there is a 1\n3 probability that it will also\nrain on the following day. On the other hand, if it does not rain on one day, there is\nonly a 1\n6 probability that it will also rain on the following day. Using this information\nwe complete the tree diagram:\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nStep 3: Compute the probability\nWe are asked what the probability is that it will rain the day after tomorrow. On the\ntree diagram above we can see that there are 2 situations where it rains on the day\nafter tomorrow. They are marked in red below.\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nTo get the probability for the first situation (that it rains tomorrow and the day after\ntomorrow) we have to multiply the probabilies along the first red line.\nP(rain tomorrow and rain day after tomorrow)\n=1\n5 × 1\n3\n= 1\n15\n427\nChapter 10.\nProbability\n\nTo get the probability for the second situation (that it does not rain tomorrow, but it\ndoes rain the day after tomorrow) we have to multiply the probabilies along the second\nred line.\nP(not rain tomorrow and rain day after tomorrow)\n=4\n5 × 1\n6\n= 2\n15\nTherefore the total probability that it will rain the day after tomorrow is the sum of the\nprobabilities along the two red paths, namely\n1\n15 + 2\n15 = 1\n5\nWorked example 11: Drawing a tree diagram\nQUESTION\nYou play the following game. You flip a coin. If it comes up tails, you get 2 points\nand your turn ends. If it comes up heads, you get only 1 point, but you can flip the\ncoin again. If you flip the coin multiple times in one turn, you add up the points. You\ncan flip the coin at most 3 times in one turn. What is the probability that you will get\nexactly 3 points in one turn? Draw a tree diagram to visualise the different possibilities.\nSOLUTION\nStep 1: Write down the events and their symbols\nEach coin toss has on of two possible outcomes, namely heads (H) and tails (T). Each\noutcome has a probability of 1\n2. We are asked to count the number of points, so we\nwill also indicate how many points we have for each outcome.\nStep 2: Draw the first level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\nThis tree diagram shows the possible outcomes after 1 flip of the coin. Remember that\nwe can have up to 3 flips, so the diagram is not complete yet. If the coin comes up\nheads, we flip the coin again. If the coin comes up tails, we stop.\n428\n10.4.\nTree diagrams\n\nStep 3: Draw the second and third level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\n1\n2\nH\n2 pts\n1\n2\nT\n3 pts\n1\n2\nH\n3 pts\n1\n2\nT\n4 pts\nIn this tree diagram you can see that we add up the points we get with each coin flip.\nAfter three coin flips, the game is over.\nStep 4: Find the relevant outcomes and compute the probability\nWe are interested in getting exactly 3 points during the game. To find these outcomes\nwe look only at the tips of the tree. We end with exactly 3 points when the coin flips\nare\n• (H; T) with probability 1\n2 × 1\n2 = 1\n4;\n• (H; H; H) with probability 1\n2 × 1\n2 × 1\n2 = 1\n8.\nNotice that we compute the probability of an outcome by multiplying all the probabil-\nities along the path from the start of the tree to the tip where the outcome is. We add\nthe above two probabilites to obtain the final probability of getting exactly 3 points as\n1\n4 + 1\n8 = 3\n8.\nWorked example 12: Drawing a tree diagram\nQUESTION\nA person takes part in a medical trial that tests the effect of a medicine on a disease.\nHalf the people are given medicine and the other half are given a sugar pill, which has\nno effect on the disease. The medicine has a 60% chance of curing someone. But,\npeople who do not get the medicine still have a 10% chance of getting well. There are\n50 people in the trial and they all have the disease. Talwar takes part in the trial, but\nwe do not know whether he got the medicine or the sugar pill. Draw a tree diagram\nof all the possible cases. What is the probability that Talwar gets cured?\nSOLUTION\nStep 1: Summarise the information in the problem\nThere are two uncertain events in this problem. Each person either receives medicine\n(probability 1\n2) or a sugar pill (probability 1\n2). Each person also gets cured (probability\n429\nChapter 10.\nProbability\n\n3\n5 with medicine and\n1\n10 without) or stays ill (probability 2\n5 with medicine and\n9\n10\nwithout).\nStep 2: Draw the tree diagram\n1\n2\nmedicine\n1\n2\nsugar pill\n3\n5\ncured\n2\n5\nnot cured\n1\n10\ncured\n9\n10\nnot cured\nIn the first level of the tree diagram we show that Talwar either gets the medicine or\nthe sugar pill. The second level of the tree diagram shows whether Talwar is cured or\nnot, depending on which one of the pills he got.\nStep 3: Compute the required probability\nWe multiply the probabilites along each path in the tree diagram that leads to Talwer\nbeing cured:\n1\n2 × 3\n5 = 3\n10\n1\n2 × 1\n10 = 1\n20\nWe then add these probabilites to get the final answer. The probability that Talwar is\ncured is 7\n20.\nExercise 10 – 5: Tree diagrams\n1. You roll a die twice and add up the dots to get a score. Draw a tree diagram to\nrepresent this experiment. What is the probability that your score is a multiple\nof 5?\n2. What is the probability of throwing at least one five in four rolls of a regular\n6-sided die? Hint: do not show all possible outcomes of each roll of the die. We\nare interested in whether the outcome is 5 or not 5 only.\n3. You flip one coin 4 times.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\n430\n10.4.\nTree diagrams\n\n4. You flip 4 different coins at the same time.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BK\n2. 23BM\n3. 23BN\n4. 23BP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.5\nContingency tables\nEMBJX\nA contingency table is another tool for keeping a record of the counts or percentages\nin a probability problem. Contingency tables are especially helpful for figuring out\nwhether events are dependent or independent.\nWe will be studying two-way contingency tables, where we count the number of out-\ncomes for 2 events and their complements, making 4 events in total. A two-way contin-\ngency table always shows the counts for the 4 possible combinations of events, as well\nas the totals for each event and its complement. We can use a contingency table to\ncompute the probabilities of various events by computing the ratios between counts,\nand to determine whether the events are dependent or independent. The example\nbelow shows a two-way contingency table, representing the outcome of a medical\nstudy.\nWorked example 13: Contingency tables\nQUESTION\nA medical trial into the effectiveness of a new medication was carried out. 120 females\nand 90 males took part in the trial. Out of those people, 50 females and 30 males\nresponded positively to the medication. Given below is a contingency table with the\ngiven information filled in.\nFemale\nMale\nTotals\nPositive\n50\n30\nNegative\nTotals\n120\n90\n1. What is the probability that the medicine gives a positive result for females?\n2. What is the probability that the medicine gives a negative result for males?\n3. Was the medication’s success independent of gender? Explain.\n431\nChapter 10.\nProbability\n\nSOLUTION\nStep 1: Complete the contingency table\nThe best place to start is always to complete the contingency table. Because the each\ncolumn has to sum up to its total, we can work out the number of females and males\nwho responded negatively to the medication. Then we can add each row to get the\ntotals on the right hand side of the table.\nFemale\nMale\nTotals\nPositive\n50\n30\n80\nNegative\n70\n60\n130\nTotals\n120\n90\n210\nStep 2: Compute the required probabilities\nThe way the first question is phrased, we need to determine the probability that a\nperson responds positively if she is female. This means that we do not include males\nin this calculation. So, the probability that the medicine gives a positive result for\nfemales is the ratio between the number of females who got a positive response and\nthe total number of females.\nP(positive if female) = n(positive and female)\nn(female)\n= 50\n120\n= 5\n12\nSimilarly, the probability that the medicine gives a negative result for males is:\nP(negative if male) = n(negative and male)\nn(male)\n= 60\n90\n= 2\n3\nStep 3: Independence\nWe need to determine whether the effect of the medicine and the gender of a par-\nticipant are dependent or independent. According to the definition, two events are\nindependent if and only if\nP(A and B) = P(A) × P(B)\nWe will look at the events that a participant is female and that the participant re-\nsponded positively to the trial.\nP(female) =\nn(female)\nn(total trials)\n= 120\n210\n= 4\n7\n432\n10.5.\nContingency tables\n\nP(positive) =\nn(positive)\nn(total trials)\n= 80\n210\n= 8\n21\nP(female and positive) = n(female and positive)\nn(total trials)\n= 50\n210\n= 5\n21\nFrom these probabilities we can see that\nP(female and positive) ̸= P(female) × P(positive)\nand therefore the gender of a participant and the outcome of a trial are dependent\nevents.\nWorked example 14: Contingency tables\nQUESTION\nUse the contingency table below to answer the following questions.\nGrade 11\nGrade 12\nTotals\nHas cellphone\n59\n50\n109\nNo cellphone\n6\n3\n9\nTotals\n65\n53\n118\n1. What is the probability that a learner from Grade 11 has a cellphone?\n2. What is the probability that a learner who does not have a cellphone is from\nGrade 11.\n3. Are the grade of a learner and whether he has a cellphone or not independent\nevents? Explain your answer.\nSOLUTION\n1. There are 65 learners in Grade 11 and 59 of them have a cellphone. Therefore\nthe probability that a learner from Grade 11 has a cellphone is 59\n65.\n2. There are 9 learners who do not have a cellphone and 6 of them are in Grade\n11. Therefore the probability that a learner who does not have a cellphone is\nfrom from Grade 11 is 6\n9 = 2\n3.\n433\nChapter 10.\nProbability\n\n3. To test for independence, we will consider whether a learner is in Grade 11 and\nwhether a learner has a cellphone. The probability that a learner is in Grade 11\nis\n65\n118. The probability that a learner has a cellphone is 109\n118. The probability that\na learner is in Grade 11 and has a cellphone is\n59\n118 = 1\n2. Since 1\n2 ̸=\n65\n118 × 109\n118\nthe grade of a learner and whether he has a cellphone are dependent.\nExercise 10 – 6: Contingency tables\n1. Use the contingency table below to answer the following questions.\nBrown eyes\nNot brown eyes\nTotals\nBlack hair\n50\n30\n80\nRed hair\n70\n80\n150\nTotals\n120\n110\n230\na) What is the probability that someone with black hair has brown eyes?\nb) What is the probability that someone has black hair?\nc) What is the probability that someone has brown eyes?\nd) Are having black hair and having brown eyes dependent or independent\nevents?\n2. Given the following contingency table, identify the events and determine\nwhether they are dependent or independent.\nLocation A\nLocation B\nTotals\nBuses left late\n15\n40\n55\nBuses left on time\n25\n20\n45\nTotals\n40\n60\n100\n3. You are given the following information.\n• Events A and B are independent.\n• P(not A) = 0,3.\n• P(B) = 0,4.\nComplete the contingency table below.\nA\nnot A\nTotals\nB\nnot B\nTotals\n50\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BQ\n2. 23BR\n3. 23BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n434\n10.5.\nContingency tables\n\n10.6\nSummary\nEMBJY\nSee presentation: 23BT at www.everythingmaths.co.za\n• Terminology:\n– Outcome: a single observation of an experiment.\n– Sample space of an experiment: the set of all possible outcomes of the\nexperiment.\n– Event: a set of outcomes of an experiment.\n– Probability of an event: a real number between 0 and 1 that describes how\nlikely it is that the event will occur.\n– Relative frequency of an event: the number of times that the event occurs\nduring experimental trials, divided by the total number of trials conducted.\n– Union of events: the set of all outcomes that occur in at least one of the\nevents, written as “A or B”.\n– Intersection of events: the set of all outcomes that occur in all of the events,\nwritten as “A and B”.\n– Mutually exclusive events: events with no outcomes in common, that is\n(A and B) = ∅.\n– Complementary events: two mutually exclusive events that together con-\ntain all the outcomes in the sample space. We write the complement as\n“not A”.\n– Independent events: two events where knowing the outcome of one event\ndoes not affect the probability of the other event. Events are independent if\nand only if P(A and B) = P(A) × P(B).\n• Identities:\n– The addition rule: P(A or B) = P(A) + P(B) −P(A and B)\n– The addition rule for 2 mutually exclusive events: P(A or B) = P(A) +\nP(B)\n– The complementary rule: P(not A) = 1 −P(A)\n• A Venn diagram is a visual tool used to show how events overlap. Each region\nin a Venn diagram represents an event and could contain either the outcomes in\nthe event, the number of outcomes in the event or the probability of the event.\n• A tree diagram is a visual tool that helps with computing probabilities for depen-\ndent events. The outcomes of each event are shown along with the probability\nof each outcome. For each event that depends on a previous event, we go one\nlevel deeper into the tree. To compute the probability of some combination of\noutcomes, we\n– find all the paths that contain the outcome of interest;\n– multiply the probabilities along each path;\n– add the probabilities between different paths.\n• A 2-way contingency table is a tool for organising data, especially when we want\nto determine whether two events, each with only two outcomes, are dependent\nor independent. The counts for each possible combination of outcomes are\nentered into the table, along with the totals of each row and column.\n435\nChapter 10.\nProbability\n\nExercise 10 – 7: End of chapter exercises\n1. Jane invested in the stock market. The probability that she will not lose all her\nmoney is 0,32. What is the probability that she will lose all her money? Explain.\n2. If D and F are mutually exclusive events, with P(not D)\n=\n0,3 and\nP(D or F) = 0,94, find P(F).\n3. A car sales person has pink, lime-green and purple models of car A and purple,\norange and multicolour models of car B. One dark night a thief steals a car.\na) What is the experiment and sample space?\nb) What is the probability of stealing either a model of A or a model of B?\nc) What is the probability of stealing both a model of A and a model of B?\n4. The probability of event X is 0,43 and the probability of event Y is 0,24. The\nprobability of both occurring together is 0,10. What is the probability that X or\nY will occur?\n5. P(H) = 0,62; P(J) = 0,39 and P(H and J) = 0,31. Calculate:\na) P(H′)\nb) P(H or J)\nc) P(H′ or J′)\nd) P(H′ or J)\ne) P(H′ and J′)\n6. The last ten letters of the alphabet are placed in a hat and people are asked to\npick one of them. Event D is picking a vowel, event E is picking a consonant\nand event F is picking one of the last four letters. Draw a Venn diagram showing\nthe outcomes in the sample space and the different events. Then calculate the\nfollowing probabilities:\na) P(not F)\nb) P(F or D)\nc) P(neither E nor F)\nd) P(D and E)\ne) P(E and F)\nf) P(E and D′)\n7. Thobeka compares three neighbourhoods (we’ll call them A, B and C) to see\nwhere the best place is to live. She interviews 80 people and asks them whether\nthey like each of the neighbourhoods, or not.\n• 40 people like neighbourhood A.\n• 35 people like neighbourhood B.\n• 40 people like neighbourhood C.\n• 21 people like both neighbourhoods A and C.\n• 18 people like both neighbourhoods B and C.\n• 68 people like at least one neighbourhood.\n• 7 people like all three neighbourhoods.\n436\n10.6.\nSummary\n\na) Use this information to draw a Venn diagram.\nb) How many people like none of the neighbourhoods?\nc) How many people like neighbourhoods A and B, but not C?\nd) What is the probability that a randomly chosen person from the survey likes\nat least one of the neighbourhoods?\n8. Let G and H be two events in a sample space.\nSuppose that P(G) = 0,4;\nP(H) = h; and P(G or H) = 0,7.\na) For what value of h are G and H mutually exclusive?\nb) For what value of h are G and H independent?\n9. The following tree diagram represents points scored by two teams in a soccer\ngame. At each level in the tree, the points are shown as (points for Team 1;\npoints for Team 2).\n0,75\n(3; 0)\n0,25\n(2; 1)\n0,5\n(2; 1)\n0,5\n(1; 2)\n0,65\n(2; 0)\n0,35\n(1; 1)\n0,4\n(1; 1)\n0,6\n(0; 2)\n0,52\n(1; 0)\n0,48\n(0; 1)\n(0; 0)\nUse this diagram to determine the probability that:\na) Team 1 will win\nb) The game will be a draw\nc) The game will end with an even number of total points\n10. A bag contains 10 orange balls and 7 black balls. You draw 3 balls from the bag\nwithout replacement. What is the probability that you will end up with exactly\n2 orange balls? Represent this experiment using a tree diagram.\n11. Complete the following contingency table and determine whether the events are\ndependent or independent.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\nDid not like living there\n140\n340\nTotals\n230\n500\n12. Summarise the following information about a medical trial with 2 types of multi-\nvitamin in a contingency table and determine whether the events are dependent\nor independent.\n• 960 people took part in the medical trial.\n• 540 people used multivitamin A for a month and 400 of those people\nshowed an improvement in their health.\n437\nChapter 10.\nProbability\n\n• 300 people showed an improvement in health when using multivitamin B\nfor a month.\nIf the events are independent, it means that the two multivitamins have the same\neffect on people. If the events are dependent, it means that one multivitamin is\nbetter than the other. Which multivitamin is better than the other, or are the both\nequally effective?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BV\n2. 23BW\n3. 23BX\n4. 23BY\n5. 23BZ\n6. 23C2\n7. 23C3\n8. 23C4\n9. 23C5\n10. 23C6\n11. 23C7\n12. 23C8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n438\n10.6.\nSummary\n\nCHAPTER\n11\nStatistics\n11.1\nRevision\n440\n11.2\nHistograms\n444\n11.3\nOgives\n451\n11.4\nVariance and standard deviation\n455\n11.5\nSymmetric and skewed data\n461\n11.6\nIdentification of outliers\n464\n11.7\nSummary\n467\n\n11\nStatistics\n11.1\nRevision\nEMBJZ\nMeasures of central tendency\nEMBK2\nThe mean and median of a data set both give an indication where the centre of the\ndata distribution is located. The mean, or average, is calculated as\nx =\nPn\ni=1 xi\nn\nwhere the xi are the data and n is the number of data. We read x as “x bar”.\nThe median is the middle value of an ordered data set. To find the median, we first\nsort the data and then pick out the value in the middle of the sorted list. If the middle\nis in between two values, the median is the average of those two values.\nSee video: 23C9 at www.everythingmaths.co.za\nWorked example 1: Computing measures of central tendency\nQUESTION\nCompute the mean and median of the following data set:\n72,5 ; 92,6 ; 15,6 ; 53,0 ; 86,4 ; 89,9 ; 90,9 ; 21,7 ; 46,0 ; 4,1 ; 51,7 ; 2,2\nSOLUTION\nStep 1: Compute the mean\nUsing the formula for the mean, we first compute the sum of the values and then divide\nby the number of values.\nx = 626,6\n12\n≈52,22\nStep 2: Compute the median\nTo find the median, we first have to sort the data:\n2,2 ; 4,1 ; 15,6 ; 21,7 ; 46,0 ; 51,7 ; 53,0 ; 72,5 ; 86,4 ; 89,9 ; 90,9 ; 92,6\nSince there are an even number of values, the median will lie between two values.\nIn this case, the two values in the middle are 51,7 and 53,0. Therefore the median is\n52,35.\n440\n11.1.\nRevision\n\nMeasures of dispersion\nEMBK3\nMeasures of dispersion tell us how spread out a data set is. If a measure of dispersion\nis small, the data are clustered in a small region. If a measure of dispersion is large,\nthe data are spread out over a large region.\nThe range is the difference between the maximum and minimum values in the data\nset.\nThe inter-quartile range is the difference between the first and third quartiles of the\ndata set. The quartiles are computed in a similar way to the median. The median is\nhalfway into the ordered data set and is sometimes also called the second quartile.\nThe first quartile is one quarter of the way into the ordered data set; whereas the third\nquartile is three quarters of the way into the ordered data set.\nSee video: 23CB at www.everythingmaths.co.za\nWorked example 2: Range and inter-quartile range\nQUESTION\nDetermine the range and the inter-quartile range of the following data set.\n14 ; 17 ; 45 ; 20 ; 19 ; 36 ; 7 ; 30 ; 8\nSOLUTION\nStep 1: Sort the values in the data set\nTo determine the range we need to find the minimum and maximum values in the\ndata set. To determine the inter-quartile range we need to compute the first and third\nquartiles of the data set. For both of these requirements, it is easier to order the data\nset first.\nThe sorted data set is\n7 ; 8 ; 14 ; 17 ; 19 ; 20 ; 30 ; 36 ; 45\nStep 2: Find the minimum, maximum and range\nThe minimum value is the first value in the ordered data set, namely 7. The maximum\nis the last value in the ordered data set, namely 45. The range is the difference between\nthe minimum and maximum: 45 −7 = 38.\nStep 3: Find the quartiles and inter-quartile range\nThe diagram below shows how we find the quartiles one quarter, one half and three\nquarters of the way into the ordered list of values.\n441\nChapter 11.\nStatistics\n\n7\n8\n14\n17\n19\n20\n30\n36\n45\n0\n1\n4\n1\n2\n3\n4\n1\nFrom this diagram we can see that the first quartile is at a value of 14, the second\nquartile (median) is at a value of 19 and the third quartile is at a value of 30.\nThe inter-quartile range is the difference between the first and third quartiles. The\nfirst quartile is 14 and the third quartile is 30. Therefore the inter-quartile range is\n30 −14 = 16.\nFive number summary\nEMBK4\nThe five number summary combines a measure of central tendency, namely the me-\ndian, with measures of dispersion, namely the range and the inter-quartile range. This\ngives a good overview of the overall data distribution. More precisely, the five number\nsummary is written in the following order:\n• minimum;\n• first quartile;\n• median;\n• third quartile;\n• maximum.\nThe five number summary is often presented visually using a box and whisker diagram.\nA box and whisker diagram is shown below, with the positions of the five relevant\nnumbers labelled. Note that this diagram is drawn vertically, but that it may also be\ndrawn horizontally.\nmaximum\nupper quartile\nmedian\nlower quartile\nminimum\ninter-quartile range\ndata range\nSee video: 23CC at www.everythingmaths.co.za\n442\n11.1.\nRevision\n\nWorked example 3: Five number summary\nQUESTION\nDraw a box and whisker diagram for the following data set:\n1,25 ; 1,5 ; 2,5 ; 2,5 ; 3,1 ; 3,2 ; 4,1 ; 4,25 ; 4,75 ; 4,8 ; 4,95 ; 5,1\nSOLUTION\nStep 1: Determine the minimum and maximum\nSince the data set is already ordered, we can read off the minimum as the first value\n(1,25) and the maximum as the last value (5,1).\nStep 2: Determine the quartiles\nThere are 12 values in the data set.\n1,25 1,5\n2,5\n2,5\n3,1\n3,2\n4,1 4,25 4,75 4,8 4,95 5,1\n0\n1\n4\n1\n2\n3\n4\n1\nUsing the figure above we can see that the median is between the sixth and seventh\nvalues, making it.\n3,2 + 4,1\n2\n= 3,65\nThe first quartile lies between the third and fourth values, making it\nQ1 = 2,5 + 2,5\n2\n= 2,5\nThe third quartile lies between the ninth and tenth values, making it\nQ3 = 4,75 + 4,8\n2\n= 4,775\nStep 3: Draw the box and whisker diagram\nWe now have the five number summary as (1,25; 2,5; 3,65; 4,775; 5,1). The box and\nwhisker diagram representing the five number summary is given below.\n1,25\n2,5\n3,65\n4,775 5,1\n443\nChapter 11.\nStatistics\n\nExercise 11 – 1: Revision\n1. For each of the following data sets, compute the mean and all the quartiles.\nRound your answers to one decimal place.\na) −3,4 ; −3,1 ; −6,1 ; −1,5 ; −7,8 ; −3,4 ; −2,7 ; −6,2\nb) −6 ; −99 ; 90 ; 81 ; 13 ; −85 ; −60 ; 65 ; −49\nc) 7 ; 45 ; 11 ; 3 ; 9 ; 35 ; 31 ; 7 ; 16 ; 40 ; 12 ; 6\n2. Use the following box and whisker diagram to determine the range and inter-\nquartile range of the data.\n−5,52\n−2,41−1,53\n0,10\n4,08\n3. Draw the box and whisker diagram for the following data.\n0,2 ; −0,2 ; −2,7 ; 2,9 ; −0,2 ; −4,2 ; −1,8 ; 0,4 ; −1,7 ; −2,5 ; 2,7 ; 0,8 ; −0,5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23CD\n1b. 23CF\n1c. 23CG\n2. 23CH\n3. 23CJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.2\nHistograms\nEMBK5\nA histogram is a graphical representation of how many times different, mutually exclu-\nsive events are observed in an experiment. To interpret a histogram, we find the events\non the x-axis and the counts on the y-axis. Each event has a rectangle that shows what\nits count (or frequency) is.\nSee video: 23CK at www.everythingmaths.co.za\nWorked example 4: Reading histograms\nQUESTION\nUse the following histogram to determine the events that were recorded and the rela-\ntive frequency of each event. Summarise your answer in a table.\n444\n11.2.\nHistograms\n\n0\n2\n4\n6\n8\n10\nnot yet\nin school\nin primary\nschool\nin high\nschool\nSOLUTION\nStep 1: Determine the events\nThe events are shown on the x-axis. In this example we have “not yet in school”, “in\nprimary school” and “in high school”.\nStep 2: Read off the count for each event\nThe counts are shown on the y-axis and the height of each rectangle shows the fre-\nquency for each event.\n• not yet in school: 2\n• in primary school: 5\n• in high school: 9\nStep 3: Calculate relative frequency\nThe relative frequency of an event in an experiment is the number of times that the\nevent occurred divided by the total number of times that the experiment was com-\npleted. In this example we add up the frequencies for all the events to get a total\nfrequency of 16. Therefore the relative frequencies are:\n• not yet in school:\n2\n16 = 1\n8\n• in primary school:\n5\n16\n• in high school:\n9\n16\nStep 4: Summarise\nEvent\nCount\nRelative frequency\nnot yet in school\n2\n1\n8\nin primary school\n5\n5\n16\nin high school\n9\n9\n16\n445\nChapter 11.\nStatistics\n\nTo draw a histogram of a data set containing numbers, the numbers first have to be\ngrouped.\nEach group is defined by an interval.\nWe then count how many times\nnumbers from each group appear in the data set and draw a histogram using the counts.\nWorked example 5: Draw a histogram\nQUESTION\nThe following data represent the heights of 16 adults in centimetres.\n162 ; 168 ; 177 ; 147 ; 189 ; 171 ; 173 ; 168\n178 ; 184 ; 165 ; 173 ; 179 ; 166 ; 168 ; 165\nDivide the data into 5 equal length intervals between 140 cm and 190 cm and draw a\nhistogram.\nSOLUTION\nStep 1: Determine intervals\nTo have 5 intervals of the same length between 140 and 190, we need and interval\nlength of 10. Therefore the intervals are (140; 150]; (150; 160]; (160; 170]; (170; 180];\nand (180; 190].\nStep 2: Count data\nThe following table summarises the number of data values in each of the intervals.\nInterval\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\n(180; 190]\nCount\n1\n0\n7\n6\n2\nStep 3: Draw the histogram\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\n446\n11.2.\nHistograms\n\nFrequency polygons\nEMBK6\nA frequency polygon is sometimes used to represent the same information as in a his-\ntogram. A frequency polygon is drawn by using line segments to connect the middle of\nthe top of each bar in the histogram. This means that the frequency polygon connects\nthe coordinates at the centre of each interval and the count in each interval.\nWorked example 6: Drawing a frequency polygon\nQUESTION\nUse the histogram from the previous example to draw a frequency polygon of the same\ndata.\nSOLUTION\nStep 1: Draw the histogram\nWe already know that the histogram looks like this:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\nStep 2: Connect the tops of the rectangles\nWhen we draw line segments between the tops of the rectangles in the histogram, we\nget the following picture:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\n447\nChapter 11.\nStatistics\n\nStep 3: Draw final frequency polygon\nFinally, we remove the histogram to show only the frequency polygon.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\nFrequency polygons are particularly useful for comparing two data sets. Comparing\ntwo histograms would be more difficult since we would have to draw the rectangles of\nthe two data sets on top of each other. Because frequency polygons are just lines, they\ndo not pose the same problem.\nWorked example 7: Drawing frequency polygons\nQUESTION\nHere is another data set of heights, this time of Grade 11 learners.\n132 ; 132 ; 156 ; 147 ; 162 ; 168 ; 152 ; 174\n141 ; 136 ; 161 ; 148 ; 140 ; 174 ; 174 ; 162\nDraw the frequency polygon for this data set using the same interval length as in the\nprevious example. Then compare the two frequency polygons on one graph to see the\ndifferences between the distributions.\nSOLUTION\nStep 1: Frequency table\nWe first create the table of counts for the new data set.\nInterval\n(130; 140]\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\nCount\n4\n3\n2\n4\n3\n448\n11.2.\nHistograms\n\nStep 2: Draw histogram and frequency polygon\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\nStep 3: Compare frequency polygons\nWe draw the two frequency polygons on the same axes. The red line indicates the\ndistribution over heights for adults and the blue line, for Grade 11 learners.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\n190\nFrom this plot we can easily see that the heights for Grade 11 learners are distributed\nmore towards the left (shorter) than adults. The learner heights also seem to be more\nevenly distributed between 130 and 180 cm, whereas the adult heights are mostly\nbetween 160 and 180 cm.\n449\nChapter 11.\nStatistics\n\nExercise 11 – 2: Histograms\n1. Use the histogram below to answer the following questions.\nThe histogram\nshows the number of people born around the world each year. The ticks on\nthe x-axis are located at the start of each year.\npeople (millions)\nyear\n79\n80\n81\n82\n83\n84\n85\n86\n87\n1994 1995 1996 1997 1998 1999 2000 2001\na) How many people were born between the beginning of 1994 and the be-\nginning of 1996?\nb) Is the number people in the world population increasing or decreasing?\n(Ignore the rate at which people are dying for this question.)\nc) How many more people were born in 1994 than in 1997?\n2. In a traffic survey, a random sample of 50 motorists were asked the distance (d)\nthey drove to work daily. The results of the survey are shown in the table below.\nDraw a histogram to represent the data.\nd\n0 < d ≤10\n10 < d ≤20\n20 < d ≤30\n30 < d ≤40\n40 < d ≤50\nf\n9\n19\n15\n5\n4\n3. Below is data for the prevalence of HIV in South Africa. HIV prevalence refers to\nthe percentage of people between the ages of 15 and 49 who are infected with\nHIV.\nyear\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nprevalence (%)\n17,7\n18,0\n18,1\n18,1\n18,1\n18,0\n17,9\n17,9\nDraw a frequency polygon of this data set.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CM\n2. 23CN\n3. 23CP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n450\n11.2.\nHistograms\n\n11.3\nOgives\nEMBK7\nCumulative histograms, also known as ogives, are graphs that can be used to deter-\nmine how many data values lie above or below a particular value in a data set. The\ncumulative frequency is calculated from a frequency table, by adding each frequency\nto the total of the frequencies of all data values before it in the data set. The last value\nfor the cumulative frequency will always be equal to the total number of data values,\nsince all frequencies will already have been added to the previous total.\nAn ogive is drawn by\n• plotting the beginning of the first interval at a y-value of zero;\n• plotting the end of every interval at the y-value equal to the cumulative count for\nthat interval; and\n• connecting the points on the plot with straight lines.\nIn this way, the end of the final interval will always be at the total number of data since\nwe will have added up across all intervals.\nWorked example 8: Cumulative frequencies and ogives\nQUESTION\nDetermine the cumulative frequencies of the following grouped data and complete the\ntable below. Use the table to draw an ogive of the data.\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n20 < n ≤30\n7\n30 < n ≤40\n12\n40 < n ≤50\n10\n50 < n ≤60\n6\nSOLUTION\nStep 1: Compute cumulative frequencies\nTo determine the cumulative frequency, we add up the frequencies going down the\ntable. The first cumulative frequency is just the same as the frequency, because we are\nadding it to zero. The final cumulative frequency is always equal to the sum of all the\nfrequencies. This gives the following table:\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n5\n20 < n ≤30\n7\n12\n30 < n ≤40\n12\n24\n40 < n ≤50\n10\n34\n50 < n ≤60\n6\n40\n451\nChapter 11.\nStatistics\n\nStep 2: Plot the ogive\nThe first coordinate in the plot always starts at a y-value of 0 because we always start\nfrom a count of zero. So, the first coordinate is at (10; 0) — at the beginning of the\nfirst interval. The second coordinate is at the end of the first interval (which is also the\nbeginning of the second interval) and at the first cumulative count, so (20; 5). The third\ncoordinate is at the end of the second interval and at the second cumulative count,\nnamely (30; 12), and so on.\nComputing all the coordinates and connecting them with straight lines gives the fol-\nlowing ogive.\nn\n0\n10\n20\n30\n40\n10\n20\n30\n40\n50\n60\n•\n•\n•\n•\n•\n•\nOgives do look similar to frequency polygons, which we saw earlier. The most impor-\ntant difference between them is that an ogive is a plot of cumulative values, whereas\na frequency polygon is a plot of the values themselves. So, to get from a frequency\npolygon to an ogive, we would add up the counts as we move from left to right in the\ngraph.\nOgives are useful for determining the median, percentiles and five number summary\nof data. Remember that the median is simply the value in the middle when we order\nthe data. A quartile is simply a quarter of the way from the beginning or the end of an\nordered data set. With an ogive we already know how many data values are above or\nbelow a certain point, so it is easy to find the middle or a quarter of the data set.\nWorked example 9: Ogives and the five number summary\nQUESTION\nUse the following ogive to compute the five number summary of the data. Remember\nthat the five number summary consists of the minimum, all the quartiles (including the\nmedian) and the maximum.\n452\n11.3.\nOgives\n\ncount\nvalue\n0\n10\n20\n30\n40\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\nSOLUTION\nStep 1: Find the minimum and maximum\nThe minimum value in the data set is 1 since this is where the ogive starts on the\nhorizontal axis. The maximum value in the data set is 10 since this is where the ogive\nstops on the horizontal axis.\nStep 2: Find the quartiles\nThe quartiles are the values that are 1\n4, 1\n2 and 3\n4 of the way into the ordered data set.\nHere the counts go up to 40, so we can find the quartiles by looking at the values\ncorresponding to counts of 10, 20 and 30. On the ogive a count of\n• 10 corresponds to a value of 3 (first quartile);\n• 20 corresponds to a value of 7 (second quartile); and\n• 30 corresponds to a value of 8 (third quartile).\nStep 3: Write down the five number summary\nThe five number summary is (1; 3; 7; 8; 10). The box-and-whisker plot of this data set\nis given below.\n1\n3\n7\n8\n10\n453\nChapter 11.\nStatistics\n\nExercise 11 – 3: Ogives\n1. Use the ogive to answer the questions below. Note that marks are given as a\npercentage.\nnumber of students\nmark\n0\n10\n20\n30\n40\n50\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n•\n•\n•\n•\n•\n•\n•\n•\n•\na) How many students got between 50% and 70%?\nb) How many students got at least 70%?\nc) Compute the average mark for this class, rounded to the nearest integer.\n2. Draw the histogram corresponding to this ogive.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n−25\n−15\n−5\n5\n15\n25\n•\n•\n•\n•\n•\n•\n3. The following data set lists the ages of 24 people.\n2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53; 38\n4; 28; 5; 73; 79; 17; 15; 5; 34; 37; 45; 56\nUse the data to answer the following questions.\na) Using an interval width of 8 construct a cumulative frequency plot.\nb) How many are below 30?\nc) How many are below 60?\nd) Giving an explanation state below what value the bottom 50% of the ages\nfall.\ne) Below what value do the bottom 40% fall?\nf) Construct a frequency polygon.\n454\n11.3.\nOgives\n\n4. The weights of bags of sand in grams is given below (rounded to the nearest\ntenth):\n50,1; 40,4; 48,5; 29,4; 50,2; 55,3; 58,1; 35,3; 54,2; 43,5\n60,1; 43,9; 45,3; 49,2; 36,6; 31,5; 63,1; 49,3; 43,4; 54,1\na) Decide on an interval width and state what you observe about your choice.\nb) Give your lowest interval.\nc) Give your highest interval.\nd) Construct a cumulative frequency graph and a frequency polygon.\ne) Below what value do 53% of the cases fall?\nf) Below what value of 60% of the cases fall?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CQ\n2. 23CR\n3. 23CS\n4. 23CT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.4\nVariance and standard deviation\nEMBK8\nMeasures of central tendency (mean, median and mode) provide information on the\ndata values at the centre of the data set. Measures of dispersion (quartiles, percentiles,\nranges) provide information on the spread of the data around the centre. In this section\nwe will look at two more measures of dispersion called the variance and the standard\ndeviation.\nSee video: 23CV at www.everythingmaths.co.za\nVariance\nEMBK9\nDEFINITION: Variance\nLet a population consist of n elements, {x1; x2; . . . ; xn}. Write the mean of the data as\nx.\nThe variance of the data is the average squared distance between the mean and each\ndata value.\nσ2 =\nPn\ni=1 (xi −x)2\nn\nNOTE:\nThe variance is written as σ2. It might seem strange that it is written in squared form,\nbut you will see why soon when we discuss the standard deviation.\n455\nChapter 11.\nStatistics\n\nThe variance has the following properties.\n• It is never negative since every term in the variance sum is squared and therefore\neither positive or zero.\n• It has squared units. For example, the variance of a set of heights measured in\ncentimetres will be given in centimeters squared. Since the population variance\nis squared, it is not directly comparable with the mean or the data themselves. In\nthe next section we will describe a different measure of dispersion, the standard\ndeviation, which has the same units as the data.\nWorked example 10: Variance\nQUESTION\nYou flip a coin 100 times and it lands on heads 44 times. You then use the same\ncoin and do another 100 flips. This time in lands on heads 49 times. You repeat this\nexperiment a total of 10 times and get the following results for the number of heads.\n{44; 49; 52; 62; 53; 48; 54; 49; 46; 51}\nCompute the mean and variance of this data set.\nSOLUTION\nStep 1: Compute the mean\nThe formula for the mean is\nx =\nPn\ni=1 xi\nn\nIn this case, we sum the data and divide by 10 to get x = 50,8.\nStep 2: Compute the variance\nThe formula for the variance is\nσ2 =\nPn\ni=1 (xi −x)2\nn\nWe first subtract the mean from each datum and then square the result.\nxi\n44\n49\n52\n62\n53\n48\n54\n49\n46\n51\nxi −x\n−6,8\n−1,8\n1,2\n11,2\n2,2\n−2,8\n3,2\n−1,8\n−4,8\n0,2\n(xi −x)2\n46,24\n3,24\n1,44\n125,44 4,84\n7,84\n10,24\n3,24\n23,04\n0,04\nThe variance is the sum of the last row in this table divided by 10, so σ2 = 22,56.\n456\n11.4.\nVariance and standard deviation\n\nStandard deviation\nEMBKB\nSince the variance is a squared quantity, it cannot be directly compared to the data val-\nues or the mean value of a data set. It is therefore more useful to have a quantity which\nis the square root of the variance. This quantity is known as the standard deviation.\nDEFINITION: Standard deviation\nLet a population consist of n elements, {x1; x2; . . . ; xn}, with a mean of x. The stan-\ndard deviation of the data is\nσ =\nsPn\ni=1 (xi −x)2\nn\nIn statistics, the standard deviation is a very common measure of dispersion. Standard\ndeviation measures how spread out the values in a data set are around the mean. More\nprecisely, it is a measure of the average distance between the values of the data in the\nset and the mean. If the data values are all similar, then the standard deviation will be\nlow (closer to zero). If the data values are highly variable, then the standard variation\nis high (further from zero).\nThe standard deviation is always a positive number and is always measured in the\nsame units as the original data. For example, if the data are distance measurements in\nkilogrammes, the standard deviation will also be measured in kilogrammes.\nThe mean and the standard deviation of a set of data are usually reported together. In\na certain sense, the standard deviation is a natural measure of dispersion if the centre\nof the data is taken as the mean.\nInvestigation: Tabulating results\nIt is often useful to set your data out in a table so that you can apply the for-\nmulae easily.\nComplete the table below to calculate the standard deviation of\n{57; 53; 58; 65; 48; 50; 66; 51}.\n• Firstly, remember to calculate the mean, x.\n• Complete the following table.\nindex: i\ndatum: xi\ndeviation: xi −x\ndeviation\nsquared: (xi −x)2\n1\n57\n2\n53\n3\n58\n4\n65\n5\n48\n6\n50\n7\n66\n8\n51\nP xi = . . .\nP(xi −x) = . . .\nP(xi −x)2 = . . .\n• The sum of the deviations is always zero. Why is this? Find out.\n• Calculate the variance using the completed table.\n• Then calculate the standard deviation.\n457\nChapter 11.\nStatistics\n\nWorked example 11: Variance and standard deviation\nQUESTION\nWhat is the variance and standard deviation of the possibilities associated with rolling\na fair die?\nSOLUTION\nStep 1: Determine all the possible outcomes\nWhen rolling a fair die, the sample space consists of 6 outcomes. The data set is\ntherefore x = {1; 2; 3; 4; 5; 6} and n = 6.\nStep 2: Calculate the mean\nThe mean is:\nx = 1\n6 (1 + 2 + 3 + 4 + 5 + 6)\n= 3,5\nStep 3: Calculate the variance\nThe variance is:\nσ2 =\nP (x −x)2\nn\n= 1\n6 (6,25 + 2,25 + 0,25 + 0,25 + 2,25 + 6,25)\n= 2,917\nStep 4: Calculate the standard deviation\nThe standard deviation is:\nσ =\np\n2,917\n= 1,708\nSee video: 23CW at www.everythingmaths.co.za\n458\n11.4.\nVariance and standard deviation\n\nInterpretation and application\nEMBKC\nA large standard deviation indicates that the data values are far from the mean and a\nsmall standard deviation indicates that they are clustered closely around the mean.\nFor example, consider the following three data sets:\n{65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\n{85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\n{43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nEach of these data sets has the same mean, namely 67. However, they have different\nstandard deviations, namely 8,97, 17,75 and 21,23. The following figures show plots\nof the data sets with the mean and standard deviation indicated on each. You can see\nhow the standard deviation is larger when the data are more spread out.\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 8,97\ndata:\n{xi} = {65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\nmean:\nx = 67\nstandard deviation:\nσ ≈8,97\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 17,75\ndata:\n{xi} = {85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\nmean:\nx = 67\nstandard deviation:\nσ ≈17,75\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 21,23\ndata:\n{xi} = {43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nmean:\nx = 67\nstandard deviation:\nσ ≈21,23\nThe standard deviation may also be thought of as a measure of uncertainty. In the phys-\nical sciences, for example, the reported standard deviation of a group of repeated mea-\nsurements represents the precision of those measurements. When deciding whether\n459\nChapter 11.\nStatistics\n\nmeasurements agree with a theoretical prediction, the standard deviation of those mea-\nsurements is very important: if the mean of the measurements is too far away from the\nprediction (with the distance measured in standard deviations), then we consider the\nmeasurements as contradicting the prediction. This makes sense since they fall outside\nthe range of values that could reasonably be expected to occur if the prediction were\ncorrect.\nExercise 11 – 4: Variance and standard deviation\n1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban.\nThe data, in rands per litre, are given below.\nCape Town\n3,96\n3,76\n4,00\n3,91\n3,69\n3,72\nDurban\n3,97\n3,81\n3,52\n4,08\n3,88\n3,68\na) Find the mean price in each city and then state which city has the lowest\nmean.\nb) Find the standard deviation of each city’s prices.\nc) Which city has the more consistently priced petrol? Give reasons for your\nanswer.\n2. Compute the mean and variance of the following set of values.\n150 ; 300 ; 250 ; 270 ; 130 ; 80 ; 700 ; 500 ; 200 ; 220 ; 110 ; 320 ; 420 ; 140\n3. Compute the mean and variance of the following set of values.\n−6,9 ; −17,3 ; 18,1 ; 1,5 ; 8,1 ; 9,6 ; −13,1 ; −14,0 ; 10,5 ; −14,8 ; −6,5 ; 1,4\n4. The times for 8 athletes who ran a 100 m sprint on the same track are shown\nbelow. All times are in seconds.\n10,2 ; 10,8 ; 10,9 ; 10,3 ; 10,2 ; 10,4 ; 10,1 ; 10,4\na) Calculate the mean time.\nb) Calculate the standard deviation for the data.\nc) How many of the athletes’ times are more than one standard deviation away\nfrom the mean?\n5. The following data set has a mean of 14,7 and a variance of 10,01.\n18 ; 11 ; 12 ; a ; 16 ; 11 ; 19 ; 14 ; b ; 13\nCompute the values of a and b.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CX\n2. 23CY\n3. 23CZ\n4. 23D2\n5. 23D3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n460\n11.4.\nVariance and standard deviation\n\n11.5\nSymmetric and skewed data\nEMBKD\nWe are now going to classify data sets into 3 categories that describe the shape of the\ndata distribution: symmetric, left skewed, right skewed. We can use this classification\nfor any data set, but here we will look only at distributions with one peak. Most of\nthe data distributions that you have seen so far have only one peak, so the plots in this\nsection should look familiar.\nDistributions with one peak are called unimodal distributions.\nUnimodal literally\nmeans having one mode. (Remember that a mode is a maximum in the distribution.)\nSymmetric distributions\nEMBKF\nA symmetric distribution is one where the left and right hand sides of the distribution\nare roughly equally balanced around the mean. The histogram below shows a typical\nsymmetric distribution.\nmean ≈median\nbalanced left and right tails\nFor symmetric distributions, the mean is approximately equal to the median. The tails\nof the distribution are the parts to the left and to the right, away from the mean. The\ntail is the part where the counts in the histogram become smaller. For a symmetric\ndistribution, the left and right tails are equally balanced, meaning that they have about\nthe same length.\nThe figure below shows the box and whisker diagram for a typical symmetric data set.\nmedian halfway\nbetween\nfirst and third quartiles\nAnother property of a symmetric distribution is that its median (second quartile) lies\nin the middle of its first and third quartiles. Note that the whiskers of the plot (the\nminimum and maximum) do not have to be equally far away from the median. In the\nnext section on outliers, you will see that the minimum and maximum values do not\nnecessarily match the rest of the data distribution well.\n461\nChapter 11.\nStatistics\n\nSkewed\nEMBKG\nA distribution that is skewed right (also known as positively skewed) is shown below.\nmean\nmedian\nmean > median\nlong right tail\nshort left tail\nNow the picture is not symmetric around the mean anymore.\nFor a right skewed\ndistribution, the mean is typically greater than the median. Also notice that the tail of\nthe distribution on the right hand (positive) side is longer than on the left hand side.\nmedian closer to first quartile\nFrom the box and whisker diagram we can also see that the median is closer to the first\nquartile than the third quartile. The fact that the right hand side tail of the distribution\nis longer than the left can also be seen.\nA distribution that is skewed left has exactly the opposite characteristics of one that is\nskewed right:\n• the mean is typically less than the median;\n• the tail of the distribution is longer on the left hand side than on the right hand\nside; and\n• the median is closer to the third quartile than to the first quartile.\nThe table below summarises the different categories visually.\nSymmetric\nSkewed right (positive)\nSkewed left (negative)\n462\n11.5.\nSymmetric and skewed data\n\nExercise 11 – 5: Symmetric and skewed data\n1. Is the following data set symmetric, skewed right or skewed left? Motivate your\nanswer.\n27 ; 28 ; 30 ; 32 ; 34 ; 38 ; 41 ; 42 ; 43 ; 44 ; 46 ; 53 ; 56 ; 62\n2. State whether each of the following data sets are symmetric, skewed right or\nskewed left.\na) A data set with this histogram:\nb) A data set with this box and whisker plot:\nc) A data set with this frequency polygon:\n• • • • • • • • •\n•\n•\n•\n• • • •\nd) The following data set:\n11,2 ; 5 ; 9,4 ; 14,9 ; 4,4 ; 18,8 ; −0,4 ; 10,5 ; 8,3 ; 17,8\n3. Two data sets have the same range and interquartile range, but one is skewed\nright and the other is skewed left. Sketch the box and whisker plot for each of\nthese data sets. Then, invent data (6 points in each data set) that matches the\ndescriptions of the two data sets.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23D4\n2a. 23D5\n2b. 23D6\n2c. 23D7\n2d. 23D8\n3. 23D9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n463\nChapter 11.\nStatistics\n\n11.6\nIdentification of outliers\nEMBKH\nAn outlier in a data set is a value that is far away from the rest of the values in the\ndata set. In a box and whisker diagram, outliers are usually close to the whiskers of\nthe diagram. This is because the centre of the diagram represents the data between\nthe first and third quartiles, which is where 50% of the data lie, while the whiskers\nrepresent the extremes — the minimum and maximum — of the data.\nWorked example 12: Identifying outliers\nQUESTION\nFind the outliers in the following data set by drawing a box and whisker diagram and\nlocating the data values on the diagram.\n0,5 ; 1 ; 1,1 ; 1,4 ; 2,4 ; 2,8 ; 3,5 ; 5,1 ; 5,2 ; 6 ; 6,5 ; 9,5\nSOLUTION\nStep 1: Determine the five number summary\nThe minimum of the data set is 0,5. The maximum of the data set is 9,5. Since there\nare 12 values in the data set, the median lies between the sixth and seventh values,\nmaking it equal to 2,8+3,5\n2\n= 3,15. The first quartile lies between the third and fourth\nvalues, making it equal to 1,1+1,4\n2\n= 1,25. The third quartile lies between the ninth\nand tenth values, making it equal to 5,2+6\n2\n= 5,6.\nStep 2: Draw the box and whisker diagram\n0,5 1,25\n3,15\n5,6\n9,5\n• •• •\n• •\n•\n••\n• •\n•\nIn the figure above, each value in the data set is shown with a black dot.\nStep 3: Find the outliers\nFrom the diagram we can see that most of the values are between 1 and 6. The only\nvalue that is very far away from this range is the maximum at 9,5. Therefore 9,5 is the\nonly outlier in the data set.\nYou should also be able to identify outliers in plots of two variables. A scatter plot\nis a graph that shows the relationship between two random variables. We call these\ndata bivariate (literally meaning two variables) and we plot the data for two different\nvariables on one set of axes. The following example shows what a typical scatter plot\nlooks like. For Grade 11 you do not need to learn how to draw these 2-dimensional\n464\n11.6.\nIdentification of outliers\n\nscatter plots, but you should be able to identify outliers on them. As before, an outlier\nis a value that is far removed from the main distribution of data.\nWorked example 13: Scatter plot\nQUESTION\nWe have a data set that relates the heights and weights of a number of people. The\nheight is the first variable and its value is plotted along the horizontal axis. The weight\nis the second variable and its value is plotted along the vertical axis. The data values\nare shown on the plot below. Identify any outliers on the scatter plot.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\nSOLUTION\nWe inspect the plot visually and notice that there are two points that lie far away from\nthe main data distribution. These two points are circled in the plot below.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\n465\nChapter 11.\nStatistics\n\nExercise 11 – 6: Outliers\n1. For each of the following data sets, draw a box and whisker diagram and deter-\nmine whether there are any outliers in the data.\na) 30 ; 21,4 ; 39,4 ; 33,4 ; 21,1 ; 29,3 ; 32,8 ; 31,6 ; 36 ;\n27,9 ; 27,3 ; 29,4 ; 29,1 ; 38,6 ; 33,8 ; 29,1 ; 37,1\nb) 198 ; 166 ; 175 ; 147 ; 125 ; 194 ; 119 ; 170 ; 142 ; 148\nc) 7,1 ; 9,6 ; 6,3 ; −5,9 ; 0,7 ; −0,1 ; 4,4 ; −11,7 ; 10 ; 2,3 ; −3,7 ; 5,8 ; −1,4\n; 1,7 ; −0,7\n2. A class’s results for a test were recorded along with the amount of time spent\nstudying for it. The results are given below. Identify any outliers in the data.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23DB\n1b. 23DC\n1c. 23DD\n2. 23DF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n466\n11.6.\nIdentification of outliers\n\n11.7\nSummary\nEMBKJ\nSee presentation: 23DG at www.everythingmaths.co.za\n• Histograms visualise how many times different events occurred. Each rectangle\nin a histogram represents one event and the height of the rectangle is relative to\nthe number of times that the event occurred.\n• Frequency polygons represent the same information as histograms, but using\nlines and points rather than rectangles. A frequency polygon connects the mid-\ndle of the top edge of each rectangle in a histogram.\n• Ogives (also known as cumulative histograms) show the total number of times\nthat a value or anything less than that value appears in the data set. To draw an\nogive you need to add up all the counts in a histogram from left to right.\n– The first count in an ogive is always zero.\n– The last count in an ogive is always the sum of all the counts in the data\nset.\n• The variance and standard deviation are measures of dispersion.\n– The standard deviation is the square root of the variance.\n– Variance: σ2 = 1\nn\nPn\ni=1(xi −x)2\n– Standard deviation: σ =\nq\n1\nn\nPn\ni=1(xi −x)2\n– The standard deviation is measured in the same units as the mean and the\ndata, but the variance is not. The variance is measured in the square of the\ndata units.\n• In a symmetric distribution\n– the mean is approximately equal to the median; and\n– the tails of the distribution are balanced.\n• In a right (positively) skewed distribution\n– the mean is greater than the median;\n– the tail on the right hand side is longer than the tail on the left hand side;\nand\n– the median is closer to the first quartile than the third quartile.\n• In a left (negatively) skewed distribution\n– the mean is less than the median;\n– the tail on the left hand side is longer than the tail on the right hand side;\nand\n– the median is closer to the third quartile than the first quartile.\n• An outlier is a value that is far away from the rest of the data.\n467\nChapter 11.\nStatistics\n\nExercise 11 – 7: End of chapter exercises\n1. Draw a histogram, frequency polygon and ogive of the following data set. To\ncount the data, use intervals with a width of 1, starting from 0.\n0,4 ; 3,1 ; 1,1 ; 2,8 ; 1,5 ; 1,3 ; 2,8 ; 3,1 ; 1,8 ; 1,3 ;\n2,6 ; 3,7 ; 3,3 ; 5,7 ; 3,7 ; 7,4 ; 4,6 ; 2,4 ; 3,5 ; 5,3\n2. Draw a box and whisker diagram of the following data set and explain whether\nit is symmetric, skewed right or skewed left.\n−4,1 ; −1,1 ; −1 ; −1,2 ; −1,5 ; −3,2 ; −4 ; −1,9 ; −4 ;\n−0,8 ; −3,3 ; −4,5 ; −2,5 ; −4,4 ; −4,6 ; −4,4 ; −3,3\n3. Eight children’s sweet consumption and sleeping habits were recorded. The data\nare given in the following table and scatter plot.\nNumber of sweets\nper week\n15\n12\n5\n3\n18\n23\n11\n4\nAverage sleeping\ntime (hours per day)\n4\n4,5\n8\n8,5\n3\n2\n5\n8\n5\n10\n15\n20\n25\nnumber of sweets\n1\n2\n3\n4\n5\n6\n7\n8\n9\nsleeping time (hours per day)\na) What is the mean and standard deviation of the number of sweets eaten per\nday?\nb) What is the mean and standard deviation of the number of hours slept per\nday?\nc) Make a list of all the outliers in the data set.\n4. The monthly incomes of eight teachers are as follows:\nR 10 050;\nR 14 300;\nR 9800;\nR 15 000;\nR 12 140;\nR 13 800;\nR 11 990;\nR 12 900.\na) What is the mean and standard deviation of their incomes?\nb) How many of the salaries are less than one standard deviation away from\nthe mean?\nc) If each teacher gets a bonus of R 500 added to their pay what is the new\nmean and standard deviation?\nd) If each teacher gets a bonus of 10% on their salary what is the new mean\nand standard deviation?\ne) Determine for both of the above, how many salaries are less than one stan-\ndard deviation away from the mean.\n468\n11.7.\nSummary\n\nf) Using the above information work out which bonus is more beneficial fi-\nnancially for the teachers.\n5. The weights of a random sample of boys in Grade 11 were recorded. The cumu-\nlative frequency graph (ogive) below represents the recorded weights.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100 110 120\n0\n10\n20\n30\n40\n50\n60\nWeight (in kilogrammes)\nCumulative frequency\nCumulative frequency curve showing weight of boys\na) How many of the boys weighed between 90 and 100 kilogrammes?\nb) Estimate the median weight of the boys.\nc) If there were 250 boys in Grade 11, estimate how many of them would\nweigh less than 80 kilogrammes?\n6. Three sets of 12 learners each had their test scores recorded. The test was out of\n50. Use the given data to answer the following questions.\nSet A\nSet B\nSet C\n25\n32\n43\n47\n34\n47\n15\n35\n16\n17\n32\n43\n16\n25\n38\n26\n16\n44\n24\n38\n42\n27\n47\n50\n22\n43\n50\n24\n29\n44\n12\n18\n43\n31\n25\n42\na) For each of the sets calculate the mean and the five number summary.\nb) Make box and whisker plots of the three data sets on the same set of axes.\nc) State, with reasons, whether each of the three data sets are symmetric or\nskewed (either right or left).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DH\n2. 23DJ\n3. 23DK\n4. 23DM\n5. 23DN\n6. 23DP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n469\nChapter 11.\nStatistics\n\n\nCHAPTER\n12\nLinear programming\n12.1\nIntroduction\n472\n\n12\nLinear programming\n12.1\nIntroduction\nEMBKK\nIn everyday life people are interested in knowing the most efficient way of carrying out\na task or achieving a goal. For example, a farmer wants to know how many hectares to\nplant during a season in order to maximise the yield (produce), a stock broker wants to\nknow how much to invest in stocks in order to maximise profit, an entrepreneur wants\nto know how many people to employ to minimise expenditure. These are optimisation\nproblems; we want to to determine either the maximum or the minimum in a specific\nsituation.\nTo describe this mathematically, we assign variables to represent the different factors\nthat influence the situation. Optimisation means finding the combination of variables\nthat gives the best result.\nSee video: 23DQ at www.everythingmaths.co.za\nWorked example 1: Mountees and Roadees\nQUESTION\nInvestigate the following situation and use your knowledge of mathematics to solve the\nproblem:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make the maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nStep 2: Organise the information given\nWrite down a summary of the information given in the problem so that we consider\n472\n12.1.\nIntroduction\n\nall the different components in the situation.\nmaximum number for M\n= 5\nmaximum number for R\n= 3\nnumber of technicians needed for M = 1\nnumber of technicians needed for R = 2\ntotal number of technicians\n= 8\nprofit per M\n= 800\nprofit per R\n= 2400\nStep 3: Draw up a table\nUse the summary to draw up a table of all the possible combinations of the number of\nMountees and Roadees that can be manufactured per day:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n(4; 3)\n5\n(5; 0)\n(5; 1)\n(5; 2)\n(5; 3)\nNote that there are 24 possible combinations.\nStep 4: Consider the limitation of the number of technicians\nIt takes 1 technician to assemble a Mountee and 2 technicians to assemble a Roadee.\nThere are a total of 8 technicians in the assembly department, therefore we can write\nthat 1(M) + 2(R) ≤8.\nWith this limitation, we are able to eliminate some of the combinations in the table\nwhere M + 2R > 8:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n\b\b\b\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n\b\b\b\n(4; 3)\n5\n(5; 0)\n(5; 1)\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nThese combinations have been excluded as possible answers. For example, (5; 3) gives\n5 + 2(3) = 11 technicians.\n473\nChapter 12.\nLinear programming\n\nStep 5: Consider the profit on the bicycles\nWe can express the profit (P) per day as: P = 800(M)+2400(R). Notice that a higher\nprofit is made on a Roadee.\nBy substituting the different combinations for M and R, we can find the values that\ngive the maximum profit:\nFor (5; 0)\nP = 800(5) + 2400(0)\n= R 4000\nFor (3; 1)\nP = 800(3) + 2400(1)\n= R 4800\nM\nR\n0\n1\n2\n3\n0\n(0; 0) ⇒R 0\n(0; 1) ⇒R 2400\n(0; 2) ⇒R 4800\n(0; 3) ⇒R 7200\n1\n(1; 0) ⇒R 800\n(1; 1) ⇒R 3200\n(1; 2) ⇒R 5600\n(1; 3) ⇒R 8000\n2\n(2; 0) ⇒R 1600\n(2; 1) ⇒R 4000\n(2; 2) ⇒R 6400\n(2; 3) ⇒R 8800\n3\n(3; 0) ⇒R 2400\n(3; 1) ⇒R 4800\n(3; 2) ⇒R 7200\n\b\b\b\n(3; 3)\n4\n(4; 0) ⇒R 3200\n(4; 1) ⇒R 5600\n(4; 2) ⇒R 8000\n\b\b\b\n(4; 3)\n5\n(5; 0) ⇒R 4000\n(5; 1) ⇒R 6400\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nStep 6: Write the final answer\nTherefore the maximum profit of R 8800 is obtained if 2 Mountees and 3 Roadees are\nmanufactured per day.\nExercise 12 – 1: Optimisation\n1. Furniture store opening special:\nAs part of their opening special, a furniture store has promised to give away at\nleast 40 prizes with a total value of at least R 4000. They intend to give away\nkettles and toasters. They decide there will be at least 10 units of each prize. A\nkettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the\ncompany. Calculate how much this combination of kettles and toasters will cost.\nUse a suitable strategy to organise the information and solve the problem.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n474\n12.1.\nIntroduction\n\nOptimisation using graphs\nA more efficient way to solve optimisation problems is using graphs.\nWe write the limitations in the situation, called constraints, as inequalities. Some con-\nstraints can be modelled by an equation, which needs to be maximised or minimized.\nWe sketch the inequalities and indicate the region above or below the line that is to be\nconsidered in determining the solution. This method of solving optimisation problems\nis called linear programming.\nSee video: 23DS at www.everythingmaths.co.za\nWorked example 2: Optimisation using graphs\nQUESTION\nConsider again the example of Mr. Hunter who manufactures Mountees and Roadees:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make a maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nNotice that the values of M and R are limited to positive integers; Mr. Hunter cannot\nsell negative numbers of bikes nor can he sell a fraction of a bike.\nStep 2: Organise the information\nWe can write these constraints as inequalities:\nnumber of Mountees: 0 ≤M ≤5\nnumber of Roadees: 0 ≤R ≤3\ntotal number of technicians: M + 2R ≤8\nWe also know that P = 800M + 2400R. This is called the objective function, some-\ntimes also referred to as the search line, because the objective (goal) is to determine\nthe maximum value of P.\n475\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nWe represent the number of Mountees manufactured daily on the horizontal axis and\nthe number of Roadees manufactured daily on the vertical axis. Since M and R are\npositive integers, we only use the first quadrant of the Cartesian plane. Note that the\ngraph only includes the integer values of M between 0 and 5 and R between 0 and 3.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nFor the number of technicians in the assembly department M + 2R ≤8. If we make\nR (represented on the y-axis) the subject of the inequality we get R ≤−1\n2M + 4.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR = −1\n2M + 4\nThe arrows indicate the region in which the solution will lie, where R ≤−1\n2M + 4.\nThis area is called the feasible region.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR ≤−1\n2M + 4\nA\n476\n12.1.\nIntroduction\n\nWe substitute the possible combinations into the profit equation P = 800M + 2400R,\nand find the combination that gives the maximum profit.\nAt A(2; 3) :\nP = 800(2) + 2400(3)\n= R 8800\nStep 4: Write the final answer\nTherefore the maximum profit is obtained if 2 Mountees and 3 Roadees are manufac-\ntured per day.\nSee video: 23DT at www.everythingmaths.co.za\nWorked example 3: Optimisation using graphs\nQUESTION\nSolve the “furniture store opening special” problem using graphs:\nAs part of their opening special, a furniture store has promised to give away at least 40\nprizes. They intend to give away kettles and toasters. They decide there will be at least\n10 units of each prize. A kettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the company.\nCalculate how much this combination of kettles and toasters will cost.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nkettles be k and the number of toasters be t, with k, t ∈Z.\nStep 2: Organise the information\nWe can write the given information as inequalities:\nnumber of kettles: k ≥10\nnumber of toasters: t ≥10\ntotal number of prizes: k + t ≥40\nWe make t the subject of the inequality:\nt ≥−k + 40\n477\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nRepresent the constraints on a set of axes:\nKettles (k)\nToasters (t)\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nt ≥−k + 40\nt ≥10\nk ≥10\nWe shade the feasible region as shown in the diagram. Remember that in this situation\nonly the points with integer coordinates inside or on the border of the feasible region\nare possible solutions. The combination giving the minimum cost will lie towards or\non the lower border of the feasible region, which gives us many points to consider. To\nfind the optimum value of C, we use the graph of the objective function\nC = 120k + 100t\nTo draw the line, we make t the subject of the formula\nt = −6\n5k + C\n100\nWe see that the gradient of the objective function is −6\n5, but we do not know the exact\nvalue of the t-intercept ( C\n100). To find the minimum value of C, we need to determine\nthe position of the objective function where it first touches the feasible region and also\ngives the lowest t-intercept.\nKettles (k)\nToasters (t)\nA\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nWe indicate the gradient of the objective function on the graph (the green search\nline). Keeping the gradient the same, we “slide” the objective function towards the\nlower border of the feasible region and find that it touches the feasible region at point\n478\n12.1.\nIntroduction\n\nA(10; 30). This optimum position of the objective function is indicated on the graph\nby the dotted line passing through point A.\nWe substitute the coordinates of A into the cost equation C = 120k + 100t:\nAt A(10; 30) :\nC = 120(10) + 100(30)\n= R 4200\nThe minimum cost can also be determined graphically by reading off the coordinates\nof the t-intercept of the objective function in the optimum position:\ntint = 42\n∴C\n100 = 42\n∴C = R 4200\nStep 4: Write the final answer\nTherefore the minimum cost to the company is R 4200 with 10 kettles and 30 toast-\ners.\nExercise 12 – 2: Optimisation\n1. You are given a test consisting of two sections. The first section is on algebra and\nthe second section is on geometry. You are not allowed to answer more than 10\nquestions from any section, but you have to answer at least 4 algebra questions.\nThe time allowed is not more than 30 minutes. An algebra problem will take 2\nminutes and a geometry problem will take 3 minutes to solve.\nLet x be the number of algebra questions and y be the number of geometry\nquestions.\na) Formulate the equations and inequalities that satisfy the above constraints.\nb) The algebra questions carry 5 marks each and the geometry questions carry\n10 marks each. If T is the total marks, write down an expression for T.\n2. A local clinic wants to produce a guide to healthy living. The clinic intends to\nproduce the guide in two formats: a short video and a printed book. The clinic\nneeds to decide how many of each format to produce for sale. Estimates show\nthat no more than 10 000 copies of both items together will be sold. At least\n4000 copies of the video and at least 2000 copies of the book could be sold,\nalthough sales of the book are not expected to exceed 4000 copies. Let x be the\nnumber of videos sold, and y the number of printed books sold.\na) Write down the constraint inequalities that can be deduced from the given\ninformation.\nb) Represent these inequalities graphically and indicate the feasible region\nclearly.\n479\nChapter 12.\nLinear programming\n\nc) The clinic is seeking to maximise the income, I, earned from the sales of\nthe two products. Each video will sell for R 50 and each book for R 30.\nWrite down the objective function for the income.\nd) What maximum income will be generated by the two guides?\n3. A certain motorcycle manufacturer produces two basic models, the Super X and\nthe Super Y. These motorcycles are sold to dealers at a profit of R 20 000 per\nSuper X and R 10 000 per Super Y. A Super X requires 150 hours for assembly,\n50 hours for painting and finishing and 10 hours for checking and testing. The\nSuper Y requires 60 hours for assembly, 40 hours for painting and finishing and\n20 hours for checking and testing. The total number of hours available per month\nis: 30 000 in the assembly department, 13 000 in the painting and finishing\ndepartment and 5000 in the checking and testing department.\nThe above information is summarised by the following table:\nDepartment\nHours for\nSuper X\nHours for\nSuper Y\nHours available\nper month\nAssembly\n150\n60\n30 000\nPainting and\nfinishing\n50\n40\n13 000\nChecking and testing\n10\n20\n5000\nLet x be the number of Super X and y be the number of Super Y models manu-\nfactured per month.\na) Write down the set of constraint inequalities.\nb) Use graph paper to represent the set of constraint inequalities.\nc) Shade the feasible region on the graph paper.\nd) Write down the profit generated in terms of x and y.\ne) How many motorcycles of each model must be produced in order to max-\nimise the monthly profit?\nf) What is the maximum monthly profit?\n4. A group of students plan to sell x hamburgers and y chicken burgers at a rugby\nmatch. They have meat for at most 300 hamburgers and at most 400 chicken\nburgers. Each burger of both types is sold in a packet. There are 500 packets\navailable. The demand is likely to be such that the number of chicken burgers\nsold is at least half the number of hamburgers sold.\na) Write the constraint inequalities and draw a graph of the feasible region.\nb) A profit of R 3 is made on each hamburger sold and R 2 on each chicken\nburger sold. Write the equation which represents the total profit P in terms\nof x and y.\nc) The objective is to maximise profit. How many of each type of burger\nshould be sold?\n5. Fashion-Cards is a small company that makes two types of cards, type X and type\nY. With the available labour and material, the company can make at most 150\ncards of type X and at most 120 cards of type Y per week. Altogether they cannot\nmake more than 200 cards per week.\n480\n12.1.\nIntroduction\n\nThere is an order for at least 40 type X cards and 10 type Y cards per week.\nFashion-Cards makes a profit of R 5 for each type X card sold and R 10 for each\ntype Y card.\nLet the number of type X cards manufactured per week be x and the number of\ntype Y cards manufactured per week be y.\na) One of the constraint inequalities which represents the restrictions above is\n0 ≤x ≤150. Write the other constraint inequalities.\nb) Represent the constraints graphically and shade the feasible region.\nc) Write the equation that represents the profit P (the objective function), in\nterms of x and y.\nd) Calculate the maximum weekly profit.\n6. To meet the requirements of a specialised diet a meal is prepared by mixing\ntwo types of cereal, Vuka and Molo. The mixture must contain x packets of\nVuka cereal and y packets of Molo cereal. The meal requires at least 15 g of\nprotein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains\n4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains\n3 g of protein and 24 g of carbohydrates. There are at most 5 packets of cereal\navailable. The feasible region is shaded on the attached graph paper.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\nNumber of packets of Vuka\nNumber of packets of Molo\na) Write down the constraint inequalities.\nb) If Vuka cereal costs R 6 per packet and Molo cereal also costs R 6 per\npacket, use the graph to determine how many packets of each cereal must\nbe used so that the total cost for the mixture is a minimum.\nc) Use the graph to determine how many packets of each cereal must be used\nso that the total cost for the mixture is a maximum (give all possibilities).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DV\n2. 23DW\n3. 23DX\n4. 23DY\n5. 23DZ\n6. 23F2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n481\nChapter 12.\nLinear programming\n\n\nSolutions to exercises\n1\nExponents and surds\nExercise 1 – 1: The number system\n1. R; Q′\n2. R; Q\n3. R; Q\n4. R; Q\n5. R; Q; Z; N0\n6. R′Q′\n7. R; Q\n8. R; Q′\n9. R′\n10. R; Q′\n11. R; Q\n12. R; Q; Z\n13. R; Q\n14. R; Q′\n15. R; Q\n16. R; Q; Z\nExercise 1 – 2: Laws of exponents\n1. 43a+3\n2. 72\n3. 9p10\n4. k2x−2\n5. 52z−2 + 5z\n6. 1\n7. x10\n8.\nb2\na2\n9.\n1\nm+n\n10. 2pts\n11.\n1\na\n12. k\n13. 2a+1\n14. h4\n15.\na4b6\nc6d2\n16. 4\n17.\nm2n2\n2\n18. 400\n19.\n1\ny7\n20. 8\n21. 26a+2\n22. 2pt\n23. 81q2sy8a+2\nExercise 1 – 3: Rational exponents and surds\n1.\na) 7\nb)\n1\n6\nc)\n1\n3√\n36\nd) −4\n3\ne) 8x3\n2.\na) s\n1\n6\nb) 16m4\nc)\n3\n2 m2\nd) 8\n3. x\n31\n16\n483\nChapter 12.\nLinear programming\n\nExercise 1 – 4: Simplification of surds\n1.\na) 4\nb) ab4c2\nc) 2\nd) xy4\n2.\na)\nab\nb−a\nb) −\n\u0010\na\n1\n2 + b\n1\n2\n\u0011\nExercise 1 – 5: Rationalising the denominator\n1. 2\n√\n5\n2.\n√\n6\n2\n3.\n√\n6\n4.\n3\n√\n5 + 3\n4\n5.\nx√y\ny\n6.\n√\n6 +\n√\n14\n2\n7.\n3p −4√p\np\n8.\n√\nt −2\n9.\n1−√m\n1−m\n10.\n√\nab\nExercise 1 – 6: Solving surd equations\n1. x = 4\n2. p = 3\n3. y = 1\n4. t = 3\n5. z = 9 or z = 1\n4\n6. x = 8 or x = −27\n7. n = −1\n4\n8. d = 3 or d = −5\n9. y = 1 or y = 81\n10. f = 5\nExercise 1 – 7: Applications of exponentials\n1. 9,7%\n2. 4 254 691\n3. 7\n4. 26 893\n484\n12.1.\nIntroduction\n\nExercise 1 – 8: End of chapter exercises\n1.\na)\n1\n4\nb) 4 1\n4\n2.\na) x4\nb) s\nc) m\n25\n3\nd) m\n8\n3\ne) −m\n8\n3\nf) 81y\n16\n3\n3.\na)\n3b\n45\n2\n(a12c\n5\n2\nb) 3a3b2\nc) a24b12\nd) x\n7\n2\ne) x\n4\n3 b\n5\n3\n4.\n1\nx\n1\n16\n5. x −2\n6.\n10√x + 10\nx −1\n7.\n3√x + 2x√x\n2x\n8.\na) 6\n√\n2\nb) 7\n√\n5\nc) 2\nd)\n1\n4\n√\n2\ne) 2\nf)\n16\n√\n15\n5\n9.\na) 6 + 4\n√\n2\nb) 6 + 5\n√\n2\nc) 4+2\n√\n2+2\n√\n3+2\n√\n6\n10.\na) 55\nb) 1\n11. 15\n√\n2x3\n12.\na) 1 + 2\n√\n5\n5\nb)\n2y + y√y −4√y −8\ny −4\nc) 2√x + 2\n√\n10\n13.\n3\n2\n15. 3\n16. −\n√\n288\n17.\na) 4\nb) −1\n3\nc) 3\nd) No solution\ne) x = 1\n8 or x = −8\n18.\nb) x = 1\n2\nEquations and inequalities\nExercise 2 – 1: Solution by factorisation\n1. t = 0 or t = −2\n2. y = −1\n3. s = ±5\n4. y = 3 or y = 2\n5. y = 4 or y = −9\n6. p = −2\n7. y = −3 or y = −8\n8. y = 6 or y = 7\n9. x = −7 or x = −2\n10. y = 4k or y = k\n11. y = 9 or y = −9\n12. y = ±\n√\n5\n13. h = ±6\n14. y = ±\n√\n14\n15. p = −2\n16. y = ±6\n√\n2\n17. f = 5\n2 or f = −3\n18. x = 1\n4\n19. y = 1\n7\n20. x ∈R, x ̸= ±3\n21. y = −13 or y = −1\n22. t = 3\n2 or t = −2\n23. m = −6\n24. t = 0 or t = 3\nExercise 2 – 2: Solution by completing the square\n1.\na) x = −5 −3\n√\n3 or x = −5 + 3\n√\n3\nb) x = −1 or x = −3\nc) p = −4 ±\n√\n21\nd) x = −3 ±\n√\n7\ne) No real solution\nf) t = −8 ± 3\n√\n6\ng) x = −1 ±\nq\n5\n3\nh) z = −4 ±\n√\n22\ni) z = 11\n2 or z = 0\nj) z = 5 or z = −1\n2. k = −3 ± √9 −a\n3. y = −q±√\nq2−4pr\n2p\n485\nChapter 12.\nLinear programming\n\nExercise 2 – 3: Solution by the quadratic formula\n1. t = 1 or t = −4\n3\n2. x = 5+\n√\n37\n2\nor t = 5−\n√\n37\n2\n3. No real solution\n4. p = 1\n2 or p = −1\n5. No real solution\n6. t = −3+\n√\n69\n10\nor t = −3−\n√\n69\n10\n7. t = 2 ±\n√\n2\n8. k = 7+\n√\n373\n18\nor k = 7−\n√\n373\n18\n9. f = 1\n2 or f = −2\n10. No real solution\nExercise 2 – 4:\n1. x = −1, x = −4, x = −2 and x = −3\n2. x = 1, x = 4 and x = −2\n3. x = −7, x = 4, x = −1 and x = −2\n4. x = −4, x = 3, x = −3 and x = 2\n5. x = 8±\n√\n40\n4\n6. x = −5, x = 3, x = −1 +\n√\n10 and\nx = −1 −\n√\n10\nExercise 2 – 5: Finding the equation\n1. x2 −x −6 = 0\n2. x2 −16 = 0\n3. 2x2 −5x −3 = 0\n4. k = 3 and x = 3\n4\n5. p = 5 and x = −1\nExercise 2 – 6: Mixed exercises\n1. y = 1\n8 or y = −8\n3\n2. x = 3\n2 or x = −7\n2\n3. t = 2\n3 or t = 2\n4. y = 1 or y = −1\n5. m = 1 or m = 4\n6. y = ± 5\n7\n7. w = 3\n2 or w = 4\n8. y = 6\n5 or y = 1\n4\n9. n = 8\n3 or n = −9\n8\n10. y = −8\n3 or y = 3\n2\n11. x = −1\n2 or x = 3\n12. y = −5\n2 or y = −5\n9\n13. y = 4\n5 or y = 1\n5\n14. g = −1\n4 or g = 1\n15. y = 2 or y = −5\n9\n16. p = 3\n7 or p = −1\n5\n17. y = −2\n9 or y = −1\n18. y = 9\n2 or y = 9\n7\n486\n12.1.\nIntroduction\n\nExercise 2 – 7: From past papers\n1.\na) Real, unequal and rational\nb) Real and equal\nc) Real, unequal and irrational\nd) Real, unequal and rational\ne) Real, unequal and irrational\nf) Non-real\ng) Real, unequal and rational\nh) Real, unequal and irrational\ni) Non-real\nj) Real and equal\n2.\nb) real and unequal\nc) k = −6 ± 2\n√\n6\n4.\na) k = 6\nb) k = 1\n3\n5.\na) k = 4 or k = 1\nb) k = 0 or k = 5\n6.\na) all real values of a, b and p\nb) a = b and p = 0\nExercise 2 – 8: Solving quadratic inequalities\n1.\na) −3 < x < 4\nb) x < −4\n3 or when x > 1\nc) no real solutions\nd) −1 < t < 3\ne) All real values of s.\nf) All real values of x.\ng) x ≤−1\n4 or x ≥0\ni) x < 3 or x > 6 with x ̸= 3\nj) −2 ≤x ≤2 and x > 7 with x ̸= 7\nk) x > 0 with x ̸= 0\n2.\na) x < −3 or x > 3\nb) −\n√\n5 ≤x ≤\n√\n5\nc) no solution\nd) All real values of x\nExercise 2 – 9: Solving simultaneous equations\n1.\na) (0; 5) and (2; 3)\nb) x = 3 ±\n√\n2 and y = 2 ±\n√\n2\nc) (−1; 0) and ( 1\n4 ; 5\n8 )\nd) b = 2 ±\n√\n88\n6\nand a = 11 ±\n√\n88\n6\ne) (−3; −20) and (2; 0)\nf) x = 6 ±\n√\n264\n2\nand y = 70 ±\n√\n264\n2\n2.\na) (−3; 8) and (2; 3)\nb) (−4; 14) and (3; 7)\nc) (3; 4) and (4; 3)\nExercise 2 – 10:\n1. b = 2 m, l = 4 m\n2. 187\n3. t = 10,5 s\n4. t = 5d; 105 minutes; 1,4 km\n5. 24 A; 70 W; 12 A\n487\nChapter 12.\nLinear programming\n\nExercise 2 – 11: End of chapter exercises\n1. x = 1,62 or x = −0,62\n2. x = ±4 or x = −1\n3. y = 0 or y = ±1\n4. x = ±2\n5.\na) x = 7 or x = 2\nb) x = 2,3 or x = −1,3\nc) x = 1,65 or x = −3,65\nd) x = 0 or x = −3\n6. x =\n√\n16+p2−2\n2\n7. a = 3; b = 10 and c = −8\n8. p = ±16\n9. x2 + 2x −15\n10. Undefined:b = −2 Zero:b = 2 or b = 3\n13. a ≥4\n14. x = −3\n2 or x = 1\n15.\na) x < 3 or x ≥7:\nb) x < 1 or x > 5:\nc) 3 < x < 7:\nd) x < −1 or x > 3\ne) 0,5 < x < 2,5\nf) x ≤−3 or 0 < x ≤5\n2\ng) x < 2\n3\nh) −1 ≤x < 0 or x ≥3\ni) −4 ≤x ≤1\nj) 2 1\n2 ≤x < 3\n16.\na) x = ±\n√\n3 and y = ±2\n√\n3\nb) a = −3 and b = −1 or a = 12 and b = 4\nc) x = −5 and y = 0 or x = 2 and y = 14\nd) p = 5\n3 and q = 2\n9 or p = −1 and q = −2\n3\ne) b = 3±\n√\n5\n2\nand a = 7±3\n√\n5\n2\nf) b = −10±\n√\n140\n4\nand a = −12±\n√\n140\n2\ng) x = 3,4 and y = 5,4 or x = 3 and y = 5\nh) b = −1,4 and a = 23,6 or\nb = 3 and a = 6\n17.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\nx\n0\ny\nb\nb\nb)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n1\n2\n3\n4\n5\n6\n−1\n−2\nx\n0\ny\nb\nb\n18. 35 m\n20.\na) y = −5\n4 or y = −9\nb) x = −9\n4 or x = 1\nc) p = −8\n7 or p = −4\n3\nd) y = −1\n4 or y = 1\n2\ne) y = −2\n9 or y = −1\nf) y = 7\n3 or y = −1\n2\ng) y = 9\n4 or y = −9\n4\nh) y = 8\n3 or y = −6\ni) y = 9\n5 or y = −7\nj) x = ±4\nk) y = ±7\n21. k = 76 and 4\n9\n22. x = 3 or x = −2 and y = 1±√−7\n2\n23. x = 4 or x = −1\n24. y = 3\n2 , y = 1\n2 and p = 9\n2 , p = 7\n2\n25.\n69\n4\n26. 7\n27.\n2±\n√\n12\n2\n28. t = 1\n2 , t = 1 or t = 3±\n√\n33\n4\n488\n12.1.\nIntroduction\n\n3\nNumber patterns\nExercise 3 – 1: Linear sequences\n1. −19; −35; −51\n2.\na) −19\nb) T2 = 15; T4 = 33\n3.\na) Tn = 10 + 3n; T10 = 40; T15 = 55;\nT30 = 100\nb) Tn = 12 + 6n; T10 = 72; T15 = 102;\nT30 = 192\nc) Tn = −5 −5n; T10 = −55; T15 = −80;\nT30 = −155\n4. T9 = 36\n5.\na) 44; 66; 121\nExercise 3 – 2: Quadratic sequences\n1.\na) 10\nb) 2\nc) 2\nd) −2\ne) 2\nf) −4\ng) 4\nh) −2\ni) 6a\nj) 6\nk) 2t\n2.\na) T4 = 53\nb) T2 = 30\nc) T1 = 17\nd) T2 = −3\ne) T4 = 63\nf) T1 = 2\n3.\na) 3; 9; 17; 27\nb) −6; −9; −14; −21\nc) 1; 8; 21; 40\nd) 0; −5; −14; −27\nExercise 3 – 3: Quadratic sequences\n1.\na) 1\nb) 2\nc) 4\nd) 8\ne) −2\n2. 12; 30; 58; 96; 144\n3. T9 = 379\n4. n = 4\n5.\na) T5 = 84; T6 = 111\nb) Tn = 2n2 + 5n + 9\n489\nChapter 12.\nLinear programming\n\nExercise 3 – 4: End of chapter exercises\n1. −4; 9; 16; 25; 36\n2.\na) Quadratic sequence\nb) Quadratic sequence\nc) Quadratic sequence\nd) Quadratic sequence\ne) Quadratic sequence\nf) Quadratic sequence\ng) Linear sequence\nh) Linear sequence\ni) Quadratic sequence\nj) Quadratic sequence\nk) Quadratic sequence\nl) Linear sequence\nm) Quadratic sequence\n3. x = 31\n4. n = 11\n5. T11 = 363\n6. n = 9\n7. T5 = 114\n8. n = 8\n9.\na) T5 = 19;\nTn = 4n −1;\nT10 = 39\nb) T5 = −3;\nTn = 22 −5n;\nT10 = −28\nc) T5 = 2 1\n2 ; Tn = 1\n2 n;\nT10 = 5\nd) T5 = a + 4b;\nTn = a −b + bn;\nT10 = a + 9b\ne) T5 = −7;\nTn = 3 −2n;\nT10 = −17\n10.\na) Tn = n2 + 3;\nT100 = 10 003\nb) Tn = 6n −4;\nT100 = 596\nc) Tn = 2n2 + 5;\nT100 = 20 005\nd) Tn = 3n2 + 2;\nT100 = 30 002\n11.\na) 2; 5; 8; 11; 14\nb) Constant difference,\nd = 3\nc) Yes\n12.\na) Tn = 4n −19\nb) n = 48\n13.\na) Incorrect\nb) Correct\n14.\nc) Linear\n15.\nb) Linear\nd) Quadratic\ne) Tn = 1\n2 n2 + 3\n2 n + 1\nf) T21 = 253\ng) 31 cm\n16.\na) −1\nb) 7\n17.\nb) 2\nc) Tn = n2 −n\nd) 210\ne) 25\n18. 4; 14; 34; 64; 104; 154\n4\nAnalytical geometry\nExercise 4 – 1: Revision\n1.\na) 2\n√\n26units\nb) 7 units\nc) x + 1units\n2. p = 6 or p = 2\n3.\na) −1\n2\nb) 3\n5. 2\n6.\na) (1; 2)\nb)\n\u0000 −1\n2 ; −1\n2\n\u0001\n7. B(4; 2)\n8.\na) y = −4x + 3 and\ny = −4x + 19\nc) AD =\n√\n17units and\nBC =\n√\n17units\nd) y = 4\n3 x −7\n3\ne) Parallelogram (one\nopposite side equal\nand parallel)\n9. N(0; 3)\n10.\na) PQ =\n√\n20 and\nSR =\n√\n20\nb) M( 3\n2 ; 1)\nd) PS: y = −2\n5 x −1\n5\nand SR: y = 1\n2 x −2\ne) No\nf) Parallelogram\nExercise 4 – 2: The two-point form of the straight line equation\n1. y = 2\n3 x + 5\n2. y = −3x + 1\n4\n3. y = x + 3\n4. y = 2x −1\n5. y = −5\n6. y = 3\n4 x + 3\n7. y = −x + (s + t)\n8. y = 5x + 2\n9. y = q\npx −q\n490\n12.1.\nIntroduction\n\nExercise 4 – 3: Gradient–point form of a straight line equation\n1. y = 2\n3 x + 4\n2. y = −x −2\n3. y = −1\n3 x\n4. y = 11\n5. y = −2x + 7\n6. x = −3\n2\n7. y = −4\n5 x + 1\n8. x = 4\n9. y = 3ax + b\nExercise 4 – 4: The gradient–intercept form of a straight line equation\n1. y = 2x + 3\n2. y = 4x −4\n3. y = −x −1\n4. y = −3\n7 x\n5. y = 1\n2 x −1\n5\n6. y = 2x −2\n7. y = −3\n2\n8. y = 3x + 4\n9. y = −5x\nExercise 4 – 5: Angle of inclination\n1.\na) 1,7\nb) −1\nc) 0\nd) 1,4\ne) Undefined\nf) 1\ng) −0,8\nh) 0\ni) 3,7\n2.\na) 36,8◦\nb) 26,6◦\nc) 45◦\nd) Horizontal line\ne) 18,4◦\nf) Vertical line\ng) 71,6◦\nh) 30◦\nExercise 4 – 6: Inclination of a straight line\n1.\na) 38,7◦\nb) 135◦\nc) 80◦\nd) 80◦\ne) 102,5◦\nf) 45◦\ng) 56,3◦\nh) 63,4◦\ni) 161,6◦\nj) Gradient undefined\n2. 85,2◦\n3. 90◦\n4. 81,8◦\nExercise 4 – 7: Parallel lines\n1.\na) Parallel\nb) Parallel\nc) Parallel\nd) Not parallel\ne) Parallel\nf) Parallel\n2. y = −2x −3\n3. y = 3x\n4. y = 3\n2 x + 1\n5. y = −7\n10 x −1\n491\nChapter 12.\nLinear programming\n\nExercise 4 – 8: Perpendicular lines\n1.\na) Perpendicular\nb) Not perpendicular\nc) Perpendicular\nd) Perpendicular\ne) Perpendicular\nf) Not perpendicular\ng) Not perpendicular\n2. y = 1\n2 x −3\n3. y = −5x + 3\n4. y = −x + 2\n5. x = −2\nExercise 4 – 9: End of chapter exercises\n1.\na) y = 1\n2 x + 7\n2\nb) y = −x + 4\nc) y = 1\n2 x + 4\nd) y = 2x + 4\ne) y = 3x\n2.\na) θ = 63,4◦\nb) θ = 18,4◦\nc) θ = 36,9◦\nd) θ = 146,3◦\ne) θ = 161,6◦\n3.\na) y = −2x + 7\nb)\n\u0000 7\n2 ; 0\n\u0001\nc) θ = 116,6◦\nd) m = 1\n2\ne) Q ˆPR = 90◦\nf) y = −2x\ng)\n\u0000 1\n2 ; −3\n2\n\u0001\nh) y = −2x −1\n2\n4.\na)\n1\n2\n3\n4\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\nb\nb\nb\ny\nx\nA(−3; 5)\nB(−7; −4)\nC(2; 0)\nD(x; y)\nb) D (6; 9)\n5.\na) (−1; −2)\nb) (8; 3)\nc) x = −1\nd) MN = 5 units\ne) M ˆ\nNP = 21,8◦\nf) y = 5\n2 x + 11\n2\n6.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nA(−2; 3)\nB(2; 4)\nC(3; 0)\ny\nx\n0\nb\nb\nb\nc) y = 1\n4 x + 7\n2\nd) D(−1; −1)\ne) E\n\u0000 5\n2 ; 2\n\u0001\n7.\na) y = 3\n2 x + 2\nb) T ˆSV = 49,6◦\n8.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nb\nb\nb\nF(−1; 3)\nH(4; 4)\nG(2; 1)\ny\nx\n0\nc) y = −5x + 11\nd) Yes\ne) y = 3\n2 x + 9\n2\n9.\na)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nb\nb\nb\nA(−1; 5)\nB(5; −3)\nC(0; −6)\nx\ny\nM\nN\n492\n12.1.\nIntroduction\n\n5\nFunctions\nExercise 5 – 1: Revision\n1.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nc)\n1\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nd)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nExercise 5 – 2: Domain and range\n1. {x : x ∈R} ; {y : y ≥−1, y ∈R}\n2. {x : x ∈R} ; {y : y ≤4, y ∈R}\n3. {x : x ∈R} ; {y : y ≥0, y ∈R}\n4. {x : x ∈R} ; {y : y ≤0, y ∈R}\n5. {x : x ∈R} ; {y : y ≤2, y ∈R}\nExercise 5 – 3: Intercepts\n1. (0; 15) and (−5; 0); (−3; 0)\n2. (0; 16) and (4; 0)\n3. (0; −3) and (1; 0); (3; 0)\n4. (0; 35) and (−7\n2 ; 0); (−5\n2 ; 0)\n5. (0; 37) and no x-intercepts\n6. (0; −4) and\n(−0,85; 0); (−2,35; 0)\nExercise 5 – 4: Turning points\n1. (3; −1)\n2. (2; 1)\n3. (−2; −1)\n4. (−1\n2 ; 1\n2 )\n5. (1; 21)\n6. (−1; −6)\nExercise 5 – 5: Axis of symmetry\n1.\na) Axis of symmetry:\nx = 5\n4\nb) Axis of symmetry:\nx = 2\nc) Axis of symmetry:\nx = 2\n2. y = ax2 + q\n493\nChapter 12.\nLinear programming\n\nExercise 5 – 6: Sketching parabolas\n1.\na)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n1\n2\n3\n4\n5\n6\n−1\ny\nx\n0\nIntercepts: (−1; 0), (5; 0), (0; 5)\nTurning point: (2; 9)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≤9, y ∈R}\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−1; 0), (0; 2)\nTurning point: (−1; 0)\nAxes of symmetry: x = −1\nDomain: {x : x ∈R}\nRange: {y : y ≥0, y ∈R}\nc)\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−0,87; 0), (1,54; 0), (0; −4)\nTurning point: (0,33; −4,33)\nAxes of symmetry: x = −0,33\nDomain: {x : x ∈R}\nRange: {y : y ≥4,33, y ∈R}\nd)\n1\n2\n3\n4\n5\n6\n−1\n1\n2\n3\n4\n−1\ny\nx\n0\nIntercepts: (0; 13) Turning point: (2; 1)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≥1, y ∈R}\n3.\na)\ny\nx\n0\nb)\ny\nx\n0\nc)\ny\nx\n0\nd)\ny\nx\n0\ne)\ny\nx\n0\nf)\ny\nx\n0\n4.\na) yshifted = 2x2 + 16x + 32\nb) yshifted = −x2 −2x\nc) yshifted = 3x2 −16x + 22\n494\n12.1.\nIntroduction\n\nExercise 5 – 7: Finding the equation\n1. y = −3(x + 1)2 + 6 or y = −3x2 −6x + 3\n2. y = 1\n2 x2 −5\n2 x\n3. y = 2\n3 (x + 2)2\n4. y = −x2 + 3x + 4\nExercise 5 – 8:\n1.\na) 11\n2.\na)\n1\n2\n3\n4\n1\n2\n−1\n−2\nf(x)\nx\n0\nA(1; 3)\nb\nb) 6\nc) y = 6x −3\n3.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n−1\n−2\ng(x)\nx\n0\nb) 1\nc) 4\nd) 0\nExercise 5 – 10: Domain and range\n1. {x : x ∈R, x ̸= 0} ; {y : y ∈R, y ̸= 1}\n2. {x : x ∈R, x ̸= 8} ; {y : y ∈R, y ̸= 4}\n3. {x : x ∈R, x ̸= −1} ; {y : y ∈R, y ̸= −3}\n4. {x : x ∈R, x ̸= 5} ; {y : y ∈R, y ̸= 3}\n5. {x : x ∈R, x ̸= −2} ; {y : y ∈R, y ̸= 2}\nExercise 5 – 11: Intercepts\n1. (0; −1 3\n4 ) and\n\u0000−3 1\n2 ; 0\n\u0001\n2.\n\u0000 5\n2 ; 0\n\u0001\n3. (0; 1) and\n\u0000 1\n3 ; 0\n\u0001\n4.\n\u00000; 3\n2\n\u0001\nand\n\u0000 1\n3 ; 0\n\u0001\n5. (0; 2) and (8; 0)\nExercise 5 – 12: Asymptotes\n1. y = −2 and x = −4\n2. y = 0 and x = 0\n3. y = 1 and x = 2\n4. y = −8 and x = 0\n5. y = 0 and x = 2\n495\nChapter 12.\nLinear programming\n\nExercise 5 – 13: Axes of symmetry\n1.\na) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (0; 1); y1 = x + 1 and\ny2 = −x + 1\nb) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (−1; 0); y1 = x + 1 and\ny2 = −x −1\nc) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (1; −1); y1 = x −2 and\ny2 = −x\n2. k(x) =\n5\nx+1 + 2\nExercise 5 – 14: Sketching graphs\n1.\na) Asymptotes: x = 0; y = 2\nIntercepts:\n\u0000−1\n2 ; 0\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= 0}\nRange: {y : y ∈R, y ̸= 2}\nb) Asymptotes: x = −4; y = −2\nIntercepts:\n\u0000−3 1\n2 ; 0\n\u0001\nand\n\u00000; −1 3\n4\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x −6\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\nc) Asymptotes: x = −1; y = 3\nIntercepts:\n\u0000−2\n3 ; 0\n\u0001\nand (0; 2)\nAxes of symmetry: y = x + 4 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 3}\nd) Asymptotes: x = −2 1\n2 ; y = −2\nIntercepts: (0; 0)\nAxes of symmetry: y = x −4 1\n2 and\ny = −x + 1\n2\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\ne) Asymptotes: x = 8; y = 4\nIntercepts: (6; 0) and (0; 3)\nAxes of symmetry: y = x −4 and\ny = −x + 12\nDomain: {x : x ∈R, x ̸= 8}\nRange: {y : y ∈R, y ̸= 4}\n2. y =\n1\nx+2 −1\n3. y = −4\nx + 2\n4.\na)\nb) Average gradient = 1\nc) Average gradient = 12\nExercise 5 – 16: Domain and range\n1. {x : x ∈R} ; {y : y > 0, y ∈R}\n2. {x : x ∈R} ; {y : y < 1, y ∈R}\n3. {x : x ∈R} ; {y : y > −3, y ∈R}\n4. {x : x ∈R} ; {y : y > n, y ∈R}\n5. {x : x ∈R} ; {y : y > 2, y ∈R}\nExercise 5 – 17: Intercepts\n1. (0; −6) and (2; 0)\n2. (0; −17 1\n3 ) and (3; 0)\n3. (0; −20) and (−1; 0)\n4. (0; 15\n16 ) and (−2; 0)\n496\n12.1.\nIntroduction\n\nExercise 5 – 18: Asymptote\n1. y = 0\n2. y = 1\n3. y = −2\n3\n4. y = −2\n5. y = −2\nExercise 5 – 19: Mixed exercises\n1.\nb)\ni. y = 3\nx + 3\nii. y =\n3\nx−3\niii. y = −3\nx\niv. y = 3\nx −1\n4\nv. y = 3\nx + 4\nvi. y =\n3\nx+2 −1\n2.\na) M(−2; 2)\nb) g(x) = −4\nx\nc) f(x) = 2(x + 1)2\nd) −2 < x < 0\ne) Range: {y : y ∈R, y ≥0}\n3.\na) For k(x) :\nIntercepts:\n(−2; 0), (1; 0) and (0; −4)\nTurning point:\n\u0000−1\n2 ; −4 1\n2\n\u0001\nAsymptote:\nnone\nFor h(x) :\nIntercepts:\n(1,41; 0)\nTurning point:\nnone\nAsymptote:\ny = 0\n6.\na) f(x) = −3\n4 (x −2)2 + 3 ;\nAxes of symmetry: x = 2 ;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≤3}\nb) g(x) = 1\n4 x2 −2;\nAxes of symmetry: x = 0;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≥−2}; h(x) = 2\nx ;\nAxes of symmetry: y = x\nDomain: {x : x ∈R, x < 0};\nRange: {y : y ∈R, y < 0};\nc) k(x) =\n\u0000 1\n2\n\u0001x + 1\n2 ;\nDomain: {x : x ∈R};\nRange:\n\b\ny : y ∈R, y > 1\n2\n\t\n7.\nb) p = 9\nc) Average gradient = −2 8\n9\nd) y =\n\u0000 1\n3\n\u0001x+2 −2\n8.\na) f(x) = 2x −3\n2 and g(x) = −1\n4 x −1\n2\nb) h(x) = −\n3\nx+2 + 1\n9.\na) AO = 2 units OB = 5 units\nOC = 10 units DE = 12,25 units\nb) DE = 12 1\n4\nc) h(x) = −2x + 10\nd) {x : x ∈R, x < −2 and x > 5}\ne) {x : x ∈R, 0 ≤x ≤5}\nf) 5,25 units\n497\nChapter 12.\nLinear programming\n\nExercise 5 – 20: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\nPeriod: = 360◦\nAmplitude: = 1\nDomain: = [0◦; 360◦]\nRange: = [−1; 1]\nx-intercepts: = (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: = (0◦; 0)\nMax. turning point: = (90◦; 1)\nMin. turning point: = (270◦; −1)\n2.\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny2 = −2 sin θ\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 2\nDomain: [0◦; 360◦]\nRange: [−2; 2]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nMax. turning point: (270◦; 2)\nMin. turning point: (90◦; −2)\n3.\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny3 = sin θ + 1\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [0; 2]\nx-intercepts: (270◦; 0)\ny-intercepts: (0◦; 1)\nMax. turning point: (90◦; 2)\nMin. turning point: (270◦; 0)\n4.\n1\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny4 = 1\n2 sin θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (90◦; 1\n2 )\nMin. turning point: (270◦; −3\n2 )\nExercise 5 – 21: Sine functions of the form y = sin kθ\n2.\na) k = 2\nb) k = −3\n4\n498\n12.1.\nIntroduction\n\nExercise 5 – 23: The sine function\n1.\na)\n1\n2\n−1\n−2\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 2 sin( θ\n2 )\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 1\n2 sin(θ −45◦)\nc)\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(θ + 90◦) + 1\nd)\n1\n−1\n60◦\n120◦\n180◦\n−60◦\n−120◦\n−180◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(−3θ\n2 )\ne)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(30◦−θ)\n2. a = 2; p = 90◦∴y = 2 sin(θ + 90◦) and\ny = 2 cos θ\nExercise 5 – 24: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\n2.\n1\n2\n3\n−1\n−2\n−3\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny2 = −3 cos θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−3; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; −3)\n3.\n1\n2\n3\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny3 = cos θ + 2\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [1; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; 1)\n4.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny4 = 1\n2 cos θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (0◦; −1\n2 ); (360◦; −1\n2 )\nMin. turning point: (180◦; −3\n2 )\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n2.\na) k = 3\n2\nb) k = 2\n3\n499\nChapter 12.\nLinear programming\n\nExercise 5 – 27: The cosine function\n1.\na)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nθ\n0◦\ny\ny = cos θ\ny = cos(θ + 15◦)\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny = cos θ\nf(θ) = 1\n3 cos(θ −60◦)\nc)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ny = −2 cos θ\nd)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(30◦−θ)\ne)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ng(θ) = 1 + cos(θ −90◦)\nf)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(2θ + 60◦)\n2.\na) a = −1\nb) p = −180◦\nc) cos(θ −180◦) = −cos θ\nExercise 5 – 28: Revision\n1.\n1\n2\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1\n2 )\nAsymptotes: 90◦; 270◦\n2.\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: 90◦; 270◦\n3.\n1\n2\n3\n4\n5\n−1\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (116,6◦; 0); (296,6◦; 0)\ny-intercepts: (0◦; 2)\nAsymptotes: 90◦; 270◦\n4.\n1\n2\n−1\n−2\n−3\n−4\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1)\nAsymptotes: 90◦; 270◦\n500\n12.1.\nIntroduction\n\nExercise 5 – 29: Tangent functions of the form y = tan kθ\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦] Range: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 240◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −120◦; 120◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n4.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 270◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; 135◦\n501\nChapter 12.\nLinear programming\n\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−225◦; 0); (−45◦; 0); (135◦; 0);\n(315◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−330◦; 0); (−150◦; 0); (30◦; 0);\n(210◦; 0)\ny-intercepts: (0◦; −0,58)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−240◦; 0); (−60◦; 0); (120◦; 0);\n(300◦; 0)\ny-intercepts: (0◦; 1.73)\nAsymptotes: −330◦; −150◦; 30◦; 210◦\nExercise 5 – 31: The tangent function\n1.\na)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n−45◦\n−90◦\nθ\ny\nb)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\nθ\ny\nc)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nd)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\ny\n2. a = −1; k = 1\n2\n502\n12.1.\nIntroduction\n\nExercise 5 – 32: Mixed exercises\n1.\na) f(θ) = 3\n2 sin 2θ and g(θ) = −3\n2 tan θ\nb) f(θ) = −2 sin θ and\ng(θ) = 2 cos(θ + 360◦\nc) y = 3 tan θ\n2\nd) y = y = 2 cos θ + 2\n2.\na)\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\nθ\n0\ng\nf\ny\nb\nb\nb\nb\n(90◦; 2)\n(270◦; −2)\nb) 360◦\nc) 1\nd) At θ = 180◦\n3.\na) a = 2, b = −1 and c = 240◦\nb) 180◦\nc) θ = 60◦; 300◦\nd) y = −tan(θ −45◦)\n4.\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny1\ny2\ny\nb\nb\nExercise 5 – 33: End of chapter exercises\n2. a = −2; k = −1\n4.\na) y = x + 22 + 2\nb) y = x −12 + 5\n5. (−1; 0)\n6.\ny\nx\n0\n4\n−4\n4\n−4\ny =\n2\nx−3 −1\n7. y =\n1\n(x−1) + 2\n9.\na) a = −1\nb) f(−15) = 0,99997\nc) x = −1\nd) h(x) = −2(x−2) + 1\n10.\na) a = 256\nb) f(x) = 256\n\u0000 3\n4\n\u0001x\nc) f(13) = 6,08\n11.\na)\n1\n−1\n90◦\n180◦\n−90◦\n−180◦\nθ\n0\ny\nb)\n1\n−1\n90◦\n180◦\nθ\n0\ny\nd)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\n0\ny\ne)\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny\nf)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\n503\nChapter 12.\nLinear programming\n\n6\nTrigonometry\nExercise 6 – 1: Revision\n1.\na) True\nb) True\nc) False\nd) True\n2.\na) 50,2◦\nb) 40,5◦\nc) 26,6◦\nd) 109,8◦\ne) No solution\nf) 17,7◦\ng) 69,4◦\n3.\na) 17,3 cm\nb) 10 cm\nc) 64,8◦\n4.\na) 10 cm\nb) 5,2 cm and 19,3 cm\nc) 50 cm2\n5.\na) 2\nb) 0\nc) −1 1\n2\nd) 1\ne) 1\n6.\na) 60◦\nb)\n1\n2\nc) 1\n7. No\nExercise 6 – 2: Trigonometric identities\n1.\na) cos α\nb) tan2 θ\nc) cos2 θ\nd) 0\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1.\na)\n√\n3\n3\nb)\n1\n8\nc) 1\n2.\na)\n1−cos2 θ\ncos θ\nb) −1\n3.\na) 2t\nb) −1\nt\nExercise 6 – 4: Using reduction formula\n1.\na) −tan θ\nb) 1\nc) 1\n2. −cos β\n3.\na)\n1\n√\n3\nb) 2\nc) 2\nd) −3\n2\ne) −4\n√\n3\n5\n5.\na) −t\nb) 1 −t2\nc) ±\nt\n√\n1−t2\nExercise 6 – 5: Co-functions\n1.\na) cos θ\nb)\n3\n2\n2.\na) p\nb)\np\n1 −p2\nc) −\np\n√\n1−p2\nd) p\n504\n12.1.\nIntroduction\n\nExercise 6 – 6: Reduction formulae\n1.\na) sin2 θ\nb) cos2 θ\nc)\ni. 1\nii. tan2 θ\n2.\na) sin 17◦\nb) cos 33◦\nc) tan 68◦\nd) −cos 33◦\n3.\na)\n√\n3\nb)\n√\n3\n2\nc)\n1\n4\nd) 1\nExercise 6 – 7: Solving trigonometric equations\n1.\na) α = 60◦; 300◦\nb) α = 220,5◦; 319,5◦\nc) α = 79,2◦; 259,2◦\nd) α = 200,1◦; 339,9◦\ne) α = 36,9◦; 143,1◦\nf) α = 109,7◦; 289,7◦\n2.\na) θ = −323,1◦; −216,9◦; 36,9◦; 143,1◦\nb) θ = −221,4◦; −138,6◦; 138,6◦; 221,4◦\nc) θ = −278,5◦; −98,5◦; 81,5◦; 261,5◦\nd) θ = −90◦; 270◦\ne) θ = −293,6◦; −66,4◦; 66,4◦; 293,6◦\nExercise 6 – 8: General solution\n1.\na) θ = −128,36◦; −101,64◦; 51,64◦\nb) θ = −80,45◦; −9,54◦; 99,55◦; 170,46◦\nc) θ = −53,27◦; 126,73◦\nd) α = 0◦\ne) θ = −180◦; 0◦; 180◦\nf) θ = −180◦; 180◦\ng) θ = 84◦\nh) θ = −120◦; 120◦\ni) θ = −60◦; −30◦; 120◦; 150◦\n2.\na) θ = −20◦+ n . 360◦\nb) α = 30◦+ n . 120◦\nc) β = 10,25◦+ n . 45◦or\nβ = 55,25◦+ n . 45◦\nd) α = 70◦+ n . 360◦or\nα = 340◦+ n . 360◦\ne) θ = 140◦+ n . 240◦or\nθ = 220◦+ n . 240◦\nf) β = 15◦+ n . 180◦\nExercise 6 – 9: Solving trigonometric equations\n1.\na) θ = 45◦+ k . 180◦or\nθ = 135◦+ k . 180◦\nb) α = 50◦+ k . 360◦or\nα = 110◦+ k . 360◦\nc) θ = 60◦+ k . 720◦or\nθ = 660◦+ k . 720◦\nd) β = 146,6◦+ k . 180◦\ne) θ = 110,27◦+ k . 360◦or\nθ = 249,73◦+ k . 360◦\nf) α = 210◦+ k . 360◦or\nα = 330◦+ k . 360◦\ng) β = 23,3◦+ k . 120◦\nh) θ = 122◦+ k . 180◦\ni) α = 21◦+ k . 180◦or\nα = 39,5◦+ k . 90◦\nj) β = 22,5◦+ k . 90◦\n2. θ = 0◦, 180◦, 210◦, 330◦or 360◦\n3.\na) θ = 120◦+ k . 360◦or\nθ = 240◦+ k . 360◦\nb) θ = 0◦+ k . 180◦or\nθ = 146,3◦+ k . 180◦\nc) α = 36,9◦+ k . 360◦or\nα = 143,1◦+ k . 360◦or\nα = 216,9◦+ k . 360◦or\nα = 323,1◦+ k . 360◦\nd) β = 15◦+ k . 120◦or\nβ = 75◦+ k . 120◦\ne) α = 48,4◦+ k . 180◦\nf) θ = 63,4◦+ k . 180◦or\nθ = 116,6◦+ k . 180◦\ng) θ = 54,8◦+ k . 180◦or\nθ = 95,25◦+ k . 180◦\n4. β = −70,5◦or β = 109,5◦\n505\nChapter 12.\nLinear programming\n\nExercise 6 – 10: The area rule\n1.\na)\nP\nQ\nR\n30◦\n10\n7\nArea △PQR = 17,5 square units\nb)\nP\nQ\nR\n110◦\n9\n8\nArea △PQR = 33,8 square units\n2. Area △XY Z = 645,6 square units\n3. Area = 106,5 square units\n4.\nˆC = 72,2◦or ˆC = 107,8◦\nExercise 6 – 11: Sine rule\n1.\na)\nˆP = 92◦, q = 6,6, p = 7,4\nb)\nˆL = 87◦, l = 1,3, k = 0,89\nc)\nˆB = 76,8◦, b = 94,3, c = 91,3\nd)\nˆY = 84◦, y = 60, z = 38,8\n2.\nˆB = 32◦, AB = 23, BC = 39\n3. ST = 78,1 km\n4. m = 26,2\n5. BC = 3,2\nExercise 6 – 12: The cosine rule\n1.\na) a = 8,5, ˆC = 83,9◦, ˆB = 26,1◦\nb)\nˆR = 120◦, ˆS = 32,2◦, ˆT = 27,8◦\nc)\nˆ\nM = 27,7◦, ˆL = 40,5◦, ˆ\nK = 111,8◦\nd) h = 19,1, ˆJ = 18,2◦, ˆ\nK = 31,8◦\ne)\nˆD = 34◦, ˆE = 44,4◦, ˆF = 101,6◦\n2.\na) x = 4,4 km\nb) y = 63,5 cm\n3.\na)\nˆ\nK = 117,3◦\nb)\nˆQ = 78,5◦\nExercise 6 – 13: Area, sine and cosine rule\n1.\na) 7,78 km\nb) 6 km\n2. XZ = 1,73 km, XY = 0,87 km\n3.\na) 1053 km\nb) 4,42◦\n4. DC = x sin a sin(b+c)\nsin(a+c) sin b\n5.\nb) 438,5 km\n6. 9,38 m2\n7. DC = x sin α\nsin β\n506\n12.1.\nIntroduction\n\nExercise 6 – 14: End of chapter exercises\n1. sin2 A\n2. 1 1\n4\n3. cos α\n4. 3\n7.\na) −1\nb) θ = 135◦or θ = 315◦\n8.\na)\nb\nx\ny\n0\nθ\n(−12; −5)\nb) −5\n13 and 12\n13\nc) θ = 202,62◦\n9.\na) a = 1 and b = −\n√\n3\nb) −\n√\n3\n2\n10.\na) x = 50,9◦or x = 309,1◦\nb) x = 127,3◦or x = 307,3◦\nc) x = 26,6◦; 153,4◦206,6◦or 333,4◦\n11.\na) x = 55◦+ k . 360◦or\nx = 175◦+ k . 360◦\nb) x = 180◦+ k . 360◦\n12.\na) x = 28,6◦+ k . 180◦or\nx = 61,4◦+ k . 180◦\nb)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nb\nb\nb\nb\nθ\n0◦\ny\ny = sin 2α\nc) 28,6◦; 61,4◦; 208,6◦; 241,4◦\n13.\na) A ˆGN = α −β\nb)\nˆ\nA = 90◦−α\nd) H = 5 m\n14.\na) AC = 9,43 m\nb) AD = 6,2 m\nc) Area = 49,25 m2\nd) Area = 49,23 m2\n7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "1.1" }, { "title": "Rational exponents and surds", "content": "", "chapter_id": "1.2" }, { "title": "Solving surd equations", "content": "", "chapter_id": "1.3" }, { "title": "Applications of exponentials", "content": "", "chapter_id": "1.4" }, { "title": "Summary", "content": "2.10\nSummary\n80\n\n2\nEquations and inequalities\n2.1\nRevision\nEMBFF\nSolving quadratic equations using factorisation\nEMBFG\nTerminology:\nExpression\nAn expression is a term or group of terms consist-\ning of numbers, variables and the basic operators\n(+, −, ×, ÷, xn).\nEquation\nA mathematical statement that asserts that two ex-\npressions are equal.\nInequality\nAn inequality states the relation between two ex-\npressions (>, <, ≥, ≤).\nSolution\nA value or set of values that satisfy the original prob-\nlem statement.\nRoot\nA root of an equation is the value of x such that\nf(x) = 0.\nA quadratic equation is an equation of the second degree; the exponent of one variable\nis 2.\nThe following are examples of quadratic equations:\n2x2 −5x = 12\na(a −3) −10 = 0\n3b\nb + 2 + 1 =\n4\nb + 1\nA quadratic equation has at most two solutions, also referred to as roots. There are\nsome situations, however, in which a quadratic equation has either one solution or no\nsolutions.\n30\n2.1.\nRevision\n\nx\ny\ny = x2 −4\n0\n−2\n2\n−4\nx\ny\ny = x2\n0\nx\ny\ny = x2 + x + 1\n0\ny = (x −2)(x + 2)\n= x2 −4\nGraph of a quadratic\nequation with two roots:\nx = −2 and x = 2.\ny = x2\nGraph of a quadratic\nequation with one root:\nx = 0.\ny = x2 + x + 1\nGraph of a quadratic\nequation with no real\nroots.\nOne method for solving quadratic equations is factorisation. The standard form of a\nquadratic equation is ax2 + bx + c = 0 and it is the starting point for solving any\nequation by factorisation.\nIt is very important to note that one side of the equation must be equal to zero.\nInvestigation: Zero product law\nSolve the following equations:\n1. 6 × 0 = ?\n2. −25 × 0 = ?\n3. 0 × 0,69 = ?\n4. 7 × ? = 0\nNow solve for the variable in each of the following:\n1. 6 × m = 0\n2. 32 × x × 2 = 0\n3. 11(z −3) = 0\n4. (k + 3)(k −4) = 0\nTo obtain the two roots we use the fact that if a × b = 0, then a = 0 and/or b = 0. This\nis called the zero product law.\n31", "chapter_id": "1.5" }, { "title": "Equations and inequalities", "content": "Chapter 2.\nEquations and inequalities\n\nMethod for solving quadratic equations\nEMBFH\n1. Rewrite the equation in the standard form ax2 + bx + c = 0.\n2. Divide the entire equation by any common factor of the coefficients to obtain a\nsimpler equation of the form ax2+bx+c = 0, where a, b and c have no common\nfactors.\n3. Factorise ax2 + bx + c = 0 to be of the form (rx + s) (ux + v) = 0.\n4. The two solutions are\n(rx + s) = 0\n(ux + v) = 0\nSo x = −s\nr\nSo x = −v\nu\n5. Always check the solution by substituting the answer back into the original equa-\ntion.\nSee video: 226R at www.everythingmaths.co.za\nWorked example 1: Solving quadratic equations using factorisation\nQUESTION\nSolve for x: x (x −3) = 10\nSOLUTION\nStep 1: Rewrite the equation in the form ax2 + bx + c = 0\nExpand the brackets and subtract 10 from both sides of the equation x2 −3x −10 = 0\nStep 2: Factorise\n(x + 2) (x −5) = 0\nStep 3: Solve for both factors\nx + 2 = 0\nx = −2\nor\nx −5 = 0\nx = 5\n32\n2.1.\nRevision\n\nThe graph shows the roots of the equation x = −2 or x = 5. This graph does not\nform part of the answer as the question did not ask for a sketch. It is shown here for\nillustration purposes only.\n2\n4\n−2\n−4\n−6\n−8\n−10\n−12\n2\n4\n6\n−2\n−4\nx\nf(x)\ny = x2 −3x −10\n0\nStep 4: Check the solution by substituting both answers back into the original equa-\ntion\nStep 5: Write the final answer\nTherefore x = −2 or x = 5.\nWorked example 2: Solving quadratic equations using factorisation\nQUESTION\nSolve the equation: 2x2 −5x −12 = 0\nSOLUTION\nStep 1: There are no common factors\nStep 2: The quadratic equation is already in the standard form ax2 + bx + c = 0\nStep 3: Factorise\nWe must determine the combination of factors of 2 and 12 that will give a middle term\ncoefficient of 5.\nWe find that 2 × 1 and 3 × 4 give a middle term coefficient of 5 so we can factorise the\nequation as\n(2x + 3)(x −4) = 0\nStep 4: Solve for both roots\n33\nChapter 2.\nEquations and inequalities\n\nWe have\n2x + 3 = 0\nx = −3\n2\nor\nx −4 = 0\nx = 4\nStep 5: Check the solution by substituting both answers back into the original equa-\ntion\nStep 6: Write the final answer\nTherefore, x = −3\n2 or x = 4.\nWorked example 3: Solving quadratic equations using factorisation\nQUESTION\nSolve for y: y2 −7 = 0\nSOLUTION\nStep 1: Factorise as a difference of two squares\nWe know that\n\u0010√\n7\n\u00112\n= 7\nWe can write the equation as\ny2 −(\n√\n7)2 = 0\nStep 2: Factorise\n(y −\n√\n7)(y +\n√\n7) = 0\nTherefore y =\n√\n7 or y = −\n√\n7\n34\n2.1.\nRevision\n\nEven though the question did not ask for a sketch, it is often very useful to draw the\ngraph. We can let f(y) = y2 −7 and draw a rough sketch of the graph to see where\nthe two roots of the equation lie.\n1\n2\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n3\n−1\n−2\n−3\ny\nf(y)\nf(y) = y2 −7\n0\nStep 3: Check the solution by substituting both answers back into the original equa-\ntion\nStep 4: Write the final answer\nTherefore y = ±\n√\n7.\nWorked example 4: Solving quadratic equations using factorisation\nQUESTION\nSolve for b:\n3b\nb + 2 + 1 =\n4\nb + 1\nSOLUTION\nStep 1: Determine the restrictions\nThe restrictions are the values for b that would result in the denominator being equal\nto 0, which would make the fraction undefined. Therefore b ̸= −2 and b ̸= −1.\nStep 2: Determine the lowest common denominator\nThe lowest common denominator is (b + 2) (b + 1).\nStep 3: Multiply each term in the equation by the lowest common denominator and\nsimplify\n35\nChapter 2.\nEquations and inequalities\n\n3b(b + 2)(b + 1)\nb + 2\n+ (b + 2)(b + 1) = 4(b + 2)(b + 1)\nb + 1\n3b(b + 1) + (b + 2)(b + 1) = 4(b + 2)\n3b2 + 3b + b2 + 3b + 2 = 4b + 8\n4b2 + 2b −6 = 0\n2b2 + b −3 = 0\nStep 4: Factorise and solve the equation\n(2b + 3)(b −1) = 0\n2b + 3 = 0 or b −1 = 0\nb = −3\n2 or b = 1\nStep 5: Check the solution by substituting both answers back into the original equa-\ntion\nStep 6: Write the final answer\nTherefore b = −1 1\n2 or b = 1.\nWorked example 5: Squaring both sides of the equation\nQUESTION\nSolve for m: m + 2 = √7 + 2m\nSOLUTION\nStep 1: Square both sides of the equation\nBefore we square both sides of the equation, we must make sure that the radical is the\nonly term on one side of the equation and all other terms are on the other, otherwise\nsquaring both sides will make the equation more complicated to solve.\n(m + 2)2 =\n\u0000√\n7 + 2m\n\u00012\nStep 2: Expand the brackets and simplify\n36\n2.1.\nRevision\n\n(m + 2)2 =\n\u0000√\n7 + 2m\n\u00012\nm2 + 4m + 4 = 7 + 2m\nm2 + 2m −3 = 0\nStep 3: Factorise and solve for m\n(m −1)(m + 3) = 0\nTherefore m = 1 or m = −3\nStep 4: Check the solution by substituting both answers back into the original equa-\ntion\nTo find the solution we squared both sides of the equation. Squaring an expression\nchanges negative values to positives and can therefore introduce invalid answers into\nthe solution. Therefore it is very important to check that the answers obtained are\nvalid. To test the answers, always substitute back into the original equation.\nIf m = 1:\nRHS =\np\n7 + 2(1)\n=\n√\n9\n= 3\nLHS = 1 + 2\n= 3\nLHS = RHS\nTherefore m = 1 is valid.\nIf m = −3:\nRHS =\np\n7 + 2(−3)\n=\n√\n1\n= 1\nLHS = −3 + 2\n= −1\nLHS ̸= RHS\nTherefore m = −3 is not valid.\nStep 5: Write the final answer\nTherefore m = 1.\nSee video: 226S at www.everythingmaths.co.za\n37\nChapter 2.\nEquations and inequalities\n\nExercise 2 – 1: Solution by factorisation\nSolve the following quadratic equations by factorisation. Answers may be left in surd\nform, where applicable.\n1. 7t2 + 14t = 0\n2. 12y2 + 24y + 12 = 0\n3. 16s2 = 400\n4. y2 −5y + 6 = 0\n5. y2 + 5y −36 = 0\n6. 4 + p = √p + 6\n7. −y2 −11y −24 = 0\n8. 13y −42 = y2\n9. (x −1)(x + 10) = −24\n10. y2 −5ky + 4k2 = 0\n11. 2y2 −61 = 101\n12. 2y2 −10 = 0\n13. −8 + h2 = 28\n14. y2 −4 = 10\n15. √5 −2p −4 = 1\n2p\n16. y2 + 28 = 100\n17. f (2f + 1) = 15\n18. 2x = √21x −5\n19.\n5y\ny −2 + 3\ny + 2 =\n−6\ny2 −2y\n20.\nx + 9\nx2 −9 +\n1\nx + 3 =\n2\nx −3\n21. y −2\ny + 1 = 2y + 1\ny −7\n22. 1 + t −2\nt −1 =\n5\nt2 −4t + 3 +\n10\n3 −t\n23.\n4\nm + 3 +\n4\n4 −m2 =\n5m −5\nm2 + m −6\n24. 5√5t + 1 −4 = 5t + 1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 226T\n2. 226V\n3. 226W\n4. 226X\n5. 226Y\n6. 226Z\n7. 2272\n8. 2273\n9. 2274\n10. 2275\n11. 2276\n12. 2277\n13. 2278\n14. 2279\n15. 227B\n16. 227C\n17. 227D\n18. 227F\n19. 227G\n20. 227H\n21. 227J\n22. 227K\n23. 227M\n24. 227N\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.2\nCompleting the square\nEMBFJ\nInvestigation: Completing the square\nCan you solve each equation using two different methods?\n1. x2 −4 = 0\n2. x2 −8 = 0\n3. x2 −4x + 4 = 0\n38\n2.2.\nCompleting the square\n\n4. x2 −4x −4 = 0\nFactorising the last equation is quite difficult. Use the previous examples as a hint and\ntry to create a difference of two squares.\nSee video: 227P at www.everythingmaths.co.za\nWe have seen that expressions of the form x2 −b2 are known as differences of squares\nand can be factorised as (x −b)(x + b). This simple factorisation leads to another\ntechnique for solving quadratic equations known as completing the square.\nConsider the equation x2 −2x −1 = 0. We cannot easily factorise this expression.\nWhen we expand the perfect square (x −1)2 and examine the terms we see that\n(x −1)2 = x2 −2x + 1.\nWe compare the two equations and notice that only the constant terms are different.\nWe can create a perfect square by adding and subtracting the same amount to the\noriginal equation.\nx2 −2x −1 = 0\n(x2 −2x + 1) −1 −1 = 0\n(x2 −2x + 1) −2 = 0\n(x −1)2 −2 = 0\nMethod 1: Take square roots on both sides of the equation to solve for x.\n(x −1)2 −2 = 0\n(x −1)2 = 2\np\n(x −1)2 = ±\n√\n2\nx −1 = ±\n√\n2\nx = 1 ±\n√\n2\nTherefore x = 1 +\n√\n2 or x = 1 −\n√\n2\nVery important: Always remember to include both a positive and a negative answer\nwhen taking the square root, since 22 = 4 and (−2)2 = 4.\nMethod 2: Factorise the expression as a difference of two squares using 2 =\n\u0000√\n2\n\u00012.\nWe can write\n(x −1)2 −2 = 0\n(x −1)2 −\n\u0010√\n2\n\u00112\n= 0\n\u0010\n(x −1) +\n√\n2\n\u0011 \u0010\n(x −1) −\n√\n2\n\u0011\n= 0\n39\nChapter 2.\nEquations and inequalities\n\nThe solution is then\n(x −1) +\n√\n2 = 0\nx = 1 −\n√\n2\nor\n(x −1) −\n√\n2 = 0\nx = 1 +\n√\n2\nMethod for solving quadratic equations by completing the square\n1. Write the equation in the standard form ax2 + bx + c = 0.\n2. Make the coefficient of the x2 term equal to 1 by dividing the entire equation by\na.\n3. Take half the coefficient of the x term and square it; then add and subtract it\nfrom the equation so that the equation remains mathematically correct. In the\nexample above, we added 1 to complete the square and then subtracted 1 so\nthat the equation remained true.\n4. Write the left hand side as a difference of two squares.\n5. Factorise the equation in terms of a difference of squares and solve for x.\nSee video: 227Q at www.everythingmaths.co.za\nWorked example 6: Solving quadratic equations by completing the square\nQUESTION\nSolve by completing the square: x2 −10x −11 = 0\nSOLUTION\nStep 1: The equation is already in the form ax2 + bx + c = 0\nStep 2: Make sure the coefficient of the x2 term is equal to 1\nx2 −10x −11 = 0\nStep 3: Take half the coefficient of the x term and square it; then add and subtract it\nfrom the equation\nThe coefficient of the x term is −10. Half of the coefficient of the x term is −5 and\nthe square of it is 25. Therefore x2 −10x + 25 −25 −11 = 0.\nStep 4: Write the trinomial as a perfect square\n40\n2.2.\nCompleting the square\n\n(x2 −10x + 25) −25 −11 = 0\n(x −5)2 −36 = 0\nStep 5: Method 1: Take square roots on both sides of the equation\n(x −5)2 −36 = 0\n(x −5)2 = 36\nx −5 = ±\n√\n36\nImportant: When taking a square root always remember that there is a positive and\nnegative answer, since (6)2 = 36 and (−6)2 = 36.\nx −5 = ±6\nStep 6: Solve for x\nx = −1 or x = 11\nStep 7: Method 2: Factorise equation as a difference of two squares\n(x −5)2 −(6)2 = 0\n[(x −5) + 6] [(x −5) −6] = 0\nStep 8: Simplify and solve for x\n(x + 1)(x −11) = 0\n∴x = −1 or x = 11\nStep 9: Write the final answer\nx = −1 or x = 11\nNotice that both methods produce the same answer. These roots are rational because\n36 is a perfect square.\n41\nChapter 2.\nEquations and inequalities\n\nWorked example 7: Solving quadratic equations by completing the square\nQUESTION\nSolve by completing the square: 2x2 −6x −10 = 0\nSOLUTION\nStep 1: The equation is already in standard form ax2 + bx + c = 0\nStep 2: Make sure that the coefficient of the x2 term is equal to 1\nThe coefficient of the x2 term is 2. Therefore divide the entire equation by 2:\nx2 −3x −5 = 0\nStep 3: Take half the coefficient of the x term, square it; then add and subtract it\nfrom the equation\nThe coefficient of the x term is −3, so then\n\u0012−3\n2\n\u00132\n= 9\n4:\n\u0012\nx2 −3x + 9\n4\n\u0013\n−9\n4 −5 = 0\nStep 4: Write the trinomial as a perfect square\n\u0012\nx −3\n2\n\u00132\n−9\n4 −20\n4 = 0\n\u0012\nx −3\n2\n\u00132\n−29\n4 = 0\nStep 5: Method 1: Take square roots on both sides of the equation\n\u0012\nx −3\n2\n\u00132\n−29\n4 = 0\n\u0012\nx −3\n2\n\u00132\n= 29\n4\nx −3\n2 = ±\nr\n29\n4\nRemember: When taking a square root there is a positive and a negative answer.\nStep 6: Solve for x\n42\n2.2.\nCompleting the square\n\nx −3\n2 = ±\nr\n29\n4\nx = 3\n2 ±\n√\n29\n2\n= 3 ±\n√\n29\n2\nStep 7: Method 2: Factorise equation as a difference of two squares\n\u0012\nx −3\n2\n\u00132\n−29\n4 = 0\n\u0012\nx −3\n2\n\u00132\n−\n r\n29\n4\n!2\n= 0\n \nx −3\n2 −\nr\n29\n4\n! \nx −3\n2 +\nr\n29\n4\n!\n= 0\nStep 8: Solve for x\n \nx −3\n2 −\n√\n29\n2\n! \nx −3\n2 +\n√\n29\n2\n!\n= 0\nTherefore x = 3\n2 +\n√\n29\n2\nor x = 3\n2 −\n√\n29\n2\nNotice that these roots are irrational since 29 is not a perfect square.\nSee video: 227R at www.everythingmaths.co.za\nExercise 2 – 2: Solution by completing the square\n1. Solve the following equations by completing the square:\na) x2 + 10x −2 = 0\nb) x2 + 4x + 3 = 0\nc) p2 −5 = −8p\nd) 2(6x + x2) = −4\ne) x2 + 5x + 9 = 0\nf) t2 + 30 = 2(10 −It)\ng) 3x2 + 6x −2 = 0\nh) z2 + 8z −6 = 0\ni) 2z2 = 11z\nj) 5 + 4z −z2 = 0\n43\nChapter 2.\nEquations and inequalities\n\n2. Solve for k in terms of a: k2 + 6k + a = 0\n3. Solve for y in terms of p, q and r: py2 + qy + r = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 227S\n1b. 227T\n1c. 227V\n1d. 227W\n1e. 227X\n1f. 227Y\n1g. 227Z\n1h. 2282\n1i. 2283\n1j. 2284\n2. 2285\n3. 2286\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.3\nQuadratic formula\nEMBFK\nIt is not always possible to solve a quadratic equation by factorisation and it can take a\nlong time to complete the square. The method of completing the square provides a way\nto derive a formula that can be used to solve any quadratic equation. The quadratic\nformula provides an easy and fast way to solve quadratic equations.\nConsider the standard form of the quadratic equation ax2 + bx + c = 0. Divide both\nsides by a (a ̸= 0) to get\nx2 + bx\na + c\na = 0\nNow using the method of completing the square, we must halve the coefficient of x\nand square it. We then add and subtract\n\u0012 b\n2a\n\u00132\nso that the equation remains true.\nx2 + bx\na + b2\n4a2 −b2\n4a2 + c\na = 0\n\u0012\nx2 + bx\na + b2\n4a2\n\u0013\n−b2\n4a2 + c\na = 0\n\u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a2\n= 0\nWe add the constant to both sides and take the square root of both sides of the equa-\ntion, being careful to include a positive and negative answer.\n44\n2.3.\nQuadratic formula\n\n\u0012\nx + b\n2a\n\u00132\n= b2 −4ac\n4a2\ns\u0012\nx + b\n2a\n\u00132\n= ±\nr\nb2 −4ac\n4a2\nx + b\n2a = ±\n√\nb2 −4ac\n2a\nx = −b\n2a ±\n√\nb2 −4ac\n2a\nx = −b ±\n√\nb2 −4ac\n2a\nTherefore, for any quadratic equation ax2 + bx + c = 0 we can determine two roots\nx = −b +\n√\nb2 −4ac\n2a\nor x = −b −\n√\nb2 −4ac\n2a\nIt is important to notice that the expression b2 −4ac must be greater than or equal to\nzero for the roots of the quadratic to be real. If the expression under the square root\nsign is less than zero, then the roots are non-real (imaginary).\nSee video: 2287 at www.everythingmaths.co.za\nWorked example 8: Using the quadratic formula\nQUESTION\nSolve for x and leave your answer in simplest surd form: 2x2 + 3x = 7\nSOLUTION\nStep 1: Check whether the expression can be factorised\nThe expression cannot be factorised, so the general quadratic formula must be used.\nStep 2: Write the equation in the standard form ax2 + bx + c = 0\n2x2 + 3x −7 = 0\nStep 3: Identify the coefficients to substitute into the formula\na = 2;\nb = 3;\nc = −7\n45\nChapter 2.\nEquations and inequalities\n\nStep 4: Apply the quadratic formula\nAlways write down the formula first and then substitute the values of a, b and c.\nx = −b ±\n√\nb2 −4ac\n2a\n=\n−(3) ±\nq\n(3)2 −4 (2) (−7)\n2 (2)\n= −3 ±\n√\n65\n4\nStep 5: Write the final answer\nThe two roots are x = −3 +\n√\n65\n4\nor x = −3 −\n√\n65\n4\n.\nWorked example 9: Using the quadratic formula\nQUESTION\nFind the roots of the function f(x) = x2 −5x + 8.\nSOLUTION\nStep 1: Finding the roots\nTo determine the roots of f(x), we let x2 −5x + 8 = 0.\nStep 2: Check whether the expression can be factorised\nThe expression cannot be factorised, so the general quadratic formula must be used.\nStep 3: Identify the coefficients to substitute into the formula\na = 1;\nb = −5;\nc = 8\nStep 4: Apply the quadratic formula\nx = −b ±\n√\nb2 −4ac\n2a\n=\n−(−5) ±\nq\n(−5)2 −4 (1) (8)\n2 (1)\n= 5 ± √−7\n2\n46\n2.3.\nQuadratic formula\n\nStep 5: Write the final answer\nThere are no real roots for f(x) = x2 −5x + 8 since the expression under the square\nroot is negative (√−7 is not a real number). This means that the graph of the quadratic\nfunction has no x-intercepts; the entire graph lies above the x-axis.\n2\n4\n6\n8\n10\n2\n4\n6\n−2\n−4\nx\nf(x)\nf(x) = x2 −5x + 8\n0\nSee video: 2288 at www.everythingmaths.co.za\nExercise 2 – 3: Solution by the quadratic formula\nSolve the following using the quadratic formula.\n1. 3t2 + t −4 = 0\n2. x2 −5x −3 = 0\n3. 2t2 + 6t + 5 = 0\n4. 2p(2p + 1) = 2\n5. −3t2 + 5t −8 = 0\n6. 5t2 + 3t −3 = 0\n7. t2 −4t + 2 = 0\n8. 9(k2 −1) = 7k\n9. 3f −2 = −2f2\n10. t2 + t + 1 = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2289\n2. 228B\n3. 228C\n4. 228D\n5. 228F\n6. 228G\n7. 228H\n8. 228J\n9. 228K\n10. 228M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n47\nChapter 2.\nEquations and inequalities\n\n2.4\nSubstitution\nEMBFM\nIt is often useful to make a substitution for a repeated expression in a quadratic equa-\ntion. This makes the equation simpler and much easier to solve.\nWorked example 10: Solving by substitution\nQUESTION\nSolve for x: x2 −2x −\n3\nx2 −2x = 2\nSOLUTION\nStep 1: Determine the restrictions for x\nThe restrictions are the values for x that would result in the denominator being equal\nto 0, which would make the fraction undefined. Therefore x ̸= 0 and x ̸= 2.\nStep 2: Substitute a single variable for the repeated expression\nWe notice that x2 −2x is a repeated expression and we therefore let k = x2 −2x so\nthat the equation becomes\nk −3\nk = 2\nStep 3: Determine the restrictions for k\nThe restrictions are the values for k that would result in the denominator being equal\nto 0, which would make the fraction undefined. Therefore k ̸= 0.\nStep 4: Solve for k\nk −3\nk = 2\nk2 −3 = 2k\nk2 −2k −3 = 0\n(k + 1)(k −3) = 0\nTherefore k = −1 or k = 3\nWe check these two roots against the restrictions for k and confirm that both are valid.\nStep 5: Use values obtained for k to solve for the original variable x\n48\n2.4.\nSubstitution\n\nFor k = −1\nx2 −2x = −1\nx2 −2x + 1 = 0\n(x −1)(x −1) = 0\nTherefore x = 1\nFor k = 3\nx2 −2x = 3\nx2 −2x −3 = 0\n(x + 1)(x −3) = 0\nTherefore x = −1 or x = 3\nWe check these roots against the restrictions for x and confirm that all three values are\nvalid.\nStep 6: Write the final answer\nThe roots of the equation are x = −1, x = 1 and x = 3.\nExercise 2 – 4:\nSolve the following quadratic equations by substitution:\n1. −24 = 10(x2 + 5x) + (x2 + 5x)2\n2. (x2 −2x)2 −8 = 7(x2 −2x)\n3. x2 + 3x −\n56\nx(x + 3) = 26\n4. x2 −18 + x +\n72\nx2 + x = 0\n5. x2 −4x + 10 −7(4x −x2) = −2\n6.\n9\nx2 + 2x −12 = x2 + 2x −12\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 228N\n2. 228P\n3. 228Q\n4. 228R\n5. 228S\n6. 228T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n49\nChapter 2.\nEquations and inequalities\n\n2.5\nFinding the equation\nEMBFN\nWe have seen that the roots are the solutions obtained from solving a quadratic equa-\ntion. Given the roots, we are also able to work backwards to determine the original\nquadratic equation.\nWorked example 11: Finding an equation when the roots are given\nQUESTION\nFind an equation with roots 13 and −5.\nSOLUTION\nStep 1: Assign a variable and write roots as two equations\nx = 13 or x = −5\nUse additive inverses to get zero on the right-hand sides\nx −13 = 0 or x + 5 = 0\nStep 2: Write down as the product of two factors\n(x −13)(x + 5) = 0\nNotice that the signs in the brackets are opposite of the given roots.\nStep 3: Expand the brackets\nx2 −8x −65 = 0\nNote that if each term in the equation is multiplied by a constant then there could be\nother possible equations which would have the same roots. For example,\nMultiply by 2:\n2x2 −16x −130 = 0\nMultiply by −3:\n−3x2 + 24x + 195 = 0\n50\n2.5.\nFinding the equation\n\nWorked example 12: Finding an equation when the roots are fractions\nQUESTION\nFind an equation with roots −3\n2 and 4.\nSOLUTION\nStep 1: Assign a variable and write roots as two equations\nx = 4 or x = −3\n2\nUse additive inverses to get zero on the right-hand sides.\nx −4 = 0 or x + 3\n2 = 0\nMultiply the second equation through by 2 to remove the fraction.\nx −4 = 0 or 2x + 3 = 0\nStep 2: Write down as the product of two factors\n(2x + 3)(x −4) = 0\nStep 3: Expand the brackets\nThe quadratic equation is 2x2 −5x −12 = 0.\nExercise 2 – 5: Finding the equation\n1. Determine a quadratic equation for a graph that has roots 3 and −2.\n2. Find a quadratic equation for a graph that has x-intercepts of (−4; 0) and (4; 0).\n3. Determine a quadratic equation of the form ax2 + bx + c = 0, where a, b and c\nare integers, that has roots −1\n2 and 3.\n4. Determine the value of k and the other root of the quadratic equation kx2 −7x+\n4 = 0 given that one of the roots is x = 1.\n5. One root of the equation 2x2 −3x = p is 21\n2. Find p and the other root.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 228V\n2. 228W\n3. 228X\n4. 228Y\n5. 228Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n51\nChapter 2.\nEquations and inequalities\n\nExercise 2 – 6: Mixed exercises\nSolve the following quadratic equations by either factorisation, using the quadratic\nformula or completing the square:\n• Always try to factorise first, then use the formula if the trinomial cannot be fac-\ntorised.\n• In a test or examination, only use the method of completing the square when\nspecifically asked.\n• Answers can be left in surd or decimal form.\n1. 24y2 + 61y −8 = 0\n2. 8x2 + 16x = 42\n3. 9t2 = 24t −12\n4. −5y2 + 0y + 5 = 0\n5. 3m2 + 12 = 15m\n6. 49y2 + 0y −25 = 0\n7. 72 = 66w −12w2\n8. −40y2 + 58y −12 = 0\n9. 37n + 72 −24n2 = 0\n10. 6y2 + 7y −24 = 0\n11. 3 = x(2x −5)\n12. −18y2 −55y −25 = 0\n13. −25y2 + 25y −4 = 0\n14. 8(1 −4g2) + 24g = 0\n15. 9y2 −13y −10 = 0\n16. (7p −3)(5p + 1) = 0\n17. −81y2 −99y −18 = 0\n18. 14y2 −81y + 81 = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2292\n2. 2293\n3. 2294\n4. 2295\n5. 2296\n6. 2297\n7. 2298\n8. 2299\n9. 229B\n10. 229C\n11. 229D\n12. 229F\n13. 229G\n14. 229H\n15. 229J\n16. 229K\n17. 229M\n18. 229N\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.6\nNature of roots\nEMBFP\nInvestigation:\n1. Use the quadratic formula to determine the roots of the quadratic equations given\nbelow and take special note of:\n• the expression under the square root sign and\n• the type of number for the final answer (rational/irrational/real/imaginary)\na) x2 −6x + 9 = 0\n52\n2.6.\nNature of roots\n\nb) x2 −4x + 3 = 0\nc) x2 −4x −3 = 0\nd) x2 −4x + 7 = 0\n2. Choose the appropriate words from the table to describe the roots obtained for\nthe equations above.\nrational\nunequal\nreal\nimaginary\nnot perfect square\nequal\nperfect square\nirrational\nundefined\n3. The expression under the square root, b2 −4ac, is called the discriminant. Can\nyou make a conjecture about the relationship between the discriminant and the\nroots of quadratic equations?\nThe discriminant\nEMBFQ\nThe discriminant is defined as ∆= b2 −4ac\nThis is the expression under the square root in the quadratic formula. The discriminant\ndetermines the nature of the roots of a quadratic equation. The word ‘nature’ refers to\nthe types of numbers the roots can be — namely real, rational, irrational or imaginary.\n∆is the Greek symbol for the letter D.\nFor a quadratic function f (x) = ax2 + bx + c, the solutions to the equation f (x) = 0\nare given by the formula x = −b ±\n√\nb2 −4ac\n2a\n= −b ±\n√\n∆\n2a\n• If ∆< 0, then roots are imaginary (non-real) and beyond the scope of this book.\n• If ∆≥0, the expression under the square root is non-negative and therefore\nroots are real. For real roots, we have the following further possibilities.\n• If ∆= 0, the roots are equal and we can say that there is only one root.\n• If ∆> 0, the roots are unequal and there are two further possibilities.\n• ∆is the square of a rational number: the roots are rational.\n• ∆is not the square of a rational number: the roots are irrational and can be\nexpressed in decimal or surd form.\n53\nChapter 2.\nEquations and inequalities\n\n∆\n∆< 0: imaginary/non-real roots\n∆≥0: real roots\n∆= 0\nequal roots\n∆> 0\nunequal roots\n∆a squared ratio-\nnal: rational roots\n∆not a squared\nrational: irrational\nroots\nNature of roots\nDiscriminant\na > 0\na < 0\nRoots are non-real\n∆< 0\nRoots are real and\nequal\n∆= 0\nRoots are real and\nunequal:\n• rational\nroots\n• irrational\nroots\n∆> 0\n• ∆\n=\nsquared\nrational\n• ∆\n=\nnot\nsquared ra-\ntional\nSee video: 229P at www.everythingmaths.co.za\nWorked example 13: Nature of roots\nQUESTION\nShow that the roots of x2 −2x −7 = 0 are irrational.\nSOLUTION\nStep 1: Interpret the question\nFor roots to be real and irrational, we need to calculate ∆and show that it is greater\nthan zero and not a perfect square.\nStep 2: Check that the equation is in standard form ax2 + bx + c = 0\n54\n2.6.\nNature of roots\n\nx2 −2x −7 = 0\nStep 3: Identify the coefficients to substitute into the formula for the discriminant\na = 1;\nb = −2;\nc = −7\nStep 4: Write down the formula and substitute values\n∆= b2 −4ac\n= (−2)2 −4(1)(−7)\n= 4 + 28\n= 32\nWe know that 32 > 0 and is not a perfect square.\nThe graph below shows the roots of the equation x2 −2x −7 = 0. Note that the graph\ndoes not form part of the answer and is included for illustration purposes only.\n2\n4\n6\n8\n−2\n−4\n−6\n−8\n2\n4\n6\n−2\n−4\nx\nf(x)\nx2 −2x −7 = 0\n0\nStep 5: Write the final answer\nWe have calculated that ∆> 0 and is not a perfect square, therefore we can conclude\nthat the roots are real, unequal and irrational.\n55\nChapter 2.\nEquations and inequalities\n\nWorked example 14: Nature of roots\nQUESTION\nFor which value(s) of k will the roots of 6x2 + 6 = 4kx be real and equal?\nSOLUTION\nStep 1: Interpret the question\nFor roots to be real and equal, we need to solve for the value(s) of k such that ∆= 0.\nStep 2: Check that the equation is in standard form ax2 + bx + c = 0\n6x2 −4kx + 6 = 0\nStep 3: Identify the coefficients to substitute into the formula for the discriminant\na = 6;\nb = −4k;\nc = 6\nStep 4: Write down the formula and substitute values\n∆= b2 −4ac\n= (−4k)2 −4(6)(6)\n= 16k2 −144\nFor roots to be real and equal, ∆= 0.\n∆= 0\n16k2 −144 = 0\n16(k2 −9) = 0\n(k −3)(k + 3) = 0\nTherefore k = 3 or k = −3.\nStep 5: Check both answers by substituting back into the original equation\n56\n2.6.\nNature of roots\n\nFor k = 3:\n6x2 −4(3)x + 6 = 0\n6x2 −12x + 6 = 0\nx2 −2x + 1 = 0\n(x −1)(x −1) = 0\n(x −1)2 = 0\nTherefore x = 1\nWe see that for k = 3 the quadratic equation has real, equal roots x = 1.\nFor k = −3:\n6x2 −4(−3)x + 6 = 0\n6x2 + 12x + 6 = 0\nx2 + 2x + 1 = 0\n(x + 1)(x + 1) = 0\n(x + 1)2 = 0\nTherefore x = −1\nWe see that for k = −3 the quadratic equation has real, equal roots x = −1.\nStep 6: Write the final answer\nFor the roots of the quadratic equation to be real and equal, k = 3 or k = −3.\nWorked example 15: Nature of roots\nQUESTION\nShow that the roots of (x + h)(x + k) = 4d2 are real for all real values of h, k and d.\nSOLUTION\nStep 1: Interpret the question\nFor roots to be real, we need to calculate ∆and show that ∆≥0 for all real values of\nh, k and d.\nStep 2: Check that the equation is in standard form ax2 + bx + c = 0\nExpand the brackets and gather like terms\n57\nChapter 2.\nEquations and inequalities\n\n(x + h)(x + k) = 4d2\nx2 + hx + kx + hk −4d2 = 0\nx2 + (h + k)x + (hk −4d2) = 0\nStep 3: Identify the coefficients to substitute into the formula for the discriminant\na = 1;\nb = h + k;\nc = hk −4d2\nStep 4: Write down the formula and substitute values\n∆= b2 −4ac\n= (h + k)2 −4(1)(hk −4d2)\n= h2 + 2hk + k2 −4hk + 16d2\n= h2 −2hk + k2 + 16d2\n= (h −k)2 + (4d)2\nFor roots to be real, ∆≥0.\nWe know that (4d)2 ≥0\nand (h −k)2 ≥0\nso then (h −k)2 + (4d)2 ≥0\ntherefore ∆≥0\nStep 5: Write the final answer\nWe have shown that ∆≥0, therefore the roots are real for all real values of h, k and\nd.\nExercise 2 – 7: From past papers\n1. Determine the nature of the roots for each of the following equations:\na) x2 + 3x = −2\nb) x2 + 9 = 6x\nc) 6y2 −6y −1 = 0\nd) 4t2 −19t −5 = 0\ne) z2 = 3\nf) 0 = p2 + 5p + 8\ng) x2 = 36\nh) 4m + m2 = 1\ni) 11 −3x + x2 = 0\nj) y2 + 1\n4 = y\n2. Given: x2 + bx −2 + k\n\u0000x2 + 3x + 2\n\u0001\n= 0, (k ̸= −1)\na) Show that the discriminant is given by: ∆= k2 + 6bk + b2 + 8\n58\n2.6.\nNature of roots\n\nb) If b = 0, discuss the nature of the roots of the equation.\nc) If b = 2, find the value(s) of k for which the roots are equal.\n[IEB, Nov. 2001, HG]\n3. Show that k2x2 + 2 = kx −x2 has non-real roots for all real values for k.\n[IEB, Nov. 2002, HG]\n4. The equation x2 + 12x = 3kx2 + 2 has real roots.\na) Find the greatest value of value k such that k ∈Z.\nb) Find one rational value of k for which the above equation has rational roots.\n[IEB, Nov. 2003, HG]\n5. Consider the equation:\nk = x2 −4\n2x −5\nwhere x ̸= 5\n2.\na) Find a value of k for which the roots are equal.\nb) Find an integer k for which the roots of the equation will be rational and\nunequal.\n[IEB, Nov. 2004, HG]\n6.\na) Prove that the roots of the equation x2 −(a + b) x + ab −p2 = 0 are real for\nall real values of a, b and p.\nb) When will the roots of the equation be equal?\n[IEB, Nov. 2005, HG]\n7. If b and c can take on only the values 1, 2 or 3, determine all pairs (b; c) such\nthat x2 + bx + c = 0 has real roots.\n[IEB, Nov. 2005, HG]\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 229Q\n1b. 229R\n1c. 229S\n1d. 229T\n1e. 229V\n1f. 229W\n1g. 229X\n1h. 229Y\n1i. 229Z\n1j. 22B2\n2. 22B3\n3. 22B4\n4. 22B5\n5. 22B6\n6. 22B7\n7. 22B8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n59\nChapter 2.\nEquations and inequalities\n\n2.7\nQuadratic inequalities\nEMBFR\nQuadratic inequalities can be of the following forms:\nax2 + bx + c > 0\nax2 + bx + c ≥0\nax2 + bx + c < 0\nax2 + bx + c ≤0\nTo solve a quadratic inequality we must determine which part of the graph of a\nquadratic function lies above or below the x-axis. An inequality can therefore be\nsolved graphically using a graph or algebraically using a table of signs to determine\nwhere the function is positive and negative.\nWorked example 16: Solving quadratic inequalities\nQUESTION\nSolve for x: x2 −5x + 6 ≥0\nSOLUTION\nStep 1: Factorise the quadratic\n(x −3)(x −2) ≥0\nStep 2: Determine the critical values of x\nFrom the factorised quadratic we see that the values for which the inequality is equal\nto zero are x = 3 and x = 2. These are called the critical values of the inequality and\nthey are used to complete a table of signs.\nStep 3: Complete a table of signs\nWe must determine where each factor of the inequality is positive and negative on the\nnumber line:\n• to the left (in the negative direction) of the critical value\n• equal to the critical value\n• to the right (in the positive direction) of the critical value\nIn the final row of the table we determine where the inequality is positive and negative\nby finding the product of the factors and their respective signs.\n60\n2.7.\nQuadratic inequalities\n\nCritical values\nx = 2\nx = 3\nx −3\n−\n−\n−\n0\n+\nx −2\n−\n0\n+\n+\n+\nf(x) = (x −3)(x −2)\n+\n0\n−\n0\n+\nFrom the table we see that f(x) is greater than or equal to zero for x ≤2 or x ≥3.\nStep 4: A rough sketch of the graph\nThe graph below does not form part of the answer and is included for illustration pur-\nposes only. A graph of the quadratic helps us determine the answer to the inequality.\nWe can find the answer graphically by seeing where the graph lies above or below the\nx-axis.\n• From the standard form, x2 −5x + 6, a > 0 and therefore the graph is a “smile”\nand has a minimum turning point.\n• From the factorised form, (x −3)(x −2), we know the x-intercepts are (2; 0) and\n(3; 0).\n1\n2\n3\n4\n5\n1\n2\n3\n4\n5\nx\ny\ny = x2 −5x + 6\nThe graph is above or on the x-axis for x ≤2 or x ≥3.\nStep 5: Write the final answer and represent on a number line\nx2 −5x + 6 ≥0 for x ≤2 or x ≥3\n1\n2\n3\n4\nb\nb\n61\nChapter 2.\nEquations and inequalities\n\nWorked example 17: Solving quadratic inequalities\nQUESTION\nSolve for x: 4x2 −4x + 1 ≤0\nSOLUTION\nStep 1: Factorise the quadratic\n(2x −1)(2x −1) ≤0\n(2x −1)2 ≤0\nStep 2: Determine the critical values of x\nFrom the factorised quadratic we see that the value for which the inequality is equal to\nzero is x = 1\n2. We know that a2 > 0 for any real number a, a ̸= 0, so then (2x −1)2\nwill never be negative.\nStep 3: A rough sketch of the graph\nThe graph below does not form part of the answer and is included for illustration\npurposes only.\n• From the standard form, 4x2 −4x + 1, a > 0 and therefore the graph is a “smile”\nand has a minimum turning point.\n• From the factorised form, (2x−1)(2x−1), we know there is only one x-intercept\nat\n\u0000 1\n2; 0\n\u0001\n.\n1\n2\n3\n4\n1\n2\nx\ny\ny = 4x2 −4x + 1\n0\nNotice that no part of the graph lies below the x-axis.\nStep 4: Write the final answer and represent on a number line\n4x2 −4x + 1 ≤0 for x = 1\n2\n−2\n−1\n0\n1\n2\nb\n62\n2.7.\nQuadratic inequalities\n\nWorked example 18: Solving quadratic inequalities\nQUESTION\nSolve for x: −x2 −3x + 5 > 0\nSOLUTION\nStep 1: Examine the form of the inequality\nNotice that the coefficient of the x2 term is −1. Remember that if we multiply or\ndivide an inequality by a negative number, then the inequality sign changes direction.\nSo we can write the same inequality in different ways and still get the same answer, as\nshown below.\n−x2 −3x + 5 > 0\nMultiply by −1 and change direction of the inequality sign\nx2 + 3x −5 < 0\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n−1\n−2\n−3\n−4\nx\ny\n−x2 −3x + 5 > 0\n0\n1\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n−1\n−2\n−3\n−4\nx\ny\nx2 + 3x −5 < 0\n0\nFrom this rough sketch, we can see that both inequalities give the same solution; the\nvalues of x that lie between the two x-intercepts.\nStep 2: Factorise the quadratic\nWe notice that −x2−3x+5 > 0 cannot be easily factorised. So we let −x2−3x+5 = 0\nand use the quadratic formula to determine the roots of the equation.\n−x2 −3x + 5 = 0\nx2 + 3x −5 = 0\n63\nChapter 2.\nEquations and inequalities\n\n∴x =\n−3 ±\nq\n(3)2 −4 (1) (−5)\n2 (1)\n= −3 ±\n√\n29\n2\nx1 = −3 −\n√\n29\n2\n≈−4,2\nx2 = −3 +\n√\n29\n2\n≈1,2\nTherefore we can write, correct to one decimal place,\nx2 + 3x −5 < 0\nas (x −1,2)(x + 4,2) < 0\nStep 3: Determine the critical values of x\nFrom the factorised quadratic we see that the critical values are x = 1,2 and x =\n−4,2.\nStep 4: Complete a table of signs\nCritical values\nx = −4,2\nx = 1,2\nx + 4,2\n−\n0\n+\n+\n+\nx −1,2\n−\n−\n−\n0\n+\nf(x) = (x + 4,2)(x −1,2)\n+\n0\n−\n0\n+\nFrom the table we see that the function is negative for −4,2 < x < 1,2.\nStep 5: A sketch of the graph\n• From the standard form, x2 + 3x −5, a > 0 and therefore the graph is a “smile”\nand has a minimum turning point.\n• From the factorised form, (x −1,2)(x + 4,2), we know the x-intercepts are\n(−4,2; 0) and (1,2; 0).\n1\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n−1\n−2\n−3\n−4\nx\ny\ny = x2 + 3x −5\n0\nFrom the graph we see that the function lies below the x-axis between −4,2 and 1,2.\n64\n2.7.\nQuadratic inequalities\n\nStep 6: Write the final answer and represent on a number line\nx2 + 3x −5 < 0 for −4,2 < x < 1,2\n−5\n−4\n−3\n−2\n−1\n0\n1\n2\n−4,2\n1,2\nImportant: When working with an inequality in which the variable is in the denomi-\nnator, a different approach is needed. Always remember to check for restrictions.\nWorked example 19: Solving quadratic inequalities with fractions\nQUESTION\nSolve for x:\n1.\n2\nx + 3 =\n1\nx −3, x ̸= ±3\n2.\n2\nx + 3 ≤\n1\nx −3, x ̸= ±3\nSOLUTION\nStep 1: Solving the equation\nTo solve this equation we multiply both sides of the equation by (x + 3)(x −3) and\nsimplfy:\n2\nx + 3 × (x + 3)(x −3) =\n1\nx −3 × (x + 3)(x −3)\n2(x −3) = x + 3\n2x −6 = x + 3\nx = 9\nStep 2: Solving the inequality\nIt is very important to recognise that we cannot use the same method as above to solve\nthe inequality. If we multiply or divide an inequality by a negative number, then the\ninequality sign changes direction. We must rather simplify the inequality to have a\nlowest common denominator and use a table of signs to determine the values that\nsatisfy the inequality.\n65\nChapter 2.\nEquations and inequalities\n\nStep 3: Subtract\n1\nx −3 from both sides of the inequality\n2\nx + 3 −\n1\nx −3 ≤0\nStep 4: Determine the lowest common denominator and simplify the fraction\n2(x −3) −(x + 3)\n(x + 3)(x −3)\n≤0\nx −9\n(x + 3)(x −3) ≤0\nKeep the denominator because it affects the final answer.\nStep 5: Determine the critical values of x\nFrom the factorised inequality we see that the critical values are x = −3, x = 3 and\nx = 9.\nStep 6: Complete a table of signs\nCritical values\nx = −3\nx = 3\nx = 9\nx + 3\n−\nundef\n+\n+\n+\n+\n+\nx −3\n−\n−\n−\nundef\n+\n+\n+\nx −9\n−\n−\n−\n−\n−\n0\n+\nf(x) =\nx −9\n(x + 3)(x −3)\n−\nundef\n+\nundef\n−\n0\n+\nFrom the table we see that the function is less than or equal to zero for x < −3 or\n3 < x ≤9. We do not include x = −3 or x = 3 in the solution because of the\nrestrictions on the denominator.\nStep 7: Write the final answer and represent on a number line\nx < −3\nor\n3 < x ≤9\n−3\n−2\n−1\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\nb\n66\n2.7.\nQuadratic inequalities\n\nExercise 2 – 8: Solving quadratic inequalities\n1. Solve the following inequalities and show each answer on a number line:\na) x2 −x < 12\nb) 3x2 > −x + 4\nc) y2 < −y −2\nd) (3 −t)(1 + t) > 0\ne) s2 −4s > −6\nf) 0 ≥7x2 −x + 8\ng) x ≥−4x2\nh) 2x2 + x + 6 ≤0\ni)\nx\nx −3 < 2, x ̸= 3\nj) x2 + 4\nx −7 ≥0, x ̸= 7\nk) x + 2\nx\n−1 ≥0, x ̸= 0\n2. Draw a sketch of the following inequalities and solve for x:\na) 2x2 −18 > 0\nb) 5 −x2 ≤0\nc) x2 < 0\nd) 0 ≥6x2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22B9\n1b. 22BB\n1c. 22BC\n1d. 22BD\n1e. 22BF\n1f. 22BG\n1g. 22BH\n1h. 22BJ\n1i. 22BK\n1j. 22BM\n1k. 22BN\n2a. 22BP\n2b. 22BQ\n2c. 22BR\n2d. 22BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.8\nSimultaneous equations\nEMBFS\nSimultaneous linear equations can be solved using three different methods: substi-\ntution, elimination or using a graph to determine where the two lines intersect. For\nsolving systems of simultaneous equations with linear and non-linear equations, we\nmostly use the substitution method. Graphical solution is useful for showing where\nthe two equations intersect.\nIn general, to solve for the values of n unknown variables requires a system of n\nindependent equations.\nAn example of a system of simultaneous equations with one linear equation and one\nquadratic equation is\ny −2x = −4\nx2 + y = 4\nSolving by substitution\nEMBFT\n• Use the simplest of the two given equations to express one of the variables in\nterms of the other.\n• Substitute into the second equation. By doing this we reduce the number of\nequations and the number of variables by one.\n67\nChapter 2.\nEquations and inequalities\n\n• We now have one equation with one unknown variable which can be solved.\n• Use the solution to substitute back into the first equation to find the value of the\nother unknown variable.\nWorked example 20: Simultaneous equations\nQUESTION\nSolve for x and y:\ny −2x = −4\n. . . (1)\nx2 + y = 4\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of the first equation\ny = 2x −4\nStep 2: Substitute into the second equation and simplify\nx2 + (2x −4) = 4\nx2 + 2x −8 = 0\nStep 3: Factorise the equation\n(x + 4) (x −2) = 0\n∴x = −4 or x = 2\nStep 4: Substitute the values of x back into the first equation to determine the corre-\nsponding y-values\nIf x = −4:\ny = 2(−4) −4\n= −12\nIf x = 2:\ny = 2(2) −4\n= 0\nStep 5: Check that the two points satisfy both original equations\n68\n2.8.\nSimultaneous equations\n\nStep 6: Write the final answer\nThe solution is x = −4 and y = −12 or x = 2 and y = 0. These are the coordinate\npairs for the points of intersection as shown below.\n2\n4\n6\n−2\n−4\n−6\n−8\n−10\n−12\n−14\n2\n4\n6\n−2\n−4\n−6\ny = 2x −4\ny = 4\n−x2\n(−4; −12)\nx\ny\nb\nb\nSolving by elimination\nEMBFV\n• Make one of the variables the subject of both equations.\n• Equate the two equations; by doing this we reduce the number of equations and\nthe number of variables by one.\n• We now have one equation with one unknown variable which can be solved.\n• Use the solution to substitute back into either original equation, to find the cor-\nresponding value of the other unknown variable.\nWorked example 21: Simultaneous equations\nQUESTION\nSolve for x and y:\ny = x2 −6x\n. . . (1)\ny + 1\n2x −3 = 0\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of the second equation\ny + 1\n2x −3 = 0\ny = −1\n2x + 3\n69\nChapter 2.\nEquations and inequalities\n\nStep 2: Equate the two equations and solve for x\nx2 −6x = −1\n2x + 3\nx2 −6x + 1\n2x −3 = 0\n2x2 −12x + x −6 = 0\n2x2 −11x −6 = 0\n(2x + 1)(x −6) = 0\nTherefore x = −1\n2 or x = 6\nStep 3: Substitute the values for x back into the second equation to calculate the\ncorresponding y-values\nIf x = −1\n2:\ny = −1\n2\n\u0012\n−1\n2\n\u0013\n+ 3\n∴y = 31\n4\nThis gives the point\n\u0012\n−1\n2; 31\n4\n\u0013\n.\nIf x = 6:\ny = −1\n2(6) + 3\n= −3 + 3\n∴y = 0\nThis gives the point (6; 0).\nStep 4: Check that the two points satisfy both original equations\nStep 5: Write the final answer\nThe solution is x = −1\n2 and y = 31\n4 or x = 6 and y = 0. These are the coordinate\npairs for the points of intersection as shown below.\n2\n4\n−2\n−4\n−6\n−8\n1\n2\n3\n4\n5\n6\n7\n−1\nx\ny\ny = x2 −6x\ny = 1\n2x + 3\n\u0000−1\n2; 3 1\n4\n\u0001\n70\n2.8.\nSimultaneous equations\n\nWorked example 22: Simultaneous equations\nQUESTION\nSolve for x and y:\ny =\n5\nx −2\n. . . (1)\ny + 1 = 2x\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of the second equation\ny + 1 = 2x\ny = 2x −1\nStep 2: Equate the two equations and solve for x\n2x −1 =\n5\nx −2\n(2x −1)(x −2) = 5\n2x2 −5x + 2 = 5\n2x2 −5x −3 = 0\n(2x + 1)(x −3) = 0\nTherefore x = −1\n2 or x = 3\nStep 3: Substitute the values for x back into the second equation to calculate the\ncorresponding y-values\nIf x −1\n2:\ny = 2(−1\n2) −1\n∴y = −2\nThis gives the point (−1\n2; −2).\nIf x = 3:\ny = 2(3) −1\n= 5\nThis gives the point (3; 5).\nStep 4: Check that the two points satisfy both original equations\n71\nChapter 2.\nEquations and inequalities\n\nStep 5: Write the final answer\nThe solution is x = −1\n2 and y = −2 or x = 3 and y = 5. These are the coordinate\npairs for the points of intersection as shown below.\n2\n4\n−2\n−4\n1\n2\n3\n4\n5\n−1\n−2\nx\ny\ny = 2x −1\ny =\n5\nx−2\nSolving graphically\nEMBFW\n• Make y the subject of each equation.\n• Draw the graph of each equation on the same system of axes.\n• The final solutions to the system of equations are the coordinates of the points\nwhere the two graphs intersect.\nWorked example 23: Simultaneous equations\nQUESTION\nSolve graphically for x and y:\ny + x2 = 1\n. . . (1)\ny −x + 5 = 0\n. . . (2)\nSOLUTION\nStep 1: Make y the subject of both equations\nFor the first equation we have\ny + x2 = 1\ny = −x2 + 1\nand for the second equation\n72\n2.8.\nSimultaneous equations\n\ny −x + 5 = 0\ny = x −5\nStep 2: Draw the straight line graph and parabola on the same system of axes\n2\n−2\n−4\n−6\n−8\n−10\n2\n4\n−2\n−4\nx\ny\ny = −x2 + 1\ny = x −5\n0\nb\nb\nStep 3: Determine where the two graphs intersect\nFrom the diagram we see that the graphs intersect at (−3; −8) and (2; −3).\nStep 4: Check that the two points satisfy both original equations\nStep 5: Write the final answer\nThe solutions to the system of simultaneous equations are (−3; −8) and (2; −3).\nExercise 2 – 9: Solving simultaneous equations\n1. Solve the following systems of equations algebraically. Leave your answer in\nsurd form, where appropriate.\na) y + x = 5\ny −x2 + 3x −5 = 0\nb) y = 6 −5x + x2\ny −x + 1 = 0\nc) y = 2x + 2\n4\ny −2x2 + 3x + 5 = 0\nd) a −2b −3 = 0; a −3b2 + 4 = 0\ne) x2 −y + 2 = 3x\n4x = 8 + y\nf) 2y + x2 + 25 = 7x\n3x = 6y + 96\n73\nChapter 2.\nEquations and inequalities\n\n2. Solve the following systems of equations graphically. Check your solutions by\nalso solving algebraically.\na) x2 −1 −y = 0\ny + x −5 = 0\nb) x + y −10 = 0\nx2 −2 −y = 0\nc) xy = 12\n7 = x + y\nd) 6 −4x −y = 0\n12 −2x2 −y = 0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22BT\n1b. 22BV\n1c. 22BW\n1d. 22BX\n1e. 22BY\n1f. 22BZ\n2a. 22C2\n2b. 22C3\n2c. 22C4\n2d. 22C5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.9\nWord problems\nEMBFX\nSolving word problems requires using mathematical language to describe real-life con-\ntexts. Problem-solving strategies are often used in the natural sciences and engineering\ndisciplines (such as physics, biology, and electrical engineering) but also in the social\nsciences (such as economics, sociology and political science). To solve word problems\nwe need to write a set of equations that describes the problem mathematically.\nExamples of real-world problem solving applications are:\n• modelling population growth;\n• modelling effects of air pollution;\n• modelling effects of global warming;\n• computer games;\n• in the sciences, to understand how the natural world works;\n• simulators that are used to train people in certain jobs, such as pilots, doctors\nand soldiers;\n• in medicine, to track the progress of a disease.\nSee video: 22C6 at www.everythingmaths.co.za\n74\n2.9.\nWord problems\n\nProblem solving strategy\nEMBFY\n1. Read the problem carefully.\n2. What is the question and what do we need to solve for?\n3. Assign variables to the unknown quantities, for example, x and y.\n4. Translate the words into algebraic expressions by rewriting the given information\nin terms of the variables.\n5. Set up a system of equations.\n6. Solve for the variables using substitution.\n7. Check the solution.\n8. Write the final answer.\nInvestigation: Simple word problems\nWrite an equation that describes the following real-world situations mathematically:\n1. Mohato and Lindiwe both have colds. Mohato sneezes twice for each sneeze of\nLindiwe’s. If Lindiwe sneezes x times, write an equation describing how many\ntimes they both sneezed.\n2. The difference of two numbers is 10 and the sum of their squares is 50. Find the\ntwo numbers.\n3. Liboko builds a rectangular storeroom. If the diagonal of the room is\n√\n1312 m\nand the perimeter is 80 m, determine the dimensions of the room.\n4. It rains half as much in July as it does in December. If it rains y mm in July, write\nan expression relating the rainfall in July and December.\n5. Zane can paint a room in 4 hours. Tlali can paint a room in 2 hours. How long\nwill it take both of them to paint a room together?\n6. 25 years ago, Arthur was 5 years more than a third of Bongani’s age. Today,\nBongani is 26 years less than twice Arthur’s age. How old is Bongani?\n7. The product of two integers is 95. Find the integers if their total is 24.\n75\nChapter 2.\nEquations and inequalities\n\nWorked example 24: Gym membership\nQUESTION\nThe annual gym subscription for a single member is R 1000, while an annual family\nmembership is R 1500. The gym is considering increasing all membership fees by\nthe same amount. If this is done then a single membership would cost 5\n7 of a family\nmembership. Determine the amount of the proposed increase.\nSOLUTION\nStep 1: Identify the unknown quantity and assign a variable\nLet the amount of the proposed increase be x.\nStep 2: Use the given information to complete a table\nnow\nafter increase\nsingle\n1000\n1000 + x\nfamily\n1500\n1500 + x\nStep 3: Set up an equation\n1000 + x = 5\n7(1500 + x)\nStep 4: Solve for x\n7000 + 7x = 7500 + 5x\n2x = 500\nx = 250\nStep 5: Write the final answer\nThe proposed increase is R 250.\nWorked example 25: Corner coffee house\nQUESTION\nErica has decided to treat her friends to coffee at the Corner Coffee House. Erica\npaid R 54,00 for four cups of cappuccino and three cups of filter coffee. If a cup of\ncappuccino costs R 3,00 more than a cup of filter coffee, calculate how much a cup of\neach type of coffee costs?\nSOLUTION\nStep 1: Method 1: identify the unknown quantities and assign two variables\nLet the cost of a cappuccino be x and the cost of a filter coffee be y.\n76\n2.9.\nWord problems\n\nStep 2: Use the given information to set up a system of equations\n4x + 3y = 54\n. . . (1)\nx = y + 3\n. . . (2)\nStep 3: Solve the equations by substituting the second equation into the first equation\n4(y + 3) + 3y = 54\n4y + 12 + 3y = 54\n7y = 42\ny = 6\nIf y = 6, then using the second equation we have\nx = y + 3\n= 6 + 3\n= 9\nStep 4: Check that the solution satisfies both original equations\nStep 5: Write the final answer\nA cup of cappuccino costs R 9 and a cup of filter coffee costs R 6.\nStep 6: Method 2: identify the unknown quantities and assign one variable\nLet the cost of a cappuccino be x and the cost of a filter coffee be x −3.\nStep 7: Use the given information to set up an equation\n4x + 3(x −3) = 54\nStep 8: Solve for x\n4x + 3(x −3) = 54\n4x + 3x −9 = 54\n7x = 63\nx = 9\nStep 9: Write the final answer\nA cup of cappuccino costs R 9 and a cup of filter coffee costs R 6.\n77\nChapter 2.\nEquations and inequalities\n\nWorked example 26: Taps filling a container\nQUESTION\nTwo taps, one more powerful than the other, are used to fill a container. Working\non its own, the less powerful tap takes 2 hours longer than the other tap to fill the\ncontainer. If both taps are opened, it takes 1 hour, 52 minutes and 30 seconds to fill\nthe container. Determine how long it takes the less powerful tap to fill the container\non its own.\nSOLUTION\nStep 1: Identify the unknown quantities and assign variables\nLet the time taken for the less powerful tap to fill the container be x and let the time\ntaken for the more powerful tap be x −2.\nStep 2: Convert all units of time to be the same\nFirst we must convert 1 hour, 52 minutes and 30 seconds to hours:\n1 + 52\n60 +\n30\n(60)2 = 1,875 hours\nStep 3: Use the given information to set up a system of equations\nWrite an equation describing the two taps working together to fill the container:\n1\nx +\n1\nx −2 =\n1\n1,875\nStep 4: Multiply the equation through by the lowest common denominator and sim-\nplify\n1,875(x −2) + 1,875x = x(x −2)\n1,875x −3,75 + 1,875x = x2 −2x\n0 = x2 −5,75x + 3,75\nMultiply the equation through by 4 to make it easier to factorise (or use the quadratic\nformula)\n0 = 4x2 −23x + 15\n0 = (4x −3)(x −5)\nTherefore x = 3\n4 or x = 5.\nWe have calculated that the less powerful tap takes 3\n4 hours or 5 hours to fill the\ncontainer, but we know that when both taps are opened it takes 1,875 hours. We can\ntherefore discard the first solution x = 3\n4 hours.\n78\n2.9.\nWord problems\n\nSo the less powerful tap fills the container in 5 hours and the more powerful tap takes\n3 hours.\nStep 5: Check that the solution satisfies the original equation\nStep 6: Write the final answer\nThe less powerful tap fills the container in 5 hours and the more powerful tap takes 3\nhours.\nExercise 2 – 10:\n1. Mr. Tsilatsila builds a fence around his rectangular vegetable garden of 8 m2.\nIf the length is twice the breadth, determine the dimensions of Mr. Tsilatsila’s\nvegetable garden.\n2. Kevin has played a few games of ten-pin bowling. In the third game, Kevin\nscored 80 more than in the second game. In the first game Kevin scored 110 less\nthan the third game. His total score for the first two games was 208. If he wants\nan average score of 146, what must he score on the fourth game?\n3. When an object is dropped or thrown downward, the distance, d, that it falls in\ntime, t, is described by the following equation:\ns = 5t2 + v0t\nIn this equation, v0 is the initial velocity, in m·s−1. Distance is measured in\nmeters and time is measured in seconds. Use the equation to find how long it\ntakes a tennis ball to reach the ground if it is thrown downward from a hot-air\nballoon that is 500 m high. The tennis ball is thrown at an initial velocity of\n5 m·s−1.\n4. The table below lists the times that Sheila takes to walk the given distances.\ntime (minutes)\n5\n10\n15\n20\n25\n30\ndistance (km)\n1\n2\n3\n4\n5\n6\nPlot the points.\nFind the equation that describes the relationship between time and distance.\nThen use the equation to answer the following questions:\na) How long will it take Sheila to walk 21 km?\nb) How far will Sheila walk in 7 minutes?\nIf Sheila were to walk half as fast as she is currently walking, what would the\ngraph of her distances and times look like?\n5. The power P (in watts) supplied to a circuit by a 12 volt battery is given by the\nformula P = 12I −0,5I2 where I is the current in amperes.\na) Since both power and current must be greater than 0, find the limits of the\ncurrent that can be drawn by the circuit.\n79\nChapter 2.\nEquations and inequalities\n\nb) Draw a graph of P = 12I −0,5I2 and use your answer to the first question\nto define the extent of the graph.\nc) What is the maximum current that can be drawn?\nd) From your graph, read off how much power is supplied to the circuit when\nthe current is 10 A. Use the equation to confirm your answer.\ne) At what value of current will the power supplied be a maximum?\n6. A wooden block is made as shown in the diagram. The ends are right-angled\ntriangles having sides 3x, 4x and 5x. The length of the block is y. The total\nsurface area of the block is 3600 cm2.\ny\n3x\n4x\n5x\nShow that\ny = 300−x2\nx\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22C7\n2. 22C8\n3. 22C9\n4. 22CB\n5. 22CC\n6. 22CD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n2.10\nSummary\nEMBFZ\nSee presentation: 22CF at www.everythingmaths.co.za\n• Zero product law: if a × b = 0, then a = 0 and/or b = 0.\n• Quadratic formula: x = −b ±\n√\nb2 −4ac\n2a\n• Discriminant: ∆= b2 −4ac\n80\n2.10.\nSummary\n\nNature of roots\nDiscriminant\nRoots are non-real\n∆< 0\nRoots are real and equal\n∆= 0\nRoots are real and unequal:\n– Rational roots\n– Irrational roots\n∆> 0\n– ∆= squared rational number\n– ∆= not squared rational num-\nber\nExercise 2 – 11: End of chapter exercises\n1. Solve: x2 −x −1 = 0. Give your answer correct to two decimal places.\n2. Solve: 16 (x + 1) = x2 (x + 1)\n3. Solve: y2 + 3 +\n12\ny2 + 3 = 7\n4. Solve for x: 2x4 −5x2 −12 = 0\n5. Solve for x:\na) x (x −9) + 14 = 0\nb) x2 −x = 3 (correct to one decimal place)\nc) x + 2 = 6\nx (correct to two decimal places)\nd)\n1\nx + 1 +\n2x\nx −1 = 1\n6. Solve for x in terms of p by completing the square: x2 −px −4 = 0\n7. The equation ax2 + bx + c = 0 has roots x = 2\n3 and x = −4. Find one set of\npossible values for a, b and c.\n8. The two roots of the equation 4x2 + px −9 = 0 differ by 5. Calculate the value\nof p.\n9. An equation of the form x2 +bx+c = 0 is written on the board. Saskia and Sven\ncopy it down incorrectly. Saskia has a mistake in the constant term and obtains\nthe solutions −4 and 2. Sven has a mistake in the coefficient of x and obtains\nthe solutions 1 and −15. Determine the correct equation that was on the board.\n10. For which values of b will the expression b2 −5b + 6\nb + 2\nbe:\na) undefined?\nb) equal to zero?\n11. Given (x2 −6)(2x + 1)\nx + 2\n= 0 solve for x if:\na) x is a real number.\nb) x is a rational number.\nc) x is an irrational number.\nd) x is an integer.\n81\nChapter 2.\nEquations and inequalities\n\n12. Given (x −6)\n1\n2\nx2 + 3 , for which value(s) of x will the expression be:\na) equal to zero?\nb) defined?\n13. Solve for a if\n√8 −2a\na −3\n≥0.\n14. Abdoul stumbled across the following formula to solve the quadratic equation\nax2 + bx + c = 0 in a foreign textbook.\nx =\n2c\n−b ±\n√\nb2 −4ac\na) Use this formula to solve the equation: 2x2 + x −3 = 0.\nb) Solve the equation again, using factorisation, to see if the formula works for\nthis equation.\nc) Trying to derive this formula to prove that it always works, Abdoul got stuck\nalong the way. His attempt is shown below:\nax2 + bx + c = 0\na + b\nx + c\nx2 = 0\nDivided by x2 where x ̸= 0\nc\nx2 + b\nx + a = 0\nRearranged\n1\nx2 + b\ncx + a\nc = 0\nDivided by c where c ̸= 0\n1\nx2 + b\ncx = −a\nc\nSubtracted a\nc from both sides\n∴1\nx2 + b\ncx + . . .\nGot stuck\nComplete his derivation.\n15. Solve for x:\na)\n4\nx −3 ≤1\nb)\n4\n(x −3)2 < 1\nc) 2x −2\nx −3 > 3\nd)\n−3\n(x −3) (x + 1) < 0\ne) (2x −3)2 < 4\nf) 2x ≤15 −x\nx\ng) x2 + 3\n3x −2 ≤0\nh) x −2 ≥3\nx\ni) x2 + 3x −4\n5 + x4\n≤0\nj) x −2\n3 −x ≥1\n82\n2.10.\nSummary\n\n16. Solve the following systems of equations algebraically. Leave your answer in\nsurd form, where appropriate.\na) y −2x = 0\ny −x2 −2x + 3 = 0\nb) a −3b = 0\na −b2 + 4 = 0\nc) y −x2 −5x = 0\n10 = y −2x\nd) p = 2p2 + q −3\np −3q = 1\ne) a −b2 = 0\na −3b + 1 = 0\nf) a −2b + 1 = 0\na −2b2 −12b + 4 = 0\ng) y + 4x −19 = 0\n8y + 5x2 −101 = 0\nh) a + 4b −18 = 0\n2a + 5b2 −57 = 0\n17. Solve the following systems of equations graphically:\na) 2y + x −2 = 0\n8y + x2 −8 = 0\nb) y + 3x −6 = 0\ny = x2 + 4 −4x\n18. A stone is thrown vertically upwards and its height (in metres) above the ground\nat time t (in seconds) is given by:\nh (t) = 35 −5t2 + 30t\nFind its initial height above the ground.\n19. After doing some research, a transport company has determined that the rate at\nwhich petrol is consumed by one of its large carriers, travelling at an average\nspeed of x km per hour, is given by:\nP(x) = 55\n2x + x\n200\nlitres per kilometre\nAssume that the petrol costs R 4,00 per litre and the driver earns R 18,00 per\nhour of travel time. Now deduce that the total cost, C, in Rands, for a 2000 km\ntrip is given by:\nC(x) = 256 000\nx\n+ 40x\n20. Solve the following quadratic equations by either factorisation, completing the\nsquare or by using the quadratic formula:\n• Always try to factorise first, then use the formula if the trinomial cannot be\nfactorised.\n• Solve some of the equations by completing the square.\na) −4y2 −41y −45 = 0\nb) 16x2 + 20x = 36\nc) 42p2 + 104p + 64 = 0\nd) 21y + 3 = 54y2\ne) 36y2 + 44y + 8 = 0\nf) 12y2 −14 = 22y\ng) 16y2 + 0y −81 = 0\nh) 3y2 + 10y −48 = 0\ni) 63 −5y2 = 26y\nj) 2x2 −30 = 2\nk) 2y2 = 98\n83\nChapter 2.\nEquations and inequalities\n\n21. One root of the equation 9y2 + 32 = ky is 8. Determine the value of k and the\nother root.\n22.\na) Solve for x in x2 −x = 6.\nb) Hence, solve for y in (y2 −y)2 −(y2 −y) −6 = 0.\n23. Solve for x: x = √8 −x + 2\n24.\na) Solve for y in −4y2 + 8y −3 = 0.\nb) Hence, solve for p in 4(p −3)2 −8(p −3) + 3 = 0.\n25. Solve for x: 2(x + 3)\n1\n2 = 9\n26.\na) Without solving the equation x + 1\nx = 3, determine the value of x2 + 1\nx2 .\nb) Now solve x + 1\nx = 3 and use the result to assess the answer obtained in\nthe question above.\n27. Solve for y: 5(y −1)2 −5 = 19 −(y −1)2\n28. Solve for t: 2t(t −3\n2) =\n3\n2t2 −3t + 2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22CG\n2. 22CH\n3. 22CJ\n4. 22CK\n5a. 22CM\n5b. 22CN\n5c. 22CP\n5d. 22CQ\n6. 22CR\n7. 22CS\n8. 22CT\n9. 22CV\n10. 22CW\n11. 22CX\n12. 22CY\n13. 22CZ\n14. 22D2\n15a. 22D3\n15b. 22D4\n15c. 22D5\n15d. 22D6\n15e. 22D7\n15f. 22D8\n15g. 22D9\n15h. 22DB\n15i. 22DC\n15j. 22DD\n16a. 22DF\n16b. 22DG\n16c. 22DH\n16d. 22DJ\n16e. 22DK\n16f. 22DM\n16g. 22DN\n16h. 22DP\n17a. 22DQ\n17b. 22DR\n18. 22DS\n19. 22DT\n20a. 22DV\n20b. 22DW\n20c. 22DX\n20d. 22DY\n20e. 22DZ\n20f. 22F2\n20g. 22F3\n20h. 22F4\n20i. 22F5\n20j. 22F6\n20k. 22F7\n21. 22F8\n22. 22F9\n23. 22FB\n24. 22FC\n25. 22FD\n26. 22FF\n27. 22FG\n28. 22FH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n84\n2.10.\nSummary\n\nCHAPTER\n3\nNumber patterns", "chapter_id": "7" }, { "title": "Revision", "content": "3.1\nRevision\n86\n3.2\nQuadratic sequences\n90\n3.3\nSummary\n99\n\n3\nNumber patterns\nIn earlier grades we learned about linear sequences, where the difference between\nconsecutive terms is constant. In this chapter, we will learn about quadratic sequences,\nwhere the difference between consecutive terms is not constant, but follows its own\npattern.\n3.1\nRevision\nEMBG2\nTerminology:\nSequence/pattern\nA sequence or pattern is an ordered set of numbers\nor variables.\nSuccessive/consecutive\nSuccessive or consecutive terms are terms that di-\nrectly follow one after another in a sequence.\nCommon difference\nThe common or constant difference (d) is the differ-\nence between any two consecutive terms in a linear\nsequence.\nGeneral term\nA mathematical expression that describes the se-\nquence and that generates any term in the pattern\nby substituting different values for n.\nConjecture\nA statement, consistent with known data, that has\nnot been proved true nor shown to be false.\nImportant: a series is not the same as a sequence or pattern. Different types of series\nare studied in Grade 12. In Grade 11 we study sequences only.\nSee video: 22FJ at www.everythingmaths.co.za\nDescribing patterns\nEMBG3\nTo describe terms in a pattern we use the following notation:\n• T1 is the first term of a sequence.\n• T4 is the fourth term of a sequence.\n• Tn is the general term and is often expressed as the nth term of a sequence.\nA sequence does not have to follow a pattern but when it does, we can write an\nequation for the general term. The general term can be used to calculate any term in\nthe sequence. For example, consider the following linear sequence: 1; 4; 7; 10; 13; . . .\nThe nth term is given by the equation Tn = 3n −2.\nYou can check this by substituting values for n:\nT1 = 3(1) −2 = 1\nT2 = 3(2) −2 = 4\nT3 = 3(3) −2 = 7\nT4 = 3(4) −2 = 10\nT5 = 3(5) −2 = 13\n86\n3.1.\nRevision\n\nIf we find the relationship between the position of a term and its value, we can describe\nthe pattern and find any term in the sequence.\nSee video: 22FK at www.everythingmaths.co.za\nLinear sequences\nEMBG4\nDEFINITION: Linear sequence\nA sequence of numbers in which there is a common difference (d) between any term\nand the term before it is called a linear sequence.\nImportant: d = T2 −T1, not T1 −T2.\nWorked example 1: Linear sequence\nQUESTION\nDetermine the common difference (d) and the general term for the following sequence:\n10; 7; 4; 1; . . .\nSOLUTION\nStep 1: Determine the common difference\nTo calculate the common difference, we find the difference between any term and the\nprevious term:\nd = Tn −Tn−1\nTherefore d = T2 −T1\n= 7 −10\n= −3\nor d = T3 −T2\n= 4 −7\n= −3\nor d = T4 −T3\n= 1 −4\n= −3\n10\n7\n4\n1\n−3\n−3\n−3\nStep 2: Determine the general term\nTo find the general term Tn, we must identify the relationship between:\n87\nChapter 3.\nNumber patterns\n\n• the value of a number in the pattern and\n• the position of a number in the pattern\nposition\n1\n2\n3\n4\nvalue\n10\n7\n4\n1\nWe start with the value of the first term in the sequence. We need to write an expres-\nsion that includes the value of the common difference (d = −3) and the position of\nthe term (n = 1).\nT1 = 10\n= 10 + (0)(−3)\n= 10 + (1 −1)(−3)\nNow we write a similar expression for the second term.\nT2 = 7\n= 10 + (1)(−3)\n= 10 + (2 −1)(−3)\nWe notice a pattern forming that links the position of a number in the sequence to its\nvalue.\nTn = 10 + (n −1)(−3)\n= 10 −3n + 3\n= −3n + 13\nStep 3: Drawing a graph of the pattern\nWe can also represent this pattern graphically, as shown below.\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n0\n1\n2\n3\n4\n5\n6\nb\nb\nb\nb\nPattern number (n)\nTerm value Tn\nNotice that the position numbers (n) can be positive integers only.\nThis pattern can also be expressed in words: “each term in the sequence can be cal-\nculated by multiplying negative three and the position number, and then adding thir-\nteen.”\n88\n3.1.\nRevision\n\nSee video: 22FM at www.everythingmaths.co.za\nExercise 3 – 1: Linear sequences\n1. Write down the next three terms in each of the following sequences:\n45; 29; 13; −3; . . .\n2. The general term is given for each sequence below. Calculate the missing terms.\na) −4; −9; −14; . . . ; −24\nTn = 1 −5n\nb) 6; . . . ; 24; . . . ; 42\nTn = 9n −3\n3. Find the general formula for the following sequences and then find T10, T15 and\nT30:\na) 13; 16; 19; 22; . . .\nb) 18; 24; 30; 36; . . .\nc) −10; −15; −20; −25; . . .\n4. The seating in a classroom is arranged so that the first row has 20 desks, the\nsecond row has 22 desks, the third row has 24 desks and so on. Calculate how\nmany desks are in the ninth row.\n5.\na) Complete the following:\n13 + 31 = . . .\n24 + 42 = . . .\n38 + 83 = . . .\nb) Look at the numbers on the left-hand side, what do you notice about the\nunit digit and the tens-digit?\nc) Investigate the pattern by trying other examples of 2-digit numbers.\nd) Make a conjecture about the pattern that you notice.\ne) Prove this conjecture.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22FN\n2a. 22FP\n2b. 22FQ\n3a. 22FR\n3b. 22FS\n3c. 22FT\n4. 22FV\n5. 22FW\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n89\nChapter 3.\nNumber patterns\n\n3.2\nQuadratic sequences\nEMBG5\nInvestigation: Quadratic sequences\nb\nb\nb\nb\nb\nb\nb\nb\nb\n1. Study the dotted-tile pattern shown and answer the following questions.\na) Complete the fourth pattern in the diagram.\nb) Complete the table below:\npattern number\n1\n2\n3\n4\n5\n20\nn\ndotted tiles\n1\n3\n5\ndifference (d)\n−\n2\nc) What do you notice about the change in number of dotted tiles?\nd) Describe the pattern in words: “The number of dotted tiles...”.\ne) Write the general term: Tn = . . .\nf) Give the mathematical name for this kind of pattern.\ng) A pattern has 819 dotted tiles. Determine the value of n.\n2. Now study the number of blank tiles (tiles without dots) and answer the following\nquestions:\na) Complete the table below:\npattern number\n1\n2\n3\n4\n5\n10\nblank tiles\n3\n6\n11\nfirst difference\n−\n3\nsecond difference\n−\n−\nb) What do you notice about the change in the number of blank tiles?\nc) Describe the pattern in words: “The number of blank tiles...”.\nd) Write the general term: Tn = . . .\ne) Give the mathematical name for this kind of pattern.\nf) A pattern has 227 blank tiles. Determine the value of n.\ng) A pattern has 79 dotted tiles. Determine the number of blank tiles.\n90\n3.2.\nQuadratic sequences\n\nDEFINITION: Quadratic sequence\nA quadratic sequence is a sequence of numbers in which the second difference be-\ntween any two consecutive terms is constant.\nConsider the following example: 1; 2; 4; 7; 11; . . .\nThe first difference is calculated by finding the difference between consecutive terms:\n1\n2\n4\n7\n11\n+1\n+2\n+3\n+4\nThe second difference is obtained by taking the difference between consecutive first\ndifferences:\n1\n2\n3\n4\n+1\n+1\n+1\nWe notice that the second differences are all equal to 1. Any sequence that has a\ncommon second difference is a quadratic sequence.\nIt is important to note that the first differences of a quadratic sequence form a sequence.\nThis sequence has a constant difference between consecutive terms. In other words, a\nlinear sequence results from taking the first differences of a quadratic sequence.\nGeneral case\nIf the sequence is quadratic, the nth term is of the form Tn = an2 + bn + c.\nn = 1\nn = 2\nn = 3\nn = 4\nTn\na + b + c\n4a + 2b + c\n9a + 3b + c\n16a + 4b + c\n1st difference\n3a + b\n5a + b\n7a + b\n2nd difference\n2a\n2a\nIn each case, the common second difference is a 2a.\nExercise 3 – 2: Quadratic sequences\n1. Determine the second difference between the terms for the following se-\nquences:\na) 5; 20; 45; 80; . . .\nb) 6; 11; 18; 27; . . .\nc) 1; 4; 9; 16; . . .\nd) 3; 0; −5; −12; . . .\ne) 1; 3; 7; 13; . . .\nf) 0; −6; −16; −30; . . .\ng) −1; 2; 9; 20; . . .\nh) 1; −3; −9; −17; . . .\ni) 3a+1; 12a+1; 27a+1; 48a+1 . . .\nj) 2; 10; 24; 44; . . .\nk) t −2; 4t −1; 9t; 16t + 1; . . .\n91\nChapter 3.\nNumber patterns\n\n2. Complete the sequence by filling in the missing term:\na) 11; 21; 35; . . . ; 75\nb) 20; . . . ; 42; 56; 72\nc) . . . ; 37; 65; 101\nd) 3; . . . ; −13; −27; −45\ne) 24; 35; 48; . . . ; 80\nf) . . . ; 11; 26; 47\n3. Use the general term to generate the first four terms in each sequence:\na) Tn = n2 + 3n −1\nb) Tn = −n2 −5\nc) Tn = 3n2 −2n\nd) Tn = −2n2 + n + 1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22FX\n1b. 22FY\n1c. 22FZ\n1d. 22G2\n1e. 22G3\n1f. 22G4\n1g. 22G5\n1h. 22G6\n1i. 22G7\n1j. 22G8\n1k. 22G9\n2a. 22GB\n2b. 22GC\n2c. 22GD\n2d. 22GF\n2e. 22GG\n2f. 22GH\n3a. 22GJ\n3b. 22GK\n3c. 22GM\n3d. 22GN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nWorked example 2: Quadratic sequences\nQUESTION\nWrite down the next two terms and determine an equation for the nth term of the\nsequence 5; 12; 23; 38; . . .\nSOLUTION\nStep 1: Find the first differences between the terms\n5\n12\n23\n38\n+7\n+11\n+15\nStep 2: Find the second differences between the terms\n7\n11\n15\n+4\n+4\nSo there is a common second difference of 4. We can therefore conclude that this is a\nquadratic sequence of the form Tn = an2 + bn + c.\nContinuing the sequence, the next first differences will be:\n...15\n19\n23...\n+4\n+4\n92\n3.2.\nQuadratic sequences\n\nStep 3: Finding the next two terms in the sequence\nThe next two terms will be:\n...38\n57\n80...\n+19\n+23\nSo the sequence will be: 5; 12; 23; 38; 57; 80; . . .\nStep 4: Determine the general term for the sequence\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve a set of simultaneous equations to determine the values of a, b and c\nWe know that T1 = 5, T2 = 12 and T3 = 23\na + b + c = 5\n4a + 2b + c = 12\n9a + 3b + c = 23\nT2 −T1 = 4a + 2b + c −(a + b + c)\n12 −5 = 4a + 2b + c −a −b −c\n7 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n23 −12 = 9a + 3b + c −4a −2b −c\n11 = 5a + b\n. . . (2)\n(2) −(1) = 5a + b −(3a + b)\n11 −7 = 5a + b −3a −b\n4 = 2a\n∴a = 2\nUsing equation (1) :\n3(2) + b = 7\n∴b = 1\nAnd using\na + b + c = 5\n2 + 1 + c = 5\n∴c = 1\nStep 5: Write the general term for the sequence\nTn = 2n2 + n + 2\n93\nChapter 3.\nNumber patterns\n\nWorked example 3: Plotting a graph of terms in a sequence\nQUESTION\nConsider the following sequence:\n3; 6; 10; 15; 21; . . .\n1. Determine the general term (Tn) for the sequence.\n2. Is this a linear or a quadratic sequence?\n3. Plot a graph of Tn vs n.\nSOLUTION\nStep 1: Determine the first and second differences\nn = 1\nn = 2\nn = 3\nn = 4\nTn\n3\n6\n10\n15\n1st difference\n3\n4\n5\n2nd difference\n1\n1\nWe see that the first differences are not constant and form the sequence 3; 4; 5; . . . and\nthat there is a common second difference of 1. Therefore the sequence is quadratic\nand has a general term of the form Tn = an2 + bn + c.\nStep 2: Determine the general term Tn\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve this set of simultaneous equations to determine the values of a, b and c. We\nknow that T1 = 3, T2 = 6 and T3 = 10.\na + b + c = 3\n4a + 2b + c = 6\n9a + 3b + c = 10\nT2 −T1 = 4a + 2b + c −(a + b + c)\n6 −3 = 4a + 2b + c −a −b −c\n3 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n10 −6 = 9a + 3b + c −4a −2b −c\n4 = 5a + b\n. . . (2)\n94\n3.2.\nQuadratic sequences\n\n(2) −(1) = 5a + b −(3a + b)\n4 −3 = 5a + b −3a −b\n1 = 2a\n∴a = 1\n2\nUsing equation (1) :\n3\n\u00121\n2\n\u0013\n+ b = 3\n∴b = 3\n2\nAnd using\na + b + c = 3\n1\n2 + 3\n2 + c = 3\n∴c = 1\nTherefore the general term for the sequence is Tn = 1\n2n2 + 3\n2n + 1.\nStep 3: Plot a graph of Tn vs n\nUse the general term for the sequence, Tn = 1\n2n2 + 3\n2n + 1, to complete the table.\nn\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nTn\n3\n6\n10\n15\n21\n28\n36\n45\n55\n66\nUse the table to plot the graph:\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nTerm value (Tn)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nPosition number (n)\nT1\nT2\nT3\nT4\nT5\nT6\nT7\nT8\nT9\nT10\nIn this case it would not be accurate to join these points, since n indicates the position\nof a term in a sequence and can therefore only be a positive integer. We can, however,\nsee that the plot of the points lies in the shape of a parabola.\n95\nChapter 3.\nNumber patterns\n\nWorked example 4: Olympic Games soccer event\nQUESTION\nIn the first stage of the soccer event at the Olympic Games, there are teams from\nfour different countries in each group. Each country in a group must play every other\ncountry in the group once.\n1. How many matches will be played in each group in the first stage of the event?\n2. How many matches would be played if there are 5 teams in each group?\n3. How many matches would be played if there are 6 teams in each group?\n4. Determine the general formula of the sequence.\nSOLUTION\nStep 1: Determine the number of matches played if there are 4 teams in a group\nLet the teams from four different countries be A, B, C and D.\nteams in a group\nmatches played\nA\nAB, AC, AD\nB\nBC, BD\nC\nCD\nD\n4\n3 + 2 + 1 = 6\nAB means that team A plays team B and BA would be the same match as AB. So if\nthere are four different teams in a group, each group plays 6 matches.\nStep 2: Determine the number of matches played if there are 5 teams in a group\nLet the teams from five different countries be A, B, C, D and E.\nteams in a group\nmatches played\nA\nAB, AC, AD, AE\nB\nBC, BD, BE\nC\nCD, CE\nD\nDE\nE\n5\n4 + 3 + 2 + 1 = 10\nSo if there are five different teams in a group, each group plays 10 matches.\nStep 3: Determine the number of matches played if there are 6 teams in a group\nLet the teams from six different countries be A, B, C, D, E and F.\n96\n3.2.\nQuadratic sequences\n\nteams in a group\nmatches to be played\nA\nAB, AC, AD, AE, AF\nB\nBC, BD, BE, BF\nC\nCD, CE, CF\nD\nDE, DF\nE\nEF\nF\n5\n5 + 4 + 3 + 2 + 1 = 15\nSo if there are six different teams in a group, each group plays 15 matches.\nWe continue to increase the number of teams in a group and find that a group of 7\nteams plays 21 matches and a group of 8 teams plays 28 matches.\nStep 4: Consider the sequence\nWe examine the sequence to determine if it is linear or quadratic:\nn = 1\nn = 2\nn = 3\nn = 4\nn = 5\nTn\n6\n10\n15\n21\n28 . . .\nfirst difference\n4\n5\n6\n7\nsecond difference\n1\n1\n1\nWe see that the first differences are not constant and that there is a common second\ndifference of 1. Therefore the sequence is quadratic and has a general term of the form\nTn = an2 + bn + c.\nStep 5: Determine the general term Tn\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve a set of simultaneous equations to determine the values of a, b and c. We\nknow that T1 = 6, T2 = 10 and T3 = 15\na + b + c = 6\n4a + 2b + c = 10\n9a + 3b + c = 15\nT2 −T1 = 4a + 2b + c −(a + b + c)\n10 −6 = 4a + 2b + c −a −b −c\n4 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n15 −10 = 9a + 3b + c −4a −2b −c\n5 = 5a + b\n. . . (2)\n97\nChapter 3.\nNumber patterns\n\n(2) −(1) = 5a + b −(3a + b)\n5 −4 = 5a + b −3a −b\n1 = 2a\n∴a = 1\n2\nUsing equation (1) :\n3\n\u00121\n2\n\u0013\n+ b = 4\n∴b = 5\n2\nAnd using a + b + c = 6\n1\n2 + 5\n2 + c = 6\n∴c = 3\nTherefore the general term for the sequence is Tn = 1\n2n2 + 5\n2n + 3.\nExercise 3 – 3: Quadratic sequences\n1. Calculate the common second difference for each of the following quadratic\nsequences:\na) 3; 6; 10; 15; 21; ...\nb) 4; 9; 16; 25; 36; ...\nc) 7; 17; 31; 49; 71; ...\nd) 2; 10; 26; 50; 82; ...\ne) 31; 30; 27; 22; 15; ...\n2. Find the first five terms of the quadratic sequence defined by: Tn = 5n2 +3n+4.\n3. Given Tn = 4n2 + 5n + 10, find T9.\n4. Given Tn = 2n2, for which value of n does Tn = 32?\n5.\na) Write down the next two terms of the quadratic sequence: 16; 27; 42; 61; . . .\nb) Find the general formula for the quadratic sequence above.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22GP\n1b. 22GQ\n1c. 22GR\n1d. 22GS\n1e. 22GT\n2. 22GV\n3. 22GW\n4. 22GX\n5. 22GY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n98\n3.2.\nQuadratic sequences\n\n3.3\nSummary\nEMBG6\nSee presentation: 22GZ at www.everythingmaths.co.za\n• Tn is the general term of a sequence.\n• Successive or consecutive terms are terms that follow one after another in a\nsequence.\n• A linear sequence has a common difference (d) between any two successive\nterms.\nd = Tn −Tn−1\n• A quadratic sequence has a common second difference between any two suc-\ncessive terms.\n• The general term for a quadratic sequence is\nTn = an2 + bn + c\n• A general quadratic sequence:\nn = 1\nn = 2\nn = 3\nn = 4\nTn\na + b + c\n4a + 2b + c\n9a + 3b + c\n16a + 4b + c\n1st difference\n3a + b\n5a + b\n7a + b\n2nd difference\n2a\n2a\nExercise 3 – 4: End of chapter exercises\n1. Find the first five terms of the quadratic sequence defined by:\nTn = n2 + 2n + 1\n2. Determine whether each of the following sequences is:\n• a linear sequence,\n• a quadratic sequence,\n• or neither.\na) 6; 9; 14; 21; 30; ...\nb) 1; 7; 17; 31; 49; ...\nc) 8; 17; 32; 53; 80; ...\nd) 9; 26; 51; 84; 125; ...\ne) 2; 20; 50; 92; 146; ...\nf) 5; 19; 41; 71; 109; ...\ng) 2; 6; 10; 14; 18; ...\nh) 3; 9; 15; 21; 27; ...\ni) 1; 2,5; 5; 8,5; 13; ...\nj) 10; 24; 44; 70; 102; ...\nk) 21\n2; 6; 101\n2; 16; 221\n2; . . .\nl) 3p2; 6p2; 9p2; 12p2; 15p2; . . .\nm) 2k; 8k; 18k; 32k; 50k; . . .\n99\nChapter 3.\nNumber patterns\n\n3. Given the pattern: 16; x; 46; . . ., determine the value of x if the pattern is linear.\n4. Given Tn = 2n2, for which value of n does Tn = 242?\n5. Given Tn = 3n2, find T11.\n6. Given Tn = n2 + 4, for which value of n does Tn = 85?\n7. Given Tn = 4n2 + 3n −1, find T5.\n8. Given Tn = 3\n2n2, for which value of n does Tn = 96?\n9. For each of the following patterns, determine:\n• the next term in the pattern,\n• and the general term,\n• the tenth term in the pattern.\na) 3; 7; 11; 15; . . .\nb) 17; 12; 7; 2; . . .\nc)\n1\n2; 1; 11\n2; 2; . . .\nd) a; a + b; a + 2b; a + 3b; . . .\ne) 1; −1; −3; −5; . . .\n10. For each of the following sequences, find the equation for the general term and\nthen use the equation to find T100.\na) 4; 7; 12; 19; 28; ...\nb) 2; 8; 14; 20; 26; ...\nc) 7; 13; 23; 37; 55; ...\nd) 5; 14; 29; 50; 77; ...\nGiven: Tn = 3n −1\n11.\na) Write down the first five terms of the sequence.\nb) What do you notice about the difference between any two consecutive\nterms?\nc) Will this always be the case for a linear sequence?\nGiven the following sequence: −15; −11; −7; . . . ; 173\n12.\na) Determine the equation for the general term.\nb) Calculate how many terms there are in the sequence.\n13. Given 3; 7; 13; 21; 31; . . .\na) Thabang determines that the general term is Tn = 4n −1. Is he correct?\nExplain.\nb) Cristina determines that the general term is Tn = n2 + n + 1. Is she correct?\nExplain.\n100\n3.3.\nSummary\n\n14. Given the following pattern of blocks:\n2\n3\n4\na) Draw pattern 5.\nb) Complete the table below:\npattern number (n)\n2\n3\n4\n5\n10\n250\nn\nnumber of white blocks (w)\n4\n8\nc) Is this a linear or a quadratic sequence?\n15. Cubes of volume 1 cm3 are stacked on top of each other to form a tower:\n1\n2\n3\na) Complete the table for the height of the tower:\ntower number (n)\n1\n2\n3\n4\n10\nn\nheight of tower (h)\n2\nb) What type of sequence is this?\nc) Now consider the number of cubes in each tower and complete the table\nbelow:\ntower number (n)\n1\n2\n3\n4\nnumber of cubes (c)\n3\nd) What type of sequence is this?\ne) Determine the general term for this sequence.\nf) How many cubes are needed for tower number 21?\ng) How high will a tower of 496 cubes be?\n16. A quadratic sequence has a second term equal to 1, a third term equal to −6 and\na fourth term equal to −14.\na) Determine the second difference for this sequence.\nb) Hence, or otherwise, calculate the first term of the pattern.\n101\nChapter 3.\nNumber patterns\n\n17. There are 15 schools competing in the U16 girls hockey championship and every\nteam must play two matches — one home match and one away match.\na) Use the given information to complete the table:\nno. of schools\nno. of matches\n1\n0\n2\n3\n4\n5\nb) Calculate the second difference.\nc) Determine a general term for the sequence.\nd) How many matches will be played if there are 15 schools competing in the\nchampionship?\ne) If 600 matches must be played, how many schools are competing in the\nchampionship?\n18. The first term of a quadratic sequence is 4, the third term is 34 and the common\nsecond difference is 10. Determine the first six terms in the sequence.\n19. Challenge question:\nGiven that the general term for a quadratic sequences is Tn = an2 + bn + c, let\nd be the first difference and D be the second common difference.\na) Show that a = D\n2 .\nb) Show that b = d −3\n2D.\nc) Show that c = T1 −d + D.\nd) Hence, show that Tn = D\n2 n2 +\n\u0012\nd −3\n2D\n\u0013\nn + (T1 −d + D).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22H2\n2a. 22H3\n2b. 22H4\n2c. 22H5\n2d. 22H6\n2e. 22H7\n2f. 22H8\n2g. 22H9\n2h. 22HB\n2i. 22HC\n2j. 22HD\n2k. 22HF\n2l. 22HG\n2m. 22HH\n3. 22HJ\n4. 22HK\n5. 22HM\n6. 22HN\n7. 22HP\n8. 22HQ\n9a. 22HR\n9b. 22HS\n9c. 22HT\n9d. 22HV\n9e. 22HW\n10a. 22HX\n10b. 22HY\n10c. 22HZ\n10d. 22J2\n11. 22J3\n12. 22J4\n13. 22J5\n14. 22J6\n15. 22J7\n16. 22J8\n17. 22J9\n18. 22JB\n19. 22JC\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n102\n3.3.\nSummary\n\nCHAPTER\n4\nAnalytical geometry\n4.1\nRevision\n104\n4.2\nEquation of a line\n113\n4.3\nInclination of a line\n124\n4.4\nParallel lines\n132\n4.5\nPerpendicular lines\n136\n4.6\nSummary\n142\n\n4\nAnalytical geometry\nAnalytical geometry, also referred to as coordinate or Cartesian geometry, is the study\nof geometric properties and relationships between points, lines and angles in the Carte-\nsian plane. Geometrical shapes are defined using a coordinate system and algebraic\nprinciples. In this chapter we deal with the equation of a straight line, parallel and\nperpendicular lines and inclination of a line.\n4.1\nRevision\nEMBG7\nPoints A(x1; y1), B(x2; y2) and C(x2; y1) are shown in the diagram below:\nb\nb\nA(x1; y1)\nC(x2; y1)\nB(x2; y2)\nx\ny\n0\nTheorem of Pythagoras\nAB2 = AC2 + BC2\nDistance formula\nDistance between two points:\nAB =\np\n(x2 −x1)2 + (y2 −y1)2\nNotice that (x1 −x2)2 = (x2 −x1)2.\nSee video: 22JD at www.everythingmaths.co.za\nGradient\nGradient (m) describes the slope or steepness of the line joining two points. The\ngradient of a line is determined by the ratio of vertical change to horizontal change.\nmAB = y2 −y1\nx2 −x1\nor\nmAB = y1 −y2\nx1 −x2\nRemember to be consistent: m ̸= y1 −y2\nx2 −x1\n.\n104\n4.1.\nRevision\n\nHorizontal lines\nx\ny\n0\nm = 0\nVertical lines\nx\ny\n0\nm is undefined\nParallel lines\nθ\nθ\nx\ny\n0\nm1 = m2\nPerpendicular lines\nθ2\nθ1\nx\ny\n0\nm1 × m2 = −1\nMid-point of a line segment\nA(x1; y1)\nM(x; y)\nB(x2; y2)\nx\ny\n0\nThe coordinates of the mid-point M(x; y) of a line between any two points A(x1; y1)\nand B(x2; y2):\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nSee video: 22JF at www.everythingmaths.co.za\n105\nChapter 4.\nAnalytical geometry\n\nPoints on a straight line\nThe diagram shows points P(x1; y1), Q(x2; y2) and R(x; y) on a straight line.\nb\nb\nx\ny\nR(x; y)\n0\nb\nQ(x2; y2)\nP(x1; y1)\nWe know that mPR = mQR = mPQ.\nUsing mPR = mPQ, we obtain the following for any point (x; y) on a straight line\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 1: Revision\nQUESTION\nGiven the points P(−5; −4) and Q(0; 6):\n1. Determine the length of the line segment PQ.\n2. Determine the mid-point T(x; y) of the line segment PQ.\n3. Show that the line passing through R(1; −3\n4) and T(x; y) is perpendicular to the\nline PQ.\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n−2\n−4\n−6\n2\n−2\n−4\n−6\nb\nb\nb\nP(−5; −4)\nT(x; y)\nQ(0; 6)\nx\ny\n0\n106\n4.1.\nRevision\n\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q (x2; y2)\nx1 = −5;\ny1 = −4;\nx2 = 0;\ny2 = 6\nWrite down the distance formula\nPQ =\np\n(x2 −x1)2 + (y2 −y1)2\n=\np\n(0 −(−5))2 + (6 −(−4))2\n=\n√\n25 + 100\n=\n√\n125\n= 5\n√\n5\nThe length of the line segment PQ is 5\n√\n5 units.\nStep 3: Write down the mid-point formula and substitute the values\nT(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nx = x1 + x2\n2\n= −5 + 0\n2\n= −5\n2\ny = y1 + y2\n2\n= −4 + 6\n2\n= 2\n2\n= 1\nThe mid-point of PQ is T(−5\n2; 1).\nStep 4: Determine the gradients of PQ and RT\nm = y2 −y1\nx2 −x1\nmPQ = 6 −(−4)\n0 −(−5)\n= 10\n5\n= 2\n107\nChapter 4.\nAnalytical geometry\n\nmRT = −3\n4 −1\n1 −(−5\n2)\n= −7\n4\n7\n2\n= −7\n4 × 2\n7\n= −1\n2\nCalculate the product of the two gradients:\nmRT × mPQ = −1\n2 × 2\n= −1\nTherefore PQ is perpendicular to RT.\nQuadrilaterals\n• A quadrilateral is a closed shape consisting of four straight line segments.\n• A parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n– Both pairs of opposite sides are equal in length.\n– Both pairs of opposite angles are equal.\n– The diagonals bisect each other.\n• A rectangle is a parallelogram that has all four angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other.\n– The diagonals are equal in length.\n108\n4.1.\nRevision\n\n• A rhombus is a parallelogram that has all four sides equal in length.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n×\n×\n××\n•\n•\n•\n•\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals of a rhombus bisect both pairs of opposite angles.\n• A square is a rhombus that has all four interior angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n•\n•\n••\n••\n••\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals are equal in length.\n– The diagonals bisect both pairs of interior opposite angles (that is, all angles\nare 45◦).\n• A trapezium is a quadrilateral with one pair of opposite sides parallel.\n• A kite is a quadrilateral with two pairs of adjacent sides equal.\nA\nB\nC\nD\nb\nb\n××\n– One pair of opposite angles are equal (the angles are between unequal\nsides).\n– The diagonal between equal sides bisects the other diagonal.\n– The diagonal between equal sides bisects the interior angles.\n– The diagonals intersect at 90◦.\n109\nChapter 4.\nAnalytical geometry\n\nWorked example 2: Quadrilaterals\nQUESTION\nPoints A (−1; 0), B (0; 3), C (8; 11) and D (x; y) are points on the Cartesian plane.\nDetermine D (x; y) if ABCD is a parallelogram.\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n−1\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\nb\nb\nx\ny\nC(8; 11)\n0\nD(x; y)\nA(−1; 0)\nb\nB(0; 3)\nM\nThe mid-point of AC will be the same as the mid-point of BD. We first find the\nmid-point of AC and then use it to determine the coordinates of point D.\nStep 2: Assign values to (x1; y1) and (x2; y2)\nLet the mid-point of AC be M(x; y)\nx1 = −1;\ny1 = 0;\nx2 = 8;\ny2 = 11\nStep 3: Write down the mid-point formula\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nStep 4: Substitute the values and calculate the coordinates of M\nM(x; y) =\n\u0012−1 + 8\n2\n; 0 + 11\n2\n\u0013\n=\n\u00127\n2; 11\n2\n\u0013\n110\n4.1.\nRevision\n\nStep 5: Use the coordinates of M to determine D\nM is also the mid-point of BD so we use M\n\u0000 7\n2; 11\n2\n\u0001\nand B (0; 3) to find D (x; y)\nStep 6: Substitute values and determine x and y\nM =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n∴\n\u00127\n2; 11\n2\n\u0013\n=\n\u00120 + x\n2\n; 3 + y\n2\n\u0013\n7\n2 = 0 + x\n2\n7 = 0 + x\n∴x = 7\n11\n2 = 3 + y\n2\n11 = 3 + y\n∴y = 8\nStep 7: Alternative method: inspection\nSince we are given that ABCD is a parallelogram, we can use the properties of a\nparallelogram and the given points to determine the coordinates of D.\nFrom the sketch we expect that point D will lie below C.\nConsider the given points A, B and C:\n• Opposite sides of a parallelogram are parallel, therefore BC must be parallel to\nAD and their gradients must be equal.\n• The vertical change from B to C is 8 units up.\n• Therefore the vertical change from A to D is also 8 units up (y = 0 + 8 = 8).\n• The horizontal change from B to C is 8 units to the right.\n• Therefore the horizontal change from A to D is also 8 units to the right (x =\n−1 + 8 = 7).\nor\n• Opposite sides of a parallelogram are parallel, therefore AB must be parallel to\nDC and their gradients must be equal.\n• The vertical change from A to B is 3 units up.\n111\nChapter 4.\nAnalytical geometry\n\n• Therefore the vertical change from C to D is 3 units down (y = 11 −3 = 8).\n• The horizontal change from A to B is 1 unit to the right.\n• Therefore the horizontal change from C to D is 1 unit to the left (x = 8 −1 = 7).\nStep 8: Write the final answer\nThe coordinates of D are (7; 8).\nExercise 4 – 1: Revision\n1. Determine the length of the line segment between the following points:\na) P(−3; 5) and Q(−1; −5)\nb) R(0,75; 3) and S(0,75; −4)\nc) T(2x; y −2) and U(3x + 1; y −2)\n2. Given Q(4; 1), T(p; 3) and length QT =\n√\n8 units, determine the value of p.\n3. Determine the gradient of the line AB if:\na) A(−5; 3) and B(−7; 4)\nb) A(3; −2) and B(1; −8)\n4. Prove that the line PQ, with P(0; 3) and Q(5; 5), is parallel to the line 5y + 5 =\n2x.\n5. Given the points A(−1; −1), B(2; 5), C(−1; −5\n2) and D(x; −4) and AB ⊥CD,\ndetermine the value of x.\n6. Calculate the coordinates of the mid-point P(x; y) of the line segment between\nthe points:\na) M(3; 5) and N(−1; −1)\nb) A(−3; −4) and B(2; 3)\n7. The line joining A(−2; 4) and B(x; y) has the mid-point C(1; 3). Determine the\nvalues of x and y.\n8. Given\nquadrilateral\nABCD\nwith\nvertices\nA(0; 3), B(4; 3), C(5; −1)\nand\nD(1; −1).\na) Determine the equation of the line AD and the line BC.\nb) Show that AD ∥BC.\nc) Calculate the lengths of AD and BC.\nd) Determine the equation of the diagonal BD.\ne) What type of quadrilateral is ABCD?\n112\n4.1.\nRevision\n\n9. MPQN is a parallelogram with points M(−5; 3), P(−1; 5) and Q(4; 5). Draw a\nsketch and determine the coordinates of N(x; y).\n10. PQRS is a quadrilateral with points P(−3; 1), Q(1; 3), R(6; 1) and S(2; −1) in\nthe Cartesian plane.\na) Determine the lengths of PQ and SR.\nb) Determine the mid-point of PR.\nc) Show that PQ ∥SR.\nd) Determine the equations of the line PS and the line SR.\ne) Is PS ⊥SR? Explain your answer.\nf) What type of quadrilateral is PQRS?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22JG\n1b. 22JH\n1c. 22JJ\n2. 22JK\n3a. 22JM\n3b. 22JN\n4. 22JP\n5. 22JQ\n6a. 22JR\n6b. 22JS\n7. 22JT\n8. 22JV\n9. 22JW\n10. 22JX\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.2\nEquation of a line\nEMBG8\nWe can derive different forms of the straight line equation. The different forms are\nused depending on the information provided in the problem:\n• The two-point form of the straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• The gradient–point form of the straight line equation: y −y1 = m(x −x1)\n• The gradient–intercept form of the straight line equation: y = mx + c\nThe two-point form of the straight line equation\nEMBG9\nb\nb\n(x1; y1)\n(x2; y2)\nx\ny\n0\n113\nChapter 4.\nAnalytical geometry\n\nGiven any two points (x1; y1) and (x2; y2), we can determine the equation of the line\npassing through the two points using the equation:\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 3: The two-point form of the straight line equation\nQUESTION\nFind the equation of the straight line passing through P (−1; −5) and Q (5; 4).\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nP(−1; −5)\nQ(5; 4)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q(x2; y2)\nx1 = −1;\ny1 = −5;\nx2 = 5;\ny2 = 4\nStep 3: Write down the two-point form of the straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\n114\n4.2.\nEquation of a line\n\nStep 4: Substitute the values and make y the subject of the equation\ny −(−5)\nx −(−1) = 4 −(−5)\n5 −(−1)\ny + 5\nx + 1 = 9\n6\ny + 5 = 3\n2(x + 1)\ny + 5 = 3\n2x + 3\n2\ny = 3\n2x −7\n2\nStep 5: Write the final answer\ny = 3\n2x −31\n2\nExercise 4 – 2: The two-point form of the straight line equation\nDetermine the equation of the straight line passing through the points:\n1. (3; 7) and (−6; 1)\n2. (1; −11\n4 ) and (2\n3; −7\n4)\n3. (−2; 1) and (3; 6)\n4. (2; 3) and (3; 5)\n5. (1; −5) and (−7; −5)\n6. (−4; 0) and (1; 15\n4 )\n7. (s; t) and (t; s)\n8. (−2; −8) and (1; 7)\n9. (2p; q) and (0; −q)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22JY\n2. 22JZ\n3. 22K2\n4. 22K3\n5. 22K4\n6. 22K5\n7. 22K6\n8. 22K7\n9. 22K8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n115\nChapter 4.\nAnalytical geometry\n\nThe gradient–point form of the straight line equation\nEMBGB\nWe derive the gradient–point form of the straight line equation using the definition of\ngradient and the two-point form of a straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nSubstitute m = y2 −y1\nx2 −x1\non the right-hand side of the equation\ny −y1\nx −x1\n= m\nMultiply both sides of the equation by (x −x1)\ny −y1 = m(x −x1)\nTo use this equation, we need to know the gradient of the line and the coordinates of\none point on the line.\nSee video: 22K9 at www.everythingmaths.co.za\nWorked example 4: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −1\n3 and passing through\nthe point (−1; 1).\nSOLUTION\nStep 1: Draw a sketch\nWe notice that m < 0, therefore the graph decreases as x increases.\n1\n2\n3\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\n(−1; 1)\nx\ny\n0\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\n116\n4.2.\nEquation of a line\n\nSubstitute the value of the gradient\ny −y1 = −1\n3(x −x1)\nSubstitute the coordinates of the given point\ny −1 = −1\n3(x −(−1))\ny −1 = −1\n3(x + 1)\ny = −1\n3x −1\n3 + 1\n= −1\n3x + 2\n3\nStep 3: Write the final answer\nThe equation of the straight line is y = −1\n3x + 2\n3.\nIf we are given two points on a straight line, we can also use the gradient–point form\nto determine the equation of a straight line. We first calculate the gradient using the\ntwo given points and then substitute either of the two points into the gradient–point\nform of the equation.\nWorked example 5: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line passing through (−3; 2) and (5; 8).\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n8\n2\n4\n6\n−2\n−4\nb\nb\n(−3; 2)\n(5; 8)\nx\ny\n0\n117\nChapter 4.\nAnalytical geometry\n\nStep 2: Assign variables to the coordinates of the given points\nx1 = −3;\ny1 = 2;\nx2 = 5;\ny2 = 8\nStep 3: Calculate the gradient using the two given points\nm = y2 −y1\nx2 −x1\n=\n8 −2\n5 −(−3)\n= 6\n8\n= 3\n4\nStep 4: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the value of the gradient\ny −y1 = 3\n4(x −x1)\nSubstitute the coordinates of a given point\ny −y1 = 3\n4(x −x1)\ny −2 = 3\n4(x −(−3))\ny −2 = 3\n4(x + 3)\ny = 3\n4x + 9\n4 + 2\n= 3\n4x + 17\n4\nStep 5: Write the final answer\nThe equation of the straight line is y = 3\n4x + 41\n4.\nSee video: 22KB at www.everythingmaths.co.za\n118\n4.2.\nEquation of a line\n\nExercise 4 – 3: Gradient–point form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (−1; 10\n3 ) and with m = 2\n3.\n2. with m = −1 and passing through the point (−2; 0).\n3. passing through the point (3; −1) and with m = −1\n3.\n4. parallel to the x-axis and passing through the point (0; 11).\n5. passing through the point (1; 5) and with m = −2.\n6. perpendicular to the x-axis and passing through the point (−3\n2; 0).\n7. with m = −0,8 and passing through the point (10; −7).\n8. with undefined gradient and passing through the point (4; 0).\n9. with m = 3a and passing through the point (−2; −6a + b).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KC\n2. 22KD\n3. 22KF\n4. 22KG\n5. 22KH\n6. 22KJ\n7. 22KK\n8. 22KM\n9. 22KN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe gradient–intercept form of a straight line equation EMBGC\nUsing the gradient–point form, we can also derive the gradient–intercept form of the\nstraight line equation.\nStarting with the equation\ny −y1 = m(x −x1)\nExpand the brackets and make y the subject of the formula\ny −y1 = mx −mx1\ny = mx −mx1 + y1\ny = mx + (y1 −mx1)\nWe define constant c such that c = y1 −mx1 so that we get the equation\ny = mx + c\nThis is also called the standard form of the straight line equation.\n119\nChapter 4.\nAnalytical geometry\n\nNotice that when x = 0, we have\ny = m(0) + c\n= c\nTherefore c is the y-intercept of the straight line.\nWorked example 6: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −2 and passing through\nthe point (−1; 7).\nSOLUTION\nStep 1: Slope of the line\nWe notice that m < 0, therefore the graph decreases as x increases.\n2\n4\n6\n8\n−2\n2\n4\n−2\n−4\nb\n(−1; 7)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\nSubstitute the value of the gradient\ny = −2x + c\n120\n4.2.\nEquation of a line\n\nSubstitute the coordinates of the given point and find c\ny = −2x + c\n7 = −2(−1) + c\n7 −2 = c\n∴c = 5\nThis gives the y-intercept (0; 5).\nStep 3: Write the final answer\nThe equation of the straight line is y = −2x + 5.\nIf we are given two points on a straight line, we can also use the gradient–intercept\nform to determine the equation of a straight line. We solve for the two unknowns m\nand c using simultaneous equations — using the methods of substitution or elimina-\ntion.\nWorked example 7: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line passing through the points (−2; −7) and\n(3; 8).\nSOLUTION\nStep 1: Draw a sketch\n4\n8\n−4\n−8\n2\n4\n−2\n−4\nb\nb\n(−2; −7)\n(3; 8)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\n121\nChapter 4.\nAnalytical geometry\n\nStep 3: Substitute the coordinates of the given points\n−7 = m(−2) + c\n−7 = −2m + c\n. . . (1)\n8 = m(3) + c\n8 = 3m + c\n. . . (2)\nWe have two equations with two unknowns; we can therefore solve using simultane-\nous equations.\nStep 4: Make the coefficient of one of the variables the same in both equations\nWe notice that the coefficient of c in both equations is 1, therefore we can subtract\none equation from the other to eliminate c:\n−7 = −2m + c\n−(8 = 3m + c)\n−15 = −5m\n∴3 = m\nSubstitute m = 3 into either of the two equations and determine c:\n−7 = −2m + c\n−7 = −2(3) + c\n∴c = −1\nor\n8 = 3m + c\n8 = 3(3) + c\n∴c = −1\nStep 5: Write the final answer\nThe equation of the straight line is y = 3x −1.\n122\n4.2.\nEquation of a line\n\nExercise 4 – 4: The gradient–intercept form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (1\n2; 4) and\nwith m = 2.\n2. passing through the points (1\n2; −2)\nand (2; 4).\n3. passing through the points (2; −3)\nand (−1; 0).\n4. passing through the point (2; −6\n7)\nand with m = −3\n7.\n5. which cuts the y-axis at y = −1\n5 and\nwith m = 1\n2.\n6.\nb\nb\n(−1; −4)\n(2; 2)\nx\ny\n0\n7.\nb −3\n2\nx\ny\n0\n8.\nb\n(−2; −2)\n4\nx\ny\n0\n9.\nb (−2; 10)\nx\ny\n0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KP\n2. 22KQ\n3. 22KR\n4. 22KS\n5. 22KT\n6. 22KV\n7. 22KW\n8. 22KX\n9. 22KY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n123\nChapter 4.\nAnalytical geometry\n\n4.3\nInclination of a line\nEMBGD\n1\n2\n3\n1\n2\n3\nθ\n∆y\n∆x\nx\ny\nThe diagram shows that a straight line makes an angle θ with the positive x-axis. This\nis called the angle of inclination of a straight line.\nWe notice that if the gradient changes, then the value of θ also changes, therefore the\nangle of inclination of a line is related to its gradient. We know that gradient is the\nratio of a change in the y-direction to a change in the x-direction:\nm = ∆y\n∆x\nFrom trigonometry we know that the tangent function is defined as the ratio:\ntan θ = opposite side\nadjacent side\nAnd from the diagram we see that\ntan θ = ∆y\n∆x\n∴m = tan θ\nfor 0◦≤θ < 180◦\nTherefore the gradient of a straight line is equal to the tangent of the angle formed\nbetween the line and the positive direction of the x-axis.\nVertical lines\n• θ = 90◦\n• Gradient is undefined since there is no change in the x-values (∆x = 0).\n• Therefore tan θ is also undefined (the graph of tan θ has an asymptote at θ =\n90◦).\n124\n4.3.\nInclination of a line\n\nHorizontal lines\n• θ = 0◦\n• Gradient is equal to 0 since there is no change in the y-values (∆y = 0).\n• Therefore tan θ is also equal to 0 (the graph of tan θ passes through the origin\n(0◦; 0).\nLines with negative gradients\nIf a straight line has a negative gradient (m < 0, tan θ < 0), then the angle formed\nbetween the line and the positive direction of the x-axis is obtuse.\nθ\nx\ny\n0\nFrom the CAST diagram in trigonometry, we know that the tangent function is negative\nin the second and fourth quadrant. If we are calculating the angle of inclination for a\nline with a negative gradient, we must add 180◦to change the negative angle in the\nfourth quadrant to an obtuse angle in the second quadrant:\nIf we are given a straight line with gradient m = −0,7, then we can determine the\nangle of inclination using a calculator:\ntan θ = m\n= −0,7\n∴θ = tan−1(−0,7)\n= −35,0◦\nThis negative angle lies in the fourth quadrant. We must add 180◦to get an obtuse\nangle in the second quadrant:\nθ = −35,0◦+ 180◦\n= 145◦\n125\nChapter 4.\nAnalytical geometry\n\nAnd we can always use our calculator to check that the obtuse angle θ = 145◦gives a\ngradient of m = −0,7.\n35◦\n180◦−35◦= 145◦\nx\ny\n0\nExercise 4 – 5: Angle of inclination\n1. Determine the gradient (correct to 1 decimal place) of each of the following\nstraight lines, given that the angle of inclination is equal to:\na) 60◦\nb) 135◦\nc) 0◦\nd) 54◦\ne) 90◦\nf) 45◦\ng) 140◦\nh) 180◦\ni) 75◦\n2. Determine the angle of inclination (correct to 1 decimal place) for each of the\nfollowing:\na) a line with m = 3\n4\nb) 2y −x = 6\nc) the line passes through the points (−4; −1) and (2; 5)\nd) y = 4\ne) x = 3y + 1\n2\nf) x = −0,25\ng) the line passes through the points (2; 5) and (2\n3; 1)\nh) a line with gradient equal to 0,577\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22KZ\n1b. 22M2\n1c. 22M3\n1d. 22M4\n1e. 22M5\n1f. 22M6\n1g. 22M7\n1h. 22M8\n1i. 22M9\n2a. 22MB\n2b. 22MC\n2c. 22MD\n2d. 22MF\n2e. 22MG\n2f. 22MH\n2g. 22MJ\n2h. 22MK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n126\n4.3.\nInclination of a line\n\nWorked example 8: Inclination of a straight line\nQUESTION\nDetermine the angle of inclination (correct to 1 decimal place) of the straight line\npassing through the points (2; 1) and (−3; −9).\nSOLUTION\nStep 1: Draw a sketch\n2\n−2\n−4\n−6\n−8\n1\n2\n−1\n−2\n−3\nb\nb\n(2; 1)\n(−3; −9)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nx1 = 2;\ny1 = 1;\nx2 = −3;\ny2 = −9\nStep 3: Determine the gradient of the line\nm = y2 −y1\nx2 −x1\n= −9 −1\n−3 −2\n= −10\n−5\n∴m = 2\nStep 4: Use the gradient to determine the angle of inclination of the line\ntan θ = m\n= 2\n∴θ = tan−1 2\n= 63,4◦\nImportant: make sure your calculator is in DEG (degrees) mode.\nStep 5: Write the final answer\nThe angle of inclination of the straight line is 63,4◦.\n127\nChapter 4.\nAnalytical geometry\n\nWorked example 9: Inclination of a straight line\nQUESTION\nDetermine the equation of the straight line passing through the point (3; 1) and with\nan angle of inclination of 135◦.\nSOLUTION\nStep 1: Use the angle of inclination to determine the gradient of the line\nm = tan θ\n= tan 135◦\n∴m = −1\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = −1\ny −y1 = −(x −x1)\nSubstitute the given point (3; 1)\ny −1 = −(x −3)\ny = −x + 3 + 1\n= −x + 4\nStep 3: Write the final answer\nThe equation of the straight line is y = −x + 4.\nWorked example 10: Inclination of a straight line\nQUESTION\nDetermine the acute angle (correct to 1 decimal place) between the line passing\nthrough the points M(−1; 13\n4) and N(4; 3) and the straight line y = −3\n2x + 4.\nSOLUTION\nStep 1: Draw a sketch\nDraw the line through points M(−1; 13\n4) and N(4; 3) and the line y = −3\n2x + 4 on a\nsuitable system of axes. Label α and β, the angles of inclination of the two lines. Label\nθ, the acute angle between the two straight lines.\n128\n4.3.\nInclination of a line\n\n2\n4\n6\n−2\n2\n4\n−2\n−4\n−6\n−8\nb\nb\nβ\nˆB1\nα\nθ\nx\ny\n0\nM(−1; 7\n4)\nN(4; 3)\nNotice that α and θ are acute angles and β is an obtuse angle.\nˆB1 = 180◦−β\n(∠on str. line)\nand θ = α + ˆB1\n(ext. ∠of △= sum int. opp)\n∴θ = α + (180◦−β)\n= 180◦+ α −β\nStep 2: Use the gradient to determine the angle of inclination β\nFrom the equation y = −3\n2x + 4 we see that m < 0, therefore β is an obtuse angle\nsuch that 90◦< β < 180◦.\ntan β = m\n= −3\n2\ntan−1\n\u0012\n−3\n2\n\u0013\n= −56,3◦\nThis negative angle lies in the fourth quadrant. We know that the angle of inclination\nβ is an obtuse angle that lies in the second quadrant, therefore\nβ = −56,3◦+ 180◦\n= 123,7◦\nStep 3: Determine the gradient and angle of inclination of the line through M and\nN\n129\nChapter 4.\nAnalytical geometry\n\nDetermine the gradient\nm = y2 −y1\nx2 −x1\n=\n3 −7\n4\n4 −(−1)\n=\n5\n4\n5\n= 1\n4\nDetermine the angle of inclination\ntan α = m\n= 1\n4\n∴α = tan−1\n\u00121\n4\n\u0013\n= 14,0◦\nStep 4: Write the final answer\nθ = 180◦+ α −β\n= 180◦+ 14,0◦−123,7◦\n= 70,3◦\nThe acute angle between the two straight lines is 70,3◦.\nExercise 4 – 6: Inclination of a straight line\n1. Determine the angle of inclination for each of the following:\na) a line with m = 4\n5\nb) x + y + 1 = 0\nc) a line with m = 5,69\nd) the line that passes through (1; 1) and (−2; 7)\ne) 3 −2y = 9x\nf) the line that passes through (−1; −6) and (−1\n2; −11\n2 )\ng) 5 = 10y −15x\n130\n4.3.\nInclination of a line\n\nh)\nb\nx\ny\n(2; 3)\n−1\n0\ni)\nb\nx\ny\n(6; 0)\n2\n0\nj)\nb\nx\ny\n(−3; 3)\n−3\n0\n2. Determine the acute angle between the line passing through the points A(−2; 1\n5)\nand B(0; 1) and the line passing through the points C(1; 0) and D(−2; 6).\n3. Determine the angle between the line y + x = 3 and the line x = y + 1\n2.\n4. Find the angle between the line y = 2x and the line passing through the points\n(−1; 7\n3) and (0; 2).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22MM\n1b. 22MN\n1c. 22MP\n1d. 22MQ\n1e. 22MR\n1f. 22MS\n1g. 22MT\n1h. 22MV\n1i. 22MW\n1j. 22MX\n2. 22MY\n3. 22MZ\n4. 22N2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n131\nChapter 4.\nAnalytical geometry\n\n4.4\nParallel lines\nEMBGF\nInvestigation: Parallel lines\n1. Draw a sketch of the line passing through the points P(−1; 0) and Q(1; 4) and\nthe line passing through the points R(1; 2) and S(2; 4).\n2. Label and measure α and β, the angles of inclination of straight lines PQ and\nRS respectively.\n3. Describe the relationship between α and β.\n4. “α and β are alternate angles, therefore PQ ∥RS.” Is this a true statement? If\nnot, provide a correct statement.\n5. Use your calculator to determine tan α and tan β.\n6. Complete the sentence: . . . . . . lines have . . . . . . angles of inclination.\n7. Determine the equations of the straight lines PQ and RS.\n8. What do you notice about mPQ and mRS?\n9. Complete the sentence: . . . . . . lines have . . . . . . gradients.\nAnother method of determining the equation of a straight line is to be given a point on\nthe unknown line, (x1; y1), and the equation of a line which is parallel to the unknown\nline.\nLet the equation of the unknown line be y = m1x + c1 and the equation of the given\nline be y = m2x + c2.\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are parallel then\nm1 = m2\n132\n4.4.\nParallel lines\n\nImportant: when determining the gradient of a line using the coefficient of x, make\nsure the given equation is written in the gradient–intercept (standard) form. y = mx+c\nSubstitute the value of m2 and the given point (x1; y1), into the gradient–intercept form\nof a straight line equation\ny −y1 = m(x −x1)\nand determine the equation of the unknown line.\nWorked example 11: Parallel lines\nQUESTION\nDetermine the equation of the line that passes through the point (−1; 1) and is parallel\nto the line y −2x + 1 = 0.\nSOLUTION\nStep 1: Write the equation in gradient–intercept form\nWe write the given equation in gradient–intercept form and determine the value of m.\ny = 2x −1\nWe know that the two lines are parallel, therefore m1 = m2 = 2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = 2\ny −y1 = 2(x −x1)\nSubstitute the given point (−1; 1)\ny −1 = 2(x −(−1))\ny −1 = 2x + 2\ny = 2x + 2 + 1\n= 2x + 3\n133\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\n(−1; 1)\ny = 2x −1\ny = 2x + 3\nx\ny\n0\nA sketch was not required, but it is always helpful and can be used to check answers.\nStep 3: Write the final answer\nThe equation of the straight line is y = 2x + 3.\nWorked example 12: Parallel lines\nQUESTION\nLine AB passes through the point A(0; 3) and has an angle of inclination of 153,4◦.\nDetermine the equation of the line CD which passes through the point C(2; −3) and\nis parallel to AB.\nSOLUTION\nStep 1: Use the given angle of inclination to determine the gradient\nmAB = tan θ\n= tan 153,4◦\n= −0,5\nStep 2: Parallel lines have equal gradients\nSince we are given AB ∥CD,\nmCD = mAB = −0,5\n134\n4.4.\nParallel lines\n\nStep 3: Write down the gradient–point form of a straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient mCD = −0,5.\ny −y1 = −1\n2(x −x1)\nSubstitute the given point (2; −3).\ny −(−3) = −1\n2(x −2)\ny + 3 = −1\n2x + 1\ny = −1\n2x −2\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\nb C(2; −3)\ny = −1\n2x −2\ny = −1\n2x + 3\nx\ny\n0\nA sketch was not required, but it is always useful.\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n2x −2.\nSee video: 22N3 at www.everythingmaths.co.za\n135\nChapter 4.\nAnalytical geometry\n\nExercise 4 – 7: Parallel lines\n1. Determine whether or not the following two lines are parallel:\na) y + 2x = 1 and −2x + 3 = y\nb)\ny\n3 + x + 5 = 0 and 2y + 6x = 1\nc) y = 2x −7 and the line passing through (1; −2) and (1\n2; −1)\nd) y + 1 = x and x + y = 3\ne) The line passing through points (−2; −1) and (−4; −3) and the line −y +\nx −4 = 0\nf) y −1 = 1\n3x and the line passing through points (−2; 4) and (1; 5)\n2. Determine the equation of the straight line that passes through the point (1; −5)\nand is parallel to the line y + 2x −1 = 0.\n3. Determine the equation of the straight line that passes through the point (−2; −6)\nand is parallel to the line 2y + 1 = 6x.\n4. Determine the equation of the straight line that passes through the point (−2; −2)\nand is parallel to the line with angle of inclination θ = 56,31◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is parallel to the line with angle of inclination θ = 145◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22N4\n1b. 22N5\n1c. 22N6\n1d. 22N7\n1e. 22N8\n1f. 22N9\n2. 22NB\n3. 22NC\n4. 22ND\n5. 22NF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.5\nPerpendicular lines\nEMBGG\nInvestigation: Perpendicular lines\n1. Draw a sketch of the line passing through the points A(−2; −3) and B(2; 5) and\nthe line passing through the points C(−1; 1\n2) and D(4; −2).\n2. Label and measure α and β, the angles of inclination of straight lines AB and\nCD respectively.\n3. Label and measure θ, the angle between the lines AB and CD.\n4. Describe the relationship between the lines AB and CD.\n5. “θ is a reflex angle, therefore AB ⊥CD.” Is this a true statement? If not, provide\na correct statement.\n136\n4.5.\nPerpendicular lines\n\n6. Determine the equation of the straight line AB and the line CD.\n7. Use your calculator to determine tan α × tan β.\n8. Determine mAB × mCD.\n9. What do you notice about these products?\n10. Complete the sentence: if two lines are . . . . . . to each other, then the product of\ntheir . . . . . . is equal . . . . . .\n11. Complete the sentence: if the gradient of a straight line is equal to the negative\n. . . . . . of the gradient of another straight line, then the two lines are . . . . . .\nDeriving the formula: m1 × m2 = −1\nb\nb\nA(4; 3)\nB(−3; 4)\nθ\n90◦+ θ\nO\ny\nx\nConsider the point A(4; 3) on the Cartesian plane with an angle of inclination A ˆOX =\nθ. Rotate through an angle of 90◦and place point B at (−3; 4) so that we have the\nangle of inclination B ˆOX = 90◦+ θ.\nWe determine the gradient of OA:\nmOA = y2 −y1\nx2 −x1\n= 3 −0\n4 −0\n= 3\n4\nAnd determine the gradient of OB:\nmOB = y2 −y1\nx2 −x1\n= 4 −0\n−3 −0\n= 4\n−3\n137\nChapter 4.\nAnalytical geometry\n\nBy rotating through an angle of 90◦we know that OB ⊥OA:\nmOA × mOB = 3\n4 × 4\n−3\n= −1\nWe can also write that\nmOA = −\n1\nmOB\nb\nb\nA(x; y)\nB(−y; x)\nθ\n90◦+ θ\nO\ny\nx\nIf we have the general point A(x; y) with an angle of inclination A ˆOX = θ and point\nB(−y; x) such that B ˆOX = 90◦+ θ, then we know that\nmOA = y\nx\nmOB = −x\ny\n∴mOA × mOB = y\nx × −x\ny\n= −1\nAnother method of determining the equation of a straight line is to be given a point on\nthe line, (x1; y1), and the equation of a line which is perpendicular to the unknown\nline. Let the equation of the unknown line be y = m1x + c1 and the equation of the\ngiven line be y = m2x + c2.\nθ2\nθ1\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are perpendicular then\nm1 × m2 = −1\nNote: this rule does not apply to vertical or horizontal lines.\n138\n4.5.\nPerpendicular lines\n\nWhen determining the gradient of a line using the coefficient of x, make sure the\ngiven equation is written in the gradient–intercept (standard) form y = mx + c. Then\nwe know that\nm1 = −1\nm2\nSubstitute the value of m1 and the given point (x1; y1), into the gradient–intercept form\nof the straight line equation y −y1 = m(x −x1) and determine the equation of the\nunknown line.\nWorked example 13: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point T(2; 2) and per-\npendicular to the line 3y + 2x −6 = 0.\nSOLUTION\nStep 1: Write the equation in standard form\nLet the gradient of the unknown line be m1 and the given gradient be m2. We write\nthe given equation in gradient–intercept form and determine the value of m2.\n3y + 2x −6 = 0\n3y = −2x + 6\ny = −2\n3x + 2\n∴m2 = −2\n3\nWe know that the two lines are perpendicular, therefore m1 × m2 = −1. Therefore\nm1 = 3\n2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m1 = 3\n2.\ny −y1 = 3\n2(x −x1)\nSubstitute the given point T(2; 2).\ny −2 = 3\n2(x −2)\ny −2 = 3\n2x −3\ny = 3\n2x −1\n139\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\nb T(2; 2)\ny = −2\n3x + 2\ny = 3\n2x −1\nx\ny\n0\nA sketch was not required, but it is useful for checking the answer.\nStep 3: Write the final answer\nThe equation of the straight line is y = 3\n2x −1.\nWorked example 14: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point (2; 1\n3) and per-\npendicular to the line with an angle of inclination of 71,57◦.\nSOLUTION\nStep 1: Use the given angle of inclination to determine gradient\nLet the gradient of the unknown line be m1 and let the given gradient be m2.\nm2 = tan θ\n= tan 71,57◦\n= 3,0\nStep 2: Determine the unknown gradient\nSince we are given that the two lines are perpendicular,\nm1 × m2 = −1\n∴m1 = −1\n3\n140\n4.5.\nPerpendicular lines\n\nStep 3: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient m1 = −1\n3.\ny −y1 = −1\n3(x −x1)\nSubstitute the given point (2; 1\n3).\ny −\n\u00121\n3\n\u0013\n= −1\n3(x −2)\ny −1\n3 = −1\n3x + 2\n3\ny = −1\n3x + 1\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n3x + 1.\nSee video: 22NG at www.everythingmaths.co.za\nExercise 4 – 8: Perpendicular lines\n1. Calculate whether or not the following two lines are perpendicular:\na) y −1 = 4x and 4y + x + 2 = 0\nb) 10x = 5y −1 and 5y −x −10 = 0\nc) x = y −5 and the line passing through (−1; 5\n4) and (3; −11\n4 )\nd) y = 2 and x = 1\ne)\ny\n3 = x and 3y + x = 9\nf) 1 −2x = y and the line passing through (2; −1) and (−1; 5)\ng) y = x + 2 and 2y + 1 = 2x\n2. Determine the equation of the straight line that passes through the point (−2; −4)\nand is perpendicular to the line y + 2x = 1.\n3. Determine the equation of the straight line that passes through the point (2; −7)\nand is perpendicular to the line 5y −x = 0.\n141\nChapter 4.\nAnalytical geometry\n\n4. Determine the equation of the straight line that passes through the point (3; −1)\nand is perpendicular to the line with angle of inclination θ = 135◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is perpendicular to the line y = 4\n3.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NH\n1b. 22NJ\n1c. 22NK\n1d. 22NM\n1e. 22NN\n1f. 22NP\n1g. 22NQ\n2. 22NR\n3. 22NS\n4. 22NT\n5. 22NV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.6\nSummary\nEMBGH\nSee presentation: 22NW at www.everythingmaths.co.za\n• Distance between two points: d =\np\n(x2 −x1)2 + (y2 −y1)2\n• Gradient of a line between two points: m = y2 −y1\nx2 −x1\n• Mid-point of a line: M(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n• Parallel lines: m1 = m2\n• Perpendicular lines: m1 × m2 = −1\n• General form of a straight line equation: ax + by + c = 0\n• Two-point form of a straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• Gradient–point form of a straight line equation: y −y1 = m(x −x1)\n• Gradient–intercept form of a straight line equation (standard form): y = mx + c\n• Angle of inclination of a straight line: θ, the angle formed between the line and\nthe positive x-axis; m = tan θ\n142\n4.6.\nSummary\n\nExercise 4 – 9: End of chapter exercises\n1. Determine the equation of the line:\na) through points (−1; 3) and (1; 4)\nb) through points (7; −3) and (0; 4)\nc) parallel to y = 1\n2x + 3 and passing through (−2; 3)\nd) perpendicular to y = −1\n2x + 3 and passing through (−1; 2)\ne) perpendicular to 3y + x = 6 and passing through the origin\n2. Determine the angle of inclination of the following lines:\na) y = 2x −3\nb) y = 1\n3x −7\nc) 4y = 3x + 8\nd) y = −2\n3x + 3\ne) 3y + x −3 = 0\n3. P(2; 3), Q(−4; 0) and R(5; −3) are the vertices of △PQR in the Cartesian plane.\nPR intersects the x-axis at S. Determine the following:\na) the equation of the line PR\nb) the coordinates of point S\nc) the angle of inclination of PR (correct to two decimal places)\nd) the gradient of line PQ\ne) Q ˆPR\nf) the equation of the line perpendicular to PQ and passing through the origin\ng) the mid-point M of QR\nh) the equation of the line parallel to PR and passing through point M\n4. Points A(−3; 5), B(−7; −4) and C(2; 0) are given.\na) Plot the points on the Cartesian plane.\nb) Determine the coordinates of D if ABCD is a parallelogram.\nc) Prove that ABCD is a rhombus.\n5.\nb\nb\nb\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\nx\ny\n0\nM\nN\nP\n143\nChapter 4.\nAnalytical geometry\n\nConsider the sketch above, with the following lines shown:\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\na) Determine the coordinates of the point N.\nb) Determine the coordinates of the point P.\nc) Determine the equation of the vertical line MN.\nd) Determine the length of the vertical line MN.\ne) Find M ˆNP.\nf) Determine the equation of the line parallel to NP and passing through the\npoint M.\n6. The following points are given: A(−2; 3), B(2; 4), C(3; 0).\na) Plot the points on the Cartesian plane.\nb) Prove that △ABC is a right-angled isosceles triangle.\nc) Determine the equation of the line AB.\nd) Determine the coordinates of D if ABCD is a square.\ne) Determine the coordinates of E, the mid-point of BC.\n7. Given points S(2; 5), T(−3; −4) and V (4; −2).\na) Determine the equation of the line ST.\nb) Determine the size of T ˆSV .\n8. Consider triangle FGH with vertices F(−1; 3), G(2; 1) and H(4; 4).\na) Sketch △FGH on the Cartesian plane.\nb) Show that △FGH is an isosceles triangle.\nc) Determine the equation of the line PQ, perpendicular bisector of FH.\nd) Does G lie on the line PQ?\ne) Determine the equation of the line parallel to GH and passing through\npoint F.\n9. Given the points A(−1; 5), B(5; −3) and C(0; −6). M is the mid-point of AB\nand N is the mid-point of AC.\na) Draw a sketch on the Cartesian plane.\nb) Show that the coordinates of M and N are (2; 1) and (−1\n2; −1\n2) respectively.\nc) Use analytical geometry methods to prove the mid-point theorem. (Prove\nthat NM ∥CB and NM = 1\n2CB.)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NX\n1b. 22NY\n1c. 22NZ\n1d. 22P2\n1e. 22P3\n2a. 22P4\n2b. 22P5\n2c. 22P6\n2d. 22P7\n2e. 22P8\n3. 22P9\n4. 22PB\n5. 22PC\n6. 22PD\n7. 22PF\n8. 22PG\n9. 22PH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n144\n4.6.\nSummary\n\nCHAPTER\n5\nFunctions\n5.1\nQuadratic functions\n146\n5.2\nAverage gradient\n164\n5.3\nHyperbolic functions\n170\n5.4\nExponential functions\n184\n5.5\nThe sine function\n197\n5.6\nThe cosine function\n209\n5.7\nThe tangent function\n222\n5.8\nSummary\n235\n\n5\nFunctions\nA function describes a specific relationship between two variables; where an indepen-\ndent (input) variable has exactly one dependent (output) variable. Every element in the\ndomain maps to only one element in the range. Functions can be one-to-one relations\nor many-to-one relations. A many-to-one relation associates two or more values of the\nindependent variable with a single value of the dependent variable. Functions allow\nus to visualise relationships in the form of graphs, which are much easier to read and\ninterpret than lists of numbers.\n5.1\nQuadratic functions\nEMBGJ\nRevision\nEMBGK\nFunctions of the form y = ax2 + q\nFunctions of the general form y = ax2 + q are called parabolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = ax2 + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\nThe turning point of f(x) is\nabove the x-axis.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\nThe turning point of f(x) is be-\nlow the x-axis.\n– q is also the y-intercept of the\nparabola.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\n• The effect of a on shape\n– For a > 0; the graph of f(x) is a “smile” and has a minimum turning point\n(0; q). As the value of a becomes larger, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n– For a < 0; the graph of f(x) is a “frown” and has a maximum turning point\n(0; q). As the value of a becomes smaller, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n146\n5.1.\nQuadratic functions\n\nExercise 5 – 1: Revision\n1. On separate axes, accurately draw each of the following functions.\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = x2\nb) y2 = 1\n2x2\nc) y3 = −x2 −1\nd) y4 = −2x2 + 4\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nyint = 0\nvalue of a\na = 1\neffect of a\nstandard\nparabola\nturning point\n(0; 0)\naxis of symmetry\nx = 0\n(y-axis)\ndomain\n{x : x ∈R}\nrange\n{y : y ≥0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22PJ\n1b. 22PK\n1c. 22PM\n1d. 22PN\n2. 22PP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22PQ at www.everythingmaths.co.za\n147\nChapter 5.\nFunctions\n\nFunctions of the form y = a(x + p)2 + q\nEMBGM\nWe now consider parabolic functions of the form y = a(x + p)2 + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a parabolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = x2\nb) y2 = (x −2)2\nc) y3 = (x −1)2\nd) y4 = (x + 1)2\ne) y5 = (x + 2)2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = x2 + 2\nb) y2 = (x −2)2 −1\nc) y3 = (x −1)2 + 1\nd) y4 = (x + 1)2 + 1\ne) y5 = (x + 2)2 −1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of q\n3. Consider the three functions given below and answer the questions that follow:\n• y1 = (x −2)2 + 1\n• y2 = 2(x −2)2 + 1\n• y3 = −1\n2(x −2)2 + 1\na) What is the value of a for y2?\nb) Does y1 have a minimum or maximum turning point?\n148\n5.1.\nQuadratic functions\n\nc) What are the coordinates of the turning point of y2?\nd) Compare the graphs of y1 and y2. Discuss the similarities and differences.\ne) What is the value of a for y3?\nf) Will the graph of y3 be narrower or wider than the graph of y1?\ng) Determine the coordinates of the turning point of y3.\nh) Compare the graphs of y1 and y3. Describe any differences.\nSee video: 22PR at www.everythingmaths.co.za\nThe effect of the parameters on y = a(x + p)2 + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects whether the turning point is to the left of the y-axis (p > 0)\nor to the right of the y-axis (p < 0). The axis of symmetry is the line x = −p.\nThe effect of q is a vertical shift. The value of q affects whether the turning point of the\ngraph is above the x-axis (q > 0) or below the x-axis (q < 0).\nThe value of a affects the shape of the graph. If a < 0, the graph is a “frown” and has\na maximum turning point. If a > 0 then the graph is a “smile” and has a minimum\nturning point. When a = 0, the graph is a horizontal line y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\nSee simulation: 22PS at www.everythingmaths.co.za\n149\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form f(x) = y = a(x + p)2 + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative. If a > 0\nwe have:\n(x + p)2 ≥0\n(perfect square is always positive)\n∴a(x + p)2 ≥0\n(a is positive)\n∴a(x + p)2 + q ≥q\n∴f(x) ≥q\nThe range is therefore {y : y ≥q, y ∈R} if a > 0. Similarly, if a < 0, the range is\n{y : y ≤q, y ∈R}.\nWorked example 1: Domain and range\nQUESTION\nState the domain and range for g(x) = −2(x −1)2 + 3.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n(x −1)2 ≥0\n−2(x −1)2 ≤0\n−2(x −1)2 + 3 ≤3\ng(x) ≤3\nTherefore the range is {g(x) : g(x) ≤3} or in interval notation (−∞; 3].\nNotice in the example above that it helps to have the function in the form y = a(x +\np)2 + q.\nWe use the method of completing the square to write a quadratic function of the\ngeneral form y = ax2 + bx + c in the form y = a(x + p)2 + q (see Chapter 2).\n150\n5.1.\nQuadratic functions\n\nExercise 5 – 2: Domain and range\nGive the domain and range for each of the following functions:\n1. f(x) = (x −4)2 −1\n2. g(x) = −(x −5)2 + 4\n3. h(x) = x2 −6x + 9\n4. j(x) = −2(x + 1)2\n5. k(x) = −x2 + 2x −3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PT\n2. 22PV\n3. 22PW\n4. 22PX\n5. 22PY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nEvery point on the y-axis has an x-coordinate of 0, therefore to calculate the y-intercept\nwe let x = 0.\nFor example, the y-intercept of g(x) = (x −1)2 + 5 is determined by setting x = 0:\ng(x) = (x −1)2 + 5\ng(0) = (0 −1)2 + 5\n= 6\nThis gives the point (0; 6).\nThe x-intercept:\nEvery point on the x-axis has a y-coordinate of 0, therefore to calculate the x-intercept\nwe let y = 0.\nFor example, the x-intercept of g(x) = (x −1)2 + 5 is determined by setting y = 0:\ng(x) = (x −1)2 + 5\n0 = (x −1)2 + 5\n−5 = (x −1)2\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n151\nChapter 5.\nFunctions\n\nExercise 5 – 3: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = (x + 4)2 −1\n2. g(x) = 16 −8x + x2\n3. h(x) = −x2 + 4x −3\n4. j(x) = 4(x −3)2 −1\n5. k(x) = 4(x −3)2 + 1\n6. l(x) = 2x2 −3x −4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PZ\n2. 22Q2\n3. 22Q3\n4. 22Q4\n5. 22Q5\n6. 22Q6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTurning point\nThe turning point of the function f(x) = a(x+p)2 +q is determined by examining the\nrange of the function:\n• If a > 0, f(x) has a minimum turning point and the range is [q; ∞):\nThe minimum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\n• If a < 0, f(x) has a maximum turning point and the range is (−∞; q]:\nThe maximum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\nTherefore the turning point of the quadratic function f(x) = a(x + p)2 + q is (−p; q).\nAlternative form for quadratic equations:\nWe can also write the quadratic equation in the form\ny = a(x −p)2 + q\nThe effect of p is still a horizontal shift, however notice that:\n• For p > 0, the graph is shifted to the right by p units.\n• For p < 0, the graph is shifted to the left by p units.\nThe turning point is (p; q) and the axis of symmetry is the line x = p.\n152\n5.1.\nQuadratic functions\n\nWorked example 2: Turning point\nQUESTION\nDetermine the turning point of g(x) = 3x2 −6x −1.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q\nWe use the method of completing the square:\ng(x) = 3x2 −6x −1\n= 3(x2 −2x) −1\n= 3\n\u0000(x −1)2 −1\n\u0001\n−1\n= 3(x −1)2 −3 −1\n= 3(x −1)2 −4\nStep 2: Determine turning point (−p; q)\nFrom the equation g(x) = 3(x −1)2 −4 we know that the turning point for g(x) is\n(1; −4).\nWorked example 3: Turning point\nQUESTION\n1. Show that the x-value for the turning point of h(x) = ax2 + bx + c is given by\nx = −b\n2a.\n2. Hence, determine the turning point of k(x) = 2 −10x + 5x2.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q and show that p =\nb\n2a\nWe use the method of completing the square:\nh(x) = ax2 + bx + c\n= a\n\u0012\nx2 + b\nax + c\na\n\u0013\nTake half the coefficient of the x term and square it; then add and subtract it from the\n153\nChapter 5.\nFunctions\n\nexpression.\nh(x) = a\n \nx2 + b\nax +\n\u0012 b\n2a\n\u00132\n−\n\u0012 b\n2a\n\u00132\n+ c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2\n4a2 + c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a2\n!\n= a\n\u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a\nFrom the above we have that the turning point is at x = −p = −b\n2a and y = q =\n−b2−4ac\n4a\n.\nStep 2: Determine the turning point of k(x)\nWrite the equation in the general form y = ax2 + bx + c.\nk(x) = 5x2 −10x + 2\nTherefore a = 5; b = −10; c = 2.\nUse the results obtained above to determine x = −b\n2a:\nx = −\n\u0012−10\n2(5)\n\u0013\n= 1\nSubstitute x = 1 to obtain the corresponding y-value :\ny = 5x2 −10x + 2\n= 5(1)2 −10(1) + 2\n= 5 −10 + 2\n= −3\nThe turning point of k(x) is (1; −3).\nExercise 5 – 4: Turning points\nDetermine the turning point of each of the following:\n1. y = x2 −6x + 8\n2. y = −x2 + 4x −3\n3. y = 1\n2(x + 2)2 −1\n4. y = 2x2 + 2x + 1\n154\n5.1.\nQuadratic functions\n\n5. y = 18 + 6x −3x2\n6. y = −2[(x + 1)2 + 3]\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22Q7\n2. 22Q8\n3. 22Q9\n4. 22QB\n5. 22QC\n6. 22QD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxis of symmetry\nThe axis of symmetry for f(x) = a(x + p)2 + q is the vertical line x = −p. The axis of\nsymmetry passes through the turning point (−p; q) and is parallel to the y-axis.\ny\nx\n0\nb\nx = −p\nf(x) = a(x + p)2 + q\nExercise 5 – 5: Axis of symmetry\n1. Determine the axis of symmetry of each of the following:\na) y = 2x2 −5x −18\nb) y = 3(x −2)2 + 1\nc) y = 4x −x2\n2. Write down the equation of a parabola where the y-axis is the axis of symmetry.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QF\n1b. 22QG\n1c. 22QH\n2. 22QJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n155\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) = a(x + p)2 + q\nIn order to sketch graphs of the form f(x) = a(x + p)2 + q, we need to determine five\ncharacteristics:\n• sign of a\n• turning point\n• y-intercept\n• x-intercept(s) (if they exist)\n• domain and range\nSee video: 22QK at www.everythingmaths.co.za\nWorked example 4: Sketching a parabola\nQUESTION\nSketch the graph of y = −1\n2(x + 1)2 −3.\nMark the intercepts, turning point and the axis of symmetry. State the domain and\nrange of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = a(x + p)2 + q\nWe notice that a < 0, therefore the graph is a “frown” and has a maximum turning\npoint.\nStep 2: Determine the turning point (−p; q)\nFrom the equation we know that the turning point is (−1; −3).\nStep 3: Determine the axis of symmetry x = −p\nFrom the equation we know that the axis of symmetry is x = −1.\nStep 4: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = −1\n2 ((0) + 1)2 −3\n= −1\n2 −3\n= −31\n2\nThis gives the point (0; −31\n2).\n156\n5.1.\nQuadratic functions\n\nStep 5: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = −1\n2 (x + 1)2 −3\n3 = −1\n2 (x + 1)2\n−6 = (x + 1)2\nwhich has no real solutions. Therefore, there are no x-intercepts and the graph lies\nbelow the x-axis.\nStep 6: Plot the points and sketch the graph\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\nb\ny\nx\n0\n(0; −3 1\n2)\n(−1; −3)\nStep 7: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≤−3, y ∈R}\nSee video: 22QM at www.everythingmaths.co.za\nWorked example 5: Sketching a parabola\nQUESTION\nSketch the graph of y = 1\n2x2 −4x + 7\n2.\nDetermine the intercepts, turning point and the axis of symmetry. Give the domain\nand range of the function.\nSOLUTION\n157\nChapter 5.\nFunctions\n\nStep 1: Examine the equation of the form y = ax2 + bx + c\nWe notice that a > 0, therefore the graph is a “smile” and has a minimum turning\npoint.\nStep 2: Determine the turning point and the axis of symmetry\nCheck that the equation is in standard form and identify the coefficients.\na = 1\n2;\nb = −4;\nc = 7\n2\nCalculate the x-value of the turning point using\nx = −b\n2a\n= −\n \n−4\n2\n\u0000 1\n2\n\u0001\n!\n= 4\nTherefore the axis of symmetry is x = 4.\nSubstitute x = 4 into the original equation to obtain the corresponding y-value.\ny = 1\n2x2 −4x + 7\n2\n= 1\n2(4)2 −4(4) + 7\n2\n= 8 −16 + 7\n2\n= −41\n2\nThis gives the point\n\u00004; −41\n2\n\u0001\n.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = 1\n2(0)2 −4(0) + 7\n2\n= 7\n2\nThis gives the point\n\u00000; 7\n2\n\u0001\n.\nStep 4: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = 1\n2x2 −4x + 7\n2\n= x2 −8x + 7\n= (x −1)(x −7)\nTherefore x = 1 or x = 7. This gives the points (1; 0) and (7; 0).\nStep 5: Plot the points and sketch the graph\n158\n5.1.\nQuadratic functions\n\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\nb\nb\ny\nx\n0\n(4; −41\n2)\n(0; 3 1\n2)\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≥−41\n2, y ∈R}\nSee video: 22QN at www.everythingmaths.co.za\nInvestigation: Shifting the equation of a parabola\nCarl and Eric are doing their Mathematics homework and decide to check each others\nanswers.\nHomework question:\nIf the parabola y = 3x2 + 1 is shifted 2 units to the right, determine the equation of the\nnew parabola.\n• Carl’s answer:\nA shift to the right means moving in the positive x direction, therefore x is re-\nplaced with x + 2 and the new equation is y = 3(x + 2)2 + 1.\n• Eric’s answer:\nWe replace x with x −2, therefore the new equation is y = 3(x −2)2 + 1.\nWork together in pairs. Discuss the two different answers and decide which one is\ncorrect. Use calculations and sketches to help explain your reasoning.\n159\nChapter 5.\nFunctions\n\nWriting an equation of a shifted parabola\nThe parabola is shifted horizontally:\n• If the parabola is shifted m units to the right, x is replaced by (x −m).\n• If the parabola is shifted m units to the left, x is replaced by (x + m).\nThe parabola is shifted vertically:\n• If the parabola is shifted n units down, y is replaced by (y + n).\n• If the parabola is shifted n units up, y is replaced by (y −n).\nWorked example 6: Shifting a parabola\nQUESTION\nGiven y = x2 −2x −3.\n1. If the parabola is shifted 1 unit to the right, determine the new equation of the\nparabola.\n2. If the parabola is shifted 3 units down, determine the new equation of the\nparabola.\nSOLUTION\nStep 1: Determine the new equation of the shifted parabola\n1. The parabola is shifted 1 unit to the right, so x must be replaced by (x −1).\ny = x2 −2x −3\n= (x −1)2 −2(x −1) −3\n= x2 −2x + 1 −2x + 2 −3\n= x2 −4x\nBe careful not to make a common error: replacing x with x + 1 for a shift to the\nright.\n2. The parabola is shifted 3 units down, so y must be replaced by (y + 3).\ny + 3 = x2 −2x −3\ny = x2 −2x −3 −3\n= x2 −2x −6\n160\n5.1.\nQuadratic functions\n\nExercise 5 – 6: Sketching parabolas\n1. Sketch graphs of the following functions and determine:\n• intercepts\n• turning point\n• axes of symmetry\n• domain and range\na) y = −x2 + 4x + 5\nb) y = 2(x + 1)2\nc) y = 3x2 −2(x + 2)\nd) y = 3(x −2)2 + 1\n2. Draw the following graphs on the same system of axes:\nf(x) = −x2 + 7\ng(x) = −(x −2)2 + 7\nh(x) = (x −2)2 −7\n3. Draw a sketch of each of the following graphs:\na) y = ax2 + bx + c if a > 0, b > 0, c < 0.\nb) y = ax2 + bx + c if a < 0, b = 0, c > 0.\nc) y = ax2 + bx + c if a < 0, b < 0, b2 −4ac < 0.\nd) y = (x + p)2 + q if p < 0, q < 0 and the x-intercepts have different signs.\ne) y = a(x + p)2 + q if a < 0, p < 0, q > 0 and one root is zero.\nf) y = a(x + p)2 + q if a > 0, p = 0, b2 −4ac > 0.\n4. Determine the new equation (in the form y = ax2 + bx + c) if:\na) y = 2x2 + 4x + 2 is shifted 3 units to the left.\nb) y = −(x + 1)2 is shifted 1 unit up.\nc) y = 3(x −1)2 + 2\n\u0000x −1\n2\n\u0001\nis shifted 2 units to the right.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QP\n1b. 22QQ\n1c. 22QR\n1d. 22QS\n2. 22QT\n3a. 22QV\n3b. 22QW\n3c. 22QX\n3d. 22QY\n3e. 22QZ\n3f. 22R2\n4a. 22R3\n4b. 22R4\n4c. 22R5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n161\nChapter 5.\nFunctions\n\nFinding the equation of a parabola from the graph\nIf the intercepts are given, use y = a(x −x1)(x −x2).\nExample:\ny\nx\n0\n(0; 2)\n(4; 0)\n(−1; 0)\nx-intercepts: (−1; 0) and (4; 0)\ny = a(x −x1)(x −x2)\n= a(x + 1)(x −4)\n= ax2 −3ax −4a\ny-intercept: (0; 2)\n−4a = 2\na = −1\n2\nEquation of the parabola:\ny = ax2 −3ax −4a\n= −1\n2x2 −3\n\u0012\n−1\n2\n\u0013\nx −4\n\u0012\n−1\n2\n\u0013\n= −1\n2x2 + 3\n2x + 2\nIf the x-intercepts and another point are given, use y = a(x −x1)(x −x2).\nExample:\nb\ny\nx\n0\n(−1; 12)\n(1; 0)\n(5; 0)\nx-intercepts: (1; 0) and (5; 0)\ny = a(x −x1)(x −x2)\n= a(x −1)(x −5)\n= ax2 −6ax + 5a\nSubstitute the point: (−1; 12)\n12 = a(−1)2 −6a(−1) + 5a\n12 = a + 6a + 5a\n12 = 12a\n1 = a\nEquation of the parabola:\ny = ax2 −6ax + 5a\n= x2 −6x + 5\nIf the turning point and another point are given, use y = a(x + p)2 + q.\nExample:\nb\nb\ny\nx\n0\n(1; 5)\n(−3; 1)\nTurning point: (−3; 1)\ny = a(x + p)2 + q\n= a(x + 3)2 + 1\n= ax2 + 6ax + 9a + 1\nSubstitute the point: (1; 5)\n5 = a(1)2 + 6a(1) + 9a + 1\n4 = 16a\n1\n4 = a\nEquation of the parabola:\ny = 1\n4(x + 3)2 + 1\n162\n5.1.\nQuadratic functions\n\nExercise 5 – 7: Finding the equation\nDetermine the equations of the following graphs. Write your answers in the form\ny = a(x + p)2 + q.\n1.\nb\nb\ny\nx\n0\n3\n(−1; 6)\n2.\nb\ny\nx\n0\n(−1; 3)\n5\n3.\nb\nb\ny\nx\n0\n(1; 6)\n−2\n4.\nb\nb\nb\ny\nx\n0\n(1; 6)\n(3; 4)\n4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22R6\n2. 22R7\n3. 22R8\n4. 22R9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n163\nChapter 5.\nFunctions\n\n5.2\nAverage gradient\nEMBGN\nWe notice that the gradient of a curve changes at every point on the curve, therefore\nwe need to work with the average gradient. The average gradient between any two\npoints on a curve is the gradient of the straight line passing through the two points.\ny\nx\n0\nb\nb\nA(−3; 7)\nC(−1; −1)\nFor the diagram above, the gradient of the line AC is\nGradient = yA −yC\nxA −xC\n= 7 −(−1)\n−3 −(−1)\n= 8\n−2\n= −4\nThis is the average gradient of the curve between the points A and C.\nWhat happens to the gradient if we fix the position of one point and move the second\npoint closer to the fixed point?\nSee video: 22RB at www.everythingmaths.co.za\nInvestigation: Gradient at a single point on a curve\nThe curve shown here is defined by y = −2x2 −5.\nPoint B is fixed at (0; −5) and the position of point\nA varies.\nComplete the table below by calculating the y-\ncoordinates of point A for the given x-coordinates\nand then calculating the average gradient between\npoints A and B.\ny\nx\n0\nb\nb\nA\nB(0; −5)\n164\n5.2.\nAverage gradient\n\nxA\nyA\nAverage gradient\n−2\n−1,5\n−1\n−0,5\n0\n0,5\n1\n1,5\n2\n1. What happens to the average gradient as A moves towards B?\n2. What happens to the average gradient as A moves away from B?\n3. What is the average gradient when A overlaps with B?\nIn the example above, the gradient of the straight line that passes through points A and\nC changes as A moves closer to C. At the point where A and C overlap, the straight\nline only passes through one point on the curve. This line is known as a tangent to the\ncurve.\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb A\nC\ny\nx\n0\nb\nb\nA\nC\nWe therefore introduce the idea of the gradient at a single point on a curve. The\ngradient at a point on a curve is the gradient of the tangent to the curve at the given\npoint.\n165\nChapter 5.\nFunctions\n\nWorked example 7: Average gradient\nQUESTION\ny\nx\n0\nb\nb\nP(a; g(a))\nQ(a + h; g(a + h))\ng(x) = x2\n1. Find the average gradient between two points P (a; g(a)) and Q (a + h; g(a + h))\non a curve g(x) = x2.\n2. Determine the average gradient between P (2; g(2)) and Q (5; g(5)).\n3. Explain what happens to the average gradient if Q moves closer to P.\nSOLUTION\nStep 1: Assign labels to the x-values for the given points\nx1 = a\nx2 = a + h\nStep 2: Determine the corresponding y-coordinates\nUsing the function g(x) = x2, we can determine:\ny1 = g(a)\n= a2\ny2 = g(a + h)\n= (a + h)2\n= a2 + 2ah + h2\n166\n5.2.\nAverage gradient\n\nStep 3: Calculate the average gradient\ny2 −y1\nx2 −x1\n=\n\u0000a2 + 2ah + h2\u0001\n−\n\u0000a2\u0001\n(a + h) −(a)\n= a2 + 2ah + h2 −a2\na + h −a\n= 2ah + h2\nh\n= h(2a + h)\nh\n= 2a + h\nThe average gradient between P (a; g(a)) and Q (a + h; g(a + h)) on the curve g(x) =\nx2 is 2a + h.\nStep 4: Calculate the average gradient between P (2; g(2)) and Q (5; g(5))\nThe x-coordinate of P is a and the x-coordinate of Q is a+h therefore if we know that\na = 2 and a + h = 5, then h = 3.\nThe average gradient is therefore 2a + h = 2 (2) + (3) = 7\nStep 5: When Q moves closer to P\nWhen point Q moves closer to point P, h gets smaller.\nWhen the point Q overlaps with the point P, h = 0 and the gradient is given by 2a.\nWe can write the equation for average gradient in another form. Given a curve f(x)\nwith two points P and Q with P (a; f(a)) and Q (a + h; f(a + h)). The average gradi-\nent between P and Q is:\nAverage gradient = yQ −yP\nxQ −xP\n= f(a + h) −f(a)\n(a + h) −(a)\n= f(a + h) −f(a)\nh\nThis result is important for calculating the gradient at a point on a curve and will be\nexplored in greater detail in Grade 12.\n167\nChapter 5.\nFunctions\n\nWorked example 8: Average gradient\nQUESTION\nGiven f(x) = −2x2.\n1. Draw a sketch of the function and determine the average gradient between the\npoints A, where x = 1, and B, where x = 3.\n2. Determine the gradient of the curve at point A.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that a < 0, therefore the graph is a “frown” and has a\nmaximum turning point. We also see that when x = 0, y = 0, therefore the graph\npasses through the origin.\nStep 2: Draw a rough sketch\ny = −2x2\nx\ny\n0\nb\nb\nA\nB\nStep 3: Calculate the average gradient between A and B\nAverage gradient = f(3) −f(1)\n3 −1\n= −2(3)2 −(−2(1)2)\n2\n= −18 + 2\n2\n= −16\n2\n= −8\n168\n5.2.\nAverage gradient\n\nStep 4: Calculate the average gradient for f(x)\nAverage gradient = f(a + h) −f(a)\n(a + h) −a\n= −2(a + h)2 −(−2a2)\nh\n= −2a2 −4ah −2h2 + 2a2\nh\n= −4ah −2h2\nh\n= h(−4a −2h)\nh\n= −4a −2h\nAt point A, h = 0 and a = 1. Therefore\nAverage gradient = −4a −2h\n= −4(1) −2(0)\n= −4\nExercise 5 – 8:\n1.\na) Determine the average gradient of the curve f(x) = x (x + 3) between\nx = 5 and x = 3.\nb) Hence, state what you can deduce about the function f between x = 5 and\nx = 3.\n2. A (1; 3) is a point on f(x) = 3x2.\na) Draw a sketch of f(x) and label point A.\nb) Determine the gradient of the curve at point A.\nc) Determine the equation of the tangent line at A.\n3. Given: g(x) = −x2 + 1.\na) Draw a sketch of g(x).\nb) Determine the average gradient of the curve between x = −2 and x = 1.\nc) Determine the gradient of g at x = 2.\nd) Determine the gradient of g at x = 0.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RC\n2. 22RD\n3. 22RF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n169\nChapter 5.\nFunctions\n\n5.3\nHyperbolic functions\nEMBGP\nRevision\nEMBGQ\nFunctions of the form y = a\nx + q\nFunctions of the general form y = a\nx + q are called hyperbolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = a\nx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted vertically upwards by q units.\n– For q < 0, f(x) is shifted vertically downwards by q units.\n– The horizontal asymptote is the line y = q.\n– The vertical asymptote is the y-axis, the line x = 0.\n• The effect of a on shape and quad-\nrants\n– For a > 0, f(x) lies in the first\nand third quadrants.\n– For a > 1, f(x) will be further\naway from both axes than y =\n1\nx.\n– For 0 < a < 1, as a tends to\n0, f(x) moves closer to the axes\nthan y = 1\nx.\n– For a < 0, f(x) lies in the sec-\nond and fourth quadrants.\n– For a < −1, f(x) will be further\naway from both axes than y =\n−1\nx.\n– For −1 < a < 0, as a tends to\n0, f(x) moves closer to the axes\nthan y = −1\nx.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\nExercise 5 – 9: Revision\n1. Consider the following hyperbolic functions:\n• y1 = 1\nx\n• y2 = −4\nx\n• y3 = 4\nx −2\n• y4 = −4\nx + 1\n170\n5.3.\nHyperbolic functions\n\nComplete the table to summarise the properties of the hyperbolic function:\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nlies in I and III quad\nasymptotes\ny-axis, x = 0\nx-axis, y = 0\naxes of symmetry\ny = x\ny = −x\ndomain\n{x : x ∈R, x ̸= 0}\nrange\n{y : y ∈R, y ̸= 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22RH at www.everythingmaths.co.za\nFunctions of the form y =\na\nx+p + q\nEMBGR\nWe now consider hyperbolic functions of the form y =\na\nx+p + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a hyperbolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 1\nx\nb) y2 =\n1\nx−2\nc) y3 =\n1\nx−1\nd) y4 =\n1\nx+1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nintercept(s)\nasymptotes\naxes of symmetry\ndomain\nrange\neffect of p\n171\nChapter 5.\nFunctions\n\n2. Complete the following sentences for functions of the form y =\na\nx+p + q:\na) A change in p causes a . . . . . . shift.\nb) If the value of p increases, the graph and the vertical asymptote . . . . . .\nc) If the value of q changes, then the . . . . . . asymptote of the hyperbola will\nshift.\nd) If the value of p decreases, the graph and the vertical asymptote . . . . . .\nThe effect of the parameters on y =\na\nx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects the vertical asymptote, the line x = −p.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position on the Cartesian plane.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n172\n5.3.\nHyperbolic functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y =\na\nx+p + q:\nDomain and range\nThe domain is {x : x ∈R, x ̸= −p}. If x = −p, the dominator is equal to zero and the\nfunction is undefined.\nWe see that\ny =\na\nx + p + q\ncan be re-written as:\ny −q =\na\nx + p\nIf x ̸= −p then:\n(y −q) (x + p) = a\nx + p =\na\ny −q\nThe range is therefore {y : y ∈R, y ̸= q}.\nThese restrictions on the domain and range determine the vertical asymptote x = −p\nand the horizontal asymptote y = q.\nWorked example 9: Domain and range\nQUESTION\nDetermine the domain and range for g(x) =\n2\nx+1 + 2.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R, x ̸= −1} since g(x) is undefined for x = −1.\nStep 2: Determine the range\nLet g(x) = y:\ny =\n2\nx + 1 + 2\ny −2 =\n2\nx + 1\n(y −2)(x + 1) = 2\nx + 1 =\n2\ny −2\nTherefore the range is {g(x) : g(x) ∈R, g(x) ̸= 2}.\n173\nChapter 5.\nFunctions\n\nExercise 5 – 10: Domain and range\nDetermine the domain and range for each of the following functions:\n1. y = 1\nx + 1\n2. g(x) =\n8\nx−8 + 4\n3. y = −\n4\nx+1 −3\n4. x =\n2\n3−y + 5\n5. (y −2)(x + 2) = 3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RJ\n2. 22RK\n3. 22RM\n4. 22RN\n5. 22RP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n2\nx+1+2\nis determined by setting x = 0:\ng(x) =\n2\nx + 1 + 2\ng(0) =\n2\n0 + 1 + 2\n= 2 + 2\n= 4\nThis gives the point (0; 4).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n2\nx+1+2\nis determined by setting y = 0:\ng(x) =\n2\nx + 1 + 2\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\n174\n5.3.\nHyperbolic functions\n\nExercise 5 – 11: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) =\n1\nx+4 −2\n2. g(x) = −5\nx + 2\n3. j(x) =\n2\nx−1 + 3\n4. h(x) =\n3\n6−x + 1\n5. k(x) =\n5\nx+2 −1\n2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RQ\n2. 22RR\n3. 22RS\n4. 22RT\n5. 22RV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptotes\nThere are two asymptotes for functions of the form y =\na\nx+p + q. The asymptotes\nindicate the values of x for which the function does not exist. In other words, the\nvalues that are excluded from the domain and the range. The horizontal asymptote is\nthe line y = q and the vertical asymptote is the line x = −p.\nExercise 5 – 12: Asymptotes\nDetermine the asymptotes for each of the following functions:\n1. y =\n1\nx+4 −2\n2. y = −5\nx\n3. y =\n3\n2−x + 1\n4. y = 1\nx −8\n5. y = −\n2\nx−2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RW\n2. 22RX\n3. 22RY\n4. 22RZ\n5. 22S2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxes of symmetry\nThere are two lines about which a hyperbola is symmetrical.\nFor the standard hyperbola y = 1\nx, we see that if we replace x ⇒y and y ⇒x, we get\ny = 1\nx. Similarly, if we replace x ⇒−y and y ⇒−x, the function remains the same.\nTherefore the function is symmetrical about the lines y = x and y = −x.\nFor the shifted hyperbola y =\na\nx+p + q, the axes of symmetry intersect at the point\n(−p; q).\n175\nChapter 5.\nFunctions\n\nTo determine the axes of symmetry we define the two straight lines y1 = m1x + c1 and\ny2 = m2x + c2. For the standard and shifted hyperbolic function, the gradient of one\nof the lines of symmetry is 1 and the gradient of the other line of symmetry is −1. The\naxes of symmetry are perpendicular to each other and the product of their gradients\nequals −1. Therefore we let y1 = x+c1 and y2 = −x+c2. We then substitute (−p; q),\nthe point of intersection of the axes of symmetry, into both equations to determine the\nvalues of c1 and c2.\nWorked example 10: Axes of symmetry\nQUESTION\nDetermine the axes of symmetry for y =\n2\nx+1 −2.\nSOLUTION\nStep 1: Determine the point of intersection (−p; q)\nFrom the equation we see that p = 1 and q = −2. So the axes of symmetry will\nintersect at (−1; −2).\nStep 2: Define two straight line equations\ny1 = x + c1\ny2 = −x + c2\nStep 3: Solve for c1 and c2\nUse (−1; −2) to solve for c1:\ny1 = x + c1\n−2 = −1 + c1\n−1 = c1\nUse (−1; −2) to solve for c2:\ny2 = −x + c2\n−2 = −(−1) + c2\n−3 = c2\nStep 4: Write the final answer\nThe axes of symmetry for y =\n2\nx+1 −2 are the lines\ny1 = x −1\ny2 = −x −3\n176\n5.3.\nHyperbolic functions\n\n1\n2\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny1 = x −1\ny2 = −x −3\nExercise 5 – 13: Axes of symmetry\n1. Complete the following for f(x) and g(x):\n• Sketch the graph.\n• Determine (−p; q).\n• Find the axes of symmetry.\nCompare f(x) and g(x) and also their axes of symmetry. What do you notice?\na) f(x) = 2\nx\ng(x) = 2\nx + 1\nb) f(x) = −3\nx\ng(x) = −\n3\nx+1\nc) f(x) = 5\nx\ng(x) =\n5\nx−1 −1\n2. A hyperbola of the form k(x) =\na\nx+p + q passes through the point (4; 3). If the\naxes of symmetry intersect at (−1; 2), determine the equation of k(x).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S3\n1b. 22S4\n1c. 22S5\n2. 22S6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n177\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) =\na\nx+p + q\nIn order to sketch graphs of functions of the form, f(x) =\na\nx+p +q, we need to calculate\nfive characteristics:\n• quadrants\n• asymptotes\n• y-intercept\n• x-intercept\n• domain and range\nWorked example 11: Sketching a hyperbola\nQUESTION\nSketch the graph of y =\n2\nx+1 + 2. Determine the intercepts, asymptotes and axes of\nsymmetry. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Determine the asymptotes\nFrom the equation we know that p = 1 and q = 2.\nTherefore the horizontal asymptote is the line y = 2 and the vertical asymptote is the\nline x = −1.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n2\n0 + 1 + 2\n= 4\nThis gives the point (0; 4).\nStep 4: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n178\n5.3.\nHyperbolic functions\n\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\nStep 5: Determine the axes of symmetry\nUsing (−1; 2) to solve for c1:\ny1 = x + c1\n2 = −1 + c1\n3 = c1\ny2 = −x + c2\n2 = −(−1) + c2\n1 = c2\nTherefore the axes of symmetry are y = x + 3 and y = −x + 1.\nStep 6: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny = x + 3\ny = −x + 1\nStep 7: State the domain and range\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 2}\n179\nChapter 5.\nFunctions\n\nWorked example 12: Sketching a hyperbola\nQUESTION\nUse horizontal and vertical shifts to sketch the graph of f(x) =\n1\nx−2 + 3.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Sketch the standard hyperbola y = 1\nx\nStart with a sketch of the standard hyperbola g(x) = 1\nx. The vertical asymptote is x = 0\nand the horizontal asymptote is y = 0.\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 3: Determine the vertical shift\nFrom the equation we see that q = 3, which means g(x) must shifted 3 units up. The\nhorizontal asymptote is also shifted 3 units up to y = 3 .\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 4: Determine the horizontal shift\nFrom the equation we see that p = −2, which means g(x) must shifted 2 units to the\nright. The vertical asymptote is also shifted 2 units to the right.\n180\n5.3.\nHyperbolic functions\n\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n5\n6\n−1\n−2\ny\nx\n0\nStep 5: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n1\n0 −2 + 3\n= 21\n2\nThis gives the point (0; 21\n2).\nStep 6: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n1\nx −2 + 3\n−3 =\n1\nx −2\n−3(x −2) = 1\n−3x + 6 = 1\n−3x = −5\nx = 5\n3\nThis gives the point (5\n3; 0).\nStep 7: Determine the domain and range\nDomain: {x : x ∈R, x ̸= 2}\nRange: {y : y ∈R, y ̸= 3}\n181\nChapter 5.\nFunctions\n\nWorked example 13: Finding the equation of a hyperbola from the graph\nQUESTION\nUse the graph below to determine the values of a, p and q for y =\na\nx+p + q.\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\nSOLUTION\nStep 1: Examine the graph and deduce the sign of a\nWe notice that the graph lies in the second and fourth quadrants, therefore a < 0.\nStep 2: Determine the asymptotes\nFrom the graph we see that the vertical asymptote is x = −1, therefore p = 1. The\nhorizontal asymptote is y = 3, and therefore q = 3.\ny =\na\nx + 1 + 3\nStep 3: Determine the value of a\nTo determine the value of a we substitute a point on the graph, namely (0; 0):\ny =\na\nx + 1 + 3\n0 =\na\n0 + 1 + 3\n∴−3 = a\nStep 4: Write the final answer\ny = −\n3\nx + 1 + 3\n182\n5.3.\nHyperbolic functions\n\nExercise 5 – 14: Sketching graphs\n1. Draw the graphs of the following functions and indicate:\n• asymptotes\n• intercepts, where applicable\n• axes of symmetry\n• domain and range\na) y = 1\nx + 2\nb) y =\n1\nx+4 −2\nc) y = −\n1\nx+1 + 3\nd) y = −\n5\nx−2 1\n2 −2\ne) y =\n8\nx−8 + 4\n2. Given the graph of the hyperbola of the form y =\n1\nx+p + q, determine the values\nof p and q.\ny\nx\n−2\n−1\n3. Given a sketch of the function of the form y =\na\nx+p + q, determine the values of\na, p and q.\ny\nx\n2\n2\n4.\na) Draw the graph of f(x) = −3\nx, x > 0.\nb) Determine the average gradient of the graph between x = 1 and x = 3.\nc) Is the gradient at (1\n2; −6) less than or greater than the average gradient be-\ntween x = 1 and x = 3? Illustrate this on your graph.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S7\n1b. 22S8\n1c. 22S9\n1d. 22SB\n1e. 22SC\n2. 22SD\n3. 22SF\n4. 22SG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n183\nChapter 5.\nFunctions\n\n5.4\nExponential functions\nEMBGS\nRevision\nEMBGT\nFunctions of the form y = abx + q\nFunctions of the general form y = abx + q, for b > 0, are called exponential functions,\nwhere a, b and q are constants.\nThe effects of a, b and q on f(x) = abx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\n– The horizontal asymptote is the\nline y = q.\n• The effect of a on shape\n– For a > 0, f(x) is increasing.\n– For a < 0, f(x) is decreasing.\nThe graph is reflected about the\nhorizontal asymptote.\n• The effect of b on direction\nAssuming a > 0:\n– If b > 1, f(x) is an increasing\nfunction.\n– If 0 < b < 1, f(x) is a decreas-\ning function.\n– If b ≤0, f(x) is not defined.\nb > 1\na < 0\na > 0\nq > 0\nq < 0\n0 < b < 1\na < 0\na > 0\nq > 0\nq < 0\nExercise 5 – 15: Revision\n1. On separate axes, accurately draw each of the following functions:\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = 3x\nb) y2 = −2 × 3x\nc) y3 = 2 × 3x + 1\nd) y4 = 3x −2\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\n184\n5.4.\nExponential functions\n\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nincreasing\nasymptote\nx-axis, y = 0\ndomain\n{x : x ∈R}\nrange\n{y : y ∈R, y > 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22SH\n1b. 22SJ\n1c. 22SK\n1d. 22SM\n2. 22SN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = ab(x+p) + q\nEMBGV\nWe now consider exponential functions of the form y = ab(x+p) + q and the effects of\nparameter p.\nSee video: 22SP at www.everythingmaths.co.za\nInvestigation: The effects of a, p and q on an exponential graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 2x\nb) y2 = 2(x−2)\nc) y3 = 2(x−1)\nd) y4 = 2(x+1)\ne) y5 = 2(x+2)\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptote\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = 2(x−1) + 2\n185\nChapter 5.\nFunctions\n\nb) y2 = 3 × 2(x−1) + 2\nc) y3 = 1\n2 × 2(x−1) + 2\nd) y4 = 0 × 2(x−1) + 2\ne) y5 = −3 × 2(x−1) + 2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptotes\ndomain\nrange\neffect of a\nThe effect of the parameters on y = abx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position relative to the horizontal\nasymptote.\n• For a > 0, the graph lies above the horizontal asymptote, y = q.\n• For a < 0, the graph lies below the horizontal asymptote, y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n186\n5.4.\nExponential functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y = ab(x+p) + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative.\nIf a > 0 we have:\nb(x+p) > 0\nab(x+p) > 0\nab(x+p) + q > q\nf(x) > q\nThe range is therefore {y : y > q, y ∈R}.\nSimilarly, if a < 0, the range is {y : y < q, y ∈R}.\nWorked example 14: Domain and range\nQUESTION\nState the domain and range for g(x) = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n3(x+1) > 0\n5 × 3(x+1) > 0\n5 × 3(x+1) −1 > −1\n∴g(x) > −1\nTherefore the range is {g(x) : g(x) > −1} or in interval notation (−1; ∞).\n187\nChapter 5.\nFunctions\n\nExercise 5 – 16: Domain and range\nGive the domain and range for each of the following functions:\n1. y =\n\u0000 3\n2\n\u0001(x+3)\n2. f(x) = −5(x−2) + 1\n3. y + 3 = 2(x+1)\n4. y = n + 3(x−m)\n5.\ny\n2 = 3(x−1) −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SQ\n2. 22SR\n3. 22SS\n4. 22ST\n5. 22SV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting x = 0:\ng(0) = 3 × 2(0+1) + 2\n= 3 × 2 + 2\n= 8\nThis gives the point (0; 8).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting y = 0:\n0 = 3 × 2(x+1) + 2\n−2 = 3 × 2(x+1)\n−2\n3 = 2(x+1)\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n188\n5.4.\nExponential functions\n\nExercise 5 – 17: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = 2(x+1) −8\n2. y = 2 × 3(x−1) −18\n3. y + 5(x+2) = 5\n4. y = 1\n2\n\u0000 3\n2\n\u0001(x+3) −0,75\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SW\n2. 22SX\n3. 22SY\n4. 22SZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptote\nExponential functions of the form y = ab(x+p) + q have a horizontal asymptote, the\nline y = q.\nWorked example 15: Asymptote\nQUESTION\nDetermine the asymptote for y = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the asymptote\nThe asymptote of g(x) can be calculated as:\n3(x+1) ̸= 0\n5 × 3(x+1) ̸= 0\n5 × 3(x+1) −1 ̸= −1\n∴y ̸= −1\nTherefore the asymptote is the line y = −1.\n189\nChapter 5.\nFunctions\n\nExercise 5 – 18: Asymptote\nGive the asymptote for each of the following functions:\n1. y = −5(x+1)\n2. y = 3(x−2) + 1\n3.\n\u0010\n3y\n2\n\u0011\n= 5(x+3) −1\n4. y = 7(x+1) −2\n5.\ny\n2 + 1 = 3(x+2)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T2\n2. 22T3\n3. 22T4\n4. 22T5\n5. 22T6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching graphs of the form f(x) = ab(x+p) + q\nIn order to sketch graphs of functions of the form, f(x) = ab(x+p) + q, we need to\ndetermine five characteristics:\n• shape\n• y-intercept\n• x-intercept\n• asymptote\n• domain and range\nWorked example 16: Sketching an exponential graph\nQUESTION\nSketch the graph of 2y = 10 × 2(x+1) −5.\nMark the intercept(s) and asymptote. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nWe notice that a > 0 and b > 1, therefore the function is increasing.\n190\n5.4.\nExponential functions\n\nStep 2: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\n2y = 10 × 2(0+1) −5\n= 10 × 2 −5\n= 15\n∴y = 71\n2\nThis gives the point (0; 71\n2).\nStep 3: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 = 10 × 2(x+1) −5\n5 = 10 × 2(x+1)\n1\n2 = 2(x+1)\n2−1 = 2(x+1)\n∴−1 = x + 1\n(same base)\n−2 = x\nThis gives the point (−2; 0).\nStep 4: Determine the asymptote\nThe horizontal asymptote is the line y = −5\n2.\nStep 5: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\n−3\n1\n2\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y > −5\n2, y ∈R}\n191\nChapter 5.\nFunctions\n\nWorked example 17: Finding the equation of an exponential function from a\ngraph\nQUESTION\nUse the given graph of y = −2 × 3(x+p) + q to determine the values of p and q.\n1\n2\n3\n4\n5\n6\n7\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nFrom the graph we see that the function is decreasing. We also note that a = −2 and\nb = 3. We need to solve for p and q.\nStep 2: Use the asymptote to determine q\nThe horizontal asymptote y = 6 is given, therefore we know that q = 6.\ny = −2 × 3(x+p) + 6\nStep 3: Use the x-intercept to determine p\nSubstitute (2; 0) into the equation and solve for p:\ny = −2 × 3(x+p) + 6\n0 = −2 × 3(2+p) + 6\n−6 = −2 × 3(2+p)\n3 = 3(2+p)\n∴1 = 2 + p\n(same base)\n∴p = −1\nStep 4: Write the final answer\ny = −2 × 3(x−1) + 6\n192\n5.4.\nExponential functions\n\nExercise 5 – 19: Mixed exercises\n1. Given the graph of the hyperbola of the form h(x) = k\nx, x < 0, which passes\nthough the point A(−1\n2; −6).\nb\ny\nx\n0\nA(−1\n2; −6)\na) Show that k = 3.\nb) Write down the equation for the new function formed if h(x):\ni. is shifted 3 units vertically upwards\nii. is shifted to the right by 3 units\niii. is reflected about the y-axis\niv. is shifted so that the asymptotes are x = 0 and y = −1\n4\nv. is shifted upwards to pass through the point (−1; 1)\nvi. is shifted to the left by 2 units and 1 unit vertically downwards (for\nx < 0)\n2. Given the graphs of f(x) = a(x + p)2 and g(x) = a\nx.\nThe axis of symmetry for f(x) is x = −1 and f(x) and g(x) intersect at point M.\nThe line y = 2 also passes through M.\nb\ny\nx\n0\nM\n−1\n2\nf\ng\n193\nChapter 5.\nFunctions\n\nDetermine:\na) the coordinates of M\nb) the equation of g(x)\nc) the equation of f(x)\nd) the values for which f(x) < g(x)\ne) the range of f(x)\n3. On the same system of axes, sketch:\na) the graphs of k(x) = 2(x + 1\n2)2 −41\n2 and h(x) = 2(x+ 1\n2 ). Determine all\nintercepts, turning point(s) and asymptotes.\nb) the reflection of h(x) about the x-axis. Label this function as j(x).\n4. Sketch the graphs of y = ax2 + bx + c for:\na) a < 0, b > 0, b2 < 4ac\nb) a > 0, b > 0, one root = 0\n5. On separate systems of axes, sketch the graphs:\ny =\n2\nx−2\ny = 2\nx −2\ny = −2(x−2)\n6. For the diagrams shown below, determine:\n• the equations of the functions; f(x) = a(x + p)2 + q, g(x) = ax2 + q,\nh(x) = a\nx, x < 0 and k(x) = bx + q\n• the axes of symmetry of each function\n• the domain and range of each function\na)\nb\ny\nx\n0\n(2; 3)\nf\n194\n5.4.\nExponential functions\n\nb)\nb\ny\nx\n0\n(−2; −1)\ng\nh\n−2\nc)\ny\nx\n0\nk\ny = 2x + 1\n2\n7. Given the graph of the function Q(x) = ax.\nb\nb\nb\ny\nx\n0\nQ = ax\n(−2; p)\n1\n(1; 1\n3)\na) Show that a = 1\n3.\nb) Find the value of p if the point (−2; p) is on Q.\nc) Calculate the average gradient of the curve between x = −2 and x = 1.\nd) Determine the equation of the new function formed if Q is shifted 2 units\nvertically downwards and 2 units to the left.\n8. Find the equation for each of the functions shown below:\na) f(x) = 2x + q\ng(x) = mx + c\n195\nChapter 5.\nFunctions\n\nb\ny\nx\n0\nf\ng\n−1\n2\n−2\nb) h(x) =\nk\nx+p + q\nb\nb\ny\nx\n0\n1\n−2\n−1\n2\nh\n9. Given: the graph of k(x) = −x2 + 3x + 10 with turning point at D. The graph of\nthe straight line h(x) = mx + c passing through points B and C is also shown.\nb\nb\nb\ny\nx\n0\nB\nA\nE\nF\nD\nC\nk\nh\nDetermine:\na) the lengths AO, OB, OC and DE\nb) the equation of DE\nc) the equation of h(x)\nd) the x-values for which k(x) < 0\n196\n5.4.\nExponential functions\n\ne) the x-values for which k(x) ≥h(x)\nf) the length of DF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T7\n2. 22T8\n3. 22T9\n4. 22TB\n5. 22TC\n6a. 22TD\n6b. 22TF\n6c. 22TG\n7. 22TH\n8a. 22TJ\n8b. 22TK\n9. 22TM\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIMPORTANT: Trigonometric functions are examined in PAPER 2.\n5.5\nThe sine function\nEMBGW\nRevision\nEMBGX\nFunctions of the form y = sin θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• Period of one complete wave is 360◦.\n• Amplitude is the maximum height of the wave above and below the x-axis and\nis always positive. Amplitude = 1.\n• Domain: [0◦; 360◦]\nFor y = sin θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Maximum turning point: (90◦; 1)\n• Minimum turning point: (270◦; −1)\n197\nChapter 5.\nFunctions\n\nFunctions of the form y = a sin θ + q\nThe effects of a and q on f(θ) = a sin θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 20: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦.\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function also determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n198\n5.5.\nThe sine function\n\n1. y1 = sin θ\n2. y2 = −2 sin θ\n3. y3 = sin θ + 1\n4. y4 = 1\n2 sin θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TN\n2. 22TP\n3. 22TQ\n4. 22TR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin kθ\nEMBGY\nWe now consider cosine functions of the form y = sin kθ and the effects of k.\nInvestigation: The effects of k on a sine graph\n1. Complete the following table for y1 = sin θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−270◦\n−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n2. Use the table of values to plot the graph of y1 = sin θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = sin(−θ)\nb) y3 = sin 2θ\nc) y4 = sin θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n199\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = sin θ and y2 = sin(−θ)?\n6. Is sin(−θ) = −sin θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = sin kθ?\nThe effect of the parameter on y = sin kθ\nThe value of k affects the period of the sine function. If k is negative, then the graph is\nreflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the sine function decreases.\nFor 0 < k < 1, the period of the sine function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\nsin(−θ) = −sin θ\nCalculating the period:\nTo determine the period of y = sin kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k (this means that k is always considered to be\npositive).\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n200\n5.5.\nThe sine function\n\nWorked example 18: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = sin θ\nb) y2 = sin 3θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin kθ\nNotice that k > 1 for y2 = sin 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\nsin θ\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\nsin 3θ\n2\n1\n0,38\n−0,71\n−0,92\n0\n0,92\n0,71\n−0,38\n−1\nStep 3: Sketch the sine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin 3\n2θ\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin 3θ\n2\nperiod\n360◦\n240◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(90◦; 1)\n(−180◦; 1) and (60◦; 1)\nminimum turning points\n(−90◦; −1)\n(−60◦; −1) and (180◦; 1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\n201\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = sin kθ\n= sin 0◦\n= 0\nThis gives the point (0◦; 0).\nExercise 5 – 21: Sine functions of the form y = sin kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦and for each graph deter-\nmine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = sin 3θ\nb) g(θ) = sin θ\n3\nc) h(θ) = sin(−2θ)\nd) k(θ) = sin 3θ\n4\n2. For each graph of the form f(θ) = sin kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n202\n5.5.\nThe sine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22TS\n1b. 22TT\n1c. 22TV\n1d. 22TW\n2a. 22TX\n2b. 22TY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin(θ + p)\nEMBGZ\nInvestigation: The effects of p on a sine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = sin θ\nb) y2 = sin(θ −90◦)\nc) y3 = sin(θ −60◦)\nd) y4 = sin(θ + 90◦)\ne) y5 = sin(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of p\n203\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = sin(θ + p)\nThe effect of p on the sine function is a horizontal shift, also called a phase shift; the\nentire graph slides to the left or to the right.\n• For p > 0, the graph of the sine function shifts to the left by p.\n• For p < 0, the graph of the sine function shifts to the right by p.\np > 0\np < 0\nWorked example 19: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = sin θ\nb) y2 = sin(θ −30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin(θ + p)\nNotice that for y1 = sin θ we have p = 0 (no phase shift) and for y2 = sin(θ −30◦),\np < 0 therefore the graph shifts to the right by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n0\n1\n0\n−1\n0\n1\n0\n−1\n0\nsin(θ −30◦)\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n204\n5.5.\nThe sine function\n\nStep 3: Sketch the sine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = sin θ\ny2 = sin(θ −30◦)\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin(θ −30◦)\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−270◦; 1) and (90◦; 1)\n(−240◦; 1) and\n(120◦; 1)\nminimum turning points\n(−90◦; −1) and\n(270◦; −1)\n(−60◦; −1) and\n(300◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; −1\n2)\nx-intercept(s)\n(−360◦; 0), (−180◦; 0),\n(0◦; 0), (180◦; 0) and\n(360◦; 0)\n(−330◦; 0), (−150◦; 0),\n(30◦; 0) and (210◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\n205\nChapter 5.\nFunctions\n\nExercise 5 – 22: Sine functions of the form y = sin(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = sin(θ + 30◦)\n2. g(θ) = sin(θ −45◦)\n3. h(θ) = sin(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TZ\n2. 22V2\n3. 22V3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching sine graphs\nEMBH2\nWorked example 20: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(45◦−θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin(θ + p).\nf(θ) = sin(45◦−θ)\n= sin(−θ + 45◦)\n= sin (−(θ −45◦))\n= −sin(θ −45◦)\nTo draw a graph of the above function, we know that the standard sine graph, y = sin θ,\n206\n5.5.\nThe sine function\n\nmust:\n• be reflected about the x-axis\n• be shifted to the right by 45◦\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n0,71\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = −sin(θ −45◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 360◦\nAmplitude: 1\nDomain: [−360◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (315◦; 1)\nMinimum turning point: (135◦; −1)\ny-intercepts: (0◦; 0,71)\nx-intercept: (45◦; 0) and (225◦; 0)\nWorked example 21: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(3θ + 60◦) for 0◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin k(θ + p).\nf(θ) = sin(3θ + 60◦)\n= sin 3(θ + 20◦)\n207\nChapter 5.\nFunctions\n\nTo draw a graph of the above equation, the standard sine graph, y = sin θ, must be\nchanged in the following ways:\n• decrease the period by a factor of 3;\n• shift to the left by 20◦.\nStep 2: Complete a table of values\nθ\n0◦\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nf(θ)\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nθ\nf(θ)\nf(θ) = sin 3(θ + 20◦)\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 180◦]\nRange: [−1; 1]\nMaximum turning point: (10◦; 1) and (130◦; 1)\nMinimum turning point: (70◦; −1)\ny-intercept: (0◦; 0,87)\nx-intercepts: (40◦; 0), (100◦; 0) and (160◦; 0)\nExercise 5 – 23: The sine function\n1. Sketch the following graphs on separate axes:\na) y = 2 sin θ\n2 for −360◦≤θ ≤360◦\nb) f(θ) = 1\n2 sin(θ −45◦) for −90◦≤θ ≤90◦\nc) y = sin(θ + 90◦) + 1 for 0◦≤θ ≤360◦\nd) y = sin(−3θ\n2 ) for −180◦≤θ ≤180◦\ne) y = sin(30◦−θ) for −360◦≤θ ≤360◦\n2. Given the graph of the function y = a sin(θ + p), determine the values of a and\np.\n208\n5.5.\nThe sine function\n\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nθ\nf(θ)\nCan you describe this graph in terms of cos θ?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22V4\n1b. 22V5\n1c. 22V6\n1d. 22V7\n1e. 22V8\n2. 22V9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n5.6\nThe cosine function\nEMBH3\nRevision\nEMBH4\nFunctions of the form y = cos θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• The period is 360◦and the amplitude is 1.\n• Domain: [0◦; 360◦]\nFor y = cos θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (90◦; 0), (270◦; 0)\n• y-intercept: (0◦; 1)\n• Maximum turning points: (0◦; 1), (360◦; 1)\n• Minimum turning point: (180◦; −1)\n209\nChapter 5.\nFunctions\n\nFunctions of the form y = a cos θ + q\nCosine functions of the general form y = a cos θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a cos θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 24: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function in the previous problem determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n210\n5.6.\nThe cosine function\n\n1. y1 = cos θ\n2. y2 = −3 cos θ\n3. y3 = cos θ + 2\n4. y4 = 1\n2 cos θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VB\n2. 22VC\n3. 22VD\n4. 22VF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos(kθ)\nEMBH5\nWe now consider cosine functions of the form y = cos kθ and the effects of k.\nInvestigation: The effects of k on a cosine graph\n1. Complete the following table for y1 = cos θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ncos θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ncos θ\n2. Use the table of values to plot the graph of y1 = cos θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = cos(−θ)\nb) y3 = cos 3θ\nc) y4 = cos 3θ\n4\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n211\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = cos θ and y2 = cos(−θ)?\n6. Is cos(−θ) = −cos θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = cos kθ?\nThe effect of the parameter k on y = cos kθ\nThe value of k affects the period of the cosine function.\n• For k > 0:\nFor k > 1, the period of the cosine function decreases.\nFor 0 < k < 1, the period of the cosine function increases.\n• For k < 0:\nFor −1 < k < 0, the period increases.\nFor k < −1, the period decreases.\nNegative angles:\ncos(−θ) = cos θ\nNotice that for negative values of θ, the graph is not reflected about the x-axis.\nCalculating the period:\nTo determine the period of y = cos kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k.\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n212\n5.6.\nThe cosine function\n\nWorked example 22: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = cos θ\nb) y2 = cos θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos kθ\nNotice that for y2 = cos θ\n2, k < 1 therefore the period of the graph increases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ncos θ\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\n−0,71\n−1\ncos θ\n2\n0\n0,38\n0,71\n0,92\n1\n0,92\n0,71\n0,38\n0\nStep 3: Sketch the cosine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = cos θ\n2\ny2 = cos θ\nStep 4: Complete the table\ny1 = cos θ\ny2 = cos θ\n2\nperiod\n360◦\n720◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[0; 1]\nmaximum turning points\n(0◦; 1)\n(0◦; 1)\nminimum turning points\n(−180◦; −1) and (180◦; −1)\nnone\ny-intercept(s)\n(0◦; 1)\n(0◦; 1)\nx-intercept(s)\n(−90◦; 0) and (90◦; 0)\n(−180◦; 0) and (180◦; 0)\n213\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos kθ\n= cos 0◦\n= 1\nThis gives the point (0◦; 1).\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = cos 2θ\nb) g(θ) = cos θ\n3\nc) h(θ) = cos(−2θ)\nd) k(θ) = cos 3θ\n4\n2. For each graph of the form f(θ) = cos kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n214\n5.6.\nThe cosine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nbA(135◦; 0)\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VG\n1b. 22VH\n1c. 22VJ\n1d. 22VK\n2a. 22VM\n2b. 22VN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos (θ + p)\nEMBH6\nWe now consider cosine functions of the form y = cos(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a cosine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = cos θ\nb) y2 = cos(θ −90◦)\nc) y3 = cos(θ −60◦)\nd) y4 = cos(θ + 90◦)\ne) y5 = cos(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning points\nminimum turning points\ny-intercept(s)\nx-intercept(s)\neffect of p\n215\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = cos(θ + p)\nThe effect of p on the cosine function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the cosine function shifts to the left by p degrees.\n• For p < 0, the graph of the cosine function shifts to the right by p degrees.\np > 0\np < 0\nWorked example 23: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = cos θ\nb) y2 = cos(θ + 30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos(θ + p)\nNotice that for y1 = cos θ we have p = 0 (no phase shift) and for y2 = cos(θ + 30◦),\np < 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\ncos θ\n1\n0\n−1\n0\n1\n0\n−1\n0\n1\ncos(θ +30◦)\n0,87\n−0,5\n−0,87\n0,5\n0,87\n−0,5\n−0,87\n0,5\n0,87\n216\n5.6.\nThe cosine function\n\nStep 3: Sketch the cosine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = cos θ\ny2 = cos(θ + 30◦)\nStep 4: Complete the table\ny1\ny2\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−360◦; 1), (0◦; 1) and\n(360◦; 1)\n(−30◦; 1) and (330◦; 1)\nminimum turning points\n(−180◦; −1) and\n(180◦; −1)\n(−210◦; −1) and\n(150◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,87)\nx-intercept(s)\n(−270◦; 0), (−90◦; 0),\n(90◦; 0) and (270◦; 0)\n(−300◦; 0), (−120◦; 0),\n(60◦; 0) and (240◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos(θ + p)\n= cos(0◦+ p)\n= cos p\nThis gives the point (0◦; cos p).\n217\nChapter 5.\nFunctions\n\nExercise 5 – 26: Cosine functions of the form y = cos(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = cos(θ + 45◦)\n2. g(θ) = cos(θ −30◦)\n3. h(θ) = cos(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VP\n2. 22VQ\n3. 22VR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching cosine graphs\nEMBH7\nWorked example 24: Sketching a cosine graph\nQUESTION\nSketch the graph of f(θ) = cos(180◦−3θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = cos k(θ + p).\nf(θ) = cos(180◦−3θ)\n= cos(−3θ + 180◦)\n= cos (−3(θ −60◦))\n= cos 3(θ −60◦)\nTo draw a graph of the above function, the standard cosine graph, y = cos θ, must be\nchanged in the following ways:\n218\n5.6.\nThe cosine function\n\n• decrease the period by a factor of 3\n• shift to the right by 60◦.\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n−1\n0,71\n0\n−0,71\n1\n−0,71\n0\n0,71\n−1\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = cos 3(θ −60◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (60◦; 1), (180◦; 1) and (300◦; 1)\nMinimum turning point: (0◦; −1), (120◦; −1), (240◦; −1) and (360◦; −1)\ny-intercepts: (0◦; −1)\nx-intercept: (30◦; 0), (90◦; 0), (150◦; 0), (210◦; 0), (270◦; 0) and (330◦; 0)\nWorked example 25: Finding the equation of a cosine graph\nQUESTION\nGiven the graph of y = a cos(kθ+p), determine the values of a, k, p and the minimum\nturning point.\nθ\ny\ny = a cos(θ + p)\nb\n(45◦; 2)\n−45◦\n315◦\n219\nChapter 5.\nFunctions\n\nSOLUTION\nStep 1: Determine the value of k\nFrom the sketch we see that the period of the graph is 360◦, therefore k = 1.\ny = a cos(θ + p)\nStep 2: Determine the value of a\nFrom the sketch we see that the maximum turning point is (45◦; 2), so we know that\nthe amplitude of the graph is 2 and therefore a = 2.\ny = 2 cos(θ + p)\nStep 3: Determine the value of p\nCompare the given graph with the standard cosine function y = cos θ and notice the\ndifference in the maximum turning points. We see that the given function has been\nshifted to the right by 45◦, therefore p = 45◦.\ny = 2 cos(θ −45◦)\nStep 4: Determine the minimum turning point\nAt the minimum turning point, y = −2:\ny = 2 cos(θ −45◦)\n−2 = 2 cos(θ −45◦)\n−1 = cos(θ −45◦)\ncos−1(−1) = θ −45◦\n180◦= θ −45◦\n225◦= θ\nThis gives the point (225◦; −2).\n220\n5.6.\nThe cosine function\n\nExercise 5 – 27: The cosine function\n1. Sketch the following graphs on separate axes:\na) y = cos(θ + 15◦) for −180◦≤θ ≤180◦\nb) f(θ) = 1\n3 cos(θ −60◦) for −90◦≤θ ≤90◦\nc) y = −2 cos θ for 0◦≤θ ≤360◦\nd) y = cos(30◦−θ) for −360◦≤θ ≤360◦\ne) g(θ) = 1 + cos(θ −90◦) for 0◦≤θ ≤360◦\nf) y = cos(2θ + 60◦) for −360◦≤θ ≤360◦\n2. Two girls are given the following graph:\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\ny\nb\na) Audrey decides that the equation for the graph is a cosine function of the\nform y = a cos θ. Determine the value of a.\nb) Megan thinks that the equation for the graph is a cosine function of the\nform y = cos(θ + p). Determine the value of p.\nc) What can they conclude?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VS\n1b. 22VT\n1c. 22VV\n1d. 22VW\n1e. 22VX\n1f. 22VY\n2. 22VZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n221\nChapter 5.\nFunctions\n\n5.7\nThe tangent function\nEMBH8\nRevision\nEMBH9\nFunctions of the form y = tan θ for 0◦≤θ ≤360◦\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nf(θ)\nθ\nThe dashed vertical lines are called the asymptotes. The asymptotes are at the values\nof θ where tan θ is not defined.\n• Period: 180◦\n• Domain: {θ : 0◦≤θ ≤360◦, θ ̸= 90◦; 270◦}\n• Range: {f(θ) : f(θ) ∈R}\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Asymptotes: the lines θ = 90◦and θ = 270◦\nFunctions of the form y = a tan θ + q\nTangent functions of the general form y = a tan θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a tan θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted vertically upwards by q units.\n– For q < 0, f(θ) is shifted vertically downwards by q units.\n• The effect of a on shape\n– For a > 1, branches of f(θ) are steeper.\n– For 0 < a < 1, branches of f(θ) are less steep and curve more.\n222\n5.7.\nThe tangent function\n\n– For a < 0, there is a reflection about the x-axis.\n– For −1 < a < 0, there is a reflection about the x-axis and the branches of\nthe graph are less steep.\n– For a < −1, there is a reflection about the x-axis and the branches of the\ngraph are steeper.\na < 0\na > 0\nq > 0\nb\n0\nb\n0\nq = 0\nb\n0\nb\n0\nq < 0\nb\n0\nb\n0\nExercise 5 – 28: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function determine the following:\n• Period\n• Domain and range\n223\nChapter 5.\nFunctions\n\n• x- and y-intercepts\n• Asymptotes\n1. y1 = tan θ −1\n2\n2. y2 = −3 tan θ\n3. y3 = tan θ + 2\n4. y4 = 2 tan θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W2\n2. 22W3\n3. 22W4\n4. 22W5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = tan(kθ)\nEMBHB\nInvestigation: The effects of k on a tangent graph\n1. Complete the following table for y1 = tan θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ntan θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ntan θ\n2. Use the table of values to plot the graph of y1 = tan θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = tan(−θ)\nb) y3 = tan 3θ\nc) y4 = tan θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of k\n224\n5.7.\nThe tangent function\n\n5. What do you notice about y1 = tan θ and y2 = tan(−θ)?\n6. Is tan(−θ) = −tan θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = tan kθ?\nThe effect of the parameter on y = tan kθ\nThe value of k affects the period of the tangent function. If k is negative, then the\ngraph is reflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the tangent function decreases.\nFor 0 < k < 1, the period of the tangent function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\ntan(−θ) = −tan θ\nCalculating the period:\nTo determine the period of y = tan kθ we use,\nPeriod = 180◦\n|k|\nwhere |k| is the absolute value of k.\nk > 0\nk < 0\n225\nChapter 5.\nFunctions\n\nWorked example 26: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan 3θ\n2\n2. For each function determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan kθ\nNotice that k > 1 for y2 = tan 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan 3θ\n2\nUNDEF\n−0,41\n1\n−2,41\n0\n2,41\n−1\n0,41\nUNDEF\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan 3θ\n2\nperiod\n180◦\n120◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦< θ < 180◦, θ ̸=\n−60◦; 60◦}\nrange\n{f(θ) : f(θ) ∈R}\n{f(θ) : f(θ) ∈R}\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −180◦; −60◦and 180◦\n226\n5.7.\nThe tangent function\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan kθ:\nDomain and range\nThe domain of one branch is {θ : −90◦\nk\n< θ < 90◦\nk , θ ∈R} because f(θ) is undefined\nfor θ = −90◦\nk and θ = 90◦\nk .\nThe range is {f(θ) : f(θ) ∈R} or (−∞; ∞).\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0 and solving for f(θ).\ny = tan kθ\n= tan 0◦\n= 0\nThis gives the point (0◦; 0).\nAsymptotes\nThese are the values of kθ for which tan kθ is undefined.\nExercise 5 – 29: Tangent functions of the form y = tan kθ\nSketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan 2θ\n2. g(θ) = tan 3θ\n4\n3. h(θ) = tan(−2θ)\n4. k(θ) = tan 2θ\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W6\n2. 22W7\n3. 22W8\n4. 22W9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n227\nChapter 5.\nFunctions\n\nFunctions of the form y = tan (θ + p)\nEMBHC\nWe now consider tangent functions of the form y = tan(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a tangent graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = tan θ\nb) y2 = tan(θ −60◦)\nc) y3 = tan(θ −90◦)\nd) y4 = tan(θ + 60◦)\ne) y5 = tan(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of p\nThe effect of the parameter on y = tan(θ + p)\nThe effect of p on the tangent function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the tangent function shifts to the left by p.\n• For p < 0, the graph of the tangent function shifts to the right by p.\np > 0\np < 0\n228\n5.7.\nThe tangent function\n\nWorked example 27: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan(θ + 30◦)\nFor each function determine the following:\n2.\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan(θ + p)\nNotice that for y1 = tan θ we have p = 0◦(no phase shift) and for y2 = tan(θ + 30◦),\np > 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−180◦−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan(θ+30◦)\n0,58\n3,73\n−1,73\n−0,27\n0,58\n3,73\n−1,73\n−0,27\n0,58\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan(θ + 30◦)\nperiod\n180◦\n180◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦≤θ ≤\n180◦, θ ̸= −120◦; 60◦}\nrange\n(−∞; ∞)\n(−∞; ∞)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,58)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and (180◦; 0)\n(−30◦; 0) and (150◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −120◦and θ = 60◦\n229\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan(θ + p):\nDomain and range\nThe domain of one branch is {θ : θ ∈(−90◦−p; 90◦−p)} because the function is\nundefined for θ = −90◦−p and θ = 90◦−p.\nThe range is {f(θ) : f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = tan(θ + p)\n= tan(0◦+ p)\n= tan p\nThis gives the point (0◦; tan p).\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan(θ + 45◦)\n2. g(θ) = tan(θ −30◦)\n3. h(θ) = tan(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WB\n2. 22WC\n3. 22WD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n230\n5.7.\nThe tangent function\n\nSketching tangent graphs\nEMBHD\nWorked example 28: Sketching a tangent graph\nQUESTION\nSketch the graph of f(θ) = tan 1\n2(θ −30◦) for −180◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that 0 < k < 1, therefore the branches of the graph will be\nless steep than the standard tangent graph y = tan θ. We also notice that p < 0 so the\ngraph will be shifted to the right on the x-axis.\nStep 2: Determine the period\nThe period for f(θ) = tan 1\n2(θ −30◦) is:\nPeriod = 180◦\n|k|\n= 180◦\n1\n2\n= 360◦\nStep 3: Determine the asymptotes\nThe standard tangent graph, y = tan θ, for −180◦≤θ ≤180◦is undefined at θ = −90◦\nand θ = 90◦. Therefore we can determine the asymptotes of f(θ) = tan 1\n2(θ −30◦):\n•\n−90◦\n0,5 + 30◦= −150◦\n•\n90◦\n0,5 + 30◦= 210◦\nThe asymptote at θ = 210◦lies outside the required interval.\n231\nChapter 5.\nFunctions\n\nStep 4: Plot the points and join with a smooth curve\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 360◦\nDomain: {θ : −180◦≤θ ≤180◦, θ ̸= −150◦}\nRange: (−∞; ∞)\ny-intercepts: (0◦; −0,27)\nx-intercept: (30◦; 0)\nAsymptotes: θ = −150◦\nExercise 5 – 31: The tangent function\n1. Sketch the following graphs on separate axes:\na) y = tan θ −1 for −90◦≤θ ≤90◦\nb) f(θ) = −tan 2θ for 0◦≤θ ≤90◦\nc) y = 1\n2 tan(θ + 45◦) for 0◦≤θ ≤360◦\nd) y = tan(30◦−θ) for −180◦≤θ ≤180◦\n2. Given the graph of y = a tan kθ, determine the values of a and k.\nθ\nf(θ)\nb\nb\n(90◦; −1)\n360◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WF\n1b. 22WG\n1c. 22WH\n1d. 22WJ\n2. 22WK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n232\n5.7.\nThe tangent function\n\nExercise 5 – 32: Mixed exercises\n1. Determine the equation for each of the following:\na) f(θ) = a sin kθ and g(θ) = a tan θ\nθ\ny\nb\nb\nf\n(45◦; −3\n2 )\ng\n(180◦; 0)\n(135◦; −1 1\n2 )\nb) f(θ) = a sin kθ and g(θ) = a cos(θ + p)\nθ\n0\ny\nb\n(−90◦; 2)\n−180◦\n180◦\nf and g\nc) y = a tan kθ\nθ\n0\ny\nb\n(90◦; 3)\n360◦\n180◦\nd) y = a cos θ + q\nθ\n0\ny\n4\n360◦\n180◦\n233\nChapter 5.\nFunctions\n\n2. Given the functions f(θ) = 2 sin θ and g(θ) = cos θ + 1:\na) Sketch the graphs of both functions on the same system of axes, for 0◦≤\nθ ≤360◦. Indicate the turning points and intercepts on the diagram.\nb) What is the period of f?\nc) What is the amplitude of g?\nd) Use your sketch to determine how many solutions there are for the equation\n2 sin θ −cos θ = 1. Give one of the solutions.\ne) Indicate on your sketch where on the graph the solution to 2 sin θ = −1 is\nfound.\n3. The sketch shows the two functions f(θ) = a cos θ and g(θ) = tan θ for 0◦≤θ ≤\n360◦. Points P(135◦; b) and Q(c; −1) lie on g(θ) and f(θ) respectively.\nθ\n0\ny\nb\nb\n360◦\n180◦\nP\nQ\ng\nf\n−1\n2\n−2\na) Determine the values of a, b and c.\nb) What is the period of g?\nc) Solve the equation cos θ = 1\n2 graphically and show your answer(s) on the\ndiagram.\nd) Determine the equation of the new graph if g is reflected about the x-axis\nand shifted to the right by 45◦.\n4. Sketch the graphs of y1 = −1\n2 sin(θ + 30◦) and y2 = cos(θ −60◦), on the same\nsystem of axes for 0◦≤θ ≤360◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WM\n1b. 22WN\n1c. 22WP\n1d. 22WQ\n2. 22WR\n3. 22WS\n4. 22WT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n234\n5.7.\nThe tangent function\n\n5.8\nSummary\nEMBHF\nSee presentation: 22WV at www.everythingmaths.co.za\n1. Parabolic functions:\nStandard form: y = ax2 + bx + c\n• y-intercept: (0; c)\n• x-intercept: x = −b±\n√\nb2−4ac\n2a\n• Turning point:\n\u0010\n−b\n2a; −b2\n4a + c\n\u0011\n• Axis of symmetry: x = −b\n2a\nCompleted square form: y = a(x + p)2 + q\n• Turning point: (−p; q)\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n2. Average gradient:\n• Average gradient = y2−y1\nx2−x1\n3. Hyperbolic functions:\nStandard form: y = k\nx\n• k > 0: first and third quadrant\n• k < 0: second and fourth quadrant\nShifted form: y =\nk\nx+p + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: x = −p and y = q\n4. Exponential functions:\nStandard form: y = abx\n• a > 0: above x-axis\n• a < 0: below x-axis\n• b > 1: increasing function if a > 0; decreasing function if a < 0\n• 0 < b < 1: decreasing function if a > 0; increasing function if a < 0\nShifted form: y = ab(x+p) + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n235\nChapter 5.\nFunctions\n\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: y = q\n5. Sine functions:\nShifted form: y = a sin(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• sin(−θ) = −sin θ\n6. Cosine functions:\nShifted form: y = a cos(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• cos(−θ) = cos θ\n7. Tangent functions:\nShifted form: y = a tan(kθ + p) + q\n• Period = 180◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• tan(−θ) = −tan θ\n• Asymptotes: 90◦−p\nk\n± 180◦n\nk\n, n ∈Z\n236\n5.8.\nSummary\n\nExercise 5 – 33: End of chapter exercises\n1. Show that if a\n<\n0,\nthen the range of f(x)\n=\na(x + p)2 + q is\n{f(x) : f(x) ∈(−∞, q]}.\n2. If (2; 7) is the turning point of f(x) = −2x2 −4ax + k, find the values of the\nconstants a and k.\n3. The following graph is represented by the equation f(x) = ax2 + bx. The coor-\ndinates of the turning point are (3; 9). Show that a = −1 and b = 6.\nb (3; 9)\nx\n0\ny\n4. Given: f(x) = x2 −2x + 3. Give the equation of the new graph originating if:\na) the graph of f is moved three units to the left.\nb) the x-axis is moved down three units.\n5. A parabola with turning point (−1; −4) is shifted vertically by 4 units upwards.\nWhat are the coordinates of the turning point of the shifted parabola?\n6. Plot the graph of the hyperbola defined by y = 2\nx for −4 ≤x ≤4. Suppose\nthe hyperbola is shifted 3 units to the right and 1 unit down. What is the new\nequation then?\n7. Based on the graph of y =\nk\n(x+p) + q, determine the equation of the graph with\nasymptotes y = 2 and x = 1 and passing through the point (2; 3).\ny\nx\n0\n2\n1\nb (2; 3)\n237\nChapter 5.\nFunctions\n\n8. The columns in the table below give the y-values for the following functions:\ny = ax, y = ax+1 and y = ax + 1. Match each function to the correct column.\nx\nA\nB\nC\n−2\n7,25\n6,25\n2,5\n−1\n3,5\n2,5\n1\n0\n2\n1\n0,4\n1\n1,4\n0,4\n0,16\n2\n1,16\n0,16\n0,064\n9. The graph of f(x) = 1 + a . 2x (a is a constant) passes through the origin.\na) Determine the value of a.\nb) Determine the value of f(−15) correct to five decimal places.\nc) Determine the value of x, if P (x; 0,5) lies on the graph of f.\nd) If the graph of f is shifted 2 units to the right to give the function h, write\ndown the equation of h.\n10. The graph of f(x) = a . bx (a ̸= 0) has the point P (2; 144) on f.\na) If b = 0,75, calculate the value of a.\nb) Hence write down the equation of f.\nc) Determine, correct to two decimal places, the value of f(13).\nd) Describe the transformation of the curve of f to h if h(x) = f(−x).\n11. Using your knowledge of the effects of p and k draw a rough sketch of the fol-\nlowing graphs without a table of values.\na) y = sin 3θ for −180◦≤θ ≤180◦\nb) y = −cos 2θ for 0◦≤θ ≤180◦\nc) y = tan 1\n2θ for 0◦≤θ ≤360◦\nd) y = sin(θ −45◦) for −360◦≤θ ≤360◦\ne) y = cos(θ + 45◦) for 0◦≤θ ≤360◦\nf) y = tan(θ −45◦) for 0◦≤θ ≤360◦\ng) y = 2 sin 2θ for −180◦≤θ ≤180◦\nh) y = sin(θ + 30◦) + 1 for −360◦≤θ ≤0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WW\n2. 22WX\n3. 22WY\n4. 22WZ\n5. 22X2\n6. 22X3\n7. 22X4\n8. 22X5\n9. 22X6\n10. 22X7\n11a. 22X8\n11b. 22X9\n11c. 22XB\n11d. 22XC\n11e. 22XD\n11f. 22XF\n11g. 22XG\n11h. 22XH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n238\n5.8.\nSummary\n\nCHAPTER\n6\nTrigonometry\n6.1\nRevision\n240\n6.2\nTrigonometric identities\n247\n6.3\nReduction formula\n253\n6.4\nTrigonometric equations\n266\n6.5\nArea, sine, and cosine rules\n280\n6.6\nSummary\n301\n\n6\nTrigonometry\n6.1\nRevision\nEMBHG\nTrigonometric ratios\nb\nb\nb\ny\nx\nP(x; y)\nQ(−x; y)\nO\nα\nβ\nr\nr\nWe plot the points P(x; y) and Q(−x; y) in the Cartesian plane and measure the angles\nfrom the positive x-axis to the terminal arms (OP and OQ).\nP(x; y) lies in the first quadrant with P ˆOX = α and Q(−x; y) lies in the second\nquadrant with Q ˆOX = β.\nUsing the theorem of Pythagoras we have that\nOP 2 = x2 + y2\nAnd OQ2 = (−x)2 + y2\n= x2 + y2\n∴OP = OQ\nLet OP = OQ = r.\nTrigonometric ratios\nsin α = y\nr\ncos α = x\nr\ntan α = y\nx\nIn the second quadrant we notice that −x < 0\nsin β = y\nr\ncos β = −x\nr\ntan β = −y\nx\n240\n6.1.\nRevision\n\nSimilarily, in the third and fourth quadrants the sign of the trigonometric ratios depends\non the signs of x and y:\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nSpecial angles\n30◦\n60◦\n1\n√\n3\n2\n45◦\n45◦\n1\n1\n√\n2\nθ\n0◦\n30◦\n45◦\n60◦\n90◦\ncos θ\n1\n√\n3\n2\n1\n√\n2\n1\n2\n0\nsin θ\n0\n1\n2\n1\n√\n2\n√\n3\n2\n1\ntan θ\n0\n1\n√\n3\n1\n√\n3\nundef\nSee video: 22XJ at www.everythingmaths.co.za\n241\nChapter 6.\nTrigonometry\n\nSolving equations\nWorked example 1: Solving equations\nQUESTION\nDetermine the values of a and b in the right-angled triangle TUW (correct to one\ndecimal place):\nT\nW\nU\n47◦\nb\n30\na\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of a\nsin θ = opposite side\nhypotenuse\nsin 47◦= 30\na\na =\n30\nsin 47◦\n∴a = 41,0\nStep 3: Determine the value of b\nAlways try to use the information that is given for calculations and not answers that you\nhave worked out in case you have made an error. For example, avoid using a = 41,0\nto determine the value of b.\ntan θ = opposite side\nadjacent side\ntan 47◦= 30\nb\nb =\n30\ntan 47◦\n∴b = 28,0\nStep 4: Write the final answer\na = 41,0 units and b = 28,0 units.\n242\n6.1.\nRevision\n\nFinding an angle\nWorked example 2: Finding an angle\nQUESTION\nCalculate the value of θ in the right-angled triangle MNP (correct to one decimal\nplace):\nP\nN\nM\n41\nθ\n24\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of θ\ntan θ = opposite side\nadjacent side\ntan θ = 41\n24\n∴θ = tan−1\n\u001241\n24\n\u0013\nθ = 59,7◦\n243\nChapter 6.\nTrigonometry\n\nWorked example 3: Finding an angle\nQUESTION\nGiven 2 sin θ\n2 = cos 43◦, for θ ∈[0◦; 90◦], determine the value of θ (correct to one\ndecimal place).\nSOLUTION\nStep 1: Simplify the equation\nAvoiding rounding off in calculations until you have determined the final answer. In\nthe calculation below, the dots indicate that the number has not been rounded so that\nthe answer is as accurate as possible.\n2 sin θ\n2 = cos 43◦\nsin θ\n2 = cos 43◦\n2\nθ\n2 = sin−1(0,365 . . .)\nθ = 2(21,449 . . .)\n∴θ = 42,9◦\nTwo-dimensional problems\nWorked example 4: Flying a kite\nQUESTION\nThelma flies a kite on a 22 m piece of string and the height of the kite above the\nground is 20,4 m. Determine the angle of inclination of the string (correct to one\ndecimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the opposite and adjacent sides and the hy-\npotenuse\nLet the angle of inclination of the string be θ.\n244\n6.1.\nRevision\n\nKite\nThelma\n20,4\nθ\n22\nStep 2: Use an appropriate trigonometric ratio to find θ\nsin θ = opposite side\nhypotenuse\n= 20,4\n22\nθ = sin−1(0,927 . . .)\n∴θ = 68,0◦\nExercise 6 – 1: Revision\n1. If p = 49◦and q = 32◦, use a calculator to determine whether the following\nstatements are true of false:\na) sin p + 3 sin p = 4 sin p\nb) sin q\ncos q = tan q\nc) cos(p −q) = cos p −cos q\nd) sin(2p) = 2 sin p cos p\n2. Determine the following angles (correct to one decimal place):\na) cos α = 0,64\nb) sin θ + 2 = 2,65\nc) 1\n2 cos 2β = 0,3\nd) tan θ\n3 = sin 48◦\ne) cos 3p = 1,03\nf) 2 sin 3β + 1 = 2,6\ng) sin θ\ncos θ = 42\n3\n3. In △ABC, A ˆCB = 30◦, AC = 20 cm and BC = 22 cm. The perpendicular\nline from A intersects BC at T.\n245\nChapter 6.\nTrigonometry\n\nDetermine:\nA\nC\nB\nT\n20 cm\n22 cm\n30◦\na) the length TC\nb) the length AT\nc) the angle B ˆAT\n4. A rhombus has a perimeter of 40 cm and one of the internal angles is 30◦.\na) Determine the length of the sides.\nb) Determine the lengths of the diagonals.\nc) Calculate the area of the rhombus.\n5. Simplify the following without using a calculator:\na) 2 sin 45◦× 2 cos 45◦\nb) cos2 30◦−sin2 60◦\nc) sin 60◦cos 30◦−cos 60◦sin 30◦−tan 45◦\nd) 4 sin 60◦cos 30◦−2 tan 45◦+ tan 60◦−2 sin 60◦\ne) sin 60◦×\n√\n2 tan 45◦+ 1 −sin 30◦\n6. Given the diagram below.\nb\nx\ny\n0\nB(2; 2\n√\n3)\nβ\nDetermine the following without using a calculator:\na) β\nb) cos β\nc) cos2 β + sin2 β\n7. The 10 m ladder of a fire truck leans against the wall of a burning building at an\nangle of 60◦. The height of an open window is 9 m from the ground. Will the\nladder reach the window?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22XK\n2a. 22XM\n2b. 22XN\n2c. 22XP\n2d. 22XQ\n2e. 22XR\n2f. 22XS\n2g. 22XT\n3. 22XV\n4. 22XW\n5a. 22XX\n5b. 22XY\n5c. 22XZ\n5d. 22Y2\n5e. 22Y3\n6. 22Y4\n7. 22Y5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n246\n6.1.\nRevision\n\n6.2\nTrigonometric identities\nEMBHH\nAn identity is a mathematical statement that equates one quantity with another. Trigono-\nmetric identities allow us to simplify a given expression so that it contains sine and co-\nsine ratios only. This enables us to solve equations and also to prove other identities.\nQuotient identity\nInvestigation: Quotient identity\n1. Complete the table without using a calculator, leaving your answer in surd form\nwhere applicable:\nθ = 45◦\nθ\n3\n5\nx\ny\n(3; 2)\nθ\nb\nsin θ\ncos θ\nsin θ\ncos θ\ntan θ\n2. Examine the last two rows of the table and make a conjecture.\n3. Are there any values of θ for which your conjecture would not be true? Explain\nyour answer.\nWe know that tan θ is defined as:\ntan θ = opposite side\nadjacent side\nUsing the diagram below and the theorem of Pythagoras, we can write the tangent\nfunction in terms of x, y and r:\nx\ny\n(x; y)\nθ\nb\nO\n247\nChapter 6.\nTrigonometry\n\ntan θ = y\nx\n= y\nx × r\nr\n= y\nr × r\nx\n= y\nr ÷ x\nr\n= sin θ ÷ cos θ\n= sin θ\ncos θ\nThis is the quotient identity:\ntan θ = sin θ\ncos θ\nNotice that tan θ is undefined if cos θ = 0, therefore θ ̸= k × 90◦, where k is an odd\ninteger.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\ntan θ\nSquare identity\nInvestigation: Square identity\n1. Use a calculator to complete the following table:\nsin2 80◦+ cos2 80◦=\ncos2 23◦+ sin2 23◦=\nsin 50◦+ cos 50◦=\nsin2 67◦−cos2 67◦=\nsin2 67◦+ cos2 67◦=\n2. What do you notice? Make a conjecture.\n3. Draw a sketch and prove your conjecture in general terms, using x, y and r.\n248\n6.2.\nTrigonometric identities\n\nx\ny\n(x; y)\nα\nb\nO\nr\nUsing the theorem of Pythagoras, we can write the sine and cosine functions in terms\nof x, y and r:\nsin2 θ + cos2 θ =\n\u0010y\nr\n\u00112\n+\n\u0010x\nr\n\u00112\n= y2\nr2 + x2\nr2\n= y2 + x2\nr2\n= r2\nr2\n= 1\nThis is the square identity:\nsin2 θ + cos2 θ = 1\nOther forms of the square identity\nComplete the following:\n1. sin2 θ = 1 −. . . . . .\n2. cos θ = ±√. . . . . .\n3. sin2 θ = (1 + . . . . . .)(1 −. . . . . .)\n4. cos2 θ −1 = . . . . . .\nHere are some useful tips for proving identities:\n• Change all trigonometric ratios to sine and cosine.\n• Choose one side of the equation to simplify and show that it is equal to the other\nside.\n• Usually it is better to choose the more complicated side to simplify.\n• Sometimes we need to simplify both sides of the equation to show that they are\nequal.\n• A square root sign often indicates that we need to use the square identity.\n• We can also add to the expression to make simplifying easier:\n– replace 1 with sin2 θ + cos2 θ.\n– multiply by 1 in the form of a suitable fraction, for example 1 + sin θ\n1 + sin θ.\n249\nChapter 6.\nTrigonometry\n\nSee video: 22Y6 at www.everythingmaths.co.za\nWorked example 5: Trigonometric identities\nQUESTION\nSimplify the following:\n1. tan2 θ × cos2 θ\n2.\n1\ncos2 θ −tan2 θ\nSOLUTION\nStep 1: Write the expression in terms of sine and cosine only\nWe use the square and quotient identities to write the given expression in terms of sine\nand cosine and then simplify as far as possible.\n1.\ntan2 θ × cos2 θ =\n\u0012 sin θ\ncos θ\n\u00132\n× cos2 θ\n= sin2 θ\ncos2 θ × cos2 θ\n= sin2 θ\n2.\n1\ncos2 θ −tan2 θ =\n1\ncos2 θ −\n\u0012 sin θ\ncos θ\n\u00132\n=\n1\ncos2 θ −sin2 θ\ncos2 θ\n= 1 −sin2 θ\ncos2 θ\n= cos2 θ\ncos2 θ\n= 1\n250\n6.2.\nTrigonometric identities\n\nWorked example 6: Trigonometric identities\nQUESTION\nProve: 1 −sin α\ncos α\n=\ncos α\n1 + sin α\nSOLUTION\nStep 1: Note restrictions\nWhen working with fractions, we must be careful that the denominator does not equal\n0. Therefore cos θ ̸= 0 for the fraction on the left-hand side and sin θ + 1 ̸= 0 for the\nfraction on the right-hand side.\nStep 2: Simplify the left-hand side\nThis is not an equation that needs to be solved. We are required to show that one side\nof the equation is equal to the other. We can choose either of the two sides to simplify.\nLHS = 1 −sin α\ncos α\n= 1 −sin α\ncos α\n× 1 + sin α\n1 + sin α\nNotice that we have not changed the equation — this is the same as multiplying by 1\nsince the numerator and the denominator are the same.\nStep 3: Determine the lowest common denominator and simplify\nLHS =\n1 −sin2 α\ncos α(1 + sin α)\n=\ncos2 α\ncos α(1 + sin α)\n=\ncos α\n1 + sin α\n= RHS\n251\nChapter 6.\nTrigonometry\n\nExercise 6 – 2: Trigonometric identities\n1. Reduce the following to one trigonometric ratio:\na) sin α\ntan α\nb) cos2 θ tan2 θ + tan2 θ sin2 θ\nc) 1 −sin θ cos θ tan θ\nd)\n\u00121 −cos2 β\ncos2 β\n\u0013\n−tan2 β\n2. Prove the following identities and state restrictions where appropriate:\na) 1 + sin θ\ncos θ\n=\ncos θ\n1 −sin θ\nb) sin2α + (cos α −tan α) (cos α + tan α) = 1 −tan2α\nc)\n1\ncos θ −cos θtan2θ\n1\n= cos θ\nd)\n2 sin θ cos θ\nsin θ + cos θ = sin θ + cos θ −\n1\nsin θ + cos θ\ne)\n\u0012cos β\nsin β + tan β\n\u0013\ncos β =\n1\nsin β\nf)\n1\n1 + sin θ +\n1\n1 −sin θ = d\n2 tan θ\nsin θ cos θ\ng) (1 + tan2 α) cos α\n(1 −tan α)\n=\n1\ncos α −sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Y7\n1b. 22Y8\n1c. 22Y9\n1d. 22YB\n2a. 22YC\n2b. 22YD\n2c. 22YF\n2d. 22YG\n2e. 22YH\n2f. 22YJ\n2g. 22YK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n252\n6.2.\nTrigonometric identities\n\n6.3\nReduction formula\nEMBHJ\nAny trigonometric function whose argument is 90◦± θ; 180◦± θ and 360◦± θ can be\nwritten simply in terms of θ.\nDeriving reduction formulae\nEMBHK\nInvestigation: Reduction formulae for function values of 180◦± θ\n1. Function values of 180◦−θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the y-axis, determine the coordi-\nnates of P ′.\nb) Write down values for sin θ, cos θ and tan θ.\nc) Use the coordinates for P ′ to determine sin(180◦−θ), cos(180◦−θ),\ntan(180◦−θ).\nd) From your results determine a relationship between the trigonometric func-\ntion values of (180◦−θ) and θ.\n253\nChapter 6.\nTrigonometry\n\n2. Function values of 180◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦+ θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the origin (the two points are sym-\nmetrical about both the x-axis and the y-axis), determine the coordinates of\nP ′.\nb) Use the coordinates for P ′ to determine sin(180◦+ θ), cos(180◦+ θ) and\ntan(180◦+ θ).\nc) From your results determine a relationship between the trigonometric func-\ntion values of (180◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(180◦−θ) = . . . . . .\nb) cos(180◦−θ) = . . . . . .\nc) tan(180◦−θ) = . . . . . .\nd) sin(180◦+ θ) = . . . . . .\ne) cos(180◦+ θ) = . . . . . .\nf) tan(180◦+ θ) = . . . . . .\n254\n6.3.\nReduction formula\n\nWorked example 7: Reduction formulae for function values of 180◦± θ\nQUESTION\nWrite the following as a single trigonometric ratio:\nsin 163◦\ncos 197◦+ tan 17◦+ cos(180◦−θ) × tan(180◦+ θ)\nSOLUTION\nStep 1: Use reduction formulae to write the trigonometric function values in terms\nof acute angles and θ\n= sin(180◦−17◦)\ncos(180◦+ 17◦) + tan 17◦+ (−cos θ) × tan θ\nStep 2: Simplify\n=\nsin 17◦\n−cos 17◦+ tan 17◦−cos θ × sin θ\ncos θ\n= −tan 17◦+ tan 17◦−sin θ\n= −sin θ\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1. Determine the value of the following expressions without using a calculator:\na) tan 150◦sin 30◦−cos 210◦\nb) (1 + cos 120◦)(1 −sin2 240◦)\nc) cos2 140◦+ sin2 220◦\n2. Write the following in terms of a single trigonometric ratio:\na) tan(180◦−θ) × sin(180◦+ θ)\nb) tan(180◦+ θ) cos(180◦−θ)\nsin(180◦−θ)\n255\nChapter 6.\nTrigonometry\n\n3. If t = tan 40◦, express the following in terms of t:\na) tan 140◦+ 3 tan 220◦\nb) cos 220◦\nsin 140◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YM\n1b. 22YN\n1c. 22YP\n2a. 22YQ\n2b. 22YR\n3. 22YS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of (360◦± θ) and (−θ)\n1. Function values of (360◦−θ) and (−θ)\nIn the Cartesian plane we measure angles from the positive x-axis to the terminal\narm, which means that an anti-clockwise rotation gives a positive angle. We can\ntherefore measure negative angles by rotating in a clockwise direction.\nFor an acute angle θ, we know that −θ will lie in the fourth quadrant.\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n360◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the x-axis (y = 0), determine the\ncoordinates of P ′.\nb) Use the coordinates of P ′ to determine sin(360◦−θ), cos(360◦−θ) and\ntan(360◦−θ).\nc) Use the coordinates of P ′ to determine sin(−θ), cos(−θ) and tan(−θ).\nd) From your results determine a relationship between the function values of\n(360◦−θ) and −θ.\n256\n6.3.\nReduction formula\n\ne) Complete the following reduction formulae:\ni. sin(360◦−θ) = . . . . . .\nii. cos(360◦−θ) = . . . . . .\niii. tan(360◦−θ) = . . . . . .\niv. sin(−θ) = . . . . . .\nv. cos(−θ) = . . . . . .\nvi. tan(−θ) = . . . . . .\n2. Function values of 360◦+ θ\nWe can also have an angle that is larger than 360◦. The angle completes a\nrevolution of 360◦and then continues to give an angle of θ.\nComplete the following reduction formulae:\na) sin(360◦+ θ) = . . . . . .\nb) cos(360◦+ θ) = . . . . . .\nc) tan(360◦+ θ) = . . . . . .\nFrom working with functions, we know that the graph of y = sin θ has a period of\n360◦. Therefore, one complete wave of a sine graph is the same as one complete\nrevolution for sin θ in the Cartesian plane.\n0\n1\n−1\n90◦\n180◦\n270◦\n360◦\n1st\n2nd\n3rd\n4th\npositive\npositive\nnegative\nnegative\n0◦/360◦\n90◦\n180◦\n270◦\n2nd\npos.\nneg.\n3rd\nneg.\n4th\n1st\npos.\nWe can also have multiple revolutions. The periodicity of the trigonometric graphs\nshows this clearly. A complete sine or cosine curve is completed in 360◦.\ny = cos θ\ny = sin θ\nθ\ny\n257\nChapter 6.\nTrigonometry\n\nIf k is any integer, then\nsin(k . 360◦+ θ) = sin θ\ncos(k . 360◦+ θ) = cos θ\ntan(k . 360◦+ θ) = tan θ\nWorked example 8: Reduction formulae for function values of 360◦± θ\nQUESTION\nIf f = tan 67◦, express the following in terms of f\nsin 293◦\ncos 427◦+ tan(−67◦) + tan 1147◦\nSOLUTION\nStep 1: Using reduction formula\n= sin(360◦−67◦)\ncos(360◦+ 67◦) −tan(67◦) + tan (3(360◦) + 67◦)\n= −sin 67◦\ncos 67◦−tan 67◦+ tan 67◦\n= −tan 67◦\n= −f\nWorked example 9: Using reduction formula\nQUESTION\nEvaluate without using a calculator:\ntan2 210◦−(1 + cos 120◦) sin2 405◦\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and special angles\n258\n6.3.\nReduction formula\n\n= tan2(180◦+ 30◦) −(1 + cos(180◦−60◦)) sin2(360◦+ 45◦)\n= tan2 30◦−(1 + (−cos 60◦)) sin2 45◦\n=\n\u0012 1\n√\n3\n\u00132\n−\n\u0012\n1 −1\n2\n\u0013 \u0012 1\n√\n2\n\u00132\n= 1\n3 −\n\u00121\n2\n\u0013 \u00121\n2\n\u0013\n= 1\n3 −1\n4\n= 1\n12\nExercise 6 – 4: Using reduction formula\n1. Simplify the following:\na) tan(180◦−θ) sin(360◦+ θ)\ncos(180◦+ θ) tan(360◦−θ)\nb) cos2(360◦+ θ) + cos(180◦+ θ) tan(360◦−θ) sin(360◦+ θ)\nc)\nsin(360◦+ α) tan(180◦+ α)\ncos(360◦−α) tan2(360◦+ α)\n2. Write the following in terms of cos β:\ncos(360◦−β) cos(−β) −1\nsin(360◦+ β) tan(360◦−β)\n3. Simplify the following without using a calculator:\na)\ncos 300◦tan 150◦\nsin 225◦cos(−45◦)\nb) 3 tan 405◦+ 2 tan 330◦cos 750◦\nc) cos 315◦cos 405◦+ sin 45◦sin 135◦\nsin 750◦\nd) tan 150◦cos 390◦−2 sin 510◦\ne) 2 sin 120◦+ 3 cos 765◦−2 sin 240◦−3 cos 45◦\n5 sin 300◦+ 3 tan 225◦−6 cos 60◦\n4. Given 90◦< α < 180◦, use a sketch to help explain why:\na) sin(−α) = −sin α\nb) cos(−α) = −cos α\n259\nChapter 6.\nTrigonometry\n\n5. If t = sin 43◦, express the following in terms of t:\na) sin 317◦\nb) cos2 403◦\nc) tan(−43◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YT\n1b. 22YV\n1c. 22YW\n2. 22YX\n3a. 22YY\n3b. 22YZ\n3c. 22Z2\n3d. 22Z3\n3e. 22Z4\n4. 22Z5\n5. 22Z6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of 90◦± θ\nIn any right-angled triangle, the two acute angles are complements of each other, ˆA +\nˆC = 90◦\nA\nB\nC\nb\na\nc\nComplete the following:\nIn △ABC\nsin ˆC = c\nb = cos . . .\ncos ˆC = a\nb = sin . . .\nComplementary angles are positive acute angles that add up to 90◦. For example 20◦\nand 70◦are complementary angles.\n260\n6.3.\nReduction formula\n\nIn the figure P(\n√\n3; 1) and P ′ lie on a circle with radius 2. OP makes an angle of\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\nP ′\n2\n2\n90◦−θ\n1. Function values of 90◦−θ\na) If points P and P ′ are symmetrical about the line y = x, determine the\ncoordinates of P ′.\nb) Use the coordinates for P ′ to determine sin(90◦−θ) and cos(90◦−θ).\nc) From your results determine a relationship between the function values of\n(90◦−θ) and θ.\n2. Function values of 90◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nθ\nb\nP\nO\nx\ny\nθ\nb\nP ′\n90◦+ θ\n2\n2\n261\nChapter 6.\nTrigonometry\n\na) If point P is rotated through 90◦to get point P ′, determine the coordinates\nof P ′.\nb) Use the coordinates for P ′ to determine sin(90◦+ θ) and cos(90◦+ θ).\nc) From your results determine a relationship between the function values of\n(90◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(90◦−θ) = . . . . . .\nb) cos(90◦−θ) = . . . . . .\nc) sin(90◦+ θ) = . . . . . .\nd) cos(90◦+ θ) = . . . . . .\nSine and cosine are known as co-functions. Two functions are called co-functions if\nf (A) = g (B) whenever A + B = 90◦(that is, A and B are complementary angles).\nThe function value of an angle is equal to the co-function of its complement.\nThus for sine and cosine we have\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nThe sine and cosine graphs illustrate this clearly: the two graphs are identical except\nthat they have a 90◦phase difference.\nθ\ny\ny = cos θ\ny = sin θ\n262\n6.3.\nReduction formula\n\nWorked example 10: Using the co-function rule\nQUESTION\nWrite each of the following in terms of sin 40◦:\n1. cos 50◦\n2. sin 320◦\n3. cos 230◦\n4. cos 130◦\nSOLUTION\n1. cos 50◦= sin(90◦−50◦) = sin 40◦\n2. sin 320◦= sin(360◦−40◦) = −sin 40◦\n3. cos 230◦= cos(180◦+ 50◦) = −cos 50◦= −cos(90◦−40◦) = −sin 40◦\n4. cos 130◦= cos(90◦+ 40◦) = −sin 40◦\nFunction values of θ −90◦\nWe can write sin(θ −90◦) as\nsin(θ −90◦) = sin [−(90◦−θ)]\n= −sin(90◦−θ)\n= −cos θ\nsimilarly, we can show that cos (θ −90◦) = sin θ\nTherefore, sin (θ −90◦) = −cos θ and cos (θ −90◦) = sin θ.\nWorked example 11: Co-functions\nQUESTION\nExpress the following in terms of t if t = sin θ:\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and co-functions\n263\nChapter 6.\nTrigonometry\n\nUse the CAST diagram to check in which quadrants the trigonometric ratios are positive\nand negative.\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\n=cos[−(90◦−θ)] cos[2(360◦) + θ] tan[−(360◦−θ)]\nsin2(360◦+ θ) cos(90◦+ θ)\n=sin θ cos θ tan θ\nsin2 θ(−sin θ)\n= −cos θ\n\u0000 sin θ\ncos θ\n\u0001\nsin2 θ\n= −\n1\nsin θ\n= −1\nt\nExercise 6 – 5: Co-functions\n1. Simplify the following:\na) cos(90◦+ θ) sin(θ + 90◦)\nsin(−θ)\nb) 2 sin(90◦−x) + sin(90◦+ x)\nsin(90◦−x) + cos(180◦+ x)\n2. Given cos 36◦= p, express the following in terms on p:\na) sin 54◦\nb) sin 36◦\nc) tan 126◦\nd) cos 324◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Z7\n1b. 22Z8\n2. 22Z9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n264\n6.3.\nReduction formula\n\nReduction formulae and co-functions:\n1. The reduction formulae hold for any angle θ. For convenience, we assume θ is\nan acute angle (0◦< θ < 90◦).\n2. When determining function values of (180◦±θ), (360◦±θ) and (−θ) the function\ndoes not change.\n3. When determining function values of (90◦±θ) and (θ±90◦) the function changes\nto its co-function.\nsecond quadrant (180◦−θ) or (90◦+ θ)\nfirst quadrant (θ) or (90◦−θ)\nsin(180◦−θ) = + sin θ\nall trig functions are positive\ncos(180◦−θ) = −cos θ\nsin(360◦+ θ) = sin θ\ntan(180◦−θ) = −tan θ\ncos(360◦+ θ) = cos θ\nsin(90◦+ θ) = + cos θ\ntan(360◦+ θ) = tan θ\ncos(90◦+ θ) = −sin θ\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nthird quadrant (180◦+ θ)\nfourth quadrant (360◦−θ)\nsin(180◦+ θ) = −sin θ\nsin(360◦−θ) = −sin θ\ncos(180◦+ θ) = −cos θ\ncos(360◦−θ) = + cos θ\ntan(180◦+ θ) = + tan θ\ntan(360◦−θ) = −tan θ\nExercise 6 – 6: Reduction formulae\n1. Write A and B as a single trigonometric ratio:\na) A = sin(360◦−θ) cos(180◦−θ) tan(360◦+ θ)\nb) B = cos(360◦+ θ) cos(−θ) sin(−θ)\ncos(90◦+ θ)\nc) Hence, determine:\ni. A + B = . . .\nii.\nA\nB = . . .\n2. Write the following as a function of an acute angle:\na) sin 163◦\nb) cos 327◦\nc) tan 248◦\nd) cos(−213◦)\n3. Determine the value of the following, without using a calculator:\na) sin(−30◦)\ntan(150◦) + cos 330◦\nb) tan 300◦cos 120◦\nc) (1 −cos 30◦)(1 −cos 210◦)\nd) cos 780◦−(sin 315◦)(cos 405◦)\n4. Prove that the following identity is true and state any restrictions:\nsin(180◦+ α) tan(360◦+ α) cos α\ncos(90◦−α)\n= sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22ZB\n2a. 22ZC\n2b. 22ZD\n2c. 22ZF\n2d. 22ZG\n3a. 22ZH\n3b. 22ZJ\n3c. 22ZK\n3d. 22ZM\n4. 22ZN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n265\nChapter 6.\nTrigonometry\n\n6.4\nTrigonometric equations\nEMBHM\nSolving trigonometric equations requires that we find the value of the angles that satisfy\nthe equation. If a specific interval for the solution is given, then we need only find the\nvalue of the angles within the given interval that satisfy the equation. If no interval is\ngiven, then we need to find the general solution. The periodic nature of trigonometric\nfunctions means that there are many values that satisfy a given equation, as shown in\nthe diagram below.\n1\n−1\n90◦180◦270◦360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nθ\n0\ny\ny = 0,5\ny = sin θ\nWorked example 12: Solving trigonometric equations\nQUESTION\nSolve for θ (correct to one decimal place), given tan θ = 5 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to solve for θ\ntan θ = 5\n∴θ = tan−1 5\n= 78,7◦\nThis value of θ is an acute angle which lies in the first quadrant and is called the\nreference angle.\nStep 2: Use the CAST diagram to determine in which quadrants tan θ is positive\nThe CAST diagram indicates that tan θ is positive in the first and third quadrants, there-\nfore we must determine the value of θ such that 180◦< θ < 270◦.\nUsing reduction formulae, we know that tan(180◦+ θ) = tan θ\nθ = 180◦+ 78,7◦\n∴θ = 258,7◦\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 78,7◦or θ = 258,7◦.\n266\n6.4.\nTrigonometric equations\n\nWorked example 13: Solving trigonometric equations\nQUESTION\nSolve for α (correct to one decimal place), given cos α = −0,7 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we do not include the negative sign. The reference\nangle must be an acute angle in the first quadrant, where all the trigonometric functions\nare positive.\nref ∠= cos−1 0,7\n= 45,6◦\nStep 2: Use the CAST diagram to determine in which quadrants cos α is negative\nThe CAST diagram indicates that cos α is negative in the second and third quadrants,\ntherefore we must determine the value of α such that 90◦< α < 270◦.\nUsing reduction formulae, we know that cos(180◦−α) = −cos α and cos(180◦+α) =\n−cos α\nIn the second quadrant:\nα = 180◦−45,6◦\n= 134,4◦\nIn the third quadrant:\nα = 180◦+ 45,6◦\n= 225,6◦\nNote: the reference angle (45,6◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nα = 134,4◦or α = 225,6◦.\n267\nChapter 6.\nTrigonometry\n\nWorked example 14: Solving trigonometric equations\nQUESTION\nSolve for β (correct to one decimal place), given sin β = −0,5 and β ∈[−360◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we use a positive value.\nref ∠= sin−1 0,5\n= 30◦\nStep 2: Use the CAST diagram to determine in which quadrants sin β is negative\nThe CAST diagram indicates that sin β is negative in the third and fourth quadrants.\nWe also need to find the values of β such that −360◦≤β ≤360◦.\nUsing reduction formulae, we know that sin(180◦+β) = −sin β and sin(360◦−β) =\n−sin β\nIn the third quadrant:\nβ = 180◦+ 30◦\n= 210◦\nor β = −180◦+ 30◦\n= −150◦\nIn the fourth quadrant:\nβ = 360◦−30◦\n= 330◦\nor β = 0◦−30◦\n= −30◦\nNotice: the reference angle (30◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nβ = −150◦, −30◦, 210◦or 330◦.\n268\n6.4.\nTrigonometric equations\n\nExercise 6 – 7: Solving trigonometric equations\n1. Determine the values of α for α ∈[0◦; 360◦] if:\na) 4 cos α = 2\nb) sin α + 3,65 = 3\nc) tan α = 51\n4\nd) cos α + 0,939 = 0\ne) 5 sin α = 3\nf)\n1\n2 tan α = −1,4\n2. Determine the values of θ for θ ∈[−360◦; 360◦] if:\na) sin θ = 0,6\nb) cos θ + 3\n4 = 0\nc) 3 tan θ = 20\nd) sin θ = cos 180◦\ne) 2 cos θ = 4\n5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22ZP\n1b. 22ZQ\n1c. 22ZR\n1d. 22ZS\n1e. 22ZT\n1f. 22ZV\n2a. 22ZW\n2b. 22ZX\n2c. 22ZY\n2d. 22ZZ\n2e. 2322\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe general solution\nEMBHN\nIn the previous worked example, the solution was restricted to a certain interval. How-\never, the periodicity of the trigonometric functions means that there are an infinite\nnumber of positive and negative angles that satisfy an equation. If we do not restrict\nthe solution, then we need to determine the general solution to the equation. We know\nthat the sine and cosine functions have a period of 360◦and the tangent function has\na period of 180◦.\nMethod for finding the general solution:\n1. Determine the reference angle (use a positive value).\n2. Use the CAST diagram to determine where the function is positive or negative\n(depending on the given equation).\n3. Find the angles in the interval [0◦; 360◦] that satisfy the equation and add multi-\nples of the period to each answer.\n4. Check answers using a calculator.\n269\nChapter 6.\nTrigonometry\n\nWorked example 15: Finding the general solution\nQUESTION\nDetermine the general solution for sin θ = 0,3 (correct to one decimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nsin θ = 0,3\n∴ref ∠= sin−1 0,3\n= 17,5◦\nStep 2: Use CAST diagram to determine in which quadrants sin θ is positive\nThe CAST diagram indicates that sin θ is positive in the first and second quadrants.\nUsing reduction formulae, we know that sin(180◦−θ) = sin θ.\nIn the first quadrant:\nθ = 17,5◦\n∴θ = 17,5◦+ k . 360◦\nIn the second quadrant:\nθ = 180◦−17,5◦\n∴θ = 162,5◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 4:\nθ = 17,5◦+ 4(360)◦\n∴θ = 1457,5◦\nAnd sin 1457,5◦= 0,3007 . . .\nThis solution is correct.\nSimilarly, if we let k = −2:\nθ = 162,5◦−2(360)◦\n∴θ = −557,5◦\nAnd sin(−557,5◦) = 0,3007 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 17,5◦+ k . 360◦or θ = 162,5◦+ k . 360◦.\n270\n6.4.\nTrigonometric equations\n\nWorked example 16: Finding the general solution\nQUESTION\nDetermine the general solution for cos 2θ = −0,6427 (give answers correct to one\ndecimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nref ∠= sin−1 0,6427\n= 50,0◦\nStep 2: Use CAST diagram to determine in which quadrants cos θ is negative\nThe CAST diagram shows that cos θ is negative in the second and third quadrants.\nTherefore we use the reduction formulae cos(180◦−θ) = −cos θ and cos(180◦+θ) =\n−cos θ.\nIn the second quadrant:\n2θ = 180◦−50◦+ k . 360◦\n= 130◦+ k . 360◦\n∴θ = 65◦+ k . 180◦\nIn the third quadrant:\n2θ = 180◦+ 50◦+ k . 360◦\n= 230◦+ k . 360◦\n∴θ = 115◦+ k . 180◦\nwhere k ∈Z.\nRemember: also divide the period (360◦) by the coefficient of θ.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 2:\nθ = 65◦+ 2(180◦)\n∴θ = 425◦\nAnd cos 2(425)◦= −0,6427 . . .\nThis solution is correct.\n271\nChapter 6.\nTrigonometry\n\nSimilarly, if we let k = −5:\nθ = 115◦−5(180◦)\n∴θ = −785◦\nAnd cos 2(−785◦) = −0,6427 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 65◦+ k . 180◦or θ = 115◦+ k . 180◦.\nWorked example 17: Finding the general solution\nQUESTION\nDetermine the general solution for tan(2α −10◦) = 2,5 such that −180◦≤α ≤180◦\n(give answers correct to one decimal place).\nSOLUTION\nStep 1: Make a substitution\nTo solve this equation, it can be useful to make a substitution: let x = 2α −10◦.\ntan(x) = 2,5\nStep 2: Use a calculator to find the reference angle\ntan x = 2,5\n∴ref ∠= tan−1 2,5\n= 68,2◦\nStep 3: Use CAST diagram to determine in which quadrants the tangent function is\npositive\nWe see that tan x is positive in the first and third quadrants, so we use the reduction\nformula tan(180◦+ x) = tan x. It is also important to remember that the period of the\ntangent function is 180◦.\n272\n6.4.\nTrigonometric equations\n\nIn the first quadrant:\nx = 68,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 68,2◦+ k . 180◦\n2α = 78,2◦+ k . 180◦\n∴α = 39,1◦+ k . 90◦\nIn the third quadrant:\nx = 180◦+ 68,2◦+ k . 180◦\n= 248,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 248,2◦+ k . 180◦\n2α = 258,2◦+ k . 180◦\n∴α = 129,1◦+ k . 90◦\nwhere k ∈Z.\nRemember: to divide the period (180◦) by the coefficient of α.\nStep 4: Find the answers within the given interval\nSubstitute suitable values of k to determine the values of α that lie within the interval\n(−180◦≤α ≤180◦).\nI: α = 39,1◦+ k . 90◦\nIII: α = 129,1◦+ k . 90◦\nk = 0\n39,1◦\n129,1◦\nk = 1\n129,1◦\n219,1◦\n(outside)\nk = 2\n219,1◦\n(outside)\nk = −1\n−50,9◦\n39,1◦\nk = −2\n−140,9◦\n−50,9◦\nk = −3\n−230,9◦\n(outside)\n−140,9◦\nk = −4\n−230,9◦\n(outside)\nNotice how some of the values repeat. This is because of the periodic nature of the\ntangent function. Therefore we need only determine the solution:\nα = 39,1◦+ k . 90◦\nfor k ∈Z.\nStep 5: Write the final answer\nα = −140,9◦; −50,9◦; 39,1◦or 129,1◦.\n273\nChapter 6.\nTrigonometry\n\nWorked example 18: Finding the general solution using co-functions\nQUESTION\nDetermine the general solution for sin(θ −20◦) = cos 2θ.\nSOLUTION\nStep 1: Use co-functions to simplify the equation\nsin(θ −20◦) = cos 2θ\n= sin(90◦−2θ)\n∴θ −20◦= 90◦−2θ + k . 360◦,\nk ∈Z\n3θ = 110◦+ k . 360◦\n∴θ = 36,7◦+ k . 120◦\nStep 2: Use the CAST diagram to determine the correct quadrants\nSince the original equation equates a sine and cosine function, we need to work in the\nquadrant where both functions are positive or in the quadrant where both functions\nare negative so that the equation holds true. We therefore determine the solution using\nthe first and third quadrants.\nIn the first quadrant: θ = 36,7◦+ k . 120◦.\nIn the third quadrant:\n3θ = 180◦+ 110◦+ k . 360◦\n= 290◦+ k . 360◦\n∴θ = 96,6◦+ k . 120◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 36,7◦+ k . 120◦or θ = 96,6◦+ k . 120◦\n274\n6.4.\nTrigonometric equations\n\nExercise 6 – 8: General solution\n1.\n• Find the general solution for each equation.\n• Hence, find all the solutions in the interval [−180◦; 180◦].\na) cos(θ + 25◦) = 0,231\nb) sin 2α = −0,327\nc) 2 tan β = −2,68\nd) cos α = 1\ne) 4 sin θ = 0\nf) cos θ = −1\ng) tan θ\n2 = 0,9\nh) 4 cos θ + 3 = 1\ni) sin 2θ = −\n√\n3\n2\n2. Find the general solution for each equation.\na) cos(θ + 20◦) = 0\nb) sin 3α = −1\nc) tan 4β = 0,866\nd) cos(α −25◦) = 0,707\ne) 2 sin 3θ\n2 = −1\nf) 5 tan(β + 15◦) =\n5\n√\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2323\n1b. 2324\n1c. 2325\n1d. 2326\n1e. 2327\n1f. 2328\n1g. 2329\n1h. 232B\n1i. 232C\n2a. 232D\n2b. 232F\n2c. 232G\n2d. 232H\n2e. 232J\n2f. 232K\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSolving quadratic trigonometric equations\nWe can use our knowledge of algebraic equations to solve quadratic trigonometric\nequations.\nWorked example 19: Quadratic trigonometric equations\nQUESTION\nFind the general solution of 4 sin2 θ = 3.\nSOLUTION\nStep 1: Simplify the equation and determine the reference angle\n4 sin2 θ = 3\nsin2 θ = 3\n4\n∴sin θ = ±\nr\n3\n4\n= ±\n√\n3\n2\n∴ref ∠= 60◦\n275\nChapter 6.\nTrigonometry\n\nStep 2: Determine in which quadrants the sine function is positive and negative\nThe CAST diagram shows that sin θ is positive in the first and second quadrants and\nnegative in the third and fourth quadrants.\nPositive in the first and second quadrants:\nθ = 60◦+ k . 360◦\nor θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nNegative in the third and fourth quadrants:\nθ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor θ = 360◦−60◦+ k . 360◦\n= 300◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 60◦+ k . 360◦or 120◦+ k . 360◦or 240◦+ k . 360◦or 300◦+ k . 360◦\nWorked example 20: Quadratic trigonometric equations\nQUESTION\nFind θ if 2 cos2 θ −cos θ −1 = 0 for θ ∈[−180◦; 180◦].\nSOLUTION\nStep 1: Factorise the equation\n2 cos2 θ −cos θ −1 = 0\n(2 cos θ + 1)(cos θ −1) = 0\n∴2 cos θ + 1 = 0 or cos θ −1 = 0\n276\n6.4.\nTrigonometric equations\n\nStep 2: Simplify the equations and solve for θ\n2 cos θ + 1 = 0\n2 cos θ = −1\ncos θ = −1\n2\n∴ref ∠= 60◦\nII quadrant: θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nIII quadrant: θ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor\ncos θ −1 = 0\ncos θ = 1\n∴ref ∠= 0◦\nII and IV quadrants: θ = k . 360◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of θ that lie within the the given interval θ ∈[−180◦; 180◦] by\nsubstituting suitable values of k.\nIf k = −1,\nθ = 240◦+ k . 360◦\n= 240◦−(360◦)\n= −120◦\nIf k = 0,\nθ = 120◦+ k . 360◦\n= 120◦+ 0(360◦)\n= 120◦\nIf k = 1,\nθ = k . 360◦\n= 0(360◦)\n= 0◦\n277\nChapter 6.\nTrigonometry\n\nStep 4: Alternative method: substitution\nWe can simplify the given equation by letting y = cos θ and then factorising as:\n2y2 −y −1 = 0\n(2y + 1)(y −1) = 0\n∴y = −1\n2 or y = 1\nWe substitute y = cos θ back into these two equations and solve for θ.\nStep 5: Write the final answer\nθ = −120◦; 0◦; 120◦\nWorked example 21: Quadratic trigonometric equations\nQUESTION\nFind α if 2 sin2 α −sin α cos α = 0 for α ∈[0◦; 360◦].\nSOLUTION\nStep 1: Factorise the equation by taking out a common factor\n2 sin2 α −sin α cos α = 0\nsin α(2 sin α −cos α) = 0\n∴sin α = 0 or 2 sin α −cos α = 0\nStep 2: Simplify the equations and solve for α\nsin α = 0\n∴ref ∠= 0◦\n∴α = 0◦+ k . 360◦\nor α = 180◦+ k . 360◦\nand since 360◦= 2 × 180◦\nwe therefore have α = k . 180◦\n278\n6.4.\nTrigonometric equations\n\nor\n2 sin α −cos α = 0\n2 sin α = cos α\nTo simplify further, we divide both sides of the equation by cos α.\n2 sin α\ncos α = cos α\ncos α\n(cos α ̸= 0)\n2 tan α = 1\ntan α = 1\n2\n∴ref ∠= 26,6◦\n∴α = 26,6◦+ k . 180◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of α that lie within the the given interval α ∈[0◦; 360◦] by\nsubstituting suitable values of k.\nIf k = 0:\nα = 0◦\nor α = 26,6◦\nIf k = 1:\nα = 180◦\nor α = 26,6◦+ 180◦\n= 206,6◦\nIf k = 2:\nα = 360◦\nStep 4: Write the final answer\nα = 0◦; 26,6◦; 180◦; 206,6◦; 360◦\n279\nChapter 6.\nTrigonometry\n\nExercise 6 – 9: Solving trigonometric equations\n1. Find the general solution for each of the following equations:\na) cos 2θ = 0\nb) sin(α + 10◦) =\n√\n3\n2\nc) 2 cos θ\n2 −\n√\n3 = 0\nd)\n1\n2 tan(β −30◦) = −1\ne) 5 cos θ = tan 300◦\nf) 3 sin α = −1,5\ng) sin 2β = cos(β + 20◦)\nh) 0,5 tan θ + 2,5 = 1,7\ni) sin(3α −10◦) = sin(α + 32◦)\nj) sin 2β = cos 2β\n2. Find θ if sin2 θ + 1\n2 sin θ = 0 for θ ∈[0◦; 360◦].\n3. Determine the general solution for each of the following:\na) 2 cos2 θ −3 cos θ = 2\nb) 3 tan2 θ + 2 tan θ = 0\nc) cos2 α = 0,64\nd) sin(4β + 35◦) = cos(10◦−β)\ne) sin(α + 15◦) = 2 cos(α + 15◦)\nf) sin2 θ −4 cos2 θ = 0\ng) cos(2θ + 30◦)\n2\n+ 0,38 = 0\n4. Find β if 1\n3 tan β = cos 200◦for β ∈[−180◦; 180◦].\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 232M\n1b. 232N\n1c. 232P\n1d. 232Q\n1e. 232R\n1f. 232S\n1g. 232T\n1h. 232V\n1i. 232W\n1j. 232X\n2. 232Y\n3a. 232Z\n3b. 2332\n3c. 2333\n3d. 2334\n3e. 2335\n3f. 2336\n3g. 2337\n4. 2338\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n6.5\nArea, sine, and cosine rules\nEMBHP\nThere are three identities relating to the trigonometric functions that make working\nwith triangles easier:\n1. the area rule\n2. the sine rule\n3. the cosine rule\n280\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nEMBHQ\nInvestigation: The area rule\n1. Consider △ABC:\nB\nA\nC\n10\n54◦\n7\nComplete the following:\na) Area △ABC = 1\n2 × . . . × AC\nb) sin ˆB = . . . and AC = . . . × . . .\nc) Therefore area △ABC = . . . × . . . × . . . × . . .\n2. Consider △A′B′C′:\nB′\nA′\nC′\n10\n54◦\n7\nComplete the following:\na) How is △A′B′C′ different from △ABC?\nb) Calculate area △A′B′C′.\n3. Use your results to write a general formula for determining the area of △PQR:\nQ\nP\nR\nr\np\nq\n281\nChapter 6.\nTrigonometry\n\nFor any △ABC with AB = c, BC = a and AC = b, we can construct a perpendicular\nheight (h) from vertex A to the line BC:\nB\nA\nC\nc\na\nb\nh\nIn △ABC:\nsin ˆB = h\nc\n∴h = c sin ˆB\nAnd we know that\nArea △ABC = 1\n2 × a × h\n= 1\n2 × a × c sin ˆB\n∴Area △ABC = 1\n2ac sin ˆB\nAlternatively, we could write that\nsin ˆC = h\nb\n∴h = b sin ˆC\nAnd then we would have that\nArea △ABC = 1\n2 × a × h\n= 1\n2ab sin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nThe area rule\nIn any △ABC:\nArea △ABC = 1\n2bc sin ˆA\n= 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n282\n6.5.\nArea, sine, and cosine rules\n\nWorked example 22: The area rule\nQUESTION\nFind the area of △ABC (correct to two decimal places):\nA\n7\nB\nC\n50◦\nSOLUTION\nStep 1: Use the given information to determine unknown angles and sides\nAB = AC = 7\n(given)\n∴ˆB = ˆC = 50◦\n(∠s opp. equal sides)\nAnd ˆA = 180◦−50◦−50◦\n(∠s sum of △ABC)\n∴ˆA = 80◦\nStep 2: Use the area rule to calculate the area of △ABC\nNotice that we do not know the length of side a and must therefore choose the form\nof the area rule that does not include this side of the triangle.\nIn △ABC:\nArea = 1\n2bc sin ˆA\n= 1\n2(7)(7) sin 80◦\n= 24,13\nStep 3: Write the final answer\nArea of △ABC = 24,13 square units.\n283\nChapter 6.\nTrigonometry\n\nWorked example 23: The area rule\nQUESTION\nShow that the area of △DEF = 1\n2df sin ˆE.\nD\nF\nE\nH\ne\nd\nf\nh\n1\n2\nSOLUTION\nStep 1: Construct a perpendicular height h\nDraw DH such that DH ⊥EF and let DH = h, D ˆEF = ˆE1 and D ˆEH = ˆE2.\nIn △DHE:\nsin ˆE2 = h\nf\nh = f sin(180◦−ˆE1)\n(∠s on str. line)\n= f sin ˆE1\nStep 2: Use the area rule to calculate the area of △DEF\nIn △DEF:\nArea = 1\n2d × h\n= 1\n2df sin ˆE1\n284\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nIn any △PQR:\nP\nP\nQ\nQ\nR\nR\nq\nq\nr\nr\np\np\nArea △PQR = 1\n2qr sin ˆP\n= 1\n2pr sin ˆQ\n= 1\n2pq sin ˆR\nThe area rule states that the area of any triangle is equal to half the product of the\nlengths of the two sides of the triangle multiplied by the sine of the angle included by\nthe two sides.\nExercise 6 – 10: The area rule\n1. Draw a sketch and calculate the area of △PQR given:\na) ˆQ = 30◦; r = 10 and p = 7\nb) ˆR = 110◦; p = 8 and q = 9\n2. Find the area of △XY Z given XZ = 52 cm, XY = 29 cm and ˆX = 58,9◦.\n3. Determine the area of a parallelogram in which two adjacent sides are 10 cm\nand 13 cm and the angle between them is 55◦.\n4. If the area of △ABC is 5000 m2 with a = 150 m and b = 70 m, what are the two\npossible sizes of ˆC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2339\n1b. 233B\n2. 233C\n3. 233D\n4. 233F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n285\nChapter 6.\nTrigonometry\n\nThe sine rule\nEMBHR\nSo far we have only applied the trigonometric ratios to right-angled triangles. We now\nexpand the application of the trigonometric ratios to triangles that do not have a right\nangle:\nInvestigation: The sine rule\nIn △ABC, AC = 15, BC = 11 and ˆA = 48◦. Find ˆB.\nA\nC\nB\nb = 15\nF\na = 11\n48◦\n1. Method 1: using the sine ratio\na) Draw a sketch of △ABC.\nb) Construct CF ⊥AB.\nc) In △CBF:\nCF\n. . . = sin ˆB\n∴CF = . . . × sin ˆB\nd) In △CAF:\nCF\n15 = . . .\n∴CF = 15 × . . .\ne) Therefore we have that:\nCF = 15 × . . .\nand CF = . . . × sin ˆB\n∴15 × . . . = . . . × sin ˆB\n∴sin ˆB = . . . . . . . . .\n∴ˆB = . . .\n2. Method 2: using the area rule\n286\n6.5.\nArea, sine, and cosine rules\n\na) In △ABC:\nArea △ABC = 1\n2AB × AC × . . .\n= 1\n2AB × . . . × . . .\nb) And we also know that\nArea △ABC = 1\n2AB × . . . × sin ˆB\nc) We can equate these two equations and solve for ˆB:\n1\n2AB × . . . × sin ˆB = 1\n2AB × . . . × . . .\n∴. . . × sin ˆB = . . . × . . .\n∴sin ˆB = . . . × . . .\n∴ˆB = . . .\n3. Use your results to write a general formula for the sine rule given △PQR:\nP\nQ\nR\nq\nr\np\nFor any triangle ABC with AB = c, BC = a and AC = b, we can construct a perpen-\ndicular height (h) at F:\nA\nC\nB\nb\nF\na\nh\nc\nMethod 1: using the sine ratio\nIn △ABF:\nsin ˆB = h\nc\n∴h = c sin ˆB\n287\nChapter 6.\nTrigonometry\n\nIn △ACF:\nsin ˆC = h\nb\n∴h = b sin ˆC\nWe can equate the two equations\nc sin ˆB = b sin ˆC\n∴sin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that:\nsin ˆA\na\n= sin ˆC\nc\nor\na\nsin ˆA\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nMethod 2: using the area rule\nIn △ABC:\nArea △ABC = 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n∴1\n2ac sin ˆB = 1\n2ab sin ˆC\nc sin ˆB = b sin ˆC\nsin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\n288\n6.5.\nArea, sine, and cosine rules\n\nThe sine rule\nIn any △ABC:\nA\nC\nB\nb\na\nc\nsin ˆA\na\n= sin ˆB\nb\n= sin ˆC\nc\na\nsin ˆA\n=\nb\nsin ˆB\n=\nc\nsin ˆC\nSee video: 233G at www.everythingmaths.co.za\nWorked example 24: The sine rule\nQUESTION\nGiven △TRS with S ˆTR = 55◦, TR = 30 and R ˆST = 40◦, determine RS, ST and\nT ˆRS.\nSOLUTION\nStep 1: Draw a sketch\nLet RS = t, ST = r and TR = s.\nR\nS\nT\n55◦\n30\n40◦\nStep 2: Find T ˆRS using angles in a triangle\nT ˆRS + R ˆST + S ˆTR = 180◦\n(∠s sum of △TRS)\n∴T ˆRS = 180◦−40◦−55◦\n= 85◦\n289\nChapter 6.\nTrigonometry\n\nStep 3: Determine t and r using the sine rule\nt\nsin ˆT\n=\ns\nsin ˆS\nt\nsin 55◦=\n30\nsin 40◦\n∴t =\n30\nsin 40◦× sin 55◦\n= 38,2\nr\nsin ˆR\n=\ns\nsin ˆS\nr\nsin 85◦=\n30\nsin 40◦\n∴r =\n30\nsin 40◦× sin 85◦\n= 46,5\nWorked example 25: The sine rule\nQUESTION\nProve the sine rule for △MNP with MS ⊥NP.\nM\nP\nN\nS\nn\nm\np\nh\n1\n2\nSOLUTION\nStep 1: Use the sine ratio to express the angles in the triangle in terms of the length\nof the sides\nIn △MSN:\nsin ˆN2 = h\np\n∴h = p sin ˆN2\nand ˆN2 = 180◦−ˆN1\n∠s on str. line\n∴h = p sin(180◦−ˆN1)\n= p sin ˆN1\n290\n6.5.\nArea, sine, and cosine rules\n\nIn △MSP:\nsin ˆP = h\nn\n∴h = n sin ˆP\nStep 2: Equate the two equations to derive the sine rule\np sin ˆN1 = n sin ˆP\n∴sin ˆN1\nn\n= sin ˆP\np\nor\nn\nsin ˆN1\n=\np\nsin ˆP\nThe ambiguous case\nIf two sides and an interior angle of a triangle are given, and the side opposite the given\nangle is the shorter of the two sides, then we can draw two different triangles (△NMP\nand △NMP ′), both having the given dimensions. We call this the ambiguous case\nbecause there are two ways of interpreting the given information and it is not certain\nwhich is the required solution.\nM\nP ′\np\nN\nP\nn\nn\nWorked example 26: The ambiguous case\nQUESTION\nIn △ABC, AB = 82, BC = 65 and ˆA = 50◦. Draw △ABC and find ˆC (correct to\none decimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the ambiguous case\nWe notice that for the given dimensions of △ABC, the side BC opposite ˆA is shorter\nthan AB. This means that we can draw two different triangles with the given dimen-\nsions.\nA\nB\nC\n50◦\n65\n82\nA′\nC′\nB′\n65\n82\n50◦\n291\nChapter 6.\nTrigonometry\n\nStep 2: Solve for unknown angle using the sine rule\nIn △ABC:\nsin ˆA\nBC = sin ˆC\nAB\nsin 50◦\n65\n= sin ˆC\n82\n∴sin 50◦\n65\n× 82 = sin ˆC\n∴ˆC = 75,1◦\nIn △A′B′C′:\nWe know that sin(180 −ˆC) = sin ˆC, which means we can also have the solution\nˆC′ = 180◦−75,1◦\n= 104,9◦\nBoth solutions are correct.\nWorked example 27: Lighthouses\nQUESTION\nThere is a coastline with two lighthouses, one on either side of a beach. The two\nlighthouses are 0,67 km apart and one is exactly due east of the other. The lighthouses\ntell how close a boat is by taking bearings to the boat (a bearing is an angle measured\nclockwise from north). These bearings are shown on the diagram below.\nCalculate how far the boat is from each lighthouse.\nˆA = 127◦\nˆB = 255◦\nC\nSOLUTION\nWe see that the two lighthouses and the boat form a triangle. Since we know the\ndistance between the lighthouses and we have two angles we can use trigonometry\n292\n6.5.\nArea, sine, and cosine rules\n\nto find the remaining two sides of the triangle, the distance of the boat from the two\nlighthouses.\nb\nA\nb B\nb\nC\n15◦\n37◦\n128◦\n0,67 km\nWe need to determine the lengths of the two sides AC and BC. We can use the sine\nrule to find the missing lengths.\nBC\nsin ˆA\n= AB\nsin ˆC\nBC = AB . sin ˆA\nsin ˆC\n= (0,67 km) sin 37◦\nsin 128◦\n= 0,51 km\nAC\nsin ˆB\n= AB\nsin ˆC\nAC = AB . sin ˆB\nsin ˆC\n= (0,67 km) sin 15◦\nsin 128◦\n= 0,22 km\nExercise 6 – 11: Sine rule\n1. Find all the unknown sides and angles of the following triangles:\na) △PQR in which ˆQ = 64◦; ˆR = 24◦and r = 3\nb) △KLM in which ˆK = 43◦; ˆ\nM = 50◦and m = 1\nc) △ABC in which ˆA = 32,7◦; ˆC = 70,5◦and a = 52,3\nd) △XY Z in which ˆX = 56◦; ˆZ = 40◦and x = 50\n2. In △ABC, ˆA = 116◦; ˆC = 32◦and AC = 23 m. Find the lengths of the sides\nAB and BC.\n3. In △RST, ˆR = 19◦; ˆS = 30◦and RT = 120 km. Find the length of the side\nST.\n4. In △KMS, ˆK = 20◦; ˆ\nM = 100◦and s = 23 cm. Find the length of the side m.\n293\nChapter 6.\nTrigonometry\n\n5. In △ABD, ˆB = 90◦, AB = 10 cm and A ˆDB = 40◦. In △BCD, ˆC = 106◦and\nC ˆDB = 15◦. Determine BC.\nA\nB\nD\nC\n10\n106◦\n15◦\n40◦\n6. In △ABC, ˆA = 33◦, AC = 21 mm and AB = 17 mm. Can you determine BC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233H\n1b. 233J\n1c. 233K\n1d. 233M\n2. 233N\n3. 233P\n4. 233Q\n5. 233R\n6. 233S\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe cosine rule\nEMBHS\nInvestigation: The cosine rule\nIf a triangle is given with two sides and the included angle known, then we can not\nsolve for the remaining unknown sides and angles using the sine rule. We therefore\ninvestigate the cosine rule:\nIn △ABC, AB = 21, AC = 17 and ˆA = 33◦. Find ˆB.\nA\nC\nB\nH\n21\nc\n17\n33◦\n1. Determine CB:\na) Construct CH ⊥AB.\nb) Let AH = c and therefore HB = . . .\n294\n6.5.\nArea, sine, and cosine rules\n\nc) Applying the theorem of Pythagoras in the right-angled triangles:\nIn△CHB:\nCB2 = BH2 + CH2\n= (. . .)2 + CH2\n= 212 −(2)(21)c + c2 + CH2 . . . . . . (1)\nIn △CHA:\nCA2 = c2 + CH2\n172 = c2 + CH2 . . . . . . (2)\nSubstitute equation (2) into equation (1):\nCB2 = 212 −(2)(21)c + 172\nNow c is the only remaining unknown. In △CHA:\nc\n17 = cos 33◦\n∴c = 17 cos 33◦\nTherefore we have that\nCB2 = 212 −(2)(21)c + 172\n= 212 −(2)(21)(17 cos 33◦) + 172\n= 212 + 172 −(2)(21)(17) cos 33◦\n= 131,189 . . .\n∴CB = 11,5\n2. Use your results to write a general formula for the cosine rule given △PQR:\nP\nQ\nR\nq\nr\np\nThe cosine rule relates the length of a side of a triangle to the angle opposite it and the\nlengths of the other two sides.\n295\nChapter 6.\nTrigonometry\n\nConsider △ABC with CD ⊥AB:\nb\nD\nb\nA\nb\nB\nbC\nh\na\nb\nc\nc −d\nd\nIn △DCB: a2 = (c −d)2 + h2 from the theorem of Pythagoras.\nIn △ACD: b2 = d2 + h2 from the theorem of Pythagoras.\nSince h2 is common to both equations we can write:\na2 = (c −d)2 + h2\n∴h2 = a2 −(c −d)2\nAnd b2 = d2 + h2\n∴h2 = b2 −d2\n∴b2 −d2 = a2 −(c −d)2\na2 = b2 + (c2 −2cd + d2) −d2\n= b2 + c2 −2cd\nIn order to eliminate d we look at △ACD, where we have: cos ˆA = d\nb. So, d = b cos ˆA.\nSubstituting back we get: a2 = b2 + c2 −2bc cos ˆA.\nThe cosine rule\nIn any △ABC:\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\na2 = b2 + c2 −2bc cos ˆA\nb2 = a2 + c2 −2ac cos ˆB\nc2 = a2 + b2 −2ab cos ˆC\nSee video: 233T at www.everythingmaths.co.za\n296\n6.5.\nArea, sine, and cosine rules\n\nWorked example 28: The cosine rule\nQUESTION\nDetermine the length of QR.\nP\nR\nQ\n13 cm\n4 cm\n70◦\nSOLUTION\nStep 1: Use the cosine rule to solve for the unknown side\nQR2 = PR2 + QP 2 −2(PR)(QP) cos ˆP\n= 42 + 132 −2(4)(13) cos 70◦\n= 149,42 . . .\n∴QR = 12,2\nStep 2: Write the final answer\nQR = 12,2 cm\nWorked example 29: The cosine rule\nQUESTION\nDetermine ˆA.\n5\n7\n8\nA\nB\nC\nSOLUTION\nApplying the cosine rule:\na2 = b2 + c2 −2bc cos ˆA\n∴cos ˆA = b2 + c2 −a2\n2bc\n= 82 + 52 −72\n2 . 8 . 5\n= 0,5\n∴ˆA = 60◦\n297\nChapter 6.\nTrigonometry\n\nIt is very important:\n• not to round off before the final answer as this will affect accuracy;\n• to take the square root;\n• to remember to give units where applicable.\nHow to determine which rule to use:\n1. Area rule:\n• if no perpendicular height is given\n2. Sine rule:\n• if no right angle is given\n• if two sides and an angle are given (not the included angle)\n• if two angles and a side are given\n3. Cosine rule:\n• if no right angle is given\n• if two sides and the included angle are given\n• if three sides are given\nExercise 6 – 12: The cosine rule\n1. Solve the following triangles (that is, find all unknown sides and angles):\na) △ABC in which ˆA = 70◦; b = 4 and c = 9\nb) △RST in which RS = 14; ST = 26 and RT = 16\nc) △KLM in which KL = 5; LM = 10 and KM = 7\nd) △JHK in which ˆH = 130◦; JH = 13 and HK = 8\ne) △DEF in which d = 4; e = 5 and f = 7\n2. Find the length of the third side of the △XY Z where:\na) ˆX = 71,4◦; y = 3,42 km and z = 4,03 km\nb) x = 103,2 cm; ˆY = 20,8◦and z = 44,59 cm\n3. Determine the largest angle in:\na) △JHK in which JH = 6; HK = 4 and JK = 3\nb) △PQR where p = 50; q = 70 and r = 60\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233V\n1b. 233W\n1c. 233X\n1d. 233Y\n1e. 233Z\n2a. 2342\n2b. 2343\n3a. 2344\n3b. 2345\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n298\n6.5.\nArea, sine, and cosine rules\n\nSee video: 2346 at www.everythingmaths.co.za\nExercise 6 – 13: Area, sine and cosine rule\n1. Q is a ship at a point 10 km due south of another ship P. R is a lighthouse on\nthe coast such that ˆP = ˆQ = 50◦.\n10 km\nP\nQ\nR\n50◦\n50◦\nDetermine:\na) the distance QR\nb) the shortest distance from the lighthouse to the line joining the two ships\n(PQ).\n2. WXY Z is a trapezium, WX ∥Y Z with WX = 3 m; Y Z = 1,5 m; ˆZ = 120◦\nand ˆW = 30◦.\nDetermine the distances XZ and XY .\n1,5 m\n3 m\n30◦\n120◦\nW\nX\nY\nZ\n3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has\nbeen flying 3◦off course. At this point the plane is 500 km from Johannesburg.\nThe direct distance between Cape Town and Johannesburg airports is 1552 km.\nDetermine, to the nearest km:\na) The distance the plane has to travel to get to Cape Town and hence the\nextra distance that the plane has had to travel due to the pilot’s error.\nb) The correction, to one hundredth of a degree, to the plane’s heading (or\ndirection).\n4. ABCD is a trapezium (meaning that AB ∥CD). AB = x; B ˆAD = a; B ˆCD = b\nand B ˆDC = c.\nFind an expression for the length of CD in terms of x, a, b and c.\nA\nB\nC\nD\na\nb\nc\nx\n299\nChapter 6.\nTrigonometry\n\n5. A surveyor is trying to determine the distance between points X and Z. However\nthe distance cannot be determined directly as a ridge lies between the two points.\nFrom a point Y which is equidistant from X and Z, he measures the angle X ˆY Z.\nY\nX\nZ\nx\nθ\na) If XY = x and X ˆY Z = θ, show that XZ = x\np\n2(1 −cos θ).\nb) Calculate XZ (to the nearest kilometre) if x = 240 km and θ = 132◦.\n6. Find the area of WXY Z (to two decimal places):\nW\nX\nY\nZ\n120◦\n3\n4\n3,5\n7. Find the area of the shaded triangle in terms of x, α, β, θ and φ:\nA\nB\nC\nD\nE\nx\nα\nβ\nθ\nφ\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2347\n2. 2348\n3. 2349\n4. 234B\n5. 234C\n6. 234D\n7. 234F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n300\n6.5.\nArea, sine, and cosine rules\n\n6.6\nSummary\nEMBHT\nSee presentation: 234G at www.everythingmaths.co.za\nsquare identity\nquotient identity\ncos2 θ + sin2 θ = 1\ntan θ = sin θ\ncos θ\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\nnegative angles\nperiodicity identities\nco-function identities\nsin(−θ) = −sin θ\nsin(θ ± 360◦) = sin θ\nsin(90◦−θ) = cos θ\ncos(−θ) = cos θ\ncos(θ ± 360◦) = cos θ\ncos(90◦−θ) = sin θ\nsine rule\narea rule\ncosine rule\nsin A\na\n= sin B\nb\n= sin C\nc\narea △ABC = 1\n2bc sin A\na2 = b2 + c2 −2bc cos A\na\nsin A =\nb\nsin B =\nc\nsin C\narea △ABC = 1\n2ac sin B\nb2 = a2 + c2 −2ac cos B\narea △ABC = 1\n2ab sin C\nc2 = a2 + b2 −2ab cos C\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nGeneral solution:\n301\nChapter 6.\nTrigonometry\n\n1.\nIf sin θ = x\nθ = sin−1 x + k . 360◦\nor θ =\n\u0000180◦−sin−1 x\n\u0001\n+ k . 360◦\n2.\nIf cos θ = x\nθ = cos−1 x + k . 360◦\nor θ =\n\u0000360◦−cos−1 x\n\u0001\n+ k . 360◦\n3.\nIf tan θ = x\nθ = tan−1 x + k . 180◦\nfor k ∈Z.\nHow to determine which rule to use:\n1. Area rule:\n• no perpendicular height is given\n2. Sine rule:\n• no right angle is given\n• two sides and an angle are given (not the included angle)\n• two angles and a side are given\n3. Cosine rule:\n• no right angle is given\n• two sides and the included angle angle are given\n• three sides are given\nExercise 6 – 14: End of chapter exercises\n1. Write the following as a single trigonometric ratio:\ncos(90◦−A) sin 20◦\nsin(180◦−A) cos 70◦+ cos(180◦+ A) sin(90◦+ A)\n2. Determine the value of the following expression without using a calculator:\nsin 240◦cos 210◦−tan2 225◦cos 300◦cos 180◦\n302\n6.6.\nSummary\n\n3. Simplify:\nsin(180◦+ θ) sin(θ + 360◦)\nsin(−θ) tan(θ −360◦)\n4. Without the use of a calculator, evaluate:\n3 sin 55◦sin2 325◦\ncos(−145◦)\n−3 cos 395◦sin 125◦\n5. Prove the following identities:\na)\n1\n(cos x −1)(cos x + 1) =\n−1\ntan2 x cos2 x\nb) (1 −tan α) cos α = sin(90 + α) + cos(90 + α)\n6.\na) Prove: tan y +\n1\ntan y =\n1\ncos2 y tan y\nb) For which values of y ∈[0◦; 360◦] is the identity above undefined?\n7.\na) Simplify: sin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\nb) Hence, solve the equation\nsin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\n= tan θ\nfor θ ∈[0◦; 360◦].\n8. Given 12 tan θ = 5 and θ > 90◦.\na) Draw a sketch.\nb) Determine without using a calculator sin θ and cos(180◦+ θ).\nc) Use a calculator to find θ (correct to two decimal places).\n9.\nθ\nP(a; b)\n2\nx\ny\nO\nb\nIn the figure, P is a point on the Cartesian plane such that OP = 2 units and\nθ = 300◦. Without the use of a calculator, determine:\na) the values of a and b\nb) the value of sin(180◦−θ)\n10. Solve for x with x ∈[−180◦; 180◦] (correct to one decimal place):\na) 2 sin x\n2 = 0,86\n303\nChapter 6.\nTrigonometry\n\nb) tan(x + 10◦) = cos 202,6◦\nc) cos2 x −4 sin2 x = 0\n11. Find the general solution for the following equations:\na)\n1\n2 sin(x −25◦) = 0,25\nb) sin2 x + 2 cos x = −2\n12. Given the equation: sin 2α = 0,84\na) Find the general solution of the equation.\nb) Illustrate how this equation could be solved graphically for α ∈[0◦; 360◦].\nc) Write down the solutions for sin 2α = 0,84 for α ∈[0◦; 360◦].\n13.\nA\nT\nG\nN\nH\nn\nα\nβ\nA is the highest point of a vertical tower AT. At point N on the tower, n metres\nfrom the top of the tower, a bird has made its nest. The angle of inclination from\nG to point A is α and the angle of inclination from G to point N is β.\na) Express A ˆGN in terms of α and β.\nb) Express ˆA in terms of α and/or β.\nc) Show that the height of the nest from the ground (H) can be determined by\nthe formula\nH = n cos α sin β\nsin(α −β)\nd) Calculate the height of the nest H if n = 10 m, α = 68◦and β = 40◦(give\nyour answer correct to the nearest metre).\n304\n6.6.\nSummary\n\n14.\nA\nD\nB\nC\n11\n8\n5\nMr. Collins wants to pave his trapezium-shaped backyard, ABCD. AB ∥DC\nand ˆB = 90◦. DC = 11 m, AB = 8 m and BC = 5 m.\na) Calculate the length of the diagonal AC.\nb) Calculate the length of the side AD.\nc) Calculate the area of the patio using geometry.\nd) Calculate the area of the patio using trigonometry.\n15.\nA\nC\nB\n2t\nF\nt\nn\nn\n2n\nα\nIn △ABC, AC = 2A, AF = BF, A ˆFB = α and FC = 2AF. Prove that\ncos α = 1\n4.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 234H\n2. 234J\n3. 234K\n4. 234M\n5a. 234N\n5b. 234P\n6. 234Q\n7. 234R\n8. 234S\n9. 234T\n10a. 234V\n10b. 234W\n10c. 234X\n11a. 234Y\n11b. 234Z\n12. 2352\n13. 2353\n14. 2354\n15. 2355\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n305\nChapter 6.\nTrigonometry\n\n\nCHAPTER\n7\nMeasurement\n7.1\nArea of a polygon\n308\n7.2\nRight prisms and cylinders\n311\n7.3\nRight pyramids, right cones and spheres\n318\n7.4\nMultiplying a dimension by a constant factor\n322\n7.5\nSummary\n326\n\n7\nMeasurement\nThis chapter is a revision of perimeters and areas of two dimensional objects and\nvolumes of three dimensional objects. We also examine different combinations of\ngeometric objects and calculate areas and volumes in a variety of real-life contexts.\nSee video: 2356 at www.everythingmaths.co.za\n7.1\nArea of a polygon\nEMBHV\nSquare\ns\ns\nArea = s2\nRectangle\nh\nb\nArea = b × h\nTriangle\nh\nb\nArea = 1\n2b × h\nSee video: 2357 at www.everythingmaths.co.za\nTrapezium\nh\nb\na\nArea = 1\n2 (a + b) × h\nParallelogram\nh\nb\nArea = b × h\nCircle\nb r\nArea = πr2\n(Circumference = 2πr)\nSee video: 2358 at www.everythingmaths.co.za\n308\n7.1.\nArea of a polygon\n\nWorked example 1: Finding the area of a polygon\nQUESTION\nABCD is a parallelogram with DC = 15 cm, h = 8 cm and BF = 9 cm.\nA\nB\nC\nD\nH\n9 cm\n15 cm\nh\nF\nCalculate:\n1. the area of ABCD\n2. the perimeter of ABCD\nSOLUTION\nStep 1: Determine the area\nThe area of a parallelogram ABCD = base × height:\nArea = 15 × 8\n= 120 cm2\nStep 2: Determine the perimeter\nThe perimeter of a parallelogram ABCD = 2DC + 2BC.\nTo find the length of BC, we use AF ⊥BC and the theorem of Pythagoras.\nIn △ABF:\nAF 2 = AB2 −BF 2\n= 152 −92\n= 144\n∴AF = 12 cm\nAreaABCD = BC × AF\n120 = BC × 12\n∴BC = 10 cm\n∴PerimeterABCD = 2(15) + 2(10)\n= 50 cm\n309\nChapter 7.\nMeasurement\n\nExercise 7 – 1: Area of a polygon\n1. Vuyo and Banele are having a competition to see who can build the best kite\nusing balsa wood (a lightweight wood) and paper. Vuyo decides to make his kite\nwith one diagonal 1 m long and the other diagonal 60 cm long. The intersection\nof the two diagonals cuts the longer diagonal in the ratio 1 : 3.\nBanele also uses diagonals of length 60 cm and 1 m, but he designs his kite to\nbe rhombus-shaped.\na) Draw a sketch of Vuyo’s kite and write down all the known measurements.\nb) Determine how much balsa wood Vuyo will need to build the outside frame\nof the kite (give answer correct to the nearest cm).\nc) Calculate how much paper he will need to cover the frame of the kite.\nd) Draw a sketch of Banele’s kite and write down all the known measure-\nments.\ne) Determine how much wood and paper Banele will need for his kite.\nf) Compare the two designs and comment on the similarities and differences.\nWhich do you think is the better design? Motivate your answer.\n2. O is the centre of the bigger semi-circle with a radius of 10 units. Two smaller\nsemi-circles are inscribed into the bigger one, as shown on the diagram. Calcu-\nlate the following (in terms of π):\nO\nb\nb\na) The area of the shaded figure.\nb) The perimeter enclosing the shaded area.\n3. Karen’s engineering textbook is 30 cm long and 20 cm wide. She notices that\nthe dimensions of her desk are in the same proportion as the dimensions of her\ntextbook.\na) If the desk is 90 cm wide, calculate the area of the top of the desk.\nb) Karen uses some cardboard to cover each corner of her desk with an isosce-\nles triangle, as shown in the diagram:\n150 mm\n150 mm\ndesk\nCalculate the new perimeter and area of the visible part of the top of her\ndesk.\n310\n7.1.\nArea of a polygon\n\nc) Use this new area to calculate the dimensions of a square desk with the\nsame desk top area.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2359\n2. 235B\n3. 235C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.2\nRight prisms and cylinders\nEMBHW\nA right prism is a geometric solid that has a polygon as its base and vertical sides\nperpendicular to the base. The base and top surface are the same shape and size. It is\ncalled a “right” prism because the angles between the base and sides are right angles.\nA triangular prism has a triangle as its base, a rectangular prism has a rectangle as its\nbase, and a cube is a rectangular prism with all its sides of equal length. A cylinder is\nanother type of right prism which has a circle as its base. Examples of right prisms are\ngiven below: a rectangular prism, a cube, a triangular prism and a cylinder.\nSurface area of prisms and cylinders\nEMBHX\nSurface area is the total area of the exposed or outer surfaces of a prism. This is easier\nto understand if we imagine the prism to be a cardboard box that we can unfold. A\nsolid that is unfolded like this is called a net. When a prism is unfolded into a net, we\ncan clearly see each of its faces. In order to calculate the surface area of the prism, we\ncan then simply calculate the area of each face, and add them all together.\nFor example, when a triangular prism is unfolded into a net, we can see that it has\ntwo faces that are triangles and three faces that are rectangles. To calculate the surface\narea of the prism, we find the area of each triangle and each rectangle, and add them\ntogether.\nIn the case of a cylinder the top and bottom faces are circles and the curved surface\nflattens into a rectangle with a length that is equal to the circumference of the circular\nbase. To calculate the surface area we therefore find the area of the two circles and the\nrectangle and add them together.\n311\nChapter 7.\nMeasurement\n\nBelow are examples of right prisms that have been unfolded into nets. A rectangular\nprism unfolded into a net is made up of six rectangles.\nA cube unfolded into a net is made up of six identical squares.\nA triangular prism unfolded into a net is made up of two triangles and three rectangles.\nThe sum of the lengths of the rectangles is equal to the perimeter of the triangles.\nA cylinder unfolded into a net is made up of two identical circles and a rectangle with\nlength equal to the circumference of the circles.\n312\n7.2.\nRight prisms and cylinders\n\nWorked example 2: Calculating surface area\nQUESTION\nA box of chocolates has the following dimensions:\nlength = 25 cm\nwidth = 20 cm\nheight = 4 cm\n25 cm\n20 cm\n4 cm\nAnd a cylindrical tin of biscuits has the following dimensions:\ndiameter = 20 cm\nheight = 20 cm\nb\n20 cm\n20 cm\n1. Calculate the area of the wrapping paper needed to cover the entire box (assume\nno overlapping at the corners).\n2. Determine if this same sheet of wrapping paper would be enough to cover the\ntin of biscuits.\nSOLUTION\nStep 1: Determine the area of the rectangular box\nSurface area = 2 × (25 × 20) + 2 × (20 × 4) + 2 × (25 × 4)\n= 1360 cm2\n313\nChapter 7.\nMeasurement\n\nStep 2: Determine the area of the cylindrical tin\nThe radius of the cylinder = 20\n2 = 10 cm.\nSurface area = 2 × π(10)2 + 2π(10)(20)\n= 1885 cm2\nStep 3: Write the final answer\nNo, the area of the sheet of wrapping paper used to cover the box is not big enough to\ncover the tin.\nExercise 7 – 2: Calculating surface area\n1. A popular chocolate container is an equilateral right triangular prism with sides\nof 34 mm. The box is 170 mm long. Calculate the surface area of the box (to the\nnearest square centimetre).\n34 mm\n34 mm\n34 mm\n170 mm\n2. Gordon buys a cylindrical water tank to catch rain water off his roof. He discov-\ners a full 2 ℓtin of green paint in his garage and decides to paint the tank (not the\nbase). If he uses 250 ml to cover 1 m2, will he have enough green paint to cover\nthe tank with one layer of paint?\nDimensions of the tank:\ndiameter = 1,1 m\nheight = 1,4 m\nb\n1,1 m\n1,4 m\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235D\n2. 235F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n314\n7.2.\nRight prisms and cylinders\n\nVolume of prisms and cylinders\nEMBHY\nVolume, sometimes also called capacity, is the three dimensional space occupied by\nan object, or the contents of an object. It is measured in cubic units.\nThe volume of a right prism is simply calculated by multiplying the area of the base of\na solid by the height of the solid.\nRectangular\nprism\nl\nb\nh\nVolume = area of base × height\n= area of rectangle × height\n= l × b × h\nTriangular\nprism\nH\nb\nh\nVolume = area of base × height\n= area of triangle × height\n=\n\u00121\n2b × h\n\u0013\n× H\nCylinder\nh\nr\nVolume = area of base × height\n= area of circle × height\n= πr2 × h\nSee video: 235G at www.everythingmaths.co.za\n315\nChapter 7.\nMeasurement\n\nWorked example 3: Calculating volume\nQUESTION\nA rectangular glass vase with dimensions 28 cm × 18 cm × 8 cm is used for flower\narrangements. A florist uses a platic cylindrical jug to pour water into the glass vase.\nThe jug has a diameter of 142 mm and a height of 28 cm.\n28 cm\n18 cm\n8 cm\n142 mm\n28 cm\njug\nvase\n1. Will the plastic jug hold 5 ℓof water?\n2. Will a full jug of water be enough to fill the glass vase?\nSOLUTION\nStep 1: Determine the volume of the plastic jug\nThe diameter of the jug is 142 mm, therefore the radius =\n142\n2×10 = 7,1 cm.\nVolume of a cylinder = area of the base × height\nVolume of the jug = πr2 × h\n= π × (7,1)2 × 28\n= 4434 cm3\nAnd 1000 cm3 = 1 ℓ\n∴Volume of the jug = 4434\n1000\n= 4,434 ℓ\nNo, the capacity of the jug is not enough to hold 5 ℓof water.\n316\n7.2.\nRight prisms and cylinders\n\nStep 2: Determine the volume of the glass vase\nVolume of a rectangular prism = area of the base × height\nVolume of the vase = l × b × h\n= 28 × 18 × 8\n= 4032 cm3\n∴Volume of the vase = 4032\n1000\n= 4,032 ℓ\nYes, the volume of the jug is greater than the volume of the vase.\nExercise 7 – 3: Calculating volume\n1. The roof of Phumza’s house is the shape of a right-angled trapezium. A cylindri-\ncal water tank is positioned next to the house so that the rain on the roof runs\ninto the tank. The diameter of the tank is 140 cm and the height is 2,2 m.\n10 m\n8 m\n7,5 m\n2,2 m\n140 cm\na) Determine the area of the roof.\nb) Determine how many litres of water the tank can hold.\n2. The length of a side of a hexagonal sweet tin is 8 cm and its height is equal to\nhalf of the side length.\nA\nB\nC\nD\nE\nF\n8 cm\nh\na) Show that the interior angles are equal to 120◦.\n317\nChapter 7.\nMeasurement\n\nb) Determine the length of the line AE.\nc) Calculate the volume of the tin.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235H\n2. 235J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.3\nRight pyramids, right cones and spheres\nEMBHZ\nA pyramid is a geometric solid that has a polygon as its base and sides that converge\nat a point called the apex. In other words the sides are not perpendicular to the base.\nb\nThe triangular pyramid and square pyramid take their names from the shape of their\nbase. We call a pyramid a “right pyramid” if the line between the apex and the centre\nof the base is perpendicular to the base. Cones are similar to pyramids except that\ntheir bases are circles instead of polygons. Spheres are solids that are perfectly round\nand look the same from any direction.\nSurface area of pyramids, cones and spheres\nEMBJ2\nSquare\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n= b2 + 4\n\u0000 1\n2bhs\n\u0001\n= b (b + 2hs)\n318\n7.3.\nRight pyramids, right cones and spheres\n\nTriangular\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n=\n\u0000 1\n2b × hb\n\u0001\n+ 3\n\u0000 1\n2b × hs\n\u0001\n= 1\n2b (hb + 3hs)\nRight cone\nh\nr\nH\nSurface area = area of base +\narea of walls\n= πr2 + 1\n2 × 2πrh\n= πr (r + h)\nSphere\nb\nr\nSurface area = 4πr2\nVolume of pyramids, cones and spheres\nEMBJ3\nSquare\npyramid\nb\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × b2 × H\nTriangular\npyramid\nb\nh\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × 1\n2bh × H\n319\nChapter 7.\nMeasurement\n\nRight cone\nr\nH\nVolume = 1\n3 × area of base ×\nheight of cone\n= 1\n3 × πr2 × H\nSphere\nb\nr\nVolume = 4\n3πr3\nSee video: 235K at www.everythingmaths.co.za\nWorked example 4: Finding surface area and volume\nQUESTION\nThe Southern African Large Telescope (SALT) is housed in a cylindrical building with\na domed roof in the shape of a hemisphere. The height of the building wall is 17 m\nand the diameter is 26 m.\n17 m\n26 m\n1. Calculate the total surface area of the building.\n2. Calculate the total volume of the building.\n320\n7.3.\nRight pyramids, right cones and spheres\n\nSOLUTION\nStep 1: Calculate the total surface area\nTotal surface area = area of the dome + area of the cylinder\nSurface area =\n\u00141\n2(4πr2)\n\u0015\n+ [2πr × h]\n= 1\n2(4π)(13)2 + 2π(13)(17)\n= 2450 m2\nStep 2: Calculate the total volume\nTotal volume = volume of the dome + volume of the cylinder\nVolume =\n\u00141\n2 ×\n\u00124\n3πr3\n\u0013\u0015\n+\n\u0002\nπr2h\n\u0003\n= 2\n3π(13)3 + π(11)2(13)\n= 9543 m3\nExercise 7 – 4: Finding surface area and volume\n1. An ice-cream cone has a diameter of 52,4 mm and a total height of 146 mm.\n52,4 mm\n146 mm\na) Calculate the surface area of the ice-cream and the cone.\nb) Calculate the total volume of the ice-cream and the cone.\nc) How many ice-cream cones can be made from a 5 ℓtub of ice-cream (as-\nsume the cone is completely filled with ice-cream)?\n321\nChapter 7.\nMeasurement\n\nd) Consider the net of the cone given below. R is the length from the tip of\nthe cone to its perimeter, P.\nP\nR\nb\nM\ni. Determine the value of R.\nii. Calculate the length of arc P.\niii. Determine the length of arc M.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.4\nMultiplying a dimension by a constant factor\nEMBJ4\nWhen one or more of the dimensions of a prism or cylinder is multiplied by a constant,\nthe surface area and volume will change. The new surface area and volume can be\ncalculated by using the formulae from the preceding section.\nIt is important to see a relationship between the change in dimensions and the resulting\nchange in surface area and volume. These relationships make it simpler to calculate\nthe new volume or surface area of an object when its dimensions are scaled up or\ndown.\nConsider a rectangular prism of dimensions l, b and h. Below we multiply one, two\nand three of its dimensions by a constant factor of 5 and calculate the new volume and\nsurface area.\n322\n7.4.\nMultiplying a dimension by a constant factor\n\nDimensions\nVolume\nSurface\nOriginal dimensions\nl\nb\nh\nV = l × b × h\n= lbh\nA\n= 2 [(l × h) + (l × b) + (b × h)]\n= 2 (lh + lb + bh)\nMultiply one\ndimension by 5\nl\nb\n5h\nV1 = l × b × 5h\n= 5 (lbh)\n= 5V\nA1\n= 2 [(l × 5h) + (l × b) + (b × 5h)]\n= 2 (5lh + lb + 5bh)\nMultiply two\ndimensions by 5\n5l\nb\n5h\nV = 5l × b × 5h\n= 5 . 5(lbh)\n= 52V\nA2\n= 2 [(5l × 5h) + (5l × b) + (b × 5h)]\n= 2 × 5(5lh + lb + bh)\nMultiply all three\ndimensions by 5\n5l\n5b\n5h\nV = 5l × 5b × 5h\n= 53(lbh)\n= 53V\nA3\n= 2 [(5l × 5h) + (5l × 5b) + (5b × 5h)]\n= 2 × (52lh + 52lb + 52bh)\n= 52 × 2(lh + lb + bh)\n= 52A\nMultiply all three\ndimensions by k\nkl\nkb\nkh\nV = kl × kb × kh\n= k3(lbh)\n= k3V\nAk\n= 2 [(kl × kh) + (kl × kb) + (kb × kh)]\n= 2 × (k2lh + k2lb + k2bh)\n= k2 × 2(lh + lb + bh)\n= k2A\n323\nChapter 7.\nMeasurement\n\nWorked example 5: The effects of k\nQUESTION\nThe Nash family wants to build a television room onto their house. The dad draws up\nthe plans for the new square room of length k metres. The mum looks at the plans and\ndecides that the area of the room needs to be doubled. To achieve this:\n• the mum suggests doubling the length of the sides of the room\n• the dad recommends adding 2 m to the length of the sides\n• the daughter suggests multiplying the length of the sides by a factor of\n√\n2\n• the son suggests doubling only the width of the room\nWho’s suggestion will double the area of the square room? Show all calculations.\nSOLUTION\nStep 1: Draw a sketch\nk\nk\n2k\n2k\nk + 2\nk + 2\n√\n2k\n√\n2k\n2k\nk\nArea O\nArea M\nArea D\nArea d\nArea s\nStep 2: Calculate and compare\nFirst calculate the area of the square room in the original plan:\nArea O = length × length\n= k2\nTherefore, double the area of the room would be 2k2.\n324\n7.4.\nMultiplying a dimension by a constant factor\n\nConsider the mum’s suggestion of doubling the length of the sides of the room:\nArea M = length × length\n= 2k × 2k\n= 4k2\nThis area would be 4 times the original area.\nThe dad suggests adding 2 m to the length of the sides of the room:\nArea D = length × length\n= (k + 2) × (k + 2)\n= k2 + 4k + 2\n̸= 2k2\nThis is not double the original area.\nThe daughter suggests multiplying the length of the sides by a factor of\n√\n2:\nArea d = length × length\n=\n√\n2k ×\n√\n2k\n= 2k2\nThe daughter’s suggestion would double the area of the room. Practically, the length\nof the room could be multiplied by\n√\n2 ≈1,41 which would given an area of 1,96 m2.\nThe son suggests doubling only the width of the room:\nArea s = length × length\n= 2k × k\n= 2k2\nThe son’s suggestion would double the area of the room, however the room would no\nlonger be a square.\nStep 3: Write the final answer\nThe daughter’s suggestion of multiplying the length of the sides of the room by a factor\nof\n√\n2 would keep the shape of the room a square and would double the area of the\nroom.\nExercise 7 – 5: The effects of k\n1. Complete the following sentences:\na) If one dimension of a cube is multiplied by a factor 1\n2, the volume of the\ncube . . .\nb) If two dimensions of a cube are multiplied by a factor 7, the volume of the\ncube . . .\n325\nChapter 7.\nMeasurement\n\nc) If three dimensions of a cube are multiplied by a factor 3, then:\ni. each side of the cube will . . .\nii. the outer surface area of the cube will . . .\niii. the volume of the cube will . . .\nd) If each side of a cube is halved, then:\ni. the outer surface area of the cube will . . .\nii. the volume of the cube will . . .\n2. The municipality intends building a swimming pool of volume W 3 cubic metres.\nHowever, they realise that it will be very expensive to fill the pool with water, so\nthey decide to make the pool smaller.\na) The length and breadth of the pool are reduced by a factor of\n7\n10. Express\nthe new volume in terms of W.\nb) The dimensions of the pool are reduced so that the volume of the pool\ndecreases by a factor of 0,8. Determine the new dimensions of the pool in\nterms of W (remember that the pool must be a cube).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235N\n2. 235P\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.5\nSummary\nEMBJ5\nSee presentation: 235Q at www.everythingmaths.co.za\n1. Area is the two dimensional space inside the boundary of a flat object.\n2. Area formulae:\n• square: s2\n• rectangle: b × h\n• triangle: 1\n2b × h\n• trapezium: 1\n2 (a + b) × h\n• parallelogram: b × h\n• circle: πr2\n3. Surface area is the total area of the exposed or outer surfaces of a prism.\n4. A net is the unfolded “plan” of a solid.\n5. Volume is the three dimensional space occupied by an object, or the contents\nof an object.\n• Volume of a rectangular prism: l × b × h\n326\n7.5.\nSummary\n\n• Volume of a triangular prism:\n\u0000 1\n2b × h\n\u0001\n× H\n• Volume of a square prism or cube: s3\n• Volume of a cylinder: πr2 × h\n6. A pyramid is a geometric solid that has a polygon as its base and sides that\nconverge at a point called the apex. The sides are not perpendicular to the base.\n7. Surface area formulae:\n• square pyramid: b (b + 2h)\n• triangular pyramid: 1\n2b (hb + 3hs)\n• right cone: πr (r + hs)\n• sphere: 4πr2\n8. Volume formulae:\n• square pyramid: 1\n3 × b2 × H\n• triangular pyramid: 1\n3 × 1\n2bh × H\n• right cone: 1\n3 × πr2 × H\n• sphere: 4\n3πr3\nExercise 7 – 6: End of chapter exercises\n1.\na) Describe this figure in terms of a prism.\nb) Draw a net of this figure.\n2. Which of the following is a net of a cube?\na)\nb)\nc)\nd)\ne)\n327\nChapter 7.\nMeasurement\n\n3. Name and draw the following figures:\na) A prism with the least number of sides.\nb) A pyramid with the least number of vertices.\nc) A right prism with a kite base.\n4.\na)\ni. Determine how much paper is needed to make a box of width 16 cm,\nheight 3 cm and length 20 cm (assume no overlapping at corners).\nii. Give a mathematical name for the shape of the box.\niii. Calculate the volume of the box.\nb) Determine how much paper is needed to make a cube with a capacity of\n1 ℓ.\nc) Compare the box and the cube. Which has the greater volume and which\nrequires the most paper to make?\n5. ABCD is a rhombus with sides of length 3\n2x millimetres. The diagonals intersect\nat O and length DO = x millimetres. Express the area of ABCD in terms of x.\nO\nB\nD\nx\nC\nA\n3\n2x\n6. The diagram shows a rectangular pyramid with a base of length 80 cm and\nbreadth 60 cm. The vertical height of the pyramid is 45 cm.\n60 cm\n80 cm\n45 cm\nb\nh\nH\na) Calculate the volume of the pyramid.\nb) Calculate H and h.\nc) Calculate the surface area of the pyramid.\n7. A group of children are playing soccer in a field. The soccer ball has a capacity\nof 5000 cc (cubic centimetres). A drain pipe in the corner of the field has a\ndiameter of 20 cm. Is it possible for the children to lose their ball down the pipe?\nShow your calculations.\n328\n7.5.\nSummary\n\n8. A litre of washing powder goes into a standard cubic container at the factory.\na) Determine the length of the sides of the container.\nb) Determine the dimensions of the cubic container required to hold double\nthe volume of washing powder.\n9. A cube has sides of length k units.\na) Describe the effect on the volume of the cube if the height is tripled.\nb) If all three dimensions of the cube are tripled, determine the effect on the\nouter surface area.\nc) If all three dimensions of the cube are tripled, determine the effect on the\nvolume.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235R\n2. 235S\n3a. 235T\n3b. 235V\n3c. 235W\n4. 235X\n5. 235Y\n6. 235Z\n7. 2362\n8. 2363\n9. 2364\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n329\nChapter 7.\nMeasurement\n\n\nCHAPTER\n8\nEuclidean geometry\n8.1\nRevision\n332\n8.2\nCircle geometry\n333\n8.3\nSummary\n363\n\n8\nEuclidean geometry\n8.1\nRevision\nEMBJ6\nParallelogram\nEMBJ7\nA parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nSummary of the properties of a parallelogram:\n• Both pairs of opposite sides are parallel.\n• Both pairs of opposite sides are equal in length.\n• Both pairs of opposite angles are equal.\n• Both diagonals bisect each other.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\nThe mid-point theorem\nEMBJ8\nThe line joining the mid-points of two sides of a triangle is parallel to the third side\nand equal to half the length of the third side.\nA\nB\nC\nD\nE\nGiven: AD = DB and AE = EC, we can conclude that DE ∥BC and DE = 1\n2BC.\n332\n8.1.\nRevision\n\n8.2\nCircle geometry\nEMBJ9\nTerminology\nThe following terms are regularly used when referring to circles:\n• Arc — a portion of the circumference of a circle.\n• Chord — a straight line joining the ends of an arc.\n• Circumference — the perimeter or boundary line of a circle.\n• Radius (r) — any straight line from the centre of the circle to a point on the\ncircumference.\n• Diameter — a special chord that passes through the centre of the circle. A di-\nameter is a straight line segment from one point on the circumference to another\npoint on the circumference that passes through the centre of the circle.\n• Segment — part of the circle that is cut off by a chord. A chord divides a circle\ninto two segments.\n• Tangent — a straight line that makes contact with a circle at only one point on\nthe circumference.\nb\nb\nA\nB\nO\nP\na\nr\nc\nchord\ntangent\ndiameter\nradius\nsegment\nSee video: 2365 at www.everythingmaths.co.za\nAxioms\nAn axiom is an established or accepted principle. For this section, the following are\naccepted as axioms.\n333\nChapter 8.\nEuclidean geometry\n\n1. The theorem of Pythagoras states that the square of the hypotenuse of a right-\nangled triangle is equal to the sum of the squares of the other two sides.\n(AC)2 = (AB)2 + (BC)2\nC\nB\nA\n(AC)2\n(AB)2\n(BC)2\n2. A tangent is perpendicular to the radius (OT ⊥ST), drawn at the point of contact\nwith the circle.\nT\nS\nb\nO\nTheorems\nEMBJB\nA theorem is a hypothesis (proposition) that can be shown to be true by accepted\nmathematical operations and arguments. A proof is the process of showing a theorem\nto be correct.\nThe converse of a theorem is the reverse of the hypothesis and the conclusion. For\nexample, given the theorem “if A, then B”, the converse is “if B, then A”.\n334\n8.2.\nCircle geometry\n\nTheorem: Perpendicular line from circle centre bisects chord\nSTATEMENT\nIf a line is drawn from the centre of a circle perpendicular to a chord, then it bisects\nthe chord.\n(Reason: ⊥from centre bisects chord)\nGiven:\nCircle with centre O and line OP perpendicular to chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = PB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA2 = OP 2 + AP 2\n(Pythagoras)\nOB2 = OP 2 + BP 2\n(Pythagoras)\nand\nOA = OB\n(equal radii)\n∴AP 2 = BP 2\n∴AP = BP\nTherefore OP bisects AB.\nAlternative proof:\nIn △OPA and in △OPB,\nO ˆPA = O ˆPB\n(given OP ⊥AB)\nOA = OB\n(equal radii)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(RHS)\n∴AP = PB\nTherefore OP bisects AB.\n335\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Line from circle centre to mid-point of\nchord is perpendicular\nSTATEMENT\nIf a line is drawn from the centre of a circle to the mid-point of a chord, then the line\nis perpendicular to the chord.\n(Reason: line from centre to mid-point ⊥)\nGiven:\nCircle with centre O and line OP to mid-point P on chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nOP ⊥AB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA = OB\n(equal radii)\nAP = PB\n(given)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(SSS)\n∴O ˆPA = O ˆPB\nand O ˆPA + O ˆPB = 180◦\n(∠on str. line)\n∴O ˆPA = O ˆPB = 90◦\nTherefore OP ⊥AB.\nSee video: 2366 at www.everythingmaths.co.za\n336\n8.2.\nCircle geometry\n\nTheorem: Perpendicular bisector of chord passes through circle centre\nSTATEMENT\nIf the perpendicular bisector of a chord is drawn, then the line will pass through the\ncentre of the circle.\n(Reason: ⊥bisector through centre)\nGiven:\nCircle with mid-point P on chord AB.\nLine QP is drawn such that Q ˆPA = Q ˆPB = 90◦.\nLine RP is drawn such that R ˆPA = R ˆPB = 90◦.\nb\nb\nA\nB\nQ\nP\nR\nRequired to prove:\nCircle centre O lies on the line PR\nPROOF\nDraw lines QA and QB.\nDraw lines RA and RB.\nIn △QPA and in △QPB,\nAP = PB\n(given)\nQP = QP\n(common side)\nQ ˆPA = Q ˆPB = 90◦\n(given)\n∴△QPA ≡△QPB\n(SAS)\n∴QA = QB\nSimilarly it can be shown that in △RPA and in △RPB, RA = RB.\nWe conclude that all the points that are equidistant from A and B will lie on the\nline PR extended. Therefore the centre O, which is equidistant to all points on the\ncircumference, must also lie on the line PR.\n337\nChapter 8.\nEuclidean geometry\n\nWorked example 1: Perpendicular line from circle centre bisects chord\nQUESTION\nGiven OQ ⊥PR and PR = 8 units, determine the value of x.\nO\nx\n5\nP\nQ\nR\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nPQ = QR = 4\n(⊥from centre bisects chord)\nStep 2: Solve for x\nIn △OQP:\nPQ = 4\n(⊥from centre bisects chord)\nOP 2 = OQ2 + QP 2\n(Pythagoras)\n52 = x2 + 42\n∴x2 = 25 −16\nx2 = 9\nx = 3\nStep 3: Write the final answer\nx = 3 units.\n338\n8.2.\nCircle geometry\n\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. In the circle with centre O, OQ ⊥PR,\nOQ = 4 units and PR = 10. Determine\nx.\nO\n4\nP\nQ\nR\nx\n2. In the circle with centre O and radius\n= 10 units, OQ ⊥PR and PR = 8. De-\ntermine x.\nO\nx\n10\nP\nQ\nR\n3. In the circle with centre O, OQ ⊥PR,\nPR = 12 units and SQ = 2 units. Deter-\nmine x.\nO\nx\nP\nQ\nR\nS\n4. In the circle with centre O, OT ⊥SQ,\nOT ⊥PR, OP = 10 units, ST = 5 units\nand PU = 8 units. Determine TU.\nO\nV\n8\nP\nR\nU\n10\n5\nT\nQ\nS\n5. In the circle with centre O, OT ⊥QP,\nOS ⊥PR, OT = 5 units, PQ = 24 units\nand PR = 25 units. Determine OS = x.\nO\nx\nP\nS\n5\nT\nQ\nR\n25\n24\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2367\n2. 2368\n3. 2369\n4. 236B\n5. 236C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n339\nChapter 8.\nEuclidean geometry\n\nInvestigation: Angles subtended by an arc at the centre and the circumference of\na circle\n1. Measure angles x and y in each of the following graphs:\nb\nx1\ny1\nb\nx2\ny2\nb\nx3\ny3\n2. Complete the table:\nx\ny\n3. Use your results to make a conjecture about the relationship between angles\nsubtended by an arc at the centre of a circle and angles at the circumference of\na circle.\n4. Now draw three of your own similar diagrams and measure the angles to check\nyour conjecture.\n340\n8.2.\nCircle geometry\n\nTheorem: Angle at the centre of a circle is twice the size of the angle at the cir-\ncumference\nSTATEMENT\nIf an arc subtends an angle at the centre of a circle and at the circumference, then the\nangle at the centre is twice the size of the angle at the circumference.\n(Reason: ∠at centre = 2∠at circum.)\nGiven:\nCircle with centre O, arc AB subtending A ˆOB at the centre of the circle, and A ˆPB at\nthe circumference.\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nRequired to prove:\nA ˆOB = 2A ˆPB\nPROOF\nDraw PO extended to Q and let A ˆOQ = ˆO1 and B ˆOQ = ˆO2.\nˆO1 = A ˆPO + P ˆAO\n(ext. ∠△= sum int. opp. ∠s)\nand A ˆPO = P ˆAO\n(equal radii, isosceles △APO)\n∴ˆO1 = A ˆPO + A ˆPO\nˆO1 = 2A ˆPO\nSimilarly, we can also show that ˆO2 = 2B ˆPO.\nFor the first two diagrams shown above we have that:\nA ˆOB = ˆO1 + ˆO2\n= 2A ˆPO + 2B ˆPO\n= 2(A ˆPO + B ˆPO)\n∴A ˆOB = 2(A ˆPB)\nAnd for the last diagram:\nA ˆOB = ˆO2 −ˆO1\n= 2B ˆPO −2A ˆPO\n= 2(B ˆPO −A ˆPO)\n∴A ˆOB = 2(A ˆPB)\n341\nChapter 8.\nEuclidean geometry\n\nWorked example 2: Angle at the centre of circle is twice angle at circumference\nQUESTION\nGiven HK, the diameter of the circle passing through centre O.\nb\nJ\nH\nK\nO\na\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nStep 2: Solve for a\nIn △HJK:\nH ˆOK = 180◦\n(∠on str. line)\n= 2a\n(∠at centre = 2∠at circum.)\n∴2a = 180◦\na = 180◦\n2\n= 90◦\nStep 3: Conclusion\nThe diameter of a circle subtends a right angle at the circumference (angles in a semi-\ncircle).\n342\n8.2.\nCircle geometry\n\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\nGiven O is the centre of the circle, determine the unknown angle in each of the fol-\nlowing diagrams:\n1.\nb\nJ\nH\nK\nO\nb\n45◦\n2.\nbO\nJ\nK\nH\n45◦\nc\n3.\nb\nO\nK\nJ\n100◦\nH\nd\n4.\nb\nO\nH\nJ\ne\nK\n35◦\n5.\nb\nO\nJ\nK\nH\n120◦\nf\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236D\n2. 236F\n3. 236G\n4. 236H\n5. 236J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n343\nChapter 8.\nEuclidean geometry\n\nInvestigation: Subtended angles in the same segment of a circle\n1. Measure angles a, b, c, d and e in the diagram below:\na\ne\nd\nb\nc\nP\nQ\n2. Choose any two points on the circumference of the circle and label them A and\nB.\n3. Draw AP and BP, and measure A ˆPB.\n4. Draw AQ and BQ, and measure A ˆQB.\n5. What do you observe? Make a conjecture about these types of angles.\nTheorem: Subtended angles in the same segment of a circle are equal\nSTATEMENT\nIf the angles subtended by a chord of the circle are on the same side of the chord, then\nthe angles are equal.\n(Reason: ∠s in same seg.)\nGiven:\nCircle with centre O, and points P and Q on the circumference of the circle. Arc AB\nsubtends A ˆPB and A ˆQB in the same segment of the circle.\n344\n8.2.\nCircle geometry\n\nbO\nA\nB\nP\nQ\nRequired to prove:\nA ˆPB = A ˆQB\nPROOF\nA ˆOB = 2A ˆPB\n(∠at centre = 2∠at circum.)\nA ˆOB = 2A ˆQB\n(∠at centre = 2∠at circum.)\n∴2A ˆPB = 2A ˆQB\nA ˆPB = A ˆQB\nEqual arcs subtend equal angles\nFrom the theorem above we can deduce that if angles at the circumference of a circle\nare subtended by arcs of equal length, then the angles are equal. In the figure below,\nnotice that if we were to move the two chords with equal length closer to each other,\nuntil they overlap, we would have the same situation as with the theorem above. This\nshows that the angles subtended by arcs of equal length are also equal.\nb\nb\n345\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Concyclic points\nSTATEMENT\nIf a line segment subtends equal angles at two other points on the same side of the line\nsegment, then these four points are concyclic (lie on a circle).\nGiven:\nLine segment AB subtending equal angles at points P and Q on the same side of the\nline segment AB.\nA\nB\nR\nQ\nP\nRequired to prove:\nA, B, P and Q lie on a circle.\nPROOF\nProof by contradiction:\nPoints on the circumference of a circle: we know that there are only two possible\noptions regarding a given point — it either lies on circumference or it does not.\nWe will assume that point P does not lie on the circumference.\nWe draw a circle that cuts AP at R and passes through A, B and Q.\nA ˆQB = A ˆRB\n(∠s in same seg.)\nbut A ˆQB = A ˆPB\n(given)\n∴A ˆRB = A ˆPB\nbut A ˆRB = A ˆPB + R ˆBP\n(ext. ∠△= sum int. opp.)\n∴R ˆBP = 0◦\nTherefore the assumption that the circle does not pass through P must be false.\nWe can conclude that A, B, Q and P lie on a circle (A, B, Q and P are concyclic).\n346\n8.2.\nCircle geometry\n\nWorked example 3: Concyclic points\nQUESTION\nGiven FH ∥EI and E ˆIF = 15◦, determine the value of b.\nE\nF\nG\nH\nI\n15◦\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nH ˆFI = 15◦\n(alt. ∠, FH ∥EI)\nand b = H ˆFI\n(∠s in same seg.)\n∴b = 15◦\nExercise 8 – 3: Subtended angles in the same segment\n1. Find the values of the unknown angles.\na)\nA\nB\nC\nD\n21◦\na\nb)\nJ\nK\nL\nM\n24◦\nc\n102◦\nd\nc)\nN\nO\nP\nQ\n17◦\nd\n347\nChapter 8.\nEuclidean geometry\n\n2.\nR\nS\nT\nU\nV\n45◦\n35◦\n15◦\ne\na) Given T ˆV S = S ˆV R, deter-\nmine the value of e.\nb) Is TV a diameter of the cir-\ncle? Explain your answer.\n3.\nb\nW\nX\nY\nZ\nO\n35◦\nf\nT\n1\n2\nGiven circle with centre O, WT =\nTY and X ˆWT = 35◦. Determine\nf.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236K\n1b. 236M\n1c. 236N\n2. 236P\n3. 236Q\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nCyclic quadrilaterals\nCyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference\nof a circle (concyclic).\nInvestigation: Cyclic quadrilaterals\nConsider the diagrams given below:\nCircle 1\nCircle 2\nCircle 3\nA\nB\nC\nD\nA\nB\nC\nD\nA\nB\nC\nD\n348\n8.2.\nCircle geometry\n\n1. Complete the following:\nABCD is a cyclic quadrilateral because . . . . . .\n2. Complete the table:\nCircle 1\nCircle 2\nCircle 3\nˆA =\nˆB =\nˆC =\nˆD =\nˆA + ˆC =\nˆB + ˆD =\n3. Use your results to make a conjecture about the relationship between angles of\ncyclic quadrilaterals.\nTheorem: Opposite angles of a cyclic quadrilateral\nSTATEMENT\nThe opposite angles of a cyclic quadrilateral are supplementary.\n(Reason: opp. ∠s cyclic quad.)\nGiven:\nCircle with centre O with points A, B, P and Q on the circumference such that ABPQ\nis a cyclic quadrilateral.\nbO\nA\nB\nP\nQ\n1\n2\nRequired to prove:\nA ˆBP + A ˆQP = 180◦and Q ˆAB + Q ˆPB = 180◦\n349\nChapter 8.\nEuclidean geometry\n\nPROOF\nDraw AO and OP. Label ˆO1 and ˆO2.\nˆO1 = 2A ˆBP\n(∠at centre = 2∠at circum.)\nˆO2 = 2A ˆQP\n(∠at centre = 2∠at circum.)\nand ˆO1 + ˆO2 = 360◦\n(∠s around a point)\n∴2A ˆBP + 2A ˆQP = 360◦\nA ˆBP + A ˆQP = 180◦\nSimilarly, we can show that Q ˆAB + Q ˆPB = 180◦.\nConverse: interior opposite angles of a quadrilateral\nIf the interior opposite angles of a quadrilateral are supplementary, then the quadrilat-\neral is cyclic.\nExterior angle of a cyclic quadrilateral\nIf a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle.\nb\nb\nWorked example 4: Opposite angles of a cyclic quadrilateral\nQUESTION\nGiven the circle with centre O and cyclic quadrilateral PQRS. SQ is drawn and\nS ˆPQ = 34◦. Determine the values of a, b and c.\nbO\nP\nQ\nR\nS\na\nb\nc\n34◦\n350\n8.2.\nCircle geometry\n\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nS ˆPQ + c = 180◦\n(opp. ∠s cyclic quad supp.)\n∴c = 180◦−34◦\n= 146◦\na = 90◦\n(∠in semi circle)\nIn △PSQ:\na + b + 34◦= 180◦\n(∠sum of △)\n∴b = 180◦−90◦−34◦\n= 56◦\nMethods for proving a quadrilateral is cyclic\nThere are three ways to prove that a quadrilateral is a cyclic quadrilateral:\nMethod of proof\nReason\nR\nQ\nS\nP\nIf ˆP + ˆR = 180◦or ˆS +\nˆQ = 180◦, then PQRS is\na cyclic quad.\nopp.\nint.\nangles\nsuppl.\nR\nQ\nS\nP\nIf ˆP = ˆQ or ˆS = ˆR, then\nPQRS is a cyclic quad.\nangles in the same\nseg.\nR\nQ\nS\nP\nT\nIf T ˆQR = ˆS, then PQRS\nis a cyclic quad.\next.\nangle equal to\nint. opp. angle\n351\nChapter 8.\nEuclidean geometry\n\nWorked example 5: Proving a quadrilateral is a cyclic quadrilateral\nQUESTION\nProve that ABDE is a cyclic quadrilateral.\nbO\nE\nC\nD\nA\nB\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Prove that ABDE is a cyclic quadrilateral\nD ˆBC = 90◦\n(∠in semi circle)\nand ˆE = 90◦\n(given)\n∴D ˆBC = ˆE\n∴ABDE is a cyclic quadrilateral\n(ext. ∠equals int. opp. ∠)\nExercise 8 – 4: Cyclic quadrilaterals\n1. Find the values of the unknown angles.\na)\nX\nY\nZ\nW\na\nb\n106◦\n87◦\nb)\nH\nI\nJ\nK\nL\n114◦\na\nc)\nU\nV\nW\nX\n57◦\na\n86◦\n352\n8.2.\nCircle geometry\n\n2. Prove that ABCD is a cyclic quadrilateral:\na) D\nC\n72◦\nB\nA\n32◦\nM\n40◦\nb) D\nC\n70◦\nB\nA\n35◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236R\n1b. 236S\n1c. 236T\n2a. 236V\n2b. 236W\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTangent line to a circle\nA tangent is a line that touches the circumference of a circle at only one place. The\nradius of a circle is perpendicular to the tangent at the point of contact.\nb\nO\n353\nChapter 8.\nEuclidean geometry\n\nTheorem: Two tangents drawn from the same point outside a circle\nSTATEMENT\nIf two tangents are drawn from the same point outside a circle, then they are equal in\nlength.\n(Reason: tangents from same point equal)\nGiven:\nCircle with centre O and tangents PA and PB, where A and B are the respective\npoints of contact for the two lines.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = BP\nPROOF\nIn △AOP and △BOP,\nO ˆAP = O ˆBP = 90◦\n(tangent ⊥radius)\nAO = BO\n(equal radii)\nOP = OP\n(common side)\n∴△AOP ≡△BOP\n(RHS)\n∴AP = BP\n354\n8.2.\nCircle geometry\n\nWorked example 6: Tangents from the same point outside a circle\nQUESTION\nIn the diagram below AE = 5 cm, AC = 8 cm and CE = 9 cm. Determine the values\nof a, b and c.\nA\nB\nC\nD\nE\nF\nAE = 5 cm\nAC = 8 cm\nCE = 9 cm\na\nb\nc\nb\nb\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for a, b and c\nAB = AF = a\n(tangents from A)\nEF = ED = c\n(tangents from E)\nCB = CD = b\n(tangents from C)\n∴AE = a + c = 5\nand AC = a + b = 8\nand CE = b + c = 9\nStep 3: Solve for the unknown variables using simultaneous equations\na + c = 5\n. . . (1)\na + b = 8\n. . . (2)\nb + c = 9\n. . . (3)\nSubtract equation (1) from equation (2) and then substitute into equation (3):\n(2) −(1)\nb −c = 8 −5\n= 3\n∴b = c + 3\nSubstitute into (3)\nc + 3 + c = 9\n2c = 6\nc = 3\n∴a = 2\nand b = 6\n355\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 5: Tangents to a circle\nFind the values of the unknown lengths.\n1.\nb\nG\nH\nI\nJ\nd\n5 cm\n8 cm\n2.\nb\nK\nL\nM\nN\nO\nP\ne\nLN = 7,5 cm\n2 cm\n6 cm\n3.\nb\nb\nR\nQ\nS\nf\n3 cm\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236X\n2. 236Y\n3. 236Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Tangent-chord theorem\nConsider the diagrams given below:\nDiagram 1\nDiagram 2\nDiagram 3\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\n1. Measure the following angles with a protractor and complete the table:\nDiagram 1\nDiagram 2\nDiagram 3\nA ˆBC =\nˆD =\nˆE =\n2. Use your results to complete the following: the angle between a tangent to a\ncircle and a chord is . . . . . . to the angle in the alternate segment.\n356\n8.2.\nCircle geometry\n\nTheorem: Tangent-chord theorem\nSTATEMENT\nThe angle between a tangent to a circle and a chord drawn at the point of contact, is\nequal to the angle which the chord subtends in the alternate segment.\n(Reason: tan. chord theorem)\nGiven:\nCircle with centre O and tangent SR touching the circle at B. Chord AB subtends ˆP1\nand ˆQ1.\nb\nO\nA\nB\nP\n1\n1\nQ\nT\n1\nS\nR\nRequired to prove:\n1. A ˆBR = A ˆPB\n2. A ˆBS = A ˆQB\nPROOF\nDraw diameter BT and join T to A.\nLet A ˆTB = T1.\nA ˆBS + A ˆBT = 90◦\n(tangent ⊥radius)\nB ˆAT = 90◦\n(∠in semi circle)\n∴A ˆBT + T1 = 90◦\n(∠sum of △BAT)\n∴A ˆBS = T1\nbut Q1 = T1\n(∠s in same segment)\n∴Q1 = A ˆBS\nA ˆBS + A ˆBR = 180◦\n(∠s on str. line)\nˆQ1 + ˆP1 = 180◦\n(opp. ∠s cyclic quad. supp.)\n∴A ˆBS + A ˆBR = Q1 + P1\nand A ˆBS = Q1\n∴A ˆBR = P1\n357\nChapter 8.\nEuclidean geometry\n\nWorked example 7: Tangent-chord theorem\nQUESTION\nDetermine the values of h and s.\nP\nO\nQ\nS\nR\nh + 20◦s\n4h\n4h −70◦\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for h\nO ˆQS = S ˆRQ\n(tangent chord theorem)\nh + 20◦= 4h −70◦\n90◦= 3h\n∴h = 30◦\nStep 3: Solve for s\nP ˆQR = Q ˆSR\n(tangent chord theorem)\ns = 4h\n= 4(30◦)\n= 120◦\n358\n8.2.\nCircle geometry\n\nExercise 8 – 6: Tangent-chord theorem\n1. Find the values of the unknown letters, stating reasons.\nQ\nR\nS\nO\nP\na\nb\n33◦\na)\nO\nP\nQ\nR\nS\nc\nd\n72◦\nb)\nO\nP\nQ\nR\nS\ng\nf\n38◦\n47◦\nc)\nR\nP\nO\nQ\nl\n1\n1\n66◦\nd)\nO\nP\nQ\nR\nS\ni\nj\nk\n39◦\n101◦\ne)\nO\nR\nQ\nS\nT\nm\nn\no\n34◦\nf)\nO\n•\nP\nR\nQ\nS\nT\np\nq\nr\n52◦\ng)\n359\nChapter 8.\nEuclidean geometry\n\n2. O is the centre of the circle and SPT is a tangent, with OP ⊥ST. Determine\na, b and c, giving reasons.\nO•\nS\nT\nP\nM\nN\na\nb\nc\n64◦\n3.\nP\nL\nA\nB\nC\n1 2\n3\n1\n2\nD\nGiven AB = AC, AP ∥BC and ˆA2 = ˆB2. Prove:\na) PAL is a tangent to the circle ABC.\nb) AB is a tangent to the circle ADP.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2372\n1b. 2373\n1c. 2374\n1d. 2375\n1e. 2376\n1f. 2377\n1g. 2378\n2. 2379\n3. 237B\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nConverse: tangent-chord theorem\nIf a line drawn through the end point of a chord forms an angle equal to the angle\nsubtended by the chord in the alternate segment, then the line is a tangent to the\ncircle.\n(Reason: ∠between line and chord = ∠in alt. seg. )\n360\n8.2.\nCircle geometry\n\nWorked example 8: Applying the theorems\nQUESTION\nA\nD\nB\nC\nO\nE\nF\nBD is a tangent to the circle with centre O, with BO ⊥AD.\nProve that:\n1. CFOE is a cyclic quadrilateral\n2. FB = BC\n3. ∠A ˆOC = 2B ˆFC\n4. Will DC be a tangent to the circle passing through C, F, O and E? Motivate your\nanswer.\nSOLUTION\nStep 1: Prove CFOE is a cyclic quadrilateral by showing opposite angles are supple-\nmentary\nBO ⊥OD\n(given)\n∴F ˆOE = 90◦\nF ˆCE = 90◦\n(∠in semi circle)\n∴CFOE is a cyclic quad.\n(opp. ∠s suppl.)\nStep 2: Prove BFC is an isosceles triangle\nTo show that FB = BC we first prove △BFC is an isosceles triangle by showing that\nB ˆFC = B ˆCF.\nB ˆCF = C ˆEO\n(tangent-chord)\nC ˆEO = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴B ˆFC = B ˆCF\n∴FB = BC\n(△BFC isosceles)\n361\nChapter 8.\nEuclidean geometry\n\nStep 3: Prove A ˆOC = 2B ˆFC\nA ˆOC = 2A ˆEC\n(∠at centre = 2∠at circum.)\nand A ˆEC = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴A ˆOC = 2B ˆFC\nStep 4: Determine if DC is a tangent to the circle through C, F, O and E\nProof by contradiction.\nLet us assume that DC is a tangent to the circle passing through the points C, F, O\nand E:\n∴D ˆCE = C ˆOE\n(tangent-chord)\nAnd using the circle with centre O and tangent BD we have that:\nD ˆCE = C ˆAE\n(tangent-chord)\nbut C ˆAE = 1\n2C ˆOE\n(∠at centre = 2∠at circum.)\n∴D ˆCE ̸= C ˆOE\nTherefore our assumption is not correct and we can conclude that DC is not a tangent\nto the circle passing through the points C, F, O and E.\nWorked example 9: Applying the theorems\nQUESTION\nA\nB\nC\nD\nE\nF\nG\nH\nFD is drawn parallel to the tangent CB\n362\n8.2.\nCircle geometry\n\nProve that:\n1. FADE is a cyclic quadrilateral\n2. F ˆEA = ˆB\nSOLUTION\nStep 1: Prove FADE is a cyclic quadrilateral using angles in the same segment\nF ˆDC = D ˆCB\n(alt. ∠s FD ∥CB)\nand D ˆCB = C ˆAE\n(tangent-chord)\n∴F ˆDC = C ˆAE\n∴FADE is a cyclic quad.\n(∠s in same seg.)\nStep 2: Prove F ˆEA = ˆB\nF ˆDA = ˆB\n(corresp. ∠s FD ∥CB)\nand F ˆEA = F ˆDA\n(∠s same seg. cyclic quad. FADE)\n∴F ˆEA = ˆB\n8.3\nSummary\nEMBJC\nSee presentation: 237C at www.everythingmaths.co.za\n• Arc An arc is a portion of the circumference of a circle.\n• Chord - a straight line joining the ends of an arc.\n• Circumference - perimeter or boundary line of a circle.\n• Radius (r) - any straight line from the centre of the circle to a point on the cir-\ncumference.\n• Diameter - a special chord that passes through the centre of the circle. A diame-\nter is the length of a straight line segment from one point on the circumference to\nanother point on the circumference, that passes through the centre of the circle.\n• Segment A segment is a part of the circle that is cut off by a chord. A chord\ndivides a circle into two segments.\n• Tangent - a straight line that makes contact with a circle at only one point on the\ncircumference.\n• A tangent line is perpendicular to the radius, drawn at the point of contact with\nthe circle.\n363\nChapter 8.\nEuclidean geometry\n\nb O\nM\nA\nB\n• If O is the centre and OM ⊥AB, then AM =\nMB.\n• If O is the centre and AM\n= MB, then\nA ˆ\nMO = B ˆ\nMO = 90◦.\n• If AM = MB and OM ⊥AB, then ⇒MO\npasses through centre O.\nb\n2x\nx\n2y\ny\nx\nIf an arc subtends an angle at the centre of a cir-\ncle and at the circumference, then the angle at the\ncentre is twice the size of the angle at the circum-\nference.\nb\nb\nAngles at the circumference subtended by the same\narc (or arcs of equal length) are equal.\nA\nB\nC\nD\n1\n2\nE\nThe four sides of a cyclic quadrilateral ABCD are\nchords of the circle with centre O.\n• ˆA + ˆC = 180◦(opp. ∠s supp.)\n• ˆB + ˆD = 180◦(opp. ∠s supp.)\n• E ˆBC = ˆD (ext. ∠cyclic quad.)\n• ˆA1 = ˆA2 = ˆC (vert. opp. ∠, ext. ∠cyclic\nquad.)\nA\nB\nC\nD\nProving a quadrilateral is cyclic: If ˆA + ˆC = 180◦or\nˆB+ ˆD = 180◦, then ABCD is a cyclic quadrilateral.\n364\n8.3.\nSummary\n\nA\nB\nC\nD\n1\n1\nIf ˆA1 = ˆC or ˆD1 = ˆB, then ABCD is a cyclic\nquadrilateral.\nA\nB\nC\nD\nIf ˆA = ˆB or ˆC = ˆD, then ABCD is a cyclic quadri-\nlateral.\nb\nA\nB\nO\nT\nIf AT and BT are tangents to circle O, then\n• OA ⊥AT (tangent ⊥radius)\n• OB ⊥BT (tangent ⊥radius)\n• TA = TB (tangents from same point equal)\nA\nB\nT\nD\nC\nx\ny\nx\ny\n• If DC is a tangent, then D ˆTA = T ˆBA and\nC ˆTB = T ˆAB\n• If D ˆTA = T ˆBA or C ˆTB = T ˆAB, then DC is\na tangent touching at T\n365\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 7: End of chapter exercises\n1.\nO\n•\nA\nB\nC\nD\nE\nF\n×\n×\nx\nAOC is a diameter of the circle with centre O. F is the mid-point of chord EC.\nB ˆOC = C ˆOD and ˆB = x. Express the following angles in terms of x, stating\nreasons:\na) ˆA\nb) C ˆOD\nc) ˆD\n2.\nM•\nD\nE\nF\nG\n1 2\n1\n2\n1\n2\n1 2\nD, E, F and G are points on circle with centre M.\nˆF1 = 7◦and ˆD2 = 51◦.\nDetermine the sizes of the following angles, stating reasons:\na)\nˆ\nM1\nb) ˆD1\nc) ˆF2\nd) ˆG\ne) ˆE1\n366\n8.3.\nSummary\n\n3.\nM\n•\nO•\nD\nA\nB\nC\n1 2\nO is a point on the circle with centre M. O is also the centre of a second circle.\nDA cuts the smaller circle at C and ˆD1 = x. Express the following angles in\nterms of x, stating reasons:\na) ˆD2\nb) O ˆAB\nc) O ˆBA\nd) A ˆOB\ne) ˆC\n4.\nO•\nA\nB\nC\nM\nO is the centre of the circle with radius 5 cm and chord BC = 8 cm. Calculate\nthe lengths of:\na) OM\nb) AM\nc) AB\n5.\nO•\nA\nB\nC\n70◦\nx\nAO ∥CB in circle with centre O. A ˆOB = 70◦and O ˆAC = x. Calculate the\nvalue of x, giving reasons.\n367\nChapter 8.\nEuclidean geometry\n\n6.\nO\n•\nP\nQ\nR\nS\nT\nx\nPQ is a diameter of the circle with centre O. SQ bisects P ˆQR and P ˆQS = x.\na) Write down two other angles that are also equal to x.\nb) Calculate P ˆOS in terms of x, giving reasons.\nc) Prove that OS is a perpendicular bisector of PR.\n7.\nO•\nA\nB\nC\nD\n35◦\nB ˆOD is a diameter of the circle with centre O. AB = AD and O ˆCD = 35◦.\nCalculate the value of the following angles, giving reasons:\na) O ˆDC\nb) C ˆOD\nc) C ˆBD\nd) B ˆAD\ne) A ˆDB\n8.\nO\n•\nR\nP\nT\nQ\nx\ny\nQP in the circle with centre O is protracted to T so that PR = PT. Express y in\nterms of x.\n368\n8.3.\nSummary\n\n9.\nO•\nA\nB\nC\nD\nE\nP\nF\nO is the centre of the circle with diameter AB. CD ⊥AB at P and chord DE\ncuts AB at F. Prove that:\na) C ˆBP = D ˆPB\nb) C ˆED = 2C ˆBA\nc) A ˆBD = 1\n2C ˆOA\n10.\nO\n•\nP\nQ\nR\nx\nS\nIn the circle with centre O, OR ⊥QP, PQ = 30 mm and RS = 9 mm. Deter-\nmine the length of OQ.\n11.\nM •\nP\nQ\nR\nS\nT\nP, Q, R and S are points on the circle with centre M. PS and QR are extended\nand meet at T. PQ = PR and P ˆQR = 70◦.\na) Determine, stating reasons, three more angles equal to 70◦.\nb) If Q ˆPS = 80◦, calculate S ˆRT, S ˆTR and P ˆQS.\nc) Explain why PQ is a tangent to the circle QST at point Q.\nd) Determine P ˆ\nMQ.\n369\nChapter 8.\nEuclidean geometry\n\n12.\nO\n•\nA\nP\nQ\nC\nB\nPOQ is a diameter of the circle with centre O. QP is protruded to A and AC is\na tangent to the circle. BA ⊥AQ and BCQ is a straight line. Prove:\na) P ˆCQ = B ˆAP\nb) BAPC is a cyclic quadrilateral\nc) AB = AC\n13.\nO•\nT\nC\nA\nB\nx\nTA and TB are tangents to the circle with centre O. C is a point on the circum-\nference and A ˆTB = x. Express the following in terms of x, giving reasons:\na) A ˆBT\nb) O ˆBA\nc) ˆC\n14.\nO•\nA\nB\nC\nE\nD\nAOB is a diameter of the circle\nAECB with centre O. OE ∥BC\nand cuts AC at D.\na) Prove AD = DC\nb) Show that A ˆBC is bisected\nby EB\nc) If O ˆEB = x, express B ˆAC\nin terms of x\nd) Calculate the radius of the\ncircle if AC = 10 cm and\nDE = 1 cm\n370\n8.3.\nSummary\n\n15.\nV\nQ\nS\nR\nP\nT\nW\nx\ny\nPQ and RS are chords of the circle and PQ ∥RS. The tangent to the circle at\nQ meets RS protruded at T. The tangent at S meets QT at V . QS and PR are\ndrawn.\nLet T ˆQS = x and Q ˆRP = y. Prove that:\na) T ˆV S = 2Q ˆRS\nb) QV SW is a cyclic quadrilateral\nc) Q ˆPS + ˆT = P ˆRT\nd) W is the centre of the circle\n16.\nF\nD\nB\nC\nE\nA\nK\nT\n1\n2\n1\n2\n1\n2\n3\n4\nThe two circles shown intersect at points F and D. BFT is a tangent to the\nsmaller circle at F. Straight line AFE is drawn such that DF = EF. CDE is a\nstraight line and chord AC and BF cut at K. Prove that:\na) BT ∥CE\nb) BCEF is a parallelogram\nc) AC = BF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237D\n2. 237F\n3. 237G\n4. 237H\n5. 237J\n6. 237K\n7. 237M\n8. 237N\n9. 237P\n10. 237Q\n11. 237R\n12. 237S\n13. 237T\n14. 237V\n15. 237W\n16. 237X\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n371\nChapter 8.\nEuclidean geometry\n\n\nCHAPTER\n9\nFinance, growth and decay\n9.1\nRevision\n374\n9.2\nSimple and compound depreciation\n377\n9.3\nTimelines\n388\n9.4\nNominal and effective interest rates\n394\n9.5\nSummary\n398\n\n9\nFinance, growth and decay\n9.1\nRevision\nEMBJD\nSimple interest is the interest calculated only on the initial amount invested, the prin-\ncipal amount. Compound interest is the interest earned on the principal amount and\non its accumulated interest. This means that interest is being earned on interest. The\naccumulated amount is the final amount; the sum of the principal amount and the\namount of interest earned.\nFormula for simple interest:\nA = P(1 + in)\nFormula for compound interest:\nA = P(1 + i)n\nwhere\nA = accumulated amount\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nWorked example 1: Simple and compound interest\nQUESTION\nSam wants to invest R 3450 for 5 years. Wise Bank offers a savings account which pays\nsimple interest at a rate of 12,5% per annum, and Grand Bank offers a savings account\npaying compound interest at a rate of 10,4% per annum. Which bank account would\ngive Sam the greatest accumulated balance at the end of the 5 year period?\nSOLUTION\nStep 1: Calculation using the simple interest formula\nWrite down the known variables and the simple interest formula\nP = 3450\ni = 0,125\nn = 5\nA = P(1 + in)\nSubstitute the values to determine the accumulated amount for the Wise Bank savings\n374\n9.1.\nRevision\n\naccount.\nA = 3450(1 + 0,125 × 5)\n= R 5606,25\nStep 2: Calculation using the compound interest formula\nWrite down the known variables and the compound interest formula.\nP = 3450\ni = 0,104\nn = 5\nA = P(1 + i)n\nSubstitute the values to determine the accumulated amount for the Grand Bank savings\naccount.\nA = 3450(1 + 0,104)5\n= R 5658,02\nStep 3: Write the final answer\nThe Grand Bank savings account would give Sam the highest accumulated balance at\nthe end of the 5 year period.\nWorked example 2: Finding i\nQUESTION\nBongani decides to put R 30 000 in an investment account. What compound interest\nrate must the investment account achieve for Bongani to double his money in 6 years?\nGive your answer correct to one decimal place.\nSOLUTION\nStep 1: Write down the known variables and the compound interest formula\nA = 60 000\nP = 30 000\nn = 6\nA = P(1 + i)n\n375\nChapter 9.\nFinance, growth and decay\n\nStep 2: Substitute the values and solve for i\n60 000 = 30 000(1 + i)6\n60 000\n30 000 = (1 + i)6\n2 = (1 + i)6\n6√\n2 = 1 + i\n6√\n2 −1 = i\n∴i = 0,122 . . .\nStep 3: Write the final answer and comment\nWe round up to a rate of 12,3% p.a. to make sure that Bongani doubles his invest-\nment.\nExercise 9 – 1: Revision\n1. Determine the value of an investment of R 10 000 at 12,1% p.a. simple interest\nfor 3 years.\n2. Calculate the value of R 8000 invested at 8,6% p.a. compound interest for 4\nyears.\n3. Calculate how much interest John will earn if he invests R 2000 for 4 years at:\na) 6,7% p.a. simple interest\nb) 5,4% p.a. compound interest\n4. The value of an investment grows from R 2200 to R 3850 in 8 years. Determine\nthe simple interest rate at which it was invested.\n5. James had R 12 000 and invested it for 5 years. If the value of his investment is\nR 15 600, what compound interest rate did it earn?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237Y\n2. 237Z\n3. 2382\n4. 2383\n5. 2384\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n376\n9.1.\nRevision\n\n9.2\nSimple and compound depreciation\nEMBJF\nAs soon as a new car leaves the dealership, its value decreases and it is considered\n“second-hand”.\nVehicles, equipment, machinery and other similar assets, all lose\nvalue over time as a result of usage and age. This loss in value is called deprecia-\ntion. Assets that have a relatively long useful lifetime, such as machines, trucks, farm-\ning equipment etc., depreciate slower than assets like office equipment, computers,\nfurniture etc. which need to be replaced more often and therefore depreciate more\nquickly.\nDepreciation is used to calculate the value of a company’s assets, which determines\nhow much tax a company must pay. Companies can take depreciation into account as\nan expense, and thereby reduce their taxable income. A lower taxable income means\nthat the company will pay less income tax to SARS (South African Revenue Service).\nWe can calculate two different kinds of depreciation: simple decay and compound\ndecay. Decay is also a term used to describe a reduction or decline in value. Simple\ndecay is also called straight-line depreciation and compound decay can also be re-\nferred to as reducing-balance depreciation. In the straight-line method the value of the\nasset is reduced by a constant amount each year, which is calculated on the principal\namount. In reducing-balance depreciation we calculate the depreciation on the re-\nduced value of the asset. This means that the value of an asset decreases by a different\namount each year.\nInvestigation: Simple and compound depreciation\n1. Mr. Sontange buys an Opel Fiesta for R 72 000. He expects that the value of the\ncar will depreciate by R 6000 every year. He draws up a table to calculate the\ndepreciated value of his Opel Fiesta.\nComplete Mr. Sontange’s table of values for the 7 year period:\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 6000\nR 66 000\n2\nR 66 000\nR 6000\n3\n4\n5\n6\n7\n2. His son, David, does not agree that the value of the car will reduce by the same\namount each year. David thinks that the car will depreciate by 10% every year.\nComplete David’s table of values:\n377\nChapter 9.\nFinance, growth and decay\n\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 7200\nR 64 800\n2\nR 64 800\nR 6480\n3\n4\n5\n6\n7\n3. Compare and discuss the results of the two different tables.\n4. Consider the graph below, which represents Mr. Sontange’s table of values:\n10 000\n20 000\n30 000\n40 000\n50 000\n60 000\n70 000\n80 000\n1\n2\n3\n4\n5\n6\n7\n8\n0\nTime (years)\nValue (Rands)\na) Draw a similar graph using David’s table of values.\nb) Interpret the two graphs and discuss the differences between them.\nc) Explain how the graphs can be used to determine the total depreciation in\neach case.\nd)\ni. Draw two new graphs by plotting the maximum value of each bar.\nii. Join the points with a line to show the general trend.\niii. Is it mathematically correct to join these points? Explain your answer.\n378\n9.2.\nSimple and compound depreciation\n\nSimple depreciation\nEMBJG\nWorked example 3: Straight-line depreciation\nQUESTION\nA new smartphone costs R 6000 and depreciates at 22% p.a. on a straight-line basis.\nDetermine the value of the smartphone at the end of each year over a 4 year period.\nSOLUTION\nStep 1: Calculate depreciation amount\nDepreciation = 6000 × 22\n100\n= 1320\nTherefore the smartphone depreciates by R 1320 every year.\nStep 2: Complete a table of values\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 6000\nR 1320\nR 4680\n2\nR 4680\nR 1320\nR 3360\n3\nR 3360\nR 1320\nR 2040\n4\nR 2040\nR 1320\nR 720\nWe notice that\nTotal depreciation = P × i × n\nwhere\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nTherefore the depreciated value of the asset (also called the book value) can be calcu-\nlated as:\nA = P(1 −in)\nNote the similarity to the simple interest formula A = P(1 + in). Interest increases the\nvalue of the principal amount, whereas with simple decay, depreciation reduces the\nvalue of the principal amount.\nImportant: to get an accurate answer do all calculations in one step on your calculator.\nDo not round off answers in your calculations until the final answer. In the worked\nexamples in this chapter, we use dots to show that the answer has not been rounded\noff. We always round the final answer to two decimal places (cents).\n379\nChapter 9.\nFinance, growth and decay\n\nWorked example 4: Straight-line depreciation method\nQUESTION\nA car is valued at R 240 000. If it depreciates at 15% p.a. using straight-line deprecia-\ntion, calculate the value of the car after 5 years.\nSOLUTION\nStep 1: Write down the known variables and the simple decay formula\nP = 240 000\ni = 0,15\nn = 5\nA = P(1 −in)\nStep 2: Substitute the values and solve for A\nA = 240 000(1 −0,15 × 5)\n= 240 000(0,25)\n= 60 000\nStep 3: Write the final answer\nAt the end of 5 years, the car is worth R 60 000.\nWorked example 5: Simple decay\nQUESTION\nA small business buys a photocopier for R 12 000. For the tax return the owner depre-\nciates this asset over 3 years using a straight-line depreciation method. What amount\nwill he fill in on his tax form at the end of each year?\nSOLUTION\nStep 1: Write down the known variables\nThe owner of the business wants the photocopier to have a book value of R 0 after 3\nyears.\nA = 0\nP = 12 000\nn = 3\n380\n9.2.\nSimple and compound depreciation\n\nTherefore we can calculate the annual depreciation as\nDepreciation = P\nn\n= 12 000\n3\n= R 4000\nStep 2: Determine the book value at the end of each year\nBook value end of first year = 12 000 −4000\n= R 8000\nBook value end of second year = 8000 −4000\n= R 4000\nBook value end of third year = 4000 −4000\n= R 0\nExercise 9 – 2: Simple decay\n1. A business buys a truck for R 560 000. Over a period of 10 years the value of\nthe truck depreciates to R 0 using the straight-line method. What is the value of\nthe truck after 8 years?\n2. Harry wants to buy his grandpa’s donkey for R 800. His grandpa is quite pleased\nwith the offer, seeing that it only depreciated at a rate of 3% per year using the\nstraight-line method. Grandpa bought the donkey 5 years ago. What did grandpa\npay for the donkey then?\n3. Seven years ago, Rocco’s drum kit cost him R 12 500. It has now been valued at\nR 2300. What rate of simple depreciation does this represent?\n4. Fiona buys a DStv satellite dish for R 3000. Due to weathering, its value depre-\nciates simply at 15% per annum. After how long will the satellite dish have a\nbook value of zero?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2385\n2. 2386\n3. 2387\n4. 2388\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n381\nChapter 9.\nFinance, growth and decay\n\nCompound depreciation\nEMBJH\nWorked example 6: Reducing-balance depreciation\nQUESTION\nA second-hand farm tractor worth R 60 000 has a limited useful life of 5 years and\ndepreciates at 20% p.a. on a reducing-balance basis. Determine the value of the\ntractor at the end of each year over the 5 year period.\nSOLUTION\nStep 1: Write down the known variables\nP = 60 000\ni = 0,2\nn = 5\nWhen we calculate depreciation using the reducing-balance method:\n1. the depreciation amount changes for each year.\n2. the depreciation amount gets smaller each year.\n3. the book value at the end of a year becomes the principal amount for the next\nyear.\n4. the asset will always have some value (the book value will never equal zero).\nStep 2: Complete a table of values\nYear\nBook value\nDepreciation\nValue at end of\nyear\n1\nR 60 000\n60 000 × 0,2 = 12 000\nR 48 000\n2\nR 48 000\n48 000 × 0,2 = 9600\nR 38 400\n3\nR 38 400\n38 400 × 0,2 = 7680\nR 30 720\n4\nR 30 720\n30 720 × 0,2 = 6144\nR 24 576\n5\nR 24 576\n24 576 × 0,2 = 4915,20\nR 19 660,80\n382\n9.2.\nSimple and compound depreciation\n\nNotice in the example above that we could also write the book value at the end of\neach year as:\nBook value end of first year\n= 60 000(1 −0,2)\nBook value end of second year = 48 000(1 −0,2) = 60 000(1 −0,2)2\nBook value end of third year\n= 38 400(1 −0,2) = 60 000(1 −0,2)3\nBook value end of fourth year = 30 720(1 −0,2) = 60 000(1 −0,2)4\nBook value end of fifth year\n= 24 576(1 −0,2) = 60 000(1 −0,2)5\nUsing the formula for simple decay and the observed pattern in the calculation above,\nwe obtain the following formula for compound decay:\nA = P(1 −i)n\nwhere\nA = book value or depreciated value\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nAgain, notice the similarity to the compound interest formula A = P(1 + i)n.\nWorked example 7: Reducing-balance depreciation\nQUESTION\nThe number of pelicans at the Berg river mouth is decreasing at a compound rate of\n12% p.a. If there are currently 3200 pelicans in the wetlands of the Berg river mouth,\nwhat will the population be in 5 years?\nSOLUTION\nStep 1: Write down the known variables and the compound decay formula\nP = 3200\ni = 0,12\nn = 5\nA = P(1 −i)n\nStep 2: Substitute the values and solve for A\nA = 3200(1 −0,12)5\n= 3200(0,88)5\n= 1688,7421 . . .\nStep 3: Write the final answer\nIn 5 years, the pelican population will be approximately 1689.\n383\nChapter 9.\nFinance, growth and decay\n\nWorked example 8: Compound decay\nQUESTION\n1. A school buys a minibus for R 950 000, which depreciates at 13,5% per annum.\nDetermine the value of the minibus after 3 years if the depreciation is calculated:\na) on a straight-line basis.\nb) on a reducing-balance basis.\n2. Which is the better option?\nSOLUTION\nStep 1: Write down known variables\nP = 950 000\ni = 0,135\nn = 3\nStep 2: Use the simple decay formula and solve for A\nA = 950 000(1 −3 × 0,135)\n= 950 000(0,865)\n= 565 250\n∴A = R 565 250\nStep 3: Use the compound decay formula and solve for A\nA = 950 000(1 −0,135)3\n= 950 000(0,865)3\n= 614 853,89\n∴A = R 614 853,89\nStep 4: Interpret the answers\nAfter a period of 3 years, the value of the minibus calculated on the straight-line\nmethod is less than the value of the minibus calculated on the reducing-balance\nmethod. The value of the minibus depreciated less on the reducing-balance basis\nbecause the amount of depreciation is calculated on a smaller amount every year,\nwhereas the straight-line method is based on the full value of the minibus every year.\n384\n9.2.\nSimple and compound depreciation\n\nWorked example 9: Compound depreciation\nQUESTION\nFarmer Jack bought a tractor and it has depreciated by 20% p.a. on a reducing-balance\nbasis. If the current value of the tractor is R 52 429, calculate how much Farmer Jack\npaid for his tractor if he bought it 7 years ago.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 52 429\ni = 0,2\nn = 7\nA = P(1 −i)n\nStep 2: Substitute the values and solve for P\n52 429 = P(1 −0,2)7\n= P(0,8)7\n∴P = 52 429\n(0,8)7\n= 250 000,95 . . .\nStep 3: Write the final answer\n7 years ago, Farmer Jack paid R 250 000 for his tractor.\nExercise 9 – 3: Compound depreciation\n1. Jwayelani buys a truck for R 89 000 and depreciates it by 9% p.a. using the\ncompound depreciation method. What is the value of the truck after 14 years?\n2. The number of cormorants at the Amanzimtoti river mouth is decreasing at a\ncompound rate of 8% p.a. If there are now 10 000 cormorants, how many will\nthere be in 18 years’ time?\n3. On January 1, 2008 the value of my Kia Sorento is R 320 000. Each year after\nthat, the car’s value will decrease 20% of the previous year’s value. What is the\nvalue of the car on January 1, 2012?\n385\nChapter 9.\nFinance, growth and decay\n\n4. The population of Bonduel decreases at a reducing-balance rate of 9,5% per\nannum as people migrate to the cities. Calculate the decrease in population over\na period of 5 years if the initial population was 2 178 000.\n5. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses\n3% of its water each day. How much does it weigh after a month of 31 days?\n6. Richard bought a car 15 years ago and it depreciated by 17% p.a. on a com-\npound depreciation basis. How much did he pay for the car if it is now worth\nR 5256?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2389\n2. 238B\n3. 238C\n4. 238D\n5. 238F\n6. 238G\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFinding i\nEMBJJ\nWorked example 10: Finding i for simple decay\nQUESTION\nAfter 4 years, the value of a computer is halved. Assuming simple decay, at what\nannual rate did it depreciate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and simple decay formula\nLet the value of the computer be x, therefore:\nA = x\n2\nP = x\nn = 4\nA = P(1 −in)\nStep 2: Substitute the values and solve for i\n386\n9.2.\nSimple and compound depreciation\n\nx\n2 = x(1 −3i)\n1\n2 = 1 −3i\n∴3i = 1 −1\n2\n∴i = 0,1667\nStep 3: Write the final answer\nThe computer depreciated at a rate of 16,67% p.a.\nWorked example 11: Finding i for compound decay\nQUESTION\nCristina bought a fridge at the beginning of 2009 for R 8999 and sold it at the end\nof 2011 for R 4500. At what rate did the value of her fridge depreciate assuming a\nreducing-balance method? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 4500\nP = 8999\nn = 3\nA = P(1 −i)n\nStep 2: Substitute the values and solve for i\n4500 = 8999(1 −i)3\n4500\n8999 = (1 −i)3\n3\nr\n4500\n8999 = 1 −i\n∴i = 1 −\n3\nr\n4500\n8999\n= 0,206\nStep 3: Write the final answer\nCristina’s fridge depreciated at a rate of 20,6% p.a.\n387\nChapter 9.\nFinance, growth and decay\n\nExercise 9 – 4: Finding i\n1. A machine costs R 45 000 and has a scrap value of R 9000 after 10 years. Deter-\nmine the annual rate of depreciation if it is calculated on the reducing balance\nmethod.\n2. After 15 years, an aeroplane is worth 1\n6 of its original value. At what annual rate\nwas depreciation compounded?\n3. Mr. Mabula buys furniture for R 20 000. After 6 years he sells the furniture for\nR 9300. Calculate the annual compound rate of depreciation of the furniture.\n4. Ayanda bought a new car 7 years ago for double what it is worth today. At what\nyearly compound rate did her car depreciate?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238H\n2. 238J\n3. 238K\n4. 238M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.3\nTimelines\nEMBJK\nInterest can be compounded more than once a year. For example, an investment can\nbe compounded monthly or quarterly. Below is a table of compounding terms and\ntheir corresponding numeric value (p). When amounts are compounded more than\nonce per annum, we multiply the number of years by p and we also divide the interest\nrate by p.\nTerm\np\nyearly / annually\n1\nhalf-yearly / bi-annually\n2\nquarterly\n4\nmonthly\n12\nweekly\n52\ndaily\n365\nWorked example 12: Timelines\nQUESTION\nR 5500 is invested for a period of 4 years in a savings account. For the first year, the\ninvestment grows at a simple interest rate of 11% p.a. and then at a rate of 12,5%\np.a. compounded quarterly for the rest of the period. Determine the value of the\ninvestment at the end of the 4 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\n388\n9.3.\nTimelines\n\nT0\nT1\nT2\nT3\nT4\n11% p.a. simple interest\n12,5% p.a. compounded quarterly\nR 5500\nIn the timeline above, the intervals are given in years. For example, T0 is the start of\nthe investment, T1 is the end of the first year and T4 is the end of the fourth year.\nStep 2: Use the simple interest formula to calculate A at T1\nA = P(1 + in)\n= 5500(1 + 0,11)\n= R 6105\nStep 3: Use the compound interest formula to calculate A at T4\nThe investment is compounded quarterly, therefore:\nn = 3 × 4\n= 12\nand i = 0,125\n4\nAlso notice that the accumulated amount at the end of the first year becomes the\nprincipal amount at the beginning of the second year.\nA = P(1 + i)n\n= 6105\n\u0012\n1 + 0,125\n4\n\u001312\n= R 8831,88\nStep 4: Write the final answer\nThe value of the investment at the end of the 4 years is R 8831,88.\n389\nChapter 9.\nFinance, growth and decay\n\nWorked example 13: Timelines\nQUESTION\nR 150 000 is deposited in an investment account for a period of 6 years at an interest\nrate of 12% p.a. compounded half-yearly for the first 4 years and then 8,5% p.a.\ncompounded yearly for the rest of the period. A deposit of R 8000 is made into the\naccount after the first year and then another deposit of R 2000 is made 5 years after\nthe initial investment. Calculate the value of the investment at the end of the 6 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n8,5% p.a. compounded yearly\nR 15 000\nT5\nT6\n12% p.a. compounded half-yearly\n+R 8000\n+R 2000\nRemember to show when the additional deposits of R 8000 and R 2000 where made\ninto the account. It is very important to note that the interest rate changes at T4.\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nBetween T0 and T4:\nWe notice that interest for the first 4 years is compounded half-yearly, therefore:\nn1 = 4 × 2\n= 8\nand i1 = 0,12\n2\nBetween T4 and T6:\nn2 = 2\nand i2 = 0,085\nTherefore the total growth of the initial deposit over the 6 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\nStep 3: The deposit at T1\nBetween T1 and T4:\n390\n9.3.\nTimelines\n\nInterest on this deposit is compounded half-yearly for 3 years, therefore:\nn3 = 3 × 2\n= 6\nand i3 = 0,12\n2\nBetween T4 and T6:\nn4 = 2\nand i4 = 0,085\nTherefore the total growth of the deposit over the 5 years is:\nA = P(1 + i3)n3(1 + i4)n4\n= 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2\nStep 4: The deposit at T5\nAccumulate interest for only 1 year:\nA = P(1 + i)n\n= 2000(1 + 0,085)1\nStep 5: Determine the total calculation\nTo get as accurate an answer as possible, we do the the calculation on the calculator\nin one step. Using the memory and answer recall function on the calculator, we avoid\nrounding off until we get the final answer.\nA = 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\n+ 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2 + 2000(1 + 0,085)1\n= R 296 977,00\nStep 6: Write the final answer\nThe value of the investment at the end of the 6 years is R 296 977,00.\n391\nChapter 9.\nFinance, growth and decay\n\nWorked example 14: Timelines\nQUESTION\nR 60 000 is invested in an account which offers interest at 7% p.a.\ncompounded\nquarterly for the first 18 months. Thereafter the interest rate changes to 5% p.a. com-\npounded monthly. Three years after the initial investment, R 5000 is withdrawn from\nthe account. How much will be in the account at the end of 5 years?\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n5% p.a. compounded monthly\nR 60 000\nT5\n7% p.a. compounded quarterly\n−R 5000\nRemember to show when the withdrawal of R 5000 was taken out of the account. It is\nalso important to note that the interest rate changes after 18 months (T1 1\n2 ).\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nInterest for the first 1,5 years is compounded quarterly, therefore:\nn1 = 1,5 × 4\n= 6\nand i1 = 0,07\n4\nInterest for the remaining 3,5 years is compounded monthly, therefore:\nn2 = 3,5 × 12\n= 42\nand i2 = 0,05\n12\nTherefore the total growth of the initial deposit over the 5 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n392\n9.3.\nTimelines\n\nStep 3: The withdrawal at T3\nWe calculate the interest that the R 5000 would have earned if it had remained in the\naccount:\nn = 2 × 12\n= 24\nand i = 0,05\n12\nTherefore we have that:\nA = P(1 + i)n\n= 5000\n\u0012\n1 + 0,05\n12\n\u001324\nStep 4: Determine the total calculation\nWe subtract the withdrawal and the interest it would have earned from the accumu-\nlated amount at the end of the 5 years:\nA = 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n−5000\n\u0012\n1 + 0,05\n12\n\u001324\n= R 73 762,19\nStep 5: Write the final answer\nThe value of the investment at the end of the 5 years is R 73 762,19.\nExercise 9 – 5: Timelines\n1. After a 20-year period Josh’s lump sum investment matures to an amount of\nR 313 550. How much did he invest if his money earned interest at a rate of\n13,65% p.a. compounded half yearly for the first 10 years, 8,4% p.a. com-\npounded quarterly for the next five years and 7,2% p.a. compounded monthly\nfor the remaining period?\n2. Sindisiwe wants to buy a motorcycle. The cost of the motorcycle is R 55 000.\nIn 1998 Sindisiwe opened an account at Sutherland Bank with R 16 000. Then\nin 2003 she added R 2000 more into the account. In 2007 Sindisiwe made\nanother change: she took R 3500 from the account. If the account pays 6% p.a.\ncompounded half-yearly, will Sindisiwe have enough money in the account at\nthe end of 2012 to buy the motorcycle?\n3. A loan has to be returned in two equal semi-annual instalments. If the rate of\ninterest is 16% per annum, compounded semi-annually and each instalment is\nR 1458, find the sum borrowed.\n393\nChapter 9.\nFinance, growth and decay\n\n4. A man named Phillip invests R 10 000 into an account at North Bank at an\ninterest rate of 7,5% p.a. compounded monthly. After 5 years the bank changes\nthe interest rate to 8% p.a. compounded quarterly. How much money will\nPhillip have in his account 9 years after the original deposit?\n5. R 75 000 is invested in an account which offers interest at 11% p.a.\ncom-\npounded monthly for the first 24 months.\nThen the interest rate changes to\n7,7% p.a. compounded half-yearly. If R 9000 is withdrawn from the account\nafter one year and then a deposit of R 3000 is made three years after the initial\ninvestment, how much will be in the account at the end of 6 years?\n6. Christopher wants to buy a computer, but right now he doesn’t have enough\nmoney. A friend told Christopher that in 5 years the computer will cost R 9150.\nHe decides to start saving money today at Durban United Bank. Christopher\ndeposits R 5000 into a savings account with an interest rate of 7,95% p.a. com-\npounded monthly.\nThen after 18 months the bank changes the interest rate\nto 6,95% p.a. compounded weekly. After another 6 months, the interest rate\nchanges again to 7,92% p.a.\ncompounded two times per year.\nHow much\nmoney will Christopher have in the account after 5 years, and will he then have\nenough money to buy the computer?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238N\n2. 238P\n3. 238Q\n4. 238R\n5. 238S\n6. 238T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.4\nNominal and effective interest rates\nEMBJM\nWe have seen that although interest is quoted as a percentage per annum it can be\ncompounded more than once a year. We therefore need a way of comparing interest\nrates. For example, is an annual interest rate of 8% compounded quarterly higher or\nlower than an interest rate of 8% p.a. compounded yearly?\nInvestigation: Nominal and effective interest rates\n1. Calculate the accumulated amount at the end of one year if R 1000 is invested\nat 8% p.a. compound interest:\nA = P(1 + i)n\n= . . . . . .\n2. Calculate the value of R 1000 if it is invested for one year at 8% p.a. com-\npounded:\n394\n9.4.\nNominal and effective interest rates\n\nFrequency\nCalculation\nAccumulated\namount\nInterest\namount\nhalf-yearly\nA = 1000\n\u0010\n1 + 0,08\n2\n\u00111×2\nR 1081,60\nR 81,60\nquarterly\nmonthly\nweekly\ndaily\n3. Use your results from the table above to calculate the effective rate that the\ninvestment of R 1000 earns in one year:\nFrequency\nAccumulated\namount\nCalculation\nEffective\ninterest\nrate\nhalf-yearly\nR 1081,60\n1081,60 = 1000(1 + i)\n1081,60\n1000\n= 1 + i\n1081,60\n1000\n−1 = i\n∴i = 0,0816\ni = 8,16%\nquarterly\nmonthly\nweekly\ndaily\n4. If you wanted to borrow R 10 000 from the bank, would it be better to pay it\nback at an interest rate of 22% p.a. compounded quarterly or 22% compounded\nmonthly? Show your calculations.\nAn interest rate compounded more than once a year is called the nominal interest rate.\nIn the investigation above, we determined that the nominal interest rate of 8% p.a.\ncompounded half-yearly is actually an effective rate of 8,16% p.a.\nGiven a nominal interest rate i(m) compounded at a frequency of m times per year\nand the effective interest rate i, the accumulated amount calculated using both interest\nrates will be equal so we can write:\nP(1 + i) = P\n \n1 + i(m)\nm\n!m\n∴1 + i =\n \n1 + i(m)\nm\n!m\n395\nChapter 9.\nFinance, growth and decay\n\nWorked example 15: Nominal and effective interest rates\nQUESTION\nInterest on a credit card is quoted as 23% p.a. compounded monthly. What is the\neffective annual interest rate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down the known variables\nInterest is being added monthly, therefore:\nm = 12\ni(12) = 0,23\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i\n1 + i =\n\u0012\n1 + 0,23\n12\n\u001312\n∴i = 1 −\n\u0012\n1 + 0,23\n12\n\u001312\n= 25,59%\nStep 3: Write the final answer\nThe effective interest rate is 25,59% per annum.\nWorked example 16: Nominal and effective interest rates\nQUESTION\nDetermine the nominal interest rate compounded quarterly if the effective interest rate\nis 9% per annum (correct to two decimal places).\nSOLUTION\nStep 1: Write down the known variables\n396\n9.4.\nNominal and effective interest rates\n\nInterest is being added quarterly, therefore:\nm = 4\ni = 0,09\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i(m)\n1 + 0,09 =\n \n1 + i(4)\n4\n!4\n4p\n1,09 = 1 + i(4)\n4\n4p\n1,09 −1 = i(4)\n4\n4\n\u0010\n4p\n1,09 −1\n\u0011\n= i(4)\n∴i(4) = 8,71%\nStep 3: Write the final answer\nThe nominal interest rate is 8,71% p.a. compounded quarterly.\nExercise 9 – 6: Nominal and effect interest rates\n1. Determine the effective annual interest rate if the nominal interest rate is:\na) 12% p.a. compounded quarterly.\nb) 14,5% p.a. compounded weekly.\nc) 20% p.a. compounded daily.\n2. Consider the following:\n• 16,8% p.a. compounded annually.\n• 16,4% p.a. compounded monthly.\n• 16,5% p.a. compounded quarterly.\na) Determine the effective annual interest rate of each of the nominal rates\nlisted above.\nb) Which is the best interest rate for an investment?\nc) Which is the best interest rate for a loan?\n397\nChapter 9.\nFinance, growth and decay\n\n3. Calculate the effective annual interest rate equivalent to a nominal interest rate\nof 8,75% p.a. compounded monthly.\n4. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every\nfour months on her investment of R 85 000.\nCalculate the effective rate per\nannum.\n5. Determine which of the following would be the better agreement for paying back\na student loan:\na) 9,1% p.a. compounded quarterly.\nb) 9% p.a. compounded monthly.\nc) 9,3% p.a. compounded half-yearly.\n6. Miranda invests R 8000 for 5 years for her son’s study fund. Determine how\nmuch money she will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 6% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 238V\n1b. 238W\n1c. 238X\n2. 238Y\n3. 238Z\n4. 2392\n5. 2393\n6. 2394\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.5\nSummary\nEMBJN\nSee presentation: 2395 at www.everythingmaths.co.za\n• Simple interest: A = P(1 + in)\n• Compound interest: A = P(1 + i)n\n• Simple depreciation: A = P(1 −in)\n• Compound depreciation: A = P(1 −i)n\n• Nominal and effective annual interest rates: 1 + i =\n\u0010\n1 + i(m)\nm\n\u0011m\n398\n9.5.\nSummary\n\nExercise 9 – 7: End of chapter exercises\n1. Thabang buys a Mercedes worth R 385 000 in 2007. What will the value of the\nMercedes be at the end of 2013 if:\na) the car depreciates at 6% p.a. straight-line depreciation.\nb) the car depreciates at 6% p.a. reducing-balance depreciation.\n2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R 8900.\nThe interest rate is quoted as 11% per annum based on simple interest. Calculate\nthe required monthly payment for this contract.\n3. A computer is purchased for R 16 000. It depreciates at 15% per annum.\na) Determine the book value of the computer after 3 years if depreciation is\ncalculated according to the straight-line method.\nb) Find the rate according to the reducing-balance method that would yield,\nafter 3 years, the same book value as calculated in the previous question.\n4. Maggie invests R 12 500 for 5 years at 12% per annum compounded monthly\nfor the first 2 years and 14% per annum compounded semi-annually for the next\n3 years. How much will Maggie receive in total after 5 years?\n5. Tintin invests R 120 000. He is quoted a nominal interest rate of 7,2% per an-\nnum compounded monthly.\na) Calculate the effective rate per annum (correct to two decimal places).\nb) Use the effective rate to calculate the value of Tintin’s investment if he\ninvested the money for 3 years.\nc) Suppose Tintin invests his money for a total period of 4 years, but after 18\nmonths makes a withdrawal of R 20 000, how much will he receive at the\nend of the 4 years?\n6. Ntombi opens accounts at a number of clothing stores and spends freely. She\ngets herself into terrible debt and she cannot pay off her accounts. She owes\nFashion World R 5000 and the shop agrees to let her pay the bill at a nominal\ninterest rate of 24% compounded monthly.\na) How much money will she owe Fashion World after two years?\nb) What is the effective rate of interest that Fashion World is charging her?\n7. John invests R 30 000 in the bank for a period of 18 months. Calculate how\nmuch money he will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 8% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\ndaily\n399\nChapter 9.\nFinance, growth and decay\n\n8. Convert an effective annual interest rate of 11,6% p.a. to a nominal interest rate\ncompounded:\na) half-yearly\nb) quarterly\nc) monthly\n9. Joseph must sell his plot on the West Coast and he needs to get R 300 000 on the\nsale of the land. If the estate agent charges him 7% commission on the selling\nprice, what must the buyer pay for the plot?\n10. Mrs. Brown retired and received a lump sum of R 200 000. She deposited the\nmoney in a fixed deposit savings account for 6 years. At the end of the 6 years\nthe value of the investment was R 265 000. If the interest on her investment was\ncompounded monthly, determine:\na) the nominal interest rate per annum\nb) the effective annual interest rate\n11. R 145 000 is invested in an account which offers interest at 9% p.a.\ncom-\npounded half-yearly for the first 2 years. Then the interest rate changes to 4%\np.a. compounded quarterly. Four years after the initial investment, R 20 000 is\nwithdrawn. 6 years after the initial investment, a deposit of R 15 000 is made.\nDetermine the balance of the account at the end of 8 years.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2396\n2. 2397\n3. 2398\n4. 2399\n5. 239B\n6. 239C\n7. 239D\n8. 239F\n9. 239G\n10. 239H\n11. 239J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n400\n9.5.\nSummary\n\nCHAPTER\n10\nProbability\n10.1\nRevision\n402\n10.2\nDependent and independent events\n411\n10.3\nMore Venn diagrams\n419\n10.4\nTree diagrams\n426\n10.5\nContingency tables\n431\n10.6\nSummary\n435\n\n10\nProbability\n10.1\nRevision\nEMBJP\nTerminology\nEMBJQ\nOutcome: a single observation of an uncertain or random process (called an experi-\nment). For example, when you accidentally drop a book, it might fall on its cover, on\nits back or on its side. Each of these options is a possible outcome.\nSample space of an experiment: the set of all possible outcomes of the experiment. For\nexample, the sample space when you roll a single 6-sided die is the set {1; 2; 3; 4; 5; 6}.\nFor a given experiment, there is exactly one sample space. The sample space is de-\nnoted by the letter S.\nEvent: a set of outcomes of an experiment. For example, during radioactive decay of\n1 gramme of uranium-234, one possible event is that the number of alpha-particles\nemitted during 1 microsecond is between 225 and 235.\nProbability of an event: a real number between 0 and 1 that describes how likely it\nis that the event will occur. A probability of 0 means the outcome of the experiment\nwill never be in the event set. A probability of 1 means the outcome of the experiment\nwill always be in the event set. When all possible outcomes of an experiment have\nequal chance of occurring, the probability of an event is the number of outcomes in\nthe event set as a fraction of the number of outcomes in the sample space.\nRelative frequency of an event: the number of times that the event occurs during\nexperimental trials, divided by the total number of trials conducted. For example, if\nwe flip a coin 10 times and it landed on heads 3 times, then the relative frequency of\nthe heads event is 3\n10 = 0,3.\nUnion of events: the set of all outcomes that occur in at least one of the events. For\n2 events called A and B, we write the union as “A or B”. Another way of writing the\nunion is using set notation: A ∪B.\nIntersection of events: the set of all outcomes that occur in all of the events. For 2\nevents called A and B, we write the intersection as “A and B”. Another way of writing\nthe intersection is using set notation: A ∩B.\nMutually exclusive events: events with no outcomes in common, that is (A and B) =\n∅. Mutually exclusive events can never occur simultaneously. For example the event\nthat a number is even and the event that the same number is odd are mutually exclu-\nsive, since a number can never be both even and odd.\nComplementary events: two mutually exclusive events that together contain all the\noutcomes in the sample space. For an event called A, we write the complement as\n“not A”. Another way of writing the complement is as A′.\nSee video: 239K at www.everythingmaths.co.za\n402\n10.1.\nRevision\n\nIdentities\nEMBJR\nThe addition rule (also called the sum rule) for any 2 events, A and B is\nP(A or B) = P(A) + P(B) −P(A and B)\nThis rule relates the probabilities of 2 events with the probabilities of their union and\nintersection.\nThe addition rule for 2 mutually exclusive events is\nP(A or B) = P(A) + P(B)\nThis rule is a special case of the previous rule. Because the events are mutually exclu-\nsive, P(A and B) = 0.\nThe complementary rule is\nP(not A) = 1 −P(A)\nThis rule is a special case of the previous rule. Since A and (not A) are mutually\nexclusive, P(A or (not A)) = 1.\nSee video: 239M at www.everythingmaths.co.za\nWorked example 1: Events\nQUESTION\nYou take all the hearts from a deck of cards. You then select a random card from the set\nof hearts. What is the sample space? What is the probability of each of the following\nevents?\n1. The card is the ace of hearts.\n2. The card has a prime number on it.\n3. The card has a letter of the alphabet on it.\nSOLUTION\nStep 1: Write down the sample space\nSince we are considering only one suit from the deck of cards (the hearts), we need to\nwrite down only the letters and numbers on the cards. Therefore the sample space is\nS = {A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K}\nStep 2: Write down the event sets\n• ace of hearts: {A}\n• prime number: {2; 3; 5; 7}\n• letter of alphabet: {A; J; Q; K}\n403\nChapter 10.\nProbability\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are 13 elements in the\nsample space. So the probability of each event is\n• ace of hearts:\n1\n13\n• prime number:\n4\n13\n• letter of alphabet:\n4\n13\nWorked example 2: Events\nQUESTION\nYou roll two 6-sided dice. Let E be the event that the total number of dots on the dice\nis 10. Let F be the event that at least one die is a 3.\n1. Write down the event sets for E and F.\n2. Determine the probabilities for E and F.\n3. Are E and F mutually exclusive? Why or why not?\nSOLUTION\nStep 1: Write down the sample space\nThe sample space of a single 6-sided die is just {1; 2; 3; 4; 5; 6}. To get the sample\nspace of two 6-sided dice, we have to take every possible pair of numbers from 1 to 6.\nS =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(1; 1)\n(1; 2)\n(1; 3)\n(1; 4)\n(1; 5)\n(1; 6)\n(2; 1)\n(2; 2)\n(2; 3)\n(2; 4)\n(2; 5)\n(2; 6)\n(3; 1)\n(3; 2)\n(3; 3)\n(3; 4)\n(3; 5)\n(3; 6)\n(4; 1)\n(4; 2)\n(4; 3)\n(4; 4)\n(4; 5)\n(4; 6)\n(5; 1)\n(5; 2)\n(5; 3)\n(5; 4)\n(5; 5)\n(5; 6)\n(6; 1)\n(6; 2)\n(6; 3)\n(6; 4)\n(6; 5)\n(6; 6)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nStep 2: Write down the events\nFor E the dice have to add to 10.\nE = {(4; 6); (5; 5); (6; 4)}\nFor F at least one die has to be 3.\nF = {(1; 3); (3; 1); (2; 3); (3; 2); (3; 3); (4; 3); (3; 4); (5; 3); (3; 5); (6; 3); (3; 6)}\n404\n10.1.\nRevision\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are\n• 6 × 6 = 36 outcomes in the sample space, S;\n• 3 outcomes in event E; and\n• 11 outcomes in event F.\nTherefore\nP(E) = 3\n36 = 1\n12\nand\nP(F) = 11\n36\nStep 4: Are they mutually exclusive\nTo test whether two events are mutually exclusive, we have to test whether their in-\ntersection is empty. Since E has no outcomes that contain a 3 on one of the dice,\nthe intersection of E and F is empty: (E and F) = ∅. This means that the events are\nmutually exclusive.\nSee video: 239N at www.everythingmaths.co.za\nExercise 10 – 1: Revision\n1. A bag contains r red balls, b blue balls and y yellow balls. What is the probability\nthat a ball drawn from the bag at random is yellow?\n2. A packet has yellow and pink sweets. The probability of taking out a pink sweet\nis 7\n12. What is the probability of taking out a yellow sweet?\n3. You flip a coin 4 times. What is the probability that you get 2 heads and 2 tails?\nWrite down the sample space and the event set to determine the probability of\nthis event.\n4. In a class of 37 children, 15 children walk to school, 20 children have pets at\nhome and 12 children who have a pet at home also walk to school. How many\nchildren walk to school and do not have a pet at home?\n5. You roll two 6-sided dice and are interested in the following two events:\n• A: the sum of the dice equals 8\n• B: at least one of the dice shows a 1\nShow that these events are mutually exclusive.\n405\nChapter 10.\nProbability\n\n6. You ask a friend to think of a number from 1 to 100. You then ask her the\nfollowing questions:\n• Is the number even?\n• Is the number divisible by 7?\nHow many possible numbers are less than 80 if she answered “yes” to both\nquestions?\n7. In a group of 42 pupils, all but 3 had a packet of chips or a Fanta or both. If 23\nhad a packet of chips and 7 of these also had a Fanta, what is the probability that\none pupil chosen at random has:\na) both chips and Fanta\nb) only Fanta\n8. Tamara has 18 loose socks in a drawer. Eight of these are orange and two are\npink. Calculate the probability that the first sock taken out at random is:\na) orange\nb) not orange\nc) pink\nd) not pink\ne) orange or pink\nf) neither orange nor pink\n9. A box contains coloured blocks. The number of blocks of each colour is given\nin the following table.\nColour\nPurple\nOrange\nWhite\nPink\nNumber of blocks\n24\n32\n41\n19\nA block is selected randomly. What is the probability that the block will be:\na) purple\nb) purple or white\nc) pink and orange\nd) not orange?\n10. The surface of a soccer ball is made up of 32 faces. 12 faces are regular pen-\ntagons, each with a surface area of about 37 cm2. The other 20 faces are regular\nhexagons, each with a surface area of about 56 cm2.\nYou roll the soccer ball. What is the probability that it stops with a pentagon\ntouching the ground?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 239P\n2. 239Q\n3. 239R\n4. 239S\n5. 239T\n6. 239V\n7. 239W\n8. 239X\n9. 239Y\n10. 239Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n406\n10.1.\nRevision\n\nVenn diagrams\nEMBJS\nA Venn diagram is used to show how events are related to one another.\nA Venn\ndiagram can be very helpful when doing calculations with probabilities. In a Venn\ndiagram each event is represented by a shape, often a circle or a rectangle. The region\ninside the shape represents the outcomes included in the event and the region outside\nthe shape represents the outcomes that are not in the event.\nS\nA\nB\nA and B\nA Venn diagram representing a sample space, S, as a square; and two events, A and\nB, as circles. The intersection of the two circles contains outcomes that are in both A\nand B.\nVenn diagrams can be used in slightly different ways and it is important to notice the\ndifferences between them. The following 3 examples show how a Venn diagram is\nused to represent\n• the outcomes included in each event;\n• the number of outcomes in each event; and\n• the probability of each event.\nWorked example 3: Venn diagrams with outcomes\nQUESTION\nChoose a number between 1 and 20. Draw a Venn diagram to answer the following\nquestions.\n1. What is the probability that the number is a multiple of 3?\n2. What is the probability that the number is a multiple of 5?\n3. What is the probability that the number is a multiple of 3 or 5?\n4. What is the probability that the number is a multiple of 3 and 5?\nSOLUTION\nStep 1: Draw a Venn diagram\nThe Venn diagram should show the sample space of all numbers from 1 to 20. It should\nalso show an event set that contains all the multiples of 3, let A = {3; 6; 9; 12; 15; 18},\n407\nChapter 10.\nProbability\n\nand another event set that contains all the multiples of 5, let B = {5; 10; 15; 20}. Note\nthat there is one shared outcome between these two events, namely 15.\n3\n18\n15\n12\n9\n6\n10\n20\n5\n1\n2\n4\n7\n8\n11\n13\n14\n16\n17\n19\nStep 2: Compute probabilities\nThe probability of an event is the number of outcomes in the event set divided by the\nnumber of outcomes in the sample space. There are 20 outcomes in the sample space.\n1. Since there are 6 outcomes in the multiples of 3 event set, the probability of a\nmultiple of 3 is P(A) = 6\n20 = 3\n10.\n2. Since there are 4 outcomes in the multiples of 5 event set, the probability of a\nmultiple of 5 is P(B) = 4\n20 = 1\n5.\n3. The event that the number is a multiple of 3 or 5 is the union of the above two\nevent sets. There are 9 elements in the union of the event sets, so the probability\nis 9\n20.\n4. The event that the number is a multiple of 3 and 5 is the intersection of the\ntwo event sets. There is 1 element in the intersection of the event sets, so the\nprobability is 1\n20.\nWorked example 4: Venn diagrams with counts\nQUESTION\nIn a group of 50 learners, 35 take Mathematics and 30 take History, while 12 take\nneither of the two subjects. Draw a Venn diagram representing this information. If a\nlearner is chosen at random from this group, what is the probability that he takes both\nMathematics and History?\nSOLUTION\nStep 1: Draw outline of Venn diagram\nThere are 2 events in this question, namely\n• M: that a learner takes Mathematics; and\n• H: that a learner takes History.\n408\n10.1.\nRevision\n\nWe need to do some calculations before drawing the full Venn diagram, but with the\ninformation above we can already draw the outline.\nS\nM\nH\nStep 2: Write down sizes of the event sets, their union and intersection\nWe are told that 12 learners take neither of the two subjects. Graphically we can\nrepresent this as:\nS\nM\nH\n12\nSince there are 50 elements in the sample space, we can see from this figure that there\nare 50 −12 = 38 elements in (M or H). So far we know\n• n(M) = 35\n• n(H) = 30\n• n(M or H) = 38\nFrom the addition rule,\nn(M or H) = n(M) + n(H) −n(M and H)\n∴n(M and H) = 35 + 30 −38\n= 27\nStep 3: Draw the final Venn diagram\nS\nM\nH\n12\n27\n8\n3\n409\nChapter 10.\nProbability\n\nWorked example 5: Venn diagrams with probabilities\nQUESTION\nDraw a Venn diagram to represent the same information as in the previous example,\nexcept showing the probabilities of the different events, rather than the counts.\nIf a learner is chosen at random from this group, what is the probability that she takes\nboth Mathematics and History?\nSOLUTION\nStep 1: Use counts to compute probabilities\nSince there are 50 elements (learners) in the sample space, we can compute the prob-\nability of any event by dividing the size of the event set by 50. This gives the following\nprobabilities:\n• P(M) = 35\n50 = 7\n10\n• P(H) = 30\n50 = 3\n5\n• P(M or H) = 38\n50 = 19\n25\n• P(M and H) = 27\n50\nStep 2: Draw the Venn diagram\nNext we replace each count from the Venn diagram in the previous example with a\nprobability.\nS\nM\nH\n6\n25\n27\n50\n4\n25\n3\n50\nStep 3: Find the answer\nThe probability that a random learner will take both Mathematics and History is\nP(M and H) = 27\n50.\nSee video: 23B2 at www.everythingmaths.co.za\n410\n10.1.\nRevision\n\nExercise 10 – 2: Venn diagram revision\n1. Given the following information:\n• P(A) = 0,3\n• P(B and A) = 0,2\n• P(B) = 0,7\nFirst draw a Venn diagram to represent this information. Then compute the value\nof P(B and (not A)).\n2. You are given the following information:\n• P(A) = 0,5\n• P(A and B) = 0,2\n• P(not B) = 0,6\nDraw a Venn diagram to represent this information and determine P(A or B).\n3. A study was undertaken to see how many people in Port Elizabeth owned either\na Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60%\nowned a Volkswagen. What percentage of people owned neither car?\n4. Let S denote the set of whole numbers from 1 to 15, X denote the set of even\nnumbers from 1 to 15 and Y denote the set of prime numbers from 1 to 15.\nDraw a Venn diagram depicting S, X and Y .\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B3\n2. 23B4\n3. 23B5\n4. 23B6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.2\nDependent and independent events\nEMBJT\nSometimes the presence or absence of one event tells us something about other events.\nWe call events dependent if knowing whether one of them happened tells us some-\nthing about whether the others happened. Independent events give us no information\nabout one another; the probability of one event occurring does not affect the probabil-\nity of the other events occurring.\nDEFINITION: Independent events\nTwo events, A and B are independent if and only if\nP(A and B) = P(A) × P(B)\nAt first it might not be clear why we should call events that satisfy the equation above\nindependent. We will explore this further using a number of examples.\n411\nChapter 10.\nProbability\n\nInvestigation: Independence\nRoll a single 6-sided die and consider the following two events:\n• E: you get an even number\n• T: you get a number that is divisible by three\nNow answer the following questions:\n• What is the probability of E?\n• What is the probability of getting an even number if you are told that the number\nwas also divisible by three?\n• Does knowing that the number was divisible by three change the probability that\nthe number was even?\nAre the events E and T dependent or independent according to the definition (hint:\ncompute the probabilities in the definition of independence)?\nSee video: 23B7 at www.everythingmaths.co.za\nSo, why do we call it independence when P(A and B) = P(A) × P(B)? For two\nevents, A and B, independence means that knowing the outcome of B does not affect\nthe probability of A.\nConsider the following Venn diagram.\nS\nA\nB\nA and B\nThe probability of A is the ratio between the number of outcomes in A and the number\nof outcomes in the sample space, S.\nP(A) = n(A)\nn(S)\n412\n10.2.\nDependent and independent events\n\nNow, let’s say that we know that event B happened. How does this affect the proba-\nbility of A? Here is how the Venn diagram changes:\nS\nA\nB\nA and B\nA lot of the possible outcomes (all of the outcomes outside B) are now out of the pic-\nture, because we know that they did not happen. Now the probability of A happening,\ngiven that we know that B happened, is the ratio between the size of the region where\nA is present (A and B) and the size of all possible events (B).\nP(A if we know B) = n(A and B)\nn(B)\nIf P(A) = P(A if we know B) we call them independent, because knowing B does\nnot change the probability of A.\nWith some algebra, we can prove that this statement of independence is the same\nas the definition of independence that we saw at the beginning of this section. For\nindependent events\nP(A and B) = P(A) × P(B)\nThis is equivalent to\nP(A) = P(A and B) ÷ P(B)\n= n(A and B)\nn(S)\n÷ n(B)\nn(S)\n= n(A and B)\nn(B)\n= P(A if we know B)\nThat is why we call events independent!\n(For enrichment only):\nThe ratio\nP(A and B)\nP(B)\nis called a conditional probability and written using the notation P(A | B). This\nnotation is read as “the probability of A given B.”\nIf (and only if) A and B are independent: P(A | B) = P(A) and P(B | A) = P(B).\nTry to prove this using the definition of independence.\n413\nChapter 10.\nProbability\n\nWorked example 6: Independent and dependent events\nQUESTION\nA bag contains 5 red and 5 blue balls. We remove a random ball from the bag, record\nits colour and put it back into the bag. We then remove another random ball from the\nbag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Probability of a red ball first\nSince there are a total of 10 balls, of which 5 are red, the probability of getting a red\nball is\nP(first ball red) = 5\n10 = 1\n2\nStep 2: Probability of a blue ball second\nThe problem states that the first ball is placed back into the bag before we take the\nsecond ball. This means that when we draw the second ball, there are again a total of\n10 balls in the bag, of which 5 are blue. Therefore the probability of drawing a blue\nball is\nP(second ball blue) = 5\n10 = 1\n2\nStep 3: Probability of red first and blue second\nWhen drawing two balls from the bag, there are 4 possibilities. We can get\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nWe want to know the probability of the second outcome, where we have to get a red\nball first. Since there are 5 red balls and 10 balls in total, there are\n5\n10 ways to get a\nred ball first. Now we put the first ball back, so there are again 5 red balls and 5 blue\nballs in the bag. Therefore there are\n5\n10 ways to get a blue ball second if the first ball\nwas red. This means that there are\n5\n10 × 5\n10 = 25\n100\n414\n10.2.\nDependent and independent events\n\nways to get a red ball first and a blue ball second. So, the probability of getting a red\nball first and a blue ball second is 1\n4.\nStep 4: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 1\n4\nSince 1\n4 = 1\n2 × 1\n2, the events are independent.\nSee video: 23B8 at www.everythingmaths.co.za\nWorked example 7: Independent and dependent events\nQUESTION\nIn the previous example, we picked a random ball and put it back into the bag before\ncontinuing. This is called sampling with replacement. In this example, we will follow\nthe same process, except that we will not put the first ball back into the bag. This is\ncalled sampling without replacement.\nSo, from a bag with 5 red and 5 blue balls, we remove a random ball and record its\ncolour. Then, without putting back the first ball, we remove another random ball from\nthe bag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Count the number of outcomes\nWe will look directly at the number of possible ways in which we can get the 4 possible\noutcomes when removing 2 balls. In the previous example, we saw that the 4 possible\noutcomes are\n415\nChapter 10.\nProbability\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nFor the first outcome, we have to get a red ball first. Since there are 5 red balls and\n10 balls in total, there are\n5\n10 ways to get a red ball first. After we have taken out a red\nball, there are now 4 red balls and 5 blue balls left. Therefore there are 4\n9 ways to get\na red ball second if the first ball was also red. This means that there are\n5\n10 × 4\n9 = 20\n90\nways to get a red ball first and a red ball second. The probability of the first outcome\nis 2\n9.\nFor the second outcome, we have to get a red ball first. As in the first outcome, there\nare\n5\n10 ways to get a red ball first; and there are now 4 red balls and 5 blue balls left.\nTherefore there are 5\n9 ways to get a blue ball second if the first ball was red. This means\nthat there are\n5\n10 × 5\n9 = 25\n90\nways to get a red ball first and a blue ball second. The probability of the second\noutcome is 5\n18.\nWe can compute the probabilities of the third and fourth outcomes in the same way as\nthe first two, but there is an easier way. Notice that there are only 2 types of ball and\nthat there are exactly equal numbers of them at the start. This means that the problem\nis completely symmetric in red and blue. We can use this symmetry to compute the\nprobabilities of the other two outcomes.\nIn the third outcome, the first ball is blue and the second ball is red. Because of\nsymmetry this outcome must have the same probability as the second outcome (when\nthe first ball is red and the second ball is blue). Therefore the probability of the third\noutcome is 5\n18.\nIn the fourth outcome, the first and second balls are both blue. From symmetry, this\noutcome must have the same probability as the first outcome (when both balls are red).\nTherefore the probability of the fourth outcome is 2\n9.\nTo summarise, these are the possible outcomes and their probabilities:\n• first ball red and second ball red: 2\n9;\n• first ball red and second ball blue:\n5\n18;\n• first ball blue and second ball red:\n5\n18;\n• first ball blue and second ball blue: 2\n9.\nStep 2: Probability of a red ball first\nTo determine the probability of getting a red ball on the first draw, we look at all of the\noutcomes that contain a red ball first. These are\n416\n10.2.\nDependent and independent events\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball.\nThe probability of the first outcome is 2\n9 and the probability of the second outcome is\n5\n18. By adding these two probabilities, we see that the probability of getting a red ball\nfirst is\nP(first ball red) = 2\n9 + 5\n18 = 1\n2\nThis is the same as in the previous exercise, which should not be too surprising since\nthe probability of the first ball being red is not affected by whether or not we put it\nback into the bag before drawing the second ball.\nStep 3: Probability of a blue ball second\nTo determine the probability of getting a blue ball on the second draw, we look at all\nof the outcomes that contain a blue ball second. These are\n• a red ball and then a blue ball;\n• a blue ball and then another blue ball.\nThe probability of the first outcome is 5\n18 and the probability of the second outcome is\n2\n9. By adding these two probabilities, we see that the probability of getting a blue ball\nsecond is\nP(second ball blue) = 5\n18 + 2\n9 = 1\n2\nThis is also the same as in the previous exercise! You might find it surprising that the\nprobability of the second ball is not affected by whether or not we replace the first ball.\nThe reason why this probability is still 1\n2 is that we are computing the probability that\nthe second ball is blue without knowing the colour of the first ball. Because there are\nonly two equal possibilities for the second ball (red and blue) and because we don’t\nknow whether the first ball is red or blue, there is an equal chance that the second ball\nwill be one colour or the other.\nStep 4: Probability of red first and blue second\nWe have already calculated the probability that the first ball is red and the second ball\nis blue. It is 5\n18.\nStep 5: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 5\n18\nSince 5\n18 ̸= 1\n2 × 1\n2, the events are dependent.\n417\nChapter 10.\nProbability\n\nWARNING!\nJust because two events are mutually exclusive does not necessarily mean that they\nare independent. To test whether events are mutually exclusive, always check that\nP(A and B) = 0. To test whether events are independent, always check that P(A and B) =\nP(A) × P(B). See the exercises below for examples of events that are mutually ex-\nclusive and independent in different combinations.\nExercise 10 – 3: Dependent and independent events\n1. Use the following Venn diagram to determine whether events X and Y are\na) mutually exclusive or not mutually exclusive;\nb) dependent or independent.\nS\nX\nY\n11\n7\n3\n14\n2. Of the 30 learners in a class 17 have black hair, 11 have brown hair and 2 have\nred hair. A learner is selected from the class at random.\na) What is the probability that the learner has black hair?\nb) What is the probability that the learner has brown hair?\nc) Are these two events mutually exclusive?\nd) Are these two events independent?\n3. P(M) = 0,45; P(N) = 0,3 and P(M or N) = 0,615. Are the events M and N\nmutually exclusive, independent or neither mutually exclusive nor independent?\n4. (For enrichment)\nProve that if event A and event B are mutually exclusive with P(A) ̸= 0 and\nP(B) ̸= 0, then A and B are always dependent.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B9\n2. 23BB\n3. 23BC\n4. 23BD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n418\n10.2.\nDependent and independent events\n\n10.3\nMore Venn diagrams\nEMBJV\nIn the rest of this chapter we will look at tools and techniques for working with proba-\nbility problems.\nWhen working with more complex problems, we can have three or more events that\nintersect in various ways.\nTo solve these problems, we usually want to count the\nnumber (or percentage) of outcomes in an event, or a combination of events. Venn\ndiagrams are a useful tool for recording and visualising the counts.\nInvestigation: Venn diagram for 3 events\nThe diagram below shows a general Venn diagram for 3 events.\nS\nA\nB\nC\nWrite down the sets corresponding to each of the three coloured regions and also\nto the shaded region. Remember that the intersections between circles represent the\nintersections between the different events.\nWhat is the event for\n• the red region;\n• the green region;\n• the blue region; and\n• the shaded region?\n419\nChapter 10.\nProbability\n\nWorked example 8: Venn diagram for 3 events\nQUESTION\nDraw a Venn diagram that shows the following sample space and events:\n• S: all the integers from 1 to 30\n• P: prime numbers\n• M: multiples of 3\n• F: factors of 30\nSOLUTION\nStep 1: Write down the sample space and event sets\nThe sample space contains all the positive integers up to 30.\nS = {1; 2; 3; . . . ; 30}\nThe prime numbers between 1 and 30 are\nP = {2; 3; 5; 7; 11; 13; 17; 19; 23; 29}\nThe multiples of 3 between 1 and 30 are\nM = {3; 6; 9; 12; 15; 18; 21; 24; 27; 30}\nThe factors of 30 are\nF = {1; 2; 3; 5; 6; 10; 15; 30}\nStep 2: Draw the outline of the Venn diagram\nThere are 3 events, namely P, M and F, and the sample space, S. Put this information\non a Venn diagram:\nS\nP\nM\nF\n420\n10.3.\nMore Venn diagrams\n\nStep 3: Place the outcomes in the appropriate event sets\nS\nP\nM\nF\n3\n2\n5\n6\n15\n30\n1\n10\n7\n11\n13\n17\n19\n23\n29\n9\n12\n21\n24\n18\n27\n4\n8\n14\n16\n20\n22\n25\n26\n28\nWorked example 9: Venn diagram for 3 events\nQUESTION\nAt Dawnview High there are 400 Grade 11 learners. 270 do Computer Science, 300\ndo English and 50 do Business studies. All those doing Computer Science do English,\n20 take Computer Science and Business studies and 35 take English and Business\nstudies. Using a Venn diagram, calculate the probability that a pupil drawn at random\nwill take:\n1. English, but not Business studies or Computer Science\n2. English but not Business studies\n3. English or Business studies but not Computer Science\n4. English or Business studies\nSOLUTION\nStep 1: Draw the outline of the Venn diagram\nWe need to be careful with this problem. In the question statement we are told that all\nthe learners who do Computer Science also do English. This means that the circle for\nComputer Science on the Venn diagram needs to be inside the circle for English.\n421\nChapter 10.\nProbability\n\nS\nE\nC\nB\nStep 2: Fill in the counts on the Venn diagram\nS\nE\nC\nB\n20\n250\n15\n15\n15\n85\nStep 3: Compute probabilities\nTo find the number of learners taking English, but not Business studies or Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 15 and there are a total of 400 learners in the grade. There-\nfore the probability that a learner will take English but not Business studies or Computer\nScience is\n15\n400 = 3\n80.\nTo find the number of learners taking English but not Business studies, we need to look\nat this region of the Venn diagram:\n422\n10.3.\nMore Venn diagrams\n\nThe count in this region is 265. Therefore the probability that a learner will take\nEnglish but not Business studies is 265\n400 = 53\n80.\nTo find the number of learners taking English or Business studies but not Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 45. Therefore the probability that a learner will take English\nor Business studies but not Computer Science is\n45\n400 = 9\n80.\nTo find the number of learners taking English or Business studies, we need to look at\nthis region of the Venn diagram:\nThe count in this region is 315. Therefore the probability that a learner will take\nEnglish or Business studies is 315\n400 = 63\n80.\n423\nChapter 10.\nProbability\n\nThere are some words that tell you which part of the Venn diagram should be filled in.\nThe following table summarises the most important ones:\nWords\nSymbols\nVenn diagram\n“all”\nA and B and C / A ∩B ∩C\n“none”\n“at least one”\nA or B or C / A ∪B ∪C\n“both A and B”\nA and B / A ∩B\n“A or B”\nA or B / A ∪B\nExercise 10 – 4: Venn diagrams\n1. Use the Venn diagram below to answer the following questions. Also given:\nn(S) = 120.\nS\nF\n8\n10\nG\n24\n15\nH\n14\n7\n2\na) Compute P(F).\nb) Compute P(G or H).\nc) Compute P(F and G).\nd) Are F and G dependent or independent?\n424\n10.3.\nMore Venn diagrams\n\n2. The Venn diagram below shows the probabilities of 3 events. Complete the Venn\ndiagram using the additional information provided.\nS\nZ\n1\n25\nY\n17\n100\nX\n17\n100\n3\n20\n• P(Z and (not Y )) =\n31\n100\n• P(Y and X) =\n23\n100\n• P(Y ) =\n39\n100\nAfter completing the Venn diagram, compute the following:\nP (Z and not (X or Y ))\n3. There are 79 Grade 10 learners at school. All of these take some combination of\nMaths, Geography and History. The number who take Geography is 41; those\nwho take History is 36; and 30 take Maths. The number who take Maths and\nHistory is 16; the number who take Geography and History is 6, and there are 8\nwho take Maths only and 16 who take History only.\na) Draw a Venn diagram to illustrate all this information.\nb) How many learners take Maths and Geography but not History?\nc) How many learners take Geography only?\nd) How many learners take all three subjects?\n4. Draw a Venn diagram with 3 mutually exclusive events. Use the diagram to\nshow that for 3 mutually exclusive events, A, B and C, the following is true:\nP(A or B or C) = P(A) + P(B) + P(C)\nThis is the addition rule for 3 mutually exclusive events.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BF\n2. 23BG\n3. 23BH\n4. 23BJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n425\nChapter 10.\nProbability\n\n10.4\nTree diagrams\nEMBJW\nTree diagrams are useful for organising and visualising the different possible outcomes\nof a sequence of events. For each possible outcome of the first event, we draw a line\nwhere we write down the probability of that outcome and the state of the world if that\noutcome happened. Then, for each possible outcome of the second event we do the\nsame thing.\nBelow is an example of a simple tree diagram, showing the possible outcomes of\nrolling a 6-sided die.\n1\n1\n6\n2\n1\n6\n3\n1\n6\n4\n1\n6\n5\n1\n6\n6\n1\n6\noutcomes\nprobabilities\nNote that each outcome (the numbers 1 to 6) is shown at the end of a line; and that\nthe probability of each outcome (all 1\n6 in this case) is shown shown on a line. The\nprobabilities have to add up to 1 in order to cover all of the possible outcomes. In the\nexamples below, we will see how to draw tree diagrams with multiple events and how\nto compute probabilities using the diagrams.\nEarlier in this chapter you learned about dependent and independent events. Tree\ndiagrams are very helpful for analysing dependent events. A tree diagram allows you\nto show how each possible outcome of one event affects the probabilities of the other\nevents.\nTree diagrams are not so useful for independent events since we can just multiply the\nprobabilities of separate events to get the probability of the combined event. Remem-\nber that for independent events:\nP(A and B) = P(A) × P(B)\nSo if you already know that events are independent, it is usually easier to solve a\nproblem without using tree diagrams. But if you are uncertain about whether events\nare independent or if you know that they are not, you should use a tree diagram.\nWorked example 10: Drawing a tree diagram\nQUESTION\nIf it rains on a given day, the probability that it rains the next day is 1\n3. If it does not rain\non a given day, the probability that it rains the next day is 1\n6. The probability that it will\nrain tomorrow is 1\n5. What is the probability that it will rain the day after tomorrow?\nDraw a tree diagram of all the possibilities to determine the answer.\nSOLUTION\nStep 1: Draw the first level of the tree diagram\nBefore we can determine what happens on the day after tomorrow, we first have to\ndetermine what might happen tomorrow. We are told that there is a 1\n5 probability that\n426\n10.4.\nTree diagrams\n\nit will rain tomorrow. Here is how to represent this information using a tree diagram:\n1\n5\nrain\n4\n5\nno rain\ntoday:\ntomorrow:\nStep 2: Draw the second level of the tree diagram\nWe are also told that if it does rain on one day, there is a 1\n3 probability that it will also\nrain on the following day. On the other hand, if it does not rain on one day, there is\nonly a 1\n6 probability that it will also rain on the following day. Using this information\nwe complete the tree diagram:\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nStep 3: Compute the probability\nWe are asked what the probability is that it will rain the day after tomorrow. On the\ntree diagram above we can see that there are 2 situations where it rains on the day\nafter tomorrow. They are marked in red below.\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nTo get the probability for the first situation (that it rains tomorrow and the day after\ntomorrow) we have to multiply the probabilies along the first red line.\nP(rain tomorrow and rain day after tomorrow)\n=1\n5 × 1\n3\n= 1\n15\n427\nChapter 10.\nProbability\n\nTo get the probability for the second situation (that it does not rain tomorrow, but it\ndoes rain the day after tomorrow) we have to multiply the probabilies along the second\nred line.\nP(not rain tomorrow and rain day after tomorrow)\n=4\n5 × 1\n6\n= 2\n15\nTherefore the total probability that it will rain the day after tomorrow is the sum of the\nprobabilities along the two red paths, namely\n1\n15 + 2\n15 = 1\n5\nWorked example 11: Drawing a tree diagram\nQUESTION\nYou play the following game. You flip a coin. If it comes up tails, you get 2 points\nand your turn ends. If it comes up heads, you get only 1 point, but you can flip the\ncoin again. If you flip the coin multiple times in one turn, you add up the points. You\ncan flip the coin at most 3 times in one turn. What is the probability that you will get\nexactly 3 points in one turn? Draw a tree diagram to visualise the different possibilities.\nSOLUTION\nStep 1: Write down the events and their symbols\nEach coin toss has on of two possible outcomes, namely heads (H) and tails (T). Each\noutcome has a probability of 1\n2. We are asked to count the number of points, so we\nwill also indicate how many points we have for each outcome.\nStep 2: Draw the first level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\nThis tree diagram shows the possible outcomes after 1 flip of the coin. Remember that\nwe can have up to 3 flips, so the diagram is not complete yet. If the coin comes up\nheads, we flip the coin again. If the coin comes up tails, we stop.\n428\n10.4.\nTree diagrams\n\nStep 3: Draw the second and third level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\n1\n2\nH\n2 pts\n1\n2\nT\n3 pts\n1\n2\nH\n3 pts\n1\n2\nT\n4 pts\nIn this tree diagram you can see that we add up the points we get with each coin flip.\nAfter three coin flips, the game is over.\nStep 4: Find the relevant outcomes and compute the probability\nWe are interested in getting exactly 3 points during the game. To find these outcomes\nwe look only at the tips of the tree. We end with exactly 3 points when the coin flips\nare\n• (H; T) with probability 1\n2 × 1\n2 = 1\n4;\n• (H; H; H) with probability 1\n2 × 1\n2 × 1\n2 = 1\n8.\nNotice that we compute the probability of an outcome by multiplying all the probabil-\nities along the path from the start of the tree to the tip where the outcome is. We add\nthe above two probabilites to obtain the final probability of getting exactly 3 points as\n1\n4 + 1\n8 = 3\n8.\nWorked example 12: Drawing a tree diagram\nQUESTION\nA person takes part in a medical trial that tests the effect of a medicine on a disease.\nHalf the people are given medicine and the other half are given a sugar pill, which has\nno effect on the disease. The medicine has a 60% chance of curing someone. But,\npeople who do not get the medicine still have a 10% chance of getting well. There are\n50 people in the trial and they all have the disease. Talwar takes part in the trial, but\nwe do not know whether he got the medicine or the sugar pill. Draw a tree diagram\nof all the possible cases. What is the probability that Talwar gets cured?\nSOLUTION\nStep 1: Summarise the information in the problem\nThere are two uncertain events in this problem. Each person either receives medicine\n(probability 1\n2) or a sugar pill (probability 1\n2). Each person also gets cured (probability\n429\nChapter 10.\nProbability\n\n3\n5 with medicine and\n1\n10 without) or stays ill (probability 2\n5 with medicine and\n9\n10\nwithout).\nStep 2: Draw the tree diagram\n1\n2\nmedicine\n1\n2\nsugar pill\n3\n5\ncured\n2\n5\nnot cured\n1\n10\ncured\n9\n10\nnot cured\nIn the first level of the tree diagram we show that Talwar either gets the medicine or\nthe sugar pill. The second level of the tree diagram shows whether Talwar is cured or\nnot, depending on which one of the pills he got.\nStep 3: Compute the required probability\nWe multiply the probabilites along each path in the tree diagram that leads to Talwer\nbeing cured:\n1\n2 × 3\n5 = 3\n10\n1\n2 × 1\n10 = 1\n20\nWe then add these probabilites to get the final answer. The probability that Talwar is\ncured is 7\n20.\nExercise 10 – 5: Tree diagrams\n1. You roll a die twice and add up the dots to get a score. Draw a tree diagram to\nrepresent this experiment. What is the probability that your score is a multiple\nof 5?\n2. What is the probability of throwing at least one five in four rolls of a regular\n6-sided die? Hint: do not show all possible outcomes of each roll of the die. We\nare interested in whether the outcome is 5 or not 5 only.\n3. You flip one coin 4 times.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\n430\n10.4.\nTree diagrams\n\n4. You flip 4 different coins at the same time.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BK\n2. 23BM\n3. 23BN\n4. 23BP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.5\nContingency tables\nEMBJX\nA contingency table is another tool for keeping a record of the counts or percentages\nin a probability problem. Contingency tables are especially helpful for figuring out\nwhether events are dependent or independent.\nWe will be studying two-way contingency tables, where we count the number of out-\ncomes for 2 events and their complements, making 4 events in total. A two-way contin-\ngency table always shows the counts for the 4 possible combinations of events, as well\nas the totals for each event and its complement. We can use a contingency table to\ncompute the probabilities of various events by computing the ratios between counts,\nand to determine whether the events are dependent or independent. The example\nbelow shows a two-way contingency table, representing the outcome of a medical\nstudy.\nWorked example 13: Contingency tables\nQUESTION\nA medical trial into the effectiveness of a new medication was carried out. 120 females\nand 90 males took part in the trial. Out of those people, 50 females and 30 males\nresponded positively to the medication. Given below is a contingency table with the\ngiven information filled in.\nFemale\nMale\nTotals\nPositive\n50\n30\nNegative\nTotals\n120\n90\n1. What is the probability that the medicine gives a positive result for females?\n2. What is the probability that the medicine gives a negative result for males?\n3. Was the medication’s success independent of gender? Explain.\n431\nChapter 10.\nProbability\n\nSOLUTION\nStep 1: Complete the contingency table\nThe best place to start is always to complete the contingency table. Because the each\ncolumn has to sum up to its total, we can work out the number of females and males\nwho responded negatively to the medication. Then we can add each row to get the\ntotals on the right hand side of the table.\nFemale\nMale\nTotals\nPositive\n50\n30\n80\nNegative\n70\n60\n130\nTotals\n120\n90\n210\nStep 2: Compute the required probabilities\nThe way the first question is phrased, we need to determine the probability that a\nperson responds positively if she is female. This means that we do not include males\nin this calculation. So, the probability that the medicine gives a positive result for\nfemales is the ratio between the number of females who got a positive response and\nthe total number of females.\nP(positive if female) = n(positive and female)\nn(female)\n= 50\n120\n= 5\n12\nSimilarly, the probability that the medicine gives a negative result for males is:\nP(negative if male) = n(negative and male)\nn(male)\n= 60\n90\n= 2\n3\nStep 3: Independence\nWe need to determine whether the effect of the medicine and the gender of a par-\nticipant are dependent or independent. According to the definition, two events are\nindependent if and only if\nP(A and B) = P(A) × P(B)\nWe will look at the events that a participant is female and that the participant re-\nsponded positively to the trial.\nP(female) =\nn(female)\nn(total trials)\n= 120\n210\n= 4\n7\n432\n10.5.\nContingency tables\n\nP(positive) =\nn(positive)\nn(total trials)\n= 80\n210\n= 8\n21\nP(female and positive) = n(female and positive)\nn(total trials)\n= 50\n210\n= 5\n21\nFrom these probabilities we can see that\nP(female and positive) ̸= P(female) × P(positive)\nand therefore the gender of a participant and the outcome of a trial are dependent\nevents.\nWorked example 14: Contingency tables\nQUESTION\nUse the contingency table below to answer the following questions.\nGrade 11\nGrade 12\nTotals\nHas cellphone\n59\n50\n109\nNo cellphone\n6\n3\n9\nTotals\n65\n53\n118\n1. What is the probability that a learner from Grade 11 has a cellphone?\n2. What is the probability that a learner who does not have a cellphone is from\nGrade 11.\n3. Are the grade of a learner and whether he has a cellphone or not independent\nevents? Explain your answer.\nSOLUTION\n1. There are 65 learners in Grade 11 and 59 of them have a cellphone. Therefore\nthe probability that a learner from Grade 11 has a cellphone is 59\n65.\n2. There are 9 learners who do not have a cellphone and 6 of them are in Grade\n11. Therefore the probability that a learner who does not have a cellphone is\nfrom from Grade 11 is 6\n9 = 2\n3.\n433\nChapter 10.\nProbability\n\n3. To test for independence, we will consider whether a learner is in Grade 11 and\nwhether a learner has a cellphone. The probability that a learner is in Grade 11\nis\n65\n118. The probability that a learner has a cellphone is 109\n118. The probability that\na learner is in Grade 11 and has a cellphone is\n59\n118 = 1\n2. Since 1\n2 ̸=\n65\n118 × 109\n118\nthe grade of a learner and whether he has a cellphone are dependent.\nExercise 10 – 6: Contingency tables\n1. Use the contingency table below to answer the following questions.\nBrown eyes\nNot brown eyes\nTotals\nBlack hair\n50\n30\n80\nRed hair\n70\n80\n150\nTotals\n120\n110\n230\na) What is the probability that someone with black hair has brown eyes?\nb) What is the probability that someone has black hair?\nc) What is the probability that someone has brown eyes?\nd) Are having black hair and having brown eyes dependent or independent\nevents?\n2. Given the following contingency table, identify the events and determine\nwhether they are dependent or independent.\nLocation A\nLocation B\nTotals\nBuses left late\n15\n40\n55\nBuses left on time\n25\n20\n45\nTotals\n40\n60\n100\n3. You are given the following information.\n• Events A and B are independent.\n• P(not A) = 0,3.\n• P(B) = 0,4.\nComplete the contingency table below.\nA\nnot A\nTotals\nB\nnot B\nTotals\n50\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BQ\n2. 23BR\n3. 23BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n434\n10.5.\nContingency tables\n\n10.6\nSummary\nEMBJY\nSee presentation: 23BT at www.everythingmaths.co.za\n• Terminology:\n– Outcome: a single observation of an experiment.\n– Sample space of an experiment: the set of all possible outcomes of the\nexperiment.\n– Event: a set of outcomes of an experiment.\n– Probability of an event: a real number between 0 and 1 that describes how\nlikely it is that the event will occur.\n– Relative frequency of an event: the number of times that the event occurs\nduring experimental trials, divided by the total number of trials conducted.\n– Union of events: the set of all outcomes that occur in at least one of the\nevents, written as “A or B”.\n– Intersection of events: the set of all outcomes that occur in all of the events,\nwritten as “A and B”.\n– Mutually exclusive events: events with no outcomes in common, that is\n(A and B) = ∅.\n– Complementary events: two mutually exclusive events that together con-\ntain all the outcomes in the sample space. We write the complement as\n“not A”.\n– Independent events: two events where knowing the outcome of one event\ndoes not affect the probability of the other event. Events are independent if\nand only if P(A and B) = P(A) × P(B).\n• Identities:\n– The addition rule: P(A or B) = P(A) + P(B) −P(A and B)\n– The addition rule for 2 mutually exclusive events: P(A or B) = P(A) +\nP(B)\n– The complementary rule: P(not A) = 1 −P(A)\n• A Venn diagram is a visual tool used to show how events overlap. Each region\nin a Venn diagram represents an event and could contain either the outcomes in\nthe event, the number of outcomes in the event or the probability of the event.\n• A tree diagram is a visual tool that helps with computing probabilities for depen-\ndent events. The outcomes of each event are shown along with the probability\nof each outcome. For each event that depends on a previous event, we go one\nlevel deeper into the tree. To compute the probability of some combination of\noutcomes, we\n– find all the paths that contain the outcome of interest;\n– multiply the probabilities along each path;\n– add the probabilities between different paths.\n• A 2-way contingency table is a tool for organising data, especially when we want\nto determine whether two events, each with only two outcomes, are dependent\nor independent. The counts for each possible combination of outcomes are\nentered into the table, along with the totals of each row and column.\n435\nChapter 10.\nProbability\n\nExercise 10 – 7: End of chapter exercises\n1. Jane invested in the stock market. The probability that she will not lose all her\nmoney is 0,32. What is the probability that she will lose all her money? Explain.\n2. If D and F are mutually exclusive events, with P(not D)\n=\n0,3 and\nP(D or F) = 0,94, find P(F).\n3. A car sales person has pink, lime-green and purple models of car A and purple,\norange and multicolour models of car B. One dark night a thief steals a car.\na) What is the experiment and sample space?\nb) What is the probability of stealing either a model of A or a model of B?\nc) What is the probability of stealing both a model of A and a model of B?\n4. The probability of event X is 0,43 and the probability of event Y is 0,24. The\nprobability of both occurring together is 0,10. What is the probability that X or\nY will occur?\n5. P(H) = 0,62; P(J) = 0,39 and P(H and J) = 0,31. Calculate:\na) P(H′)\nb) P(H or J)\nc) P(H′ or J′)\nd) P(H′ or J)\ne) P(H′ and J′)\n6. The last ten letters of the alphabet are placed in a hat and people are asked to\npick one of them. Event D is picking a vowel, event E is picking a consonant\nand event F is picking one of the last four letters. Draw a Venn diagram showing\nthe outcomes in the sample space and the different events. Then calculate the\nfollowing probabilities:\na) P(not F)\nb) P(F or D)\nc) P(neither E nor F)\nd) P(D and E)\ne) P(E and F)\nf) P(E and D′)\n7. Thobeka compares three neighbourhoods (we’ll call them A, B and C) to see\nwhere the best place is to live. She interviews 80 people and asks them whether\nthey like each of the neighbourhoods, or not.\n• 40 people like neighbourhood A.\n• 35 people like neighbourhood B.\n• 40 people like neighbourhood C.\n• 21 people like both neighbourhoods A and C.\n• 18 people like both neighbourhoods B and C.\n• 68 people like at least one neighbourhood.\n• 7 people like all three neighbourhoods.\n436\n10.6.\nSummary\n\na) Use this information to draw a Venn diagram.\nb) How many people like none of the neighbourhoods?\nc) How many people like neighbourhoods A and B, but not C?\nd) What is the probability that a randomly chosen person from the survey likes\nat least one of the neighbourhoods?\n8. Let G and H be two events in a sample space.\nSuppose that P(G) = 0,4;\nP(H) = h; and P(G or H) = 0,7.\na) For what value of h are G and H mutually exclusive?\nb) For what value of h are G and H independent?\n9. The following tree diagram represents points scored by two teams in a soccer\ngame. At each level in the tree, the points are shown as (points for Team 1;\npoints for Team 2).\n0,75\n(3; 0)\n0,25\n(2; 1)\n0,5\n(2; 1)\n0,5\n(1; 2)\n0,65\n(2; 0)\n0,35\n(1; 1)\n0,4\n(1; 1)\n0,6\n(0; 2)\n0,52\n(1; 0)\n0,48\n(0; 1)\n(0; 0)\nUse this diagram to determine the probability that:\na) Team 1 will win\nb) The game will be a draw\nc) The game will end with an even number of total points\n10. A bag contains 10 orange balls and 7 black balls. You draw 3 balls from the bag\nwithout replacement. What is the probability that you will end up with exactly\n2 orange balls? Represent this experiment using a tree diagram.\n11. Complete the following contingency table and determine whether the events are\ndependent or independent.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\nDid not like living there\n140\n340\nTotals\n230\n500\n12. Summarise the following information about a medical trial with 2 types of multi-\nvitamin in a contingency table and determine whether the events are dependent\nor independent.\n• 960 people took part in the medical trial.\n• 540 people used multivitamin A for a month and 400 of those people\nshowed an improvement in their health.\n437\nChapter 10.\nProbability\n\n• 300 people showed an improvement in health when using multivitamin B\nfor a month.\nIf the events are independent, it means that the two multivitamins have the same\neffect on people. If the events are dependent, it means that one multivitamin is\nbetter than the other. Which multivitamin is better than the other, or are the both\nequally effective?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BV\n2. 23BW\n3. 23BX\n4. 23BY\n5. 23BZ\n6. 23C2\n7. 23C3\n8. 23C4\n9. 23C5\n10. 23C6\n11. 23C7\n12. 23C8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n438\n10.6.\nSummary\n\nCHAPTER\n11\nStatistics\n11.1\nRevision\n440\n11.2\nHistograms\n444\n11.3\nOgives\n451\n11.4\nVariance and standard deviation\n455\n11.5\nSymmetric and skewed data\n461\n11.6\nIdentification of outliers\n464\n11.7\nSummary\n467\n\n11\nStatistics\n11.1\nRevision\nEMBJZ\nMeasures of central tendency\nEMBK2\nThe mean and median of a data set both give an indication where the centre of the\ndata distribution is located. The mean, or average, is calculated as\nx =\nPn\ni=1 xi\nn\nwhere the xi are the data and n is the number of data. We read x as “x bar”.\nThe median is the middle value of an ordered data set. To find the median, we first\nsort the data and then pick out the value in the middle of the sorted list. If the middle\nis in between two values, the median is the average of those two values.\nSee video: 23C9 at www.everythingmaths.co.za\nWorked example 1: Computing measures of central tendency\nQUESTION\nCompute the mean and median of the following data set:\n72,5 ; 92,6 ; 15,6 ; 53,0 ; 86,4 ; 89,9 ; 90,9 ; 21,7 ; 46,0 ; 4,1 ; 51,7 ; 2,2\nSOLUTION\nStep 1: Compute the mean\nUsing the formula for the mean, we first compute the sum of the values and then divide\nby the number of values.\nx = 626,6\n12\n≈52,22\nStep 2: Compute the median\nTo find the median, we first have to sort the data:\n2,2 ; 4,1 ; 15,6 ; 21,7 ; 46,0 ; 51,7 ; 53,0 ; 72,5 ; 86,4 ; 89,9 ; 90,9 ; 92,6\nSince there are an even number of values, the median will lie between two values.\nIn this case, the two values in the middle are 51,7 and 53,0. Therefore the median is\n52,35.\n440\n11.1.\nRevision\n\nMeasures of dispersion\nEMBK3\nMeasures of dispersion tell us how spread out a data set is. If a measure of dispersion\nis small, the data are clustered in a small region. If a measure of dispersion is large,\nthe data are spread out over a large region.\nThe range is the difference between the maximum and minimum values in the data\nset.\nThe inter-quartile range is the difference between the first and third quartiles of the\ndata set. The quartiles are computed in a similar way to the median. The median is\nhalfway into the ordered data set and is sometimes also called the second quartile.\nThe first quartile is one quarter of the way into the ordered data set; whereas the third\nquartile is three quarters of the way into the ordered data set.\nSee video: 23CB at www.everythingmaths.co.za\nWorked example 2: Range and inter-quartile range\nQUESTION\nDetermine the range and the inter-quartile range of the following data set.\n14 ; 17 ; 45 ; 20 ; 19 ; 36 ; 7 ; 30 ; 8\nSOLUTION\nStep 1: Sort the values in the data set\nTo determine the range we need to find the minimum and maximum values in the\ndata set. To determine the inter-quartile range we need to compute the first and third\nquartiles of the data set. For both of these requirements, it is easier to order the data\nset first.\nThe sorted data set is\n7 ; 8 ; 14 ; 17 ; 19 ; 20 ; 30 ; 36 ; 45\nStep 2: Find the minimum, maximum and range\nThe minimum value is the first value in the ordered data set, namely 7. The maximum\nis the last value in the ordered data set, namely 45. The range is the difference between\nthe minimum and maximum: 45 −7 = 38.\nStep 3: Find the quartiles and inter-quartile range\nThe diagram below shows how we find the quartiles one quarter, one half and three\nquarters of the way into the ordered list of values.\n441\nChapter 11.\nStatistics\n\n7\n8\n14\n17\n19\n20\n30\n36\n45\n0\n1\n4\n1\n2\n3\n4\n1\nFrom this diagram we can see that the first quartile is at a value of 14, the second\nquartile (median) is at a value of 19 and the third quartile is at a value of 30.\nThe inter-quartile range is the difference between the first and third quartiles. The\nfirst quartile is 14 and the third quartile is 30. Therefore the inter-quartile range is\n30 −14 = 16.\nFive number summary\nEMBK4\nThe five number summary combines a measure of central tendency, namely the me-\ndian, with measures of dispersion, namely the range and the inter-quartile range. This\ngives a good overview of the overall data distribution. More precisely, the five number\nsummary is written in the following order:\n• minimum;\n• first quartile;\n• median;\n• third quartile;\n• maximum.\nThe five number summary is often presented visually using a box and whisker diagram.\nA box and whisker diagram is shown below, with the positions of the five relevant\nnumbers labelled. Note that this diagram is drawn vertically, but that it may also be\ndrawn horizontally.\nmaximum\nupper quartile\nmedian\nlower quartile\nminimum\ninter-quartile range\ndata range\nSee video: 23CC at www.everythingmaths.co.za\n442\n11.1.\nRevision\n\nWorked example 3: Five number summary\nQUESTION\nDraw a box and whisker diagram for the following data set:\n1,25 ; 1,5 ; 2,5 ; 2,5 ; 3,1 ; 3,2 ; 4,1 ; 4,25 ; 4,75 ; 4,8 ; 4,95 ; 5,1\nSOLUTION\nStep 1: Determine the minimum and maximum\nSince the data set is already ordered, we can read off the minimum as the first value\n(1,25) and the maximum as the last value (5,1).\nStep 2: Determine the quartiles\nThere are 12 values in the data set.\n1,25 1,5\n2,5\n2,5\n3,1\n3,2\n4,1 4,25 4,75 4,8 4,95 5,1\n0\n1\n4\n1\n2\n3\n4\n1\nUsing the figure above we can see that the median is between the sixth and seventh\nvalues, making it.\n3,2 + 4,1\n2\n= 3,65\nThe first quartile lies between the third and fourth values, making it\nQ1 = 2,5 + 2,5\n2\n= 2,5\nThe third quartile lies between the ninth and tenth values, making it\nQ3 = 4,75 + 4,8\n2\n= 4,775\nStep 3: Draw the box and whisker diagram\nWe now have the five number summary as (1,25; 2,5; 3,65; 4,775; 5,1). The box and\nwhisker diagram representing the five number summary is given below.\n1,25\n2,5\n3,65\n4,775 5,1\n443\nChapter 11.\nStatistics\n\nExercise 11 – 1: Revision\n1. For each of the following data sets, compute the mean and all the quartiles.\nRound your answers to one decimal place.\na) −3,4 ; −3,1 ; −6,1 ; −1,5 ; −7,8 ; −3,4 ; −2,7 ; −6,2\nb) −6 ; −99 ; 90 ; 81 ; 13 ; −85 ; −60 ; 65 ; −49\nc) 7 ; 45 ; 11 ; 3 ; 9 ; 35 ; 31 ; 7 ; 16 ; 40 ; 12 ; 6\n2. Use the following box and whisker diagram to determine the range and inter-\nquartile range of the data.\n−5,52\n−2,41−1,53\n0,10\n4,08\n3. Draw the box and whisker diagram for the following data.\n0,2 ; −0,2 ; −2,7 ; 2,9 ; −0,2 ; −4,2 ; −1,8 ; 0,4 ; −1,7 ; −2,5 ; 2,7 ; 0,8 ; −0,5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23CD\n1b. 23CF\n1c. 23CG\n2. 23CH\n3. 23CJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.2\nHistograms\nEMBK5\nA histogram is a graphical representation of how many times different, mutually exclu-\nsive events are observed in an experiment. To interpret a histogram, we find the events\non the x-axis and the counts on the y-axis. Each event has a rectangle that shows what\nits count (or frequency) is.\nSee video: 23CK at www.everythingmaths.co.za\nWorked example 4: Reading histograms\nQUESTION\nUse the following histogram to determine the events that were recorded and the rela-\ntive frequency of each event. Summarise your answer in a table.\n444\n11.2.\nHistograms\n\n0\n2\n4\n6\n8\n10\nnot yet\nin school\nin primary\nschool\nin high\nschool\nSOLUTION\nStep 1: Determine the events\nThe events are shown on the x-axis. In this example we have “not yet in school”, “in\nprimary school” and “in high school”.\nStep 2: Read off the count for each event\nThe counts are shown on the y-axis and the height of each rectangle shows the fre-\nquency for each event.\n• not yet in school: 2\n• in primary school: 5\n• in high school: 9\nStep 3: Calculate relative frequency\nThe relative frequency of an event in an experiment is the number of times that the\nevent occurred divided by the total number of times that the experiment was com-\npleted. In this example we add up the frequencies for all the events to get a total\nfrequency of 16. Therefore the relative frequencies are:\n• not yet in school:\n2\n16 = 1\n8\n• in primary school:\n5\n16\n• in high school:\n9\n16\nStep 4: Summarise\nEvent\nCount\nRelative frequency\nnot yet in school\n2\n1\n8\nin primary school\n5\n5\n16\nin high school\n9\n9\n16\n445\nChapter 11.\nStatistics\n\nTo draw a histogram of a data set containing numbers, the numbers first have to be\ngrouped.\nEach group is defined by an interval.\nWe then count how many times\nnumbers from each group appear in the data set and draw a histogram using the counts.\nWorked example 5: Draw a histogram\nQUESTION\nThe following data represent the heights of 16 adults in centimetres.\n162 ; 168 ; 177 ; 147 ; 189 ; 171 ; 173 ; 168\n178 ; 184 ; 165 ; 173 ; 179 ; 166 ; 168 ; 165\nDivide the data into 5 equal length intervals between 140 cm and 190 cm and draw a\nhistogram.\nSOLUTION\nStep 1: Determine intervals\nTo have 5 intervals of the same length between 140 and 190, we need and interval\nlength of 10. Therefore the intervals are (140; 150]; (150; 160]; (160; 170]; (170; 180];\nand (180; 190].\nStep 2: Count data\nThe following table summarises the number of data values in each of the intervals.\nInterval\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\n(180; 190]\nCount\n1\n0\n7\n6\n2\nStep 3: Draw the histogram\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\n446\n11.2.\nHistograms\n\nFrequency polygons\nEMBK6\nA frequency polygon is sometimes used to represent the same information as in a his-\ntogram. A frequency polygon is drawn by using line segments to connect the middle of\nthe top of each bar in the histogram. This means that the frequency polygon connects\nthe coordinates at the centre of each interval and the count in each interval.\nWorked example 6: Drawing a frequency polygon\nQUESTION\nUse the histogram from the previous example to draw a frequency polygon of the same\ndata.\nSOLUTION\nStep 1: Draw the histogram\nWe already know that the histogram looks like this:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\nStep 2: Connect the tops of the rectangles\nWhen we draw line segments between the tops of the rectangles in the histogram, we\nget the following picture:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\n447\nChapter 11.\nStatistics\n\nStep 3: Draw final frequency polygon\nFinally, we remove the histogram to show only the frequency polygon.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\nFrequency polygons are particularly useful for comparing two data sets. Comparing\ntwo histograms would be more difficult since we would have to draw the rectangles of\nthe two data sets on top of each other. Because frequency polygons are just lines, they\ndo not pose the same problem.\nWorked example 7: Drawing frequency polygons\nQUESTION\nHere is another data set of heights, this time of Grade 11 learners.\n132 ; 132 ; 156 ; 147 ; 162 ; 168 ; 152 ; 174\n141 ; 136 ; 161 ; 148 ; 140 ; 174 ; 174 ; 162\nDraw the frequency polygon for this data set using the same interval length as in the\nprevious example. Then compare the two frequency polygons on one graph to see the\ndifferences between the distributions.\nSOLUTION\nStep 1: Frequency table\nWe first create the table of counts for the new data set.\nInterval\n(130; 140]\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\nCount\n4\n3\n2\n4\n3\n448\n11.2.\nHistograms\n\nStep 2: Draw histogram and frequency polygon\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\nStep 3: Compare frequency polygons\nWe draw the two frequency polygons on the same axes. The red line indicates the\ndistribution over heights for adults and the blue line, for Grade 11 learners.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\n190\nFrom this plot we can easily see that the heights for Grade 11 learners are distributed\nmore towards the left (shorter) than adults. The learner heights also seem to be more\nevenly distributed between 130 and 180 cm, whereas the adult heights are mostly\nbetween 160 and 180 cm.\n449\nChapter 11.\nStatistics\n\nExercise 11 – 2: Histograms\n1. Use the histogram below to answer the following questions.\nThe histogram\nshows the number of people born around the world each year. The ticks on\nthe x-axis are located at the start of each year.\npeople (millions)\nyear\n79\n80\n81\n82\n83\n84\n85\n86\n87\n1994 1995 1996 1997 1998 1999 2000 2001\na) How many people were born between the beginning of 1994 and the be-\nginning of 1996?\nb) Is the number people in the world population increasing or decreasing?\n(Ignore the rate at which people are dying for this question.)\nc) How many more people were born in 1994 than in 1997?\n2. In a traffic survey, a random sample of 50 motorists were asked the distance (d)\nthey drove to work daily. The results of the survey are shown in the table below.\nDraw a histogram to represent the data.\nd\n0 < d ≤10\n10 < d ≤20\n20 < d ≤30\n30 < d ≤40\n40 < d ≤50\nf\n9\n19\n15\n5\n4\n3. Below is data for the prevalence of HIV in South Africa. HIV prevalence refers to\nthe percentage of people between the ages of 15 and 49 who are infected with\nHIV.\nyear\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nprevalence (%)\n17,7\n18,0\n18,1\n18,1\n18,1\n18,0\n17,9\n17,9\nDraw a frequency polygon of this data set.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CM\n2. 23CN\n3. 23CP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n450\n11.2.\nHistograms\n\n11.3\nOgives\nEMBK7\nCumulative histograms, also known as ogives, are graphs that can be used to deter-\nmine how many data values lie above or below a particular value in a data set. The\ncumulative frequency is calculated from a frequency table, by adding each frequency\nto the total of the frequencies of all data values before it in the data set. The last value\nfor the cumulative frequency will always be equal to the total number of data values,\nsince all frequencies will already have been added to the previous total.\nAn ogive is drawn by\n• plotting the beginning of the first interval at a y-value of zero;\n• plotting the end of every interval at the y-value equal to the cumulative count for\nthat interval; and\n• connecting the points on the plot with straight lines.\nIn this way, the end of the final interval will always be at the total number of data since\nwe will have added up across all intervals.\nWorked example 8: Cumulative frequencies and ogives\nQUESTION\nDetermine the cumulative frequencies of the following grouped data and complete the\ntable below. Use the table to draw an ogive of the data.\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n20 < n ≤30\n7\n30 < n ≤40\n12\n40 < n ≤50\n10\n50 < n ≤60\n6\nSOLUTION\nStep 1: Compute cumulative frequencies\nTo determine the cumulative frequency, we add up the frequencies going down the\ntable. The first cumulative frequency is just the same as the frequency, because we are\nadding it to zero. The final cumulative frequency is always equal to the sum of all the\nfrequencies. This gives the following table:\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n5\n20 < n ≤30\n7\n12\n30 < n ≤40\n12\n24\n40 < n ≤50\n10\n34\n50 < n ≤60\n6\n40\n451\nChapter 11.\nStatistics\n\nStep 2: Plot the ogive\nThe first coordinate in the plot always starts at a y-value of 0 because we always start\nfrom a count of zero. So, the first coordinate is at (10; 0) — at the beginning of the\nfirst interval. The second coordinate is at the end of the first interval (which is also the\nbeginning of the second interval) and at the first cumulative count, so (20; 5). The third\ncoordinate is at the end of the second interval and at the second cumulative count,\nnamely (30; 12), and so on.\nComputing all the coordinates and connecting them with straight lines gives the fol-\nlowing ogive.\nn\n0\n10\n20\n30\n40\n10\n20\n30\n40\n50\n60\n•\n•\n•\n•\n•\n•\nOgives do look similar to frequency polygons, which we saw earlier. The most impor-\ntant difference between them is that an ogive is a plot of cumulative values, whereas\na frequency polygon is a plot of the values themselves. So, to get from a frequency\npolygon to an ogive, we would add up the counts as we move from left to right in the\ngraph.\nOgives are useful for determining the median, percentiles and five number summary\nof data. Remember that the median is simply the value in the middle when we order\nthe data. A quartile is simply a quarter of the way from the beginning or the end of an\nordered data set. With an ogive we already know how many data values are above or\nbelow a certain point, so it is easy to find the middle or a quarter of the data set.\nWorked example 9: Ogives and the five number summary\nQUESTION\nUse the following ogive to compute the five number summary of the data. Remember\nthat the five number summary consists of the minimum, all the quartiles (including the\nmedian) and the maximum.\n452\n11.3.\nOgives\n\ncount\nvalue\n0\n10\n20\n30\n40\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\nSOLUTION\nStep 1: Find the minimum and maximum\nThe minimum value in the data set is 1 since this is where the ogive starts on the\nhorizontal axis. The maximum value in the data set is 10 since this is where the ogive\nstops on the horizontal axis.\nStep 2: Find the quartiles\nThe quartiles are the values that are 1\n4, 1\n2 and 3\n4 of the way into the ordered data set.\nHere the counts go up to 40, so we can find the quartiles by looking at the values\ncorresponding to counts of 10, 20 and 30. On the ogive a count of\n• 10 corresponds to a value of 3 (first quartile);\n• 20 corresponds to a value of 7 (second quartile); and\n• 30 corresponds to a value of 8 (third quartile).\nStep 3: Write down the five number summary\nThe five number summary is (1; 3; 7; 8; 10). The box-and-whisker plot of this data set\nis given below.\n1\n3\n7\n8\n10\n453\nChapter 11.\nStatistics\n\nExercise 11 – 3: Ogives\n1. Use the ogive to answer the questions below. Note that marks are given as a\npercentage.\nnumber of students\nmark\n0\n10\n20\n30\n40\n50\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n•\n•\n•\n•\n•\n•\n•\n•\n•\na) How many students got between 50% and 70%?\nb) How many students got at least 70%?\nc) Compute the average mark for this class, rounded to the nearest integer.\n2. Draw the histogram corresponding to this ogive.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n−25\n−15\n−5\n5\n15\n25\n•\n•\n•\n•\n•\n•\n3. The following data set lists the ages of 24 people.\n2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53; 38\n4; 28; 5; 73; 79; 17; 15; 5; 34; 37; 45; 56\nUse the data to answer the following questions.\na) Using an interval width of 8 construct a cumulative frequency plot.\nb) How many are below 30?\nc) How many are below 60?\nd) Giving an explanation state below what value the bottom 50% of the ages\nfall.\ne) Below what value do the bottom 40% fall?\nf) Construct a frequency polygon.\n454\n11.3.\nOgives\n\n4. The weights of bags of sand in grams is given below (rounded to the nearest\ntenth):\n50,1; 40,4; 48,5; 29,4; 50,2; 55,3; 58,1; 35,3; 54,2; 43,5\n60,1; 43,9; 45,3; 49,2; 36,6; 31,5; 63,1; 49,3; 43,4; 54,1\na) Decide on an interval width and state what you observe about your choice.\nb) Give your lowest interval.\nc) Give your highest interval.\nd) Construct a cumulative frequency graph and a frequency polygon.\ne) Below what value do 53% of the cases fall?\nf) Below what value of 60% of the cases fall?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CQ\n2. 23CR\n3. 23CS\n4. 23CT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.4\nVariance and standard deviation\nEMBK8\nMeasures of central tendency (mean, median and mode) provide information on the\ndata values at the centre of the data set. Measures of dispersion (quartiles, percentiles,\nranges) provide information on the spread of the data around the centre. In this section\nwe will look at two more measures of dispersion called the variance and the standard\ndeviation.\nSee video: 23CV at www.everythingmaths.co.za\nVariance\nEMBK9\nDEFINITION: Variance\nLet a population consist of n elements, {x1; x2; . . . ; xn}. Write the mean of the data as\nx.\nThe variance of the data is the average squared distance between the mean and each\ndata value.\nσ2 =\nPn\ni=1 (xi −x)2\nn\nNOTE:\nThe variance is written as σ2. It might seem strange that it is written in squared form,\nbut you will see why soon when we discuss the standard deviation.\n455\nChapter 11.\nStatistics\n\nThe variance has the following properties.\n• It is never negative since every term in the variance sum is squared and therefore\neither positive or zero.\n• It has squared units. For example, the variance of a set of heights measured in\ncentimetres will be given in centimeters squared. Since the population variance\nis squared, it is not directly comparable with the mean or the data themselves. In\nthe next section we will describe a different measure of dispersion, the standard\ndeviation, which has the same units as the data.\nWorked example 10: Variance\nQUESTION\nYou flip a coin 100 times and it lands on heads 44 times. You then use the same\ncoin and do another 100 flips. This time in lands on heads 49 times. You repeat this\nexperiment a total of 10 times and get the following results for the number of heads.\n{44; 49; 52; 62; 53; 48; 54; 49; 46; 51}\nCompute the mean and variance of this data set.\nSOLUTION\nStep 1: Compute the mean\nThe formula for the mean is\nx =\nPn\ni=1 xi\nn\nIn this case, we sum the data and divide by 10 to get x = 50,8.\nStep 2: Compute the variance\nThe formula for the variance is\nσ2 =\nPn\ni=1 (xi −x)2\nn\nWe first subtract the mean from each datum and then square the result.\nxi\n44\n49\n52\n62\n53\n48\n54\n49\n46\n51\nxi −x\n−6,8\n−1,8\n1,2\n11,2\n2,2\n−2,8\n3,2\n−1,8\n−4,8\n0,2\n(xi −x)2\n46,24\n3,24\n1,44\n125,44 4,84\n7,84\n10,24\n3,24\n23,04\n0,04\nThe variance is the sum of the last row in this table divided by 10, so σ2 = 22,56.\n456\n11.4.\nVariance and standard deviation\n\nStandard deviation\nEMBKB\nSince the variance is a squared quantity, it cannot be directly compared to the data val-\nues or the mean value of a data set. It is therefore more useful to have a quantity which\nis the square root of the variance. This quantity is known as the standard deviation.\nDEFINITION: Standard deviation\nLet a population consist of n elements, {x1; x2; . . . ; xn}, with a mean of x. The stan-\ndard deviation of the data is\nσ =\nsPn\ni=1 (xi −x)2\nn\nIn statistics, the standard deviation is a very common measure of dispersion. Standard\ndeviation measures how spread out the values in a data set are around the mean. More\nprecisely, it is a measure of the average distance between the values of the data in the\nset and the mean. If the data values are all similar, then the standard deviation will be\nlow (closer to zero). If the data values are highly variable, then the standard variation\nis high (further from zero).\nThe standard deviation is always a positive number and is always measured in the\nsame units as the original data. For example, if the data are distance measurements in\nkilogrammes, the standard deviation will also be measured in kilogrammes.\nThe mean and the standard deviation of a set of data are usually reported together. In\na certain sense, the standard deviation is a natural measure of dispersion if the centre\nof the data is taken as the mean.\nInvestigation: Tabulating results\nIt is often useful to set your data out in a table so that you can apply the for-\nmulae easily.\nComplete the table below to calculate the standard deviation of\n{57; 53; 58; 65; 48; 50; 66; 51}.\n• Firstly, remember to calculate the mean, x.\n• Complete the following table.\nindex: i\ndatum: xi\ndeviation: xi −x\ndeviation\nsquared: (xi −x)2\n1\n57\n2\n53\n3\n58\n4\n65\n5\n48\n6\n50\n7\n66\n8\n51\nP xi = . . .\nP(xi −x) = . . .\nP(xi −x)2 = . . .\n• The sum of the deviations is always zero. Why is this? Find out.\n• Calculate the variance using the completed table.\n• Then calculate the standard deviation.\n457\nChapter 11.\nStatistics\n\nWorked example 11: Variance and standard deviation\nQUESTION\nWhat is the variance and standard deviation of the possibilities associated with rolling\na fair die?\nSOLUTION\nStep 1: Determine all the possible outcomes\nWhen rolling a fair die, the sample space consists of 6 outcomes. The data set is\ntherefore x = {1; 2; 3; 4; 5; 6} and n = 6.\nStep 2: Calculate the mean\nThe mean is:\nx = 1\n6 (1 + 2 + 3 + 4 + 5 + 6)\n= 3,5\nStep 3: Calculate the variance\nThe variance is:\nσ2 =\nP (x −x)2\nn\n= 1\n6 (6,25 + 2,25 + 0,25 + 0,25 + 2,25 + 6,25)\n= 2,917\nStep 4: Calculate the standard deviation\nThe standard deviation is:\nσ =\np\n2,917\n= 1,708\nSee video: 23CW at www.everythingmaths.co.za\n458\n11.4.\nVariance and standard deviation\n\nInterpretation and application\nEMBKC\nA large standard deviation indicates that the data values are far from the mean and a\nsmall standard deviation indicates that they are clustered closely around the mean.\nFor example, consider the following three data sets:\n{65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\n{85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\n{43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nEach of these data sets has the same mean, namely 67. However, they have different\nstandard deviations, namely 8,97, 17,75 and 21,23. The following figures show plots\nof the data sets with the mean and standard deviation indicated on each. You can see\nhow the standard deviation is larger when the data are more spread out.\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 8,97\ndata:\n{xi} = {65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\nmean:\nx = 67\nstandard deviation:\nσ ≈8,97\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 17,75\ndata:\n{xi} = {85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\nmean:\nx = 67\nstandard deviation:\nσ ≈17,75\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 21,23\ndata:\n{xi} = {43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nmean:\nx = 67\nstandard deviation:\nσ ≈21,23\nThe standard deviation may also be thought of as a measure of uncertainty. In the phys-\nical sciences, for example, the reported standard deviation of a group of repeated mea-\nsurements represents the precision of those measurements. When deciding whether\n459\nChapter 11.\nStatistics\n\nmeasurements agree with a theoretical prediction, the standard deviation of those mea-\nsurements is very important: if the mean of the measurements is too far away from the\nprediction (with the distance measured in standard deviations), then we consider the\nmeasurements as contradicting the prediction. This makes sense since they fall outside\nthe range of values that could reasonably be expected to occur if the prediction were\ncorrect.\nExercise 11 – 4: Variance and standard deviation\n1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban.\nThe data, in rands per litre, are given below.\nCape Town\n3,96\n3,76\n4,00\n3,91\n3,69\n3,72\nDurban\n3,97\n3,81\n3,52\n4,08\n3,88\n3,68\na) Find the mean price in each city and then state which city has the lowest\nmean.\nb) Find the standard deviation of each city’s prices.\nc) Which city has the more consistently priced petrol? Give reasons for your\nanswer.\n2. Compute the mean and variance of the following set of values.\n150 ; 300 ; 250 ; 270 ; 130 ; 80 ; 700 ; 500 ; 200 ; 220 ; 110 ; 320 ; 420 ; 140\n3. Compute the mean and variance of the following set of values.\n−6,9 ; −17,3 ; 18,1 ; 1,5 ; 8,1 ; 9,6 ; −13,1 ; −14,0 ; 10,5 ; −14,8 ; −6,5 ; 1,4\n4. The times for 8 athletes who ran a 100 m sprint on the same track are shown\nbelow. All times are in seconds.\n10,2 ; 10,8 ; 10,9 ; 10,3 ; 10,2 ; 10,4 ; 10,1 ; 10,4\na) Calculate the mean time.\nb) Calculate the standard deviation for the data.\nc) How many of the athletes’ times are more than one standard deviation away\nfrom the mean?\n5. The following data set has a mean of 14,7 and a variance of 10,01.\n18 ; 11 ; 12 ; a ; 16 ; 11 ; 19 ; 14 ; b ; 13\nCompute the values of a and b.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CX\n2. 23CY\n3. 23CZ\n4. 23D2\n5. 23D3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n460\n11.4.\nVariance and standard deviation\n\n11.5\nSymmetric and skewed data\nEMBKD\nWe are now going to classify data sets into 3 categories that describe the shape of the\ndata distribution: symmetric, left skewed, right skewed. We can use this classification\nfor any data set, but here we will look only at distributions with one peak. Most of\nthe data distributions that you have seen so far have only one peak, so the plots in this\nsection should look familiar.\nDistributions with one peak are called unimodal distributions.\nUnimodal literally\nmeans having one mode. (Remember that a mode is a maximum in the distribution.)\nSymmetric distributions\nEMBKF\nA symmetric distribution is one where the left and right hand sides of the distribution\nare roughly equally balanced around the mean. The histogram below shows a typical\nsymmetric distribution.\nmean ≈median\nbalanced left and right tails\nFor symmetric distributions, the mean is approximately equal to the median. The tails\nof the distribution are the parts to the left and to the right, away from the mean. The\ntail is the part where the counts in the histogram become smaller. For a symmetric\ndistribution, the left and right tails are equally balanced, meaning that they have about\nthe same length.\nThe figure below shows the box and whisker diagram for a typical symmetric data set.\nmedian halfway\nbetween\nfirst and third quartiles\nAnother property of a symmetric distribution is that its median (second quartile) lies\nin the middle of its first and third quartiles. Note that the whiskers of the plot (the\nminimum and maximum) do not have to be equally far away from the median. In the\nnext section on outliers, you will see that the minimum and maximum values do not\nnecessarily match the rest of the data distribution well.\n461\nChapter 11.\nStatistics\n\nSkewed\nEMBKG\nA distribution that is skewed right (also known as positively skewed) is shown below.\nmean\nmedian\nmean > median\nlong right tail\nshort left tail\nNow the picture is not symmetric around the mean anymore.\nFor a right skewed\ndistribution, the mean is typically greater than the median. Also notice that the tail of\nthe distribution on the right hand (positive) side is longer than on the left hand side.\nmedian closer to first quartile\nFrom the box and whisker diagram we can also see that the median is closer to the first\nquartile than the third quartile. The fact that the right hand side tail of the distribution\nis longer than the left can also be seen.\nA distribution that is skewed left has exactly the opposite characteristics of one that is\nskewed right:\n• the mean is typically less than the median;\n• the tail of the distribution is longer on the left hand side than on the right hand\nside; and\n• the median is closer to the third quartile than to the first quartile.\nThe table below summarises the different categories visually.\nSymmetric\nSkewed right (positive)\nSkewed left (negative)\n462\n11.5.\nSymmetric and skewed data\n\nExercise 11 – 5: Symmetric and skewed data\n1. Is the following data set symmetric, skewed right or skewed left? Motivate your\nanswer.\n27 ; 28 ; 30 ; 32 ; 34 ; 38 ; 41 ; 42 ; 43 ; 44 ; 46 ; 53 ; 56 ; 62\n2. State whether each of the following data sets are symmetric, skewed right or\nskewed left.\na) A data set with this histogram:\nb) A data set with this box and whisker plot:\nc) A data set with this frequency polygon:\n• • • • • • • • •\n•\n•\n•\n• • • •\nd) The following data set:\n11,2 ; 5 ; 9,4 ; 14,9 ; 4,4 ; 18,8 ; −0,4 ; 10,5 ; 8,3 ; 17,8\n3. Two data sets have the same range and interquartile range, but one is skewed\nright and the other is skewed left. Sketch the box and whisker plot for each of\nthese data sets. Then, invent data (6 points in each data set) that matches the\ndescriptions of the two data sets.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23D4\n2a. 23D5\n2b. 23D6\n2c. 23D7\n2d. 23D8\n3. 23D9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n463\nChapter 11.\nStatistics\n\n11.6\nIdentification of outliers\nEMBKH\nAn outlier in a data set is a value that is far away from the rest of the values in the\ndata set. In a box and whisker diagram, outliers are usually close to the whiskers of\nthe diagram. This is because the centre of the diagram represents the data between\nthe first and third quartiles, which is where 50% of the data lie, while the whiskers\nrepresent the extremes — the minimum and maximum — of the data.\nWorked example 12: Identifying outliers\nQUESTION\nFind the outliers in the following data set by drawing a box and whisker diagram and\nlocating the data values on the diagram.\n0,5 ; 1 ; 1,1 ; 1,4 ; 2,4 ; 2,8 ; 3,5 ; 5,1 ; 5,2 ; 6 ; 6,5 ; 9,5\nSOLUTION\nStep 1: Determine the five number summary\nThe minimum of the data set is 0,5. The maximum of the data set is 9,5. Since there\nare 12 values in the data set, the median lies between the sixth and seventh values,\nmaking it equal to 2,8+3,5\n2\n= 3,15. The first quartile lies between the third and fourth\nvalues, making it equal to 1,1+1,4\n2\n= 1,25. The third quartile lies between the ninth\nand tenth values, making it equal to 5,2+6\n2\n= 5,6.\nStep 2: Draw the box and whisker diagram\n0,5 1,25\n3,15\n5,6\n9,5\n• •• •\n• •\n•\n••\n• •\n•\nIn the figure above, each value in the data set is shown with a black dot.\nStep 3: Find the outliers\nFrom the diagram we can see that most of the values are between 1 and 6. The only\nvalue that is very far away from this range is the maximum at 9,5. Therefore 9,5 is the\nonly outlier in the data set.\nYou should also be able to identify outliers in plots of two variables. A scatter plot\nis a graph that shows the relationship between two random variables. We call these\ndata bivariate (literally meaning two variables) and we plot the data for two different\nvariables on one set of axes. The following example shows what a typical scatter plot\nlooks like. For Grade 11 you do not need to learn how to draw these 2-dimensional\n464\n11.6.\nIdentification of outliers\n\nscatter plots, but you should be able to identify outliers on them. As before, an outlier\nis a value that is far removed from the main distribution of data.\nWorked example 13: Scatter plot\nQUESTION\nWe have a data set that relates the heights and weights of a number of people. The\nheight is the first variable and its value is plotted along the horizontal axis. The weight\nis the second variable and its value is plotted along the vertical axis. The data values\nare shown on the plot below. Identify any outliers on the scatter plot.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\nSOLUTION\nWe inspect the plot visually and notice that there are two points that lie far away from\nthe main data distribution. These two points are circled in the plot below.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\n465\nChapter 11.\nStatistics\n\nExercise 11 – 6: Outliers\n1. For each of the following data sets, draw a box and whisker diagram and deter-\nmine whether there are any outliers in the data.\na) 30 ; 21,4 ; 39,4 ; 33,4 ; 21,1 ; 29,3 ; 32,8 ; 31,6 ; 36 ;\n27,9 ; 27,3 ; 29,4 ; 29,1 ; 38,6 ; 33,8 ; 29,1 ; 37,1\nb) 198 ; 166 ; 175 ; 147 ; 125 ; 194 ; 119 ; 170 ; 142 ; 148\nc) 7,1 ; 9,6 ; 6,3 ; −5,9 ; 0,7 ; −0,1 ; 4,4 ; −11,7 ; 10 ; 2,3 ; −3,7 ; 5,8 ; −1,4\n; 1,7 ; −0,7\n2. A class’s results for a test were recorded along with the amount of time spent\nstudying for it. The results are given below. Identify any outliers in the data.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23DB\n1b. 23DC\n1c. 23DD\n2. 23DF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n466\n11.6.\nIdentification of outliers\n\n11.7\nSummary\nEMBKJ\nSee presentation: 23DG at www.everythingmaths.co.za\n• Histograms visualise how many times different events occurred. Each rectangle\nin a histogram represents one event and the height of the rectangle is relative to\nthe number of times that the event occurred.\n• Frequency polygons represent the same information as histograms, but using\nlines and points rather than rectangles. A frequency polygon connects the mid-\ndle of the top edge of each rectangle in a histogram.\n• Ogives (also known as cumulative histograms) show the total number of times\nthat a value or anything less than that value appears in the data set. To draw an\nogive you need to add up all the counts in a histogram from left to right.\n– The first count in an ogive is always zero.\n– The last count in an ogive is always the sum of all the counts in the data\nset.\n• The variance and standard deviation are measures of dispersion.\n– The standard deviation is the square root of the variance.\n– Variance: σ2 = 1\nn\nPn\ni=1(xi −x)2\n– Standard deviation: σ =\nq\n1\nn\nPn\ni=1(xi −x)2\n– The standard deviation is measured in the same units as the mean and the\ndata, but the variance is not. The variance is measured in the square of the\ndata units.\n• In a symmetric distribution\n– the mean is approximately equal to the median; and\n– the tails of the distribution are balanced.\n• In a right (positively) skewed distribution\n– the mean is greater than the median;\n– the tail on the right hand side is longer than the tail on the left hand side;\nand\n– the median is closer to the first quartile than the third quartile.\n• In a left (negatively) skewed distribution\n– the mean is less than the median;\n– the tail on the left hand side is longer than the tail on the right hand side;\nand\n– the median is closer to the third quartile than the first quartile.\n• An outlier is a value that is far away from the rest of the data.\n467\nChapter 11.\nStatistics\n\nExercise 11 – 7: End of chapter exercises\n1. Draw a histogram, frequency polygon and ogive of the following data set. To\ncount the data, use intervals with a width of 1, starting from 0.\n0,4 ; 3,1 ; 1,1 ; 2,8 ; 1,5 ; 1,3 ; 2,8 ; 3,1 ; 1,8 ; 1,3 ;\n2,6 ; 3,7 ; 3,3 ; 5,7 ; 3,7 ; 7,4 ; 4,6 ; 2,4 ; 3,5 ; 5,3\n2. Draw a box and whisker diagram of the following data set and explain whether\nit is symmetric, skewed right or skewed left.\n−4,1 ; −1,1 ; −1 ; −1,2 ; −1,5 ; −3,2 ; −4 ; −1,9 ; −4 ;\n−0,8 ; −3,3 ; −4,5 ; −2,5 ; −4,4 ; −4,6 ; −4,4 ; −3,3\n3. Eight children’s sweet consumption and sleeping habits were recorded. The data\nare given in the following table and scatter plot.\nNumber of sweets\nper week\n15\n12\n5\n3\n18\n23\n11\n4\nAverage sleeping\ntime (hours per day)\n4\n4,5\n8\n8,5\n3\n2\n5\n8\n5\n10\n15\n20\n25\nnumber of sweets\n1\n2\n3\n4\n5\n6\n7\n8\n9\nsleeping time (hours per day)\na) What is the mean and standard deviation of the number of sweets eaten per\nday?\nb) What is the mean and standard deviation of the number of hours slept per\nday?\nc) Make a list of all the outliers in the data set.\n4. The monthly incomes of eight teachers are as follows:\nR 10 050;\nR 14 300;\nR 9800;\nR 15 000;\nR 12 140;\nR 13 800;\nR 11 990;\nR 12 900.\na) What is the mean and standard deviation of their incomes?\nb) How many of the salaries are less than one standard deviation away from\nthe mean?\nc) If each teacher gets a bonus of R 500 added to their pay what is the new\nmean and standard deviation?\nd) If each teacher gets a bonus of 10% on their salary what is the new mean\nand standard deviation?\ne) Determine for both of the above, how many salaries are less than one stan-\ndard deviation away from the mean.\n468\n11.7.\nSummary\n\nf) Using the above information work out which bonus is more beneficial fi-\nnancially for the teachers.\n5. The weights of a random sample of boys in Grade 11 were recorded. The cumu-\nlative frequency graph (ogive) below represents the recorded weights.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100 110 120\n0\n10\n20\n30\n40\n50\n60\nWeight (in kilogrammes)\nCumulative frequency\nCumulative frequency curve showing weight of boys\na) How many of the boys weighed between 90 and 100 kilogrammes?\nb) Estimate the median weight of the boys.\nc) If there were 250 boys in Grade 11, estimate how many of them would\nweigh less than 80 kilogrammes?\n6. Three sets of 12 learners each had their test scores recorded. The test was out of\n50. Use the given data to answer the following questions.\nSet A\nSet B\nSet C\n25\n32\n43\n47\n34\n47\n15\n35\n16\n17\n32\n43\n16\n25\n38\n26\n16\n44\n24\n38\n42\n27\n47\n50\n22\n43\n50\n24\n29\n44\n12\n18\n43\n31\n25\n42\na) For each of the sets calculate the mean and the five number summary.\nb) Make box and whisker plots of the three data sets on the same set of axes.\nc) State, with reasons, whether each of the three data sets are symmetric or\nskewed (either right or left).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DH\n2. 23DJ\n3. 23DK\n4. 23DM\n5. 23DN\n6. 23DP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n469\nChapter 11.\nStatistics\n\n\nCHAPTER\n12\nLinear programming\n12.1\nIntroduction\n472\n\n12\nLinear programming\n12.1\nIntroduction\nEMBKK\nIn everyday life people are interested in knowing the most efficient way of carrying out\na task or achieving a goal. For example, a farmer wants to know how many hectares to\nplant during a season in order to maximise the yield (produce), a stock broker wants to\nknow how much to invest in stocks in order to maximise profit, an entrepreneur wants\nto know how many people to employ to minimise expenditure. These are optimisation\nproblems; we want to to determine either the maximum or the minimum in a specific\nsituation.\nTo describe this mathematically, we assign variables to represent the different factors\nthat influence the situation. Optimisation means finding the combination of variables\nthat gives the best result.\nSee video: 23DQ at www.everythingmaths.co.za\nWorked example 1: Mountees and Roadees\nQUESTION\nInvestigate the following situation and use your knowledge of mathematics to solve the\nproblem:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make the maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nStep 2: Organise the information given\nWrite down a summary of the information given in the problem so that we consider\n472\n12.1.\nIntroduction\n\nall the different components in the situation.\nmaximum number for M\n= 5\nmaximum number for R\n= 3\nnumber of technicians needed for M = 1\nnumber of technicians needed for R = 2\ntotal number of technicians\n= 8\nprofit per M\n= 800\nprofit per R\n= 2400\nStep 3: Draw up a table\nUse the summary to draw up a table of all the possible combinations of the number of\nMountees and Roadees that can be manufactured per day:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n(4; 3)\n5\n(5; 0)\n(5; 1)\n(5; 2)\n(5; 3)\nNote that there are 24 possible combinations.\nStep 4: Consider the limitation of the number of technicians\nIt takes 1 technician to assemble a Mountee and 2 technicians to assemble a Roadee.\nThere are a total of 8 technicians in the assembly department, therefore we can write\nthat 1(M) + 2(R) ≤8.\nWith this limitation, we are able to eliminate some of the combinations in the table\nwhere M + 2R > 8:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n\b\b\b\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n\b\b\b\n(4; 3)\n5\n(5; 0)\n(5; 1)\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nThese combinations have been excluded as possible answers. For example, (5; 3) gives\n5 + 2(3) = 11 technicians.\n473\nChapter 12.\nLinear programming\n\nStep 5: Consider the profit on the bicycles\nWe can express the profit (P) per day as: P = 800(M)+2400(R). Notice that a higher\nprofit is made on a Roadee.\nBy substituting the different combinations for M and R, we can find the values that\ngive the maximum profit:\nFor (5; 0)\nP = 800(5) + 2400(0)\n= R 4000\nFor (3; 1)\nP = 800(3) + 2400(1)\n= R 4800\nM\nR\n0\n1\n2\n3\n0\n(0; 0) ⇒R 0\n(0; 1) ⇒R 2400\n(0; 2) ⇒R 4800\n(0; 3) ⇒R 7200\n1\n(1; 0) ⇒R 800\n(1; 1) ⇒R 3200\n(1; 2) ⇒R 5600\n(1; 3) ⇒R 8000\n2\n(2; 0) ⇒R 1600\n(2; 1) ⇒R 4000\n(2; 2) ⇒R 6400\n(2; 3) ⇒R 8800\n3\n(3; 0) ⇒R 2400\n(3; 1) ⇒R 4800\n(3; 2) ⇒R 7200\n\b\b\b\n(3; 3)\n4\n(4; 0) ⇒R 3200\n(4; 1) ⇒R 5600\n(4; 2) ⇒R 8000\n\b\b\b\n(4; 3)\n5\n(5; 0) ⇒R 4000\n(5; 1) ⇒R 6400\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nStep 6: Write the final answer\nTherefore the maximum profit of R 8800 is obtained if 2 Mountees and 3 Roadees are\nmanufactured per day.\nExercise 12 – 1: Optimisation\n1. Furniture store opening special:\nAs part of their opening special, a furniture store has promised to give away at\nleast 40 prizes with a total value of at least R 4000. They intend to give away\nkettles and toasters. They decide there will be at least 10 units of each prize. A\nkettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the\ncompany. Calculate how much this combination of kettles and toasters will cost.\nUse a suitable strategy to organise the information and solve the problem.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n474\n12.1.\nIntroduction\n\nOptimisation using graphs\nA more efficient way to solve optimisation problems is using graphs.\nWe write the limitations in the situation, called constraints, as inequalities. Some con-\nstraints can be modelled by an equation, which needs to be maximised or minimized.\nWe sketch the inequalities and indicate the region above or below the line that is to be\nconsidered in determining the solution. This method of solving optimisation problems\nis called linear programming.\nSee video: 23DS at www.everythingmaths.co.za\nWorked example 2: Optimisation using graphs\nQUESTION\nConsider again the example of Mr. Hunter who manufactures Mountees and Roadees:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make a maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nNotice that the values of M and R are limited to positive integers; Mr. Hunter cannot\nsell negative numbers of bikes nor can he sell a fraction of a bike.\nStep 2: Organise the information\nWe can write these constraints as inequalities:\nnumber of Mountees: 0 ≤M ≤5\nnumber of Roadees: 0 ≤R ≤3\ntotal number of technicians: M + 2R ≤8\nWe also know that P = 800M + 2400R. This is called the objective function, some-\ntimes also referred to as the search line, because the objective (goal) is to determine\nthe maximum value of P.\n475\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nWe represent the number of Mountees manufactured daily on the horizontal axis and\nthe number of Roadees manufactured daily on the vertical axis. Since M and R are\npositive integers, we only use the first quadrant of the Cartesian plane. Note that the\ngraph only includes the integer values of M between 0 and 5 and R between 0 and 3.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nFor the number of technicians in the assembly department M + 2R ≤8. If we make\nR (represented on the y-axis) the subject of the inequality we get R ≤−1\n2M + 4.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR = −1\n2M + 4\nThe arrows indicate the region in which the solution will lie, where R ≤−1\n2M + 4.\nThis area is called the feasible region.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR ≤−1\n2M + 4\nA\n476\n12.1.\nIntroduction\n\nWe substitute the possible combinations into the profit equation P = 800M + 2400R,\nand find the combination that gives the maximum profit.\nAt A(2; 3) :\nP = 800(2) + 2400(3)\n= R 8800\nStep 4: Write the final answer\nTherefore the maximum profit is obtained if 2 Mountees and 3 Roadees are manufac-\ntured per day.\nSee video: 23DT at www.everythingmaths.co.za\nWorked example 3: Optimisation using graphs\nQUESTION\nSolve the “furniture store opening special” problem using graphs:\nAs part of their opening special, a furniture store has promised to give away at least 40\nprizes. They intend to give away kettles and toasters. They decide there will be at least\n10 units of each prize. A kettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the company.\nCalculate how much this combination of kettles and toasters will cost.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nkettles be k and the number of toasters be t, with k, t ∈Z.\nStep 2: Organise the information\nWe can write the given information as inequalities:\nnumber of kettles: k ≥10\nnumber of toasters: t ≥10\ntotal number of prizes: k + t ≥40\nWe make t the subject of the inequality:\nt ≥−k + 40\n477\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nRepresent the constraints on a set of axes:\nKettles (k)\nToasters (t)\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nt ≥−k + 40\nt ≥10\nk ≥10\nWe shade the feasible region as shown in the diagram. Remember that in this situation\nonly the points with integer coordinates inside or on the border of the feasible region\nare possible solutions. The combination giving the minimum cost will lie towards or\non the lower border of the feasible region, which gives us many points to consider. To\nfind the optimum value of C, we use the graph of the objective function\nC = 120k + 100t\nTo draw the line, we make t the subject of the formula\nt = −6\n5k + C\n100\nWe see that the gradient of the objective function is −6\n5, but we do not know the exact\nvalue of the t-intercept ( C\n100). To find the minimum value of C, we need to determine\nthe position of the objective function where it first touches the feasible region and also\ngives the lowest t-intercept.\nKettles (k)\nToasters (t)\nA\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nWe indicate the gradient of the objective function on the graph (the green search\nline). Keeping the gradient the same, we “slide” the objective function towards the\nlower border of the feasible region and find that it touches the feasible region at point\n478\n12.1.\nIntroduction\n\nA(10; 30). This optimum position of the objective function is indicated on the graph\nby the dotted line passing through point A.\nWe substitute the coordinates of A into the cost equation C = 120k + 100t:\nAt A(10; 30) :\nC = 120(10) + 100(30)\n= R 4200\nThe minimum cost can also be determined graphically by reading off the coordinates\nof the t-intercept of the objective function in the optimum position:\ntint = 42\n∴C\n100 = 42\n∴C = R 4200\nStep 4: Write the final answer\nTherefore the minimum cost to the company is R 4200 with 10 kettles and 30 toast-\ners.\nExercise 12 – 2: Optimisation\n1. You are given a test consisting of two sections. The first section is on algebra and\nthe second section is on geometry. You are not allowed to answer more than 10\nquestions from any section, but you have to answer at least 4 algebra questions.\nThe time allowed is not more than 30 minutes. An algebra problem will take 2\nminutes and a geometry problem will take 3 minutes to solve.\nLet x be the number of algebra questions and y be the number of geometry\nquestions.\na) Formulate the equations and inequalities that satisfy the above constraints.\nb) The algebra questions carry 5 marks each and the geometry questions carry\n10 marks each. If T is the total marks, write down an expression for T.\n2. A local clinic wants to produce a guide to healthy living. The clinic intends to\nproduce the guide in two formats: a short video and a printed book. The clinic\nneeds to decide how many of each format to produce for sale. Estimates show\nthat no more than 10 000 copies of both items together will be sold. At least\n4000 copies of the video and at least 2000 copies of the book could be sold,\nalthough sales of the book are not expected to exceed 4000 copies. Let x be the\nnumber of videos sold, and y the number of printed books sold.\na) Write down the constraint inequalities that can be deduced from the given\ninformation.\nb) Represent these inequalities graphically and indicate the feasible region\nclearly.\n479\nChapter 12.\nLinear programming\n\nc) The clinic is seeking to maximise the income, I, earned from the sales of\nthe two products. Each video will sell for R 50 and each book for R 30.\nWrite down the objective function for the income.\nd) What maximum income will be generated by the two guides?\n3. A certain motorcycle manufacturer produces two basic models, the Super X and\nthe Super Y. These motorcycles are sold to dealers at a profit of R 20 000 per\nSuper X and R 10 000 per Super Y. A Super X requires 150 hours for assembly,\n50 hours for painting and finishing and 10 hours for checking and testing. The\nSuper Y requires 60 hours for assembly, 40 hours for painting and finishing and\n20 hours for checking and testing. The total number of hours available per month\nis: 30 000 in the assembly department, 13 000 in the painting and finishing\ndepartment and 5000 in the checking and testing department.\nThe above information is summarised by the following table:\nDepartment\nHours for\nSuper X\nHours for\nSuper Y\nHours available\nper month\nAssembly\n150\n60\n30 000\nPainting and\nfinishing\n50\n40\n13 000\nChecking and testing\n10\n20\n5000\nLet x be the number of Super X and y be the number of Super Y models manu-\nfactured per month.\na) Write down the set of constraint inequalities.\nb) Use graph paper to represent the set of constraint inequalities.\nc) Shade the feasible region on the graph paper.\nd) Write down the profit generated in terms of x and y.\ne) How many motorcycles of each model must be produced in order to max-\nimise the monthly profit?\nf) What is the maximum monthly profit?\n4. A group of students plan to sell x hamburgers and y chicken burgers at a rugby\nmatch. They have meat for at most 300 hamburgers and at most 400 chicken\nburgers. Each burger of both types is sold in a packet. There are 500 packets\navailable. The demand is likely to be such that the number of chicken burgers\nsold is at least half the number of hamburgers sold.\na) Write the constraint inequalities and draw a graph of the feasible region.\nb) A profit of R 3 is made on each hamburger sold and R 2 on each chicken\nburger sold. Write the equation which represents the total profit P in terms\nof x and y.\nc) The objective is to maximise profit. How many of each type of burger\nshould be sold?\n5. Fashion-Cards is a small company that makes two types of cards, type X and type\nY. With the available labour and material, the company can make at most 150\ncards of type X and at most 120 cards of type Y per week. Altogether they cannot\nmake more than 200 cards per week.\n480\n12.1.\nIntroduction\n\nThere is an order for at least 40 type X cards and 10 type Y cards per week.\nFashion-Cards makes a profit of R 5 for each type X card sold and R 10 for each\ntype Y card.\nLet the number of type X cards manufactured per week be x and the number of\ntype Y cards manufactured per week be y.\na) One of the constraint inequalities which represents the restrictions above is\n0 ≤x ≤150. Write the other constraint inequalities.\nb) Represent the constraints graphically and shade the feasible region.\nc) Write the equation that represents the profit P (the objective function), in\nterms of x and y.\nd) Calculate the maximum weekly profit.\n6. To meet the requirements of a specialised diet a meal is prepared by mixing\ntwo types of cereal, Vuka and Molo. The mixture must contain x packets of\nVuka cereal and y packets of Molo cereal. The meal requires at least 15 g of\nprotein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains\n4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains\n3 g of protein and 24 g of carbohydrates. There are at most 5 packets of cereal\navailable. The feasible region is shaded on the attached graph paper.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\nNumber of packets of Vuka\nNumber of packets of Molo\na) Write down the constraint inequalities.\nb) If Vuka cereal costs R 6 per packet and Molo cereal also costs R 6 per\npacket, use the graph to determine how many packets of each cereal must\nbe used so that the total cost for the mixture is a minimum.\nc) Use the graph to determine how many packets of each cereal must be used\nso that the total cost for the mixture is a maximum (give all possibilities).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DV\n2. 23DW\n3. 23DX\n4. 23DY\n5. 23DZ\n6. 23F2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n481\nChapter 12.\nLinear programming\n\n\nSolutions to exercises\n1\nExponents and surds\nExercise 1 – 1: The number system\n1. R; Q′\n2. R; Q\n3. R; Q\n4. R; Q\n5. R; Q; Z; N0\n6. R′Q′\n7. R; Q\n8. R; Q′\n9. R′\n10. R; Q′\n11. R; Q\n12. R; Q; Z\n13. R; Q\n14. R; Q′\n15. R; Q\n16. R; Q; Z\nExercise 1 – 2: Laws of exponents\n1. 43a+3\n2. 72\n3. 9p10\n4. k2x−2\n5. 52z−2 + 5z\n6. 1\n7. x10\n8.\nb2\na2\n9.\n1\nm+n\n10. 2pts\n11.\n1\na\n12. k\n13. 2a+1\n14. h4\n15.\na4b6\nc6d2\n16. 4\n17.\nm2n2\n2\n18. 400\n19.\n1\ny7\n20. 8\n21. 26a+2\n22. 2pt\n23. 81q2sy8a+2\nExercise 1 – 3: Rational exponents and surds\n1.\na) 7\nb)\n1\n6\nc)\n1\n3√\n36\nd) −4\n3\ne) 8x3\n2.\na) s\n1\n6\nb) 16m4\nc)\n3\n2 m2\nd) 8\n3. x\n31\n16\n483\nChapter 12.\nLinear programming\n\nExercise 1 – 4: Simplification of surds\n1.\na) 4\nb) ab4c2\nc) 2\nd) xy4\n2.\na)\nab\nb−a\nb) −\n\u0010\na\n1\n2 + b\n1\n2\n\u0011\nExercise 1 – 5: Rationalising the denominator\n1. 2\n√\n5\n2.\n√\n6\n2\n3.\n√\n6\n4.\n3\n√\n5 + 3\n4\n5.\nx√y\ny\n6.\n√\n6 +\n√\n14\n2\n7.\n3p −4√p\np\n8.\n√\nt −2\n9.\n1−√m\n1−m\n10.\n√\nab\nExercise 1 – 6: Solving surd equations\n1. x = 4\n2. p = 3\n3. y = 1\n4. t = 3\n5. z = 9 or z = 1\n4\n6. x = 8 or x = −27\n7. n = −1\n4\n8. d = 3 or d = −5\n9. y = 1 or y = 81\n10. f = 5\nExercise 1 – 7: Applications of exponentials\n1. 9,7%\n2. 4 254 691\n3. 7\n4. 26 893\n484\n12.1.\nIntroduction\n\nExercise 1 – 8: End of chapter exercises\n1.\na)\n1\n4\nb) 4 1\n4\n2.\na) x4\nb) s\nc) m\n25\n3\nd) m\n8\n3\ne) −m\n8\n3\nf) 81y\n16\n3\n3.\na)\n3b\n45\n2\n(a12c\n5\n2\nb) 3a3b2\nc) a24b12\nd) x\n7\n2\ne) x\n4\n3 b\n5\n3\n4.\n1\nx\n1\n16\n5. x −2\n6.\n10√x + 10\nx −1\n7.\n3√x + 2x√x\n2x\n8.\na) 6\n√\n2\nb) 7\n√\n5\nc) 2\nd)\n1\n4\n√\n2\ne) 2\nf)\n16\n√\n15\n5\n9.\na) 6 + 4\n√\n2\nb) 6 + 5\n√\n2\nc) 4+2\n√\n2+2\n√\n3+2\n√\n6\n10.\na) 55\nb) 1\n11. 15\n√\n2x3\n12.\na) 1 + 2\n√\n5\n5\nb)\n2y + y√y −4√y −8\ny −4\nc) 2√x + 2\n√\n10\n13.\n3\n2\n15. 3\n16. −\n√\n288\n17.\na) 4\nb) −1\n3\nc) 3\nd) No solution\ne) x = 1\n8 or x = −8\n18.\nb) x = 1\n2\nEquations and inequalities\nExercise 2 – 1: Solution by factorisation\n1. t = 0 or t = −2\n2. y = −1\n3. s = ±5\n4. y = 3 or y = 2\n5. y = 4 or y = −9\n6. p = −2\n7. y = −3 or y = −8\n8. y = 6 or y = 7\n9. x = −7 or x = −2\n10. y = 4k or y = k\n11. y = 9 or y = −9\n12. y = ±\n√\n5\n13. h = ±6\n14. y = ±\n√\n14\n15. p = −2\n16. y = ±6\n√\n2\n17. f = 5\n2 or f = −3\n18. x = 1\n4\n19. y = 1\n7\n20. x ∈R, x ̸= ±3\n21. y = −13 or y = −1\n22. t = 3\n2 or t = −2\n23. m = −6\n24. t = 0 or t = 3\nExercise 2 – 2: Solution by completing the square\n1.\na) x = −5 −3\n√\n3 or x = −5 + 3\n√\n3\nb) x = −1 or x = −3\nc) p = −4 ±\n√\n21\nd) x = −3 ±\n√\n7\ne) No real solution\nf) t = −8 ± 3\n√\n6\ng) x = −1 ±\nq\n5\n3\nh) z = −4 ±\n√\n22\ni) z = 11\n2 or z = 0\nj) z = 5 or z = −1\n2. k = −3 ± √9 −a\n3. y = −q±√\nq2−4pr\n2p\n485\nChapter 12.\nLinear programming\n\nExercise 2 – 3: Solution by the quadratic formula\n1. t = 1 or t = −4\n3\n2. x = 5+\n√\n37\n2\nor t = 5−\n√\n37\n2\n3. No real solution\n4. p = 1\n2 or p = −1\n5. No real solution\n6. t = −3+\n√\n69\n10\nor t = −3−\n√\n69\n10\n7. t = 2 ±\n√\n2\n8. k = 7+\n√\n373\n18\nor k = 7−\n√\n373\n18\n9. f = 1\n2 or f = −2\n10. No real solution\nExercise 2 – 4:\n1. x = −1, x = −4, x = −2 and x = −3\n2. x = 1, x = 4 and x = −2\n3. x = −7, x = 4, x = −1 and x = −2\n4. x = −4, x = 3, x = −3 and x = 2\n5. x = 8±\n√\n40\n4\n6. x = −5, x = 3, x = −1 +\n√\n10 and\nx = −1 −\n√\n10\nExercise 2 – 5: Finding the equation\n1. x2 −x −6 = 0\n2. x2 −16 = 0\n3. 2x2 −5x −3 = 0\n4. k = 3 and x = 3\n4\n5. p = 5 and x = −1\nExercise 2 – 6: Mixed exercises\n1. y = 1\n8 or y = −8\n3\n2. x = 3\n2 or x = −7\n2\n3. t = 2\n3 or t = 2\n4. y = 1 or y = −1\n5. m = 1 or m = 4\n6. y = ± 5\n7\n7. w = 3\n2 or w = 4\n8. y = 6\n5 or y = 1\n4\n9. n = 8\n3 or n = −9\n8\n10. y = −8\n3 or y = 3\n2\n11. x = −1\n2 or x = 3\n12. y = −5\n2 or y = −5\n9\n13. y = 4\n5 or y = 1\n5\n14. g = −1\n4 or g = 1\n15. y = 2 or y = −5\n9\n16. p = 3\n7 or p = −1\n5\n17. y = −2\n9 or y = −1\n18. y = 9\n2 or y = 9\n7\n486\n12.1.\nIntroduction\n\nExercise 2 – 7: From past papers\n1.\na) Real, unequal and rational\nb) Real and equal\nc) Real, unequal and irrational\nd) Real, unequal and rational\ne) Real, unequal and irrational\nf) Non-real\ng) Real, unequal and rational\nh) Real, unequal and irrational\ni) Non-real\nj) Real and equal\n2.\nb) real and unequal\nc) k = −6 ± 2\n√\n6\n4.\na) k = 6\nb) k = 1\n3\n5.\na) k = 4 or k = 1\nb) k = 0 or k = 5\n6.\na) all real values of a, b and p\nb) a = b and p = 0\nExercise 2 – 8: Solving quadratic inequalities\n1.\na) −3 < x < 4\nb) x < −4\n3 or when x > 1\nc) no real solutions\nd) −1 < t < 3\ne) All real values of s.\nf) All real values of x.\ng) x ≤−1\n4 or x ≥0\ni) x < 3 or x > 6 with x ̸= 3\nj) −2 ≤x ≤2 and x > 7 with x ̸= 7\nk) x > 0 with x ̸= 0\n2.\na) x < −3 or x > 3\nb) −\n√\n5 ≤x ≤\n√\n5\nc) no solution\nd) All real values of x\nExercise 2 – 9: Solving simultaneous equations\n1.\na) (0; 5) and (2; 3)\nb) x = 3 ±\n√\n2 and y = 2 ±\n√\n2\nc) (−1; 0) and ( 1\n4 ; 5\n8 )\nd) b = 2 ±\n√\n88\n6\nand a = 11 ±\n√\n88\n6\ne) (−3; −20) and (2; 0)\nf) x = 6 ±\n√\n264\n2\nand y = 70 ±\n√\n264\n2\n2.\na) (−3; 8) and (2; 3)\nb) (−4; 14) and (3; 7)\nc) (3; 4) and (4; 3)\nExercise 2 – 10:\n1. b = 2 m, l = 4 m\n2. 187\n3. t = 10,5 s\n4. t = 5d; 105 minutes; 1,4 km\n5. 24 A; 70 W; 12 A\n487\nChapter 12.\nLinear programming\n\nExercise 2 – 11: End of chapter exercises\n1. x = 1,62 or x = −0,62\n2. x = ±4 or x = −1\n3. y = 0 or y = ±1\n4. x = ±2\n5.\na) x = 7 or x = 2\nb) x = 2,3 or x = −1,3\nc) x = 1,65 or x = −3,65\nd) x = 0 or x = −3\n6. x =\n√\n16+p2−2\n2\n7. a = 3; b = 10 and c = −8\n8. p = ±16\n9. x2 + 2x −15\n10. Undefined:b = −2 Zero:b = 2 or b = 3\n13. a ≥4\n14. x = −3\n2 or x = 1\n15.\na) x < 3 or x ≥7:\nb) x < 1 or x > 5:\nc) 3 < x < 7:\nd) x < −1 or x > 3\ne) 0,5 < x < 2,5\nf) x ≤−3 or 0 < x ≤5\n2\ng) x < 2\n3\nh) −1 ≤x < 0 or x ≥3\ni) −4 ≤x ≤1\nj) 2 1\n2 ≤x < 3\n16.\na) x = ±\n√\n3 and y = ±2\n√\n3\nb) a = −3 and b = −1 or a = 12 and b = 4\nc) x = −5 and y = 0 or x = 2 and y = 14\nd) p = 5\n3 and q = 2\n9 or p = −1 and q = −2\n3\ne) b = 3±\n√\n5\n2\nand a = 7±3\n√\n5\n2\nf) b = −10±\n√\n140\n4\nand a = −12±\n√\n140\n2\ng) x = 3,4 and y = 5,4 or x = 3 and y = 5\nh) b = −1,4 and a = 23,6 or\nb = 3 and a = 6\n17.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\nx\n0\ny\nb\nb\nb)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n1\n2\n3\n4\n5\n6\n−1\n−2\nx\n0\ny\nb\nb\n18. 35 m\n20.\na) y = −5\n4 or y = −9\nb) x = −9\n4 or x = 1\nc) p = −8\n7 or p = −4\n3\nd) y = −1\n4 or y = 1\n2\ne) y = −2\n9 or y = −1\nf) y = 7\n3 or y = −1\n2\ng) y = 9\n4 or y = −9\n4\nh) y = 8\n3 or y = −6\ni) y = 9\n5 or y = −7\nj) x = ±4\nk) y = ±7\n21. k = 76 and 4\n9\n22. x = 3 or x = −2 and y = 1±√−7\n2\n23. x = 4 or x = −1\n24. y = 3\n2 , y = 1\n2 and p = 9\n2 , p = 7\n2\n25.\n69\n4\n26. 7\n27.\n2±\n√\n12\n2\n28. t = 1\n2 , t = 1 or t = 3±\n√\n33\n4\n488\n12.1.\nIntroduction\n\n3\nNumber patterns\nExercise 3 – 1: Linear sequences\n1. −19; −35; −51\n2.\na) −19\nb) T2 = 15; T4 = 33\n3.\na) Tn = 10 + 3n; T10 = 40; T15 = 55;\nT30 = 100\nb) Tn = 12 + 6n; T10 = 72; T15 = 102;\nT30 = 192\nc) Tn = −5 −5n; T10 = −55; T15 = −80;\nT30 = −155\n4. T9 = 36\n5.\na) 44; 66; 121\nExercise 3 – 2: Quadratic sequences\n1.\na) 10\nb) 2\nc) 2\nd) −2\ne) 2\nf) −4\ng) 4\nh) −2\ni) 6a\nj) 6\nk) 2t\n2.\na) T4 = 53\nb) T2 = 30\nc) T1 = 17\nd) T2 = −3\ne) T4 = 63\nf) T1 = 2\n3.\na) 3; 9; 17; 27\nb) −6; −9; −14; −21\nc) 1; 8; 21; 40\nd) 0; −5; −14; −27\nExercise 3 – 3: Quadratic sequences\n1.\na) 1\nb) 2\nc) 4\nd) 8\ne) −2\n2. 12; 30; 58; 96; 144\n3. T9 = 379\n4. n = 4\n5.\na) T5 = 84; T6 = 111\nb) Tn = 2n2 + 5n + 9\n489\nChapter 12.\nLinear programming\n\nExercise 3 – 4: End of chapter exercises\n1. −4; 9; 16; 25; 36\n2.\na) Quadratic sequence\nb) Quadratic sequence\nc) Quadratic sequence\nd) Quadratic sequence\ne) Quadratic sequence\nf) Quadratic sequence\ng) Linear sequence\nh) Linear sequence\ni) Quadratic sequence\nj) Quadratic sequence\nk) Quadratic sequence\nl) Linear sequence\nm) Quadratic sequence\n3. x = 31\n4. n = 11\n5. T11 = 363\n6. n = 9\n7. T5 = 114\n8. n = 8\n9.\na) T5 = 19;\nTn = 4n −1;\nT10 = 39\nb) T5 = −3;\nTn = 22 −5n;\nT10 = −28\nc) T5 = 2 1\n2 ; Tn = 1\n2 n;\nT10 = 5\nd) T5 = a + 4b;\nTn = a −b + bn;\nT10 = a + 9b\ne) T5 = −7;\nTn = 3 −2n;\nT10 = −17\n10.\na) Tn = n2 + 3;\nT100 = 10 003\nb) Tn = 6n −4;\nT100 = 596\nc) Tn = 2n2 + 5;\nT100 = 20 005\nd) Tn = 3n2 + 2;\nT100 = 30 002\n11.\na) 2; 5; 8; 11; 14\nb) Constant difference,\nd = 3\nc) Yes\n12.\na) Tn = 4n −19\nb) n = 48\n13.\na) Incorrect\nb) Correct\n14.\nc) Linear\n15.\nb) Linear\nd) Quadratic\ne) Tn = 1\n2 n2 + 3\n2 n + 1\nf) T21 = 253\ng) 31 cm\n16.\na) −1\nb) 7\n17.\nb) 2\nc) Tn = n2 −n\nd) 210\ne) 25\n18. 4; 14; 34; 64; 104; 154\n4\nAnalytical geometry\nExercise 4 – 1: Revision\n1.\na) 2\n√\n26units\nb) 7 units\nc) x + 1units\n2. p = 6 or p = 2\n3.\na) −1\n2\nb) 3\n5. 2\n6.\na) (1; 2)\nb)\n\u0000 −1\n2 ; −1\n2\n\u0001\n7. B(4; 2)\n8.\na) y = −4x + 3 and\ny = −4x + 19\nc) AD =\n√\n17units and\nBC =\n√\n17units\nd) y = 4\n3 x −7\n3\ne) Parallelogram (one\nopposite side equal\nand parallel)\n9. N(0; 3)\n10.\na) PQ =\n√\n20 and\nSR =\n√\n20\nb) M( 3\n2 ; 1)\nd) PS: y = −2\n5 x −1\n5\nand SR: y = 1\n2 x −2\ne) No\nf) Parallelogram\nExercise 4 – 2: The two-point form of the straight line equation\n1. y = 2\n3 x + 5\n2. y = −3x + 1\n4\n3. y = x + 3\n4. y = 2x −1\n5. y = −5\n6. y = 3\n4 x + 3\n7. y = −x + (s + t)\n8. y = 5x + 2\n9. y = q\npx −q\n490\n12.1.\nIntroduction\n\nExercise 4 – 3: Gradient–point form of a straight line equation\n1. y = 2\n3 x + 4\n2. y = −x −2\n3. y = −1\n3 x\n4. y = 11\n5. y = −2x + 7\n6. x = −3\n2\n7. y = −4\n5 x + 1\n8. x = 4\n9. y = 3ax + b\nExercise 4 – 4: The gradient–intercept form of a straight line equation\n1. y = 2x + 3\n2. y = 4x −4\n3. y = −x −1\n4. y = −3\n7 x\n5. y = 1\n2 x −1\n5\n6. y = 2x −2\n7. y = −3\n2\n8. y = 3x + 4\n9. y = −5x\nExercise 4 – 5: Angle of inclination\n1.\na) 1,7\nb) −1\nc) 0\nd) 1,4\ne) Undefined\nf) 1\ng) −0,8\nh) 0\ni) 3,7\n2.\na) 36,8◦\nb) 26,6◦\nc) 45◦\nd) Horizontal line\ne) 18,4◦\nf) Vertical line\ng) 71,6◦\nh) 30◦\nExercise 4 – 6: Inclination of a straight line\n1.\na) 38,7◦\nb) 135◦\nc) 80◦\nd) 80◦\ne) 102,5◦\nf) 45◦\ng) 56,3◦\nh) 63,4◦\ni) 161,6◦\nj) Gradient undefined\n2. 85,2◦\n3. 90◦\n4. 81,8◦\nExercise 4 – 7: Parallel lines\n1.\na) Parallel\nb) Parallel\nc) Parallel\nd) Not parallel\ne) Parallel\nf) Parallel\n2. y = −2x −3\n3. y = 3x\n4. y = 3\n2 x + 1\n5. y = −7\n10 x −1\n491\nChapter 12.\nLinear programming\n\nExercise 4 – 8: Perpendicular lines\n1.\na) Perpendicular\nb) Not perpendicular\nc) Perpendicular\nd) Perpendicular\ne) Perpendicular\nf) Not perpendicular\ng) Not perpendicular\n2. y = 1\n2 x −3\n3. y = −5x + 3\n4. y = −x + 2\n5. x = −2\nExercise 4 – 9: End of chapter exercises\n1.\na) y = 1\n2 x + 7\n2\nb) y = −x + 4\nc) y = 1\n2 x + 4\nd) y = 2x + 4\ne) y = 3x\n2.\na) θ = 63,4◦\nb) θ = 18,4◦\nc) θ = 36,9◦\nd) θ = 146,3◦\ne) θ = 161,6◦\n3.\na) y = −2x + 7\nb)\n\u0000 7\n2 ; 0\n\u0001\nc) θ = 116,6◦\nd) m = 1\n2\ne) Q ˆPR = 90◦\nf) y = −2x\ng)\n\u0000 1\n2 ; −3\n2\n\u0001\nh) y = −2x −1\n2\n4.\na)\n1\n2\n3\n4\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\nb\nb\nb\ny\nx\nA(−3; 5)\nB(−7; −4)\nC(2; 0)\nD(x; y)\nb) D (6; 9)\n5.\na) (−1; −2)\nb) (8; 3)\nc) x = −1\nd) MN = 5 units\ne) M ˆ\nNP = 21,8◦\nf) y = 5\n2 x + 11\n2\n6.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nA(−2; 3)\nB(2; 4)\nC(3; 0)\ny\nx\n0\nb\nb\nb\nc) y = 1\n4 x + 7\n2\nd) D(−1; −1)\ne) E\n\u0000 5\n2 ; 2\n\u0001\n7.\na) y = 3\n2 x + 2\nb) T ˆSV = 49,6◦\n8.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nb\nb\nb\nF(−1; 3)\nH(4; 4)\nG(2; 1)\ny\nx\n0\nc) y = −5x + 11\nd) Yes\ne) y = 3\n2 x + 9\n2\n9.\na)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nb\nb\nb\nA(−1; 5)\nB(5; −3)\nC(0; −6)\nx\ny\nM\nN\n492\n12.1.\nIntroduction\n\n5\nFunctions\nExercise 5 – 1: Revision\n1.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nc)\n1\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nd)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nExercise 5 – 2: Domain and range\n1. {x : x ∈R} ; {y : y ≥−1, y ∈R}\n2. {x : x ∈R} ; {y : y ≤4, y ∈R}\n3. {x : x ∈R} ; {y : y ≥0, y ∈R}\n4. {x : x ∈R} ; {y : y ≤0, y ∈R}\n5. {x : x ∈R} ; {y : y ≤2, y ∈R}\nExercise 5 – 3: Intercepts\n1. (0; 15) and (−5; 0); (−3; 0)\n2. (0; 16) and (4; 0)\n3. (0; −3) and (1; 0); (3; 0)\n4. (0; 35) and (−7\n2 ; 0); (−5\n2 ; 0)\n5. (0; 37) and no x-intercepts\n6. (0; −4) and\n(−0,85; 0); (−2,35; 0)\nExercise 5 – 4: Turning points\n1. (3; −1)\n2. (2; 1)\n3. (−2; −1)\n4. (−1\n2 ; 1\n2 )\n5. (1; 21)\n6. (−1; −6)\nExercise 5 – 5: Axis of symmetry\n1.\na) Axis of symmetry:\nx = 5\n4\nb) Axis of symmetry:\nx = 2\nc) Axis of symmetry:\nx = 2\n2. y = ax2 + q\n493\nChapter 12.\nLinear programming\n\nExercise 5 – 6: Sketching parabolas\n1.\na)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n1\n2\n3\n4\n5\n6\n−1\ny\nx\n0\nIntercepts: (−1; 0), (5; 0), (0; 5)\nTurning point: (2; 9)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≤9, y ∈R}\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−1; 0), (0; 2)\nTurning point: (−1; 0)\nAxes of symmetry: x = −1\nDomain: {x : x ∈R}\nRange: {y : y ≥0, y ∈R}\nc)\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−0,87; 0), (1,54; 0), (0; −4)\nTurning point: (0,33; −4,33)\nAxes of symmetry: x = −0,33\nDomain: {x : x ∈R}\nRange: {y : y ≥4,33, y ∈R}\nd)\n1\n2\n3\n4\n5\n6\n−1\n1\n2\n3\n4\n−1\ny\nx\n0\nIntercepts: (0; 13) Turning point: (2; 1)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≥1, y ∈R}\n3.\na)\ny\nx\n0\nb)\ny\nx\n0\nc)\ny\nx\n0\nd)\ny\nx\n0\ne)\ny\nx\n0\nf)\ny\nx\n0\n4.\na) yshifted = 2x2 + 16x + 32\nb) yshifted = −x2 −2x\nc) yshifted = 3x2 −16x + 22\n494\n12.1.\nIntroduction\n\nExercise 5 – 7: Finding the equation\n1. y = −3(x + 1)2 + 6 or y = −3x2 −6x + 3\n2. y = 1\n2 x2 −5\n2 x\n3. y = 2\n3 (x + 2)2\n4. y = −x2 + 3x + 4\nExercise 5 – 8:\n1.\na) 11\n2.\na)\n1\n2\n3\n4\n1\n2\n−1\n−2\nf(x)\nx\n0\nA(1; 3)\nb\nb) 6\nc) y = 6x −3\n3.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n−1\n−2\ng(x)\nx\n0\nb) 1\nc) 4\nd) 0\nExercise 5 – 10: Domain and range\n1. {x : x ∈R, x ̸= 0} ; {y : y ∈R, y ̸= 1}\n2. {x : x ∈R, x ̸= 8} ; {y : y ∈R, y ̸= 4}\n3. {x : x ∈R, x ̸= −1} ; {y : y ∈R, y ̸= −3}\n4. {x : x ∈R, x ̸= 5} ; {y : y ∈R, y ̸= 3}\n5. {x : x ∈R, x ̸= −2} ; {y : y ∈R, y ̸= 2}\nExercise 5 – 11: Intercepts\n1. (0; −1 3\n4 ) and\n\u0000−3 1\n2 ; 0\n\u0001\n2.\n\u0000 5\n2 ; 0\n\u0001\n3. (0; 1) and\n\u0000 1\n3 ; 0\n\u0001\n4.\n\u00000; 3\n2\n\u0001\nand\n\u0000 1\n3 ; 0\n\u0001\n5. (0; 2) and (8; 0)\nExercise 5 – 12: Asymptotes\n1. y = −2 and x = −4\n2. y = 0 and x = 0\n3. y = 1 and x = 2\n4. y = −8 and x = 0\n5. y = 0 and x = 2\n495\nChapter 12.\nLinear programming\n\nExercise 5 – 13: Axes of symmetry\n1.\na) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (0; 1); y1 = x + 1 and\ny2 = −x + 1\nb) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (−1; 0); y1 = x + 1 and\ny2 = −x −1\nc) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (1; −1); y1 = x −2 and\ny2 = −x\n2. k(x) =\n5\nx+1 + 2\nExercise 5 – 14: Sketching graphs\n1.\na) Asymptotes: x = 0; y = 2\nIntercepts:\n\u0000−1\n2 ; 0\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= 0}\nRange: {y : y ∈R, y ̸= 2}\nb) Asymptotes: x = −4; y = −2\nIntercepts:\n\u0000−3 1\n2 ; 0\n\u0001\nand\n\u00000; −1 3\n4\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x −6\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\nc) Asymptotes: x = −1; y = 3\nIntercepts:\n\u0000−2\n3 ; 0\n\u0001\nand (0; 2)\nAxes of symmetry: y = x + 4 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 3}\nd) Asymptotes: x = −2 1\n2 ; y = −2\nIntercepts: (0; 0)\nAxes of symmetry: y = x −4 1\n2 and\ny = −x + 1\n2\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\ne) Asymptotes: x = 8; y = 4\nIntercepts: (6; 0) and (0; 3)\nAxes of symmetry: y = x −4 and\ny = −x + 12\nDomain: {x : x ∈R, x ̸= 8}\nRange: {y : y ∈R, y ̸= 4}\n2. y =\n1\nx+2 −1\n3. y = −4\nx + 2\n4.\na)\nb) Average gradient = 1\nc) Average gradient = 12\nExercise 5 – 16: Domain and range\n1. {x : x ∈R} ; {y : y > 0, y ∈R}\n2. {x : x ∈R} ; {y : y < 1, y ∈R}\n3. {x : x ∈R} ; {y : y > −3, y ∈R}\n4. {x : x ∈R} ; {y : y > n, y ∈R}\n5. {x : x ∈R} ; {y : y > 2, y ∈R}\nExercise 5 – 17: Intercepts\n1. (0; −6) and (2; 0)\n2. (0; −17 1\n3 ) and (3; 0)\n3. (0; −20) and (−1; 0)\n4. (0; 15\n16 ) and (−2; 0)\n496\n12.1.\nIntroduction\n\nExercise 5 – 18: Asymptote\n1. y = 0\n2. y = 1\n3. y = −2\n3\n4. y = −2\n5. y = −2\nExercise 5 – 19: Mixed exercises\n1.\nb)\ni. y = 3\nx + 3\nii. y =\n3\nx−3\niii. y = −3\nx\niv. y = 3\nx −1\n4\nv. y = 3\nx + 4\nvi. y =\n3\nx+2 −1\n2.\na) M(−2; 2)\nb) g(x) = −4\nx\nc) f(x) = 2(x + 1)2\nd) −2 < x < 0\ne) Range: {y : y ∈R, y ≥0}\n3.\na) For k(x) :\nIntercepts:\n(−2; 0), (1; 0) and (0; −4)\nTurning point:\n\u0000−1\n2 ; −4 1\n2\n\u0001\nAsymptote:\nnone\nFor h(x) :\nIntercepts:\n(1,41; 0)\nTurning point:\nnone\nAsymptote:\ny = 0\n6.\na) f(x) = −3\n4 (x −2)2 + 3 ;\nAxes of symmetry: x = 2 ;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≤3}\nb) g(x) = 1\n4 x2 −2;\nAxes of symmetry: x = 0;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≥−2}; h(x) = 2\nx ;\nAxes of symmetry: y = x\nDomain: {x : x ∈R, x < 0};\nRange: {y : y ∈R, y < 0};\nc) k(x) =\n\u0000 1\n2\n\u0001x + 1\n2 ;\nDomain: {x : x ∈R};\nRange:\n\b\ny : y ∈R, y > 1\n2\n\t\n7.\nb) p = 9\nc) Average gradient = −2 8\n9\nd) y =\n\u0000 1\n3\n\u0001x+2 −2\n8.\na) f(x) = 2x −3\n2 and g(x) = −1\n4 x −1\n2\nb) h(x) = −\n3\nx+2 + 1\n9.\na) AO = 2 units OB = 5 units\nOC = 10 units DE = 12,25 units\nb) DE = 12 1\n4\nc) h(x) = −2x + 10\nd) {x : x ∈R, x < −2 and x > 5}\ne) {x : x ∈R, 0 ≤x ≤5}\nf) 5,25 units\n497\nChapter 12.\nLinear programming\n\nExercise 5 – 20: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\nPeriod: = 360◦\nAmplitude: = 1\nDomain: = [0◦; 360◦]\nRange: = [−1; 1]\nx-intercepts: = (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: = (0◦; 0)\nMax. turning point: = (90◦; 1)\nMin. turning point: = (270◦; −1)\n2.\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny2 = −2 sin θ\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 2\nDomain: [0◦; 360◦]\nRange: [−2; 2]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nMax. turning point: (270◦; 2)\nMin. turning point: (90◦; −2)\n3.\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny3 = sin θ + 1\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [0; 2]\nx-intercepts: (270◦; 0)\ny-intercepts: (0◦; 1)\nMax. turning point: (90◦; 2)\nMin. turning point: (270◦; 0)\n4.\n1\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny4 = 1\n2 sin θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (90◦; 1\n2 )\nMin. turning point: (270◦; −3\n2 )\nExercise 5 – 21: Sine functions of the form y = sin kθ\n2.\na) k = 2\nb) k = −3\n4\n498\n12.1.\nIntroduction\n\nExercise 5 – 23: The sine function\n1.\na)\n1\n2\n−1\n−2\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 2 sin( θ\n2 )\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 1\n2 sin(θ −45◦)\nc)\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(θ + 90◦) + 1\nd)\n1\n−1\n60◦\n120◦\n180◦\n−60◦\n−120◦\n−180◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(−3θ\n2 )\ne)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(30◦−θ)\n2. a = 2; p = 90◦∴y = 2 sin(θ + 90◦) and\ny = 2 cos θ\nExercise 5 – 24: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\n2.\n1\n2\n3\n−1\n−2\n−3\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny2 = −3 cos θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−3; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; −3)\n3.\n1\n2\n3\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny3 = cos θ + 2\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [1; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; 1)\n4.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny4 = 1\n2 cos θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (0◦; −1\n2 ); (360◦; −1\n2 )\nMin. turning point: (180◦; −3\n2 )\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n2.\na) k = 3\n2\nb) k = 2\n3\n499\nChapter 12.\nLinear programming\n\nExercise 5 – 27: The cosine function\n1.\na)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nθ\n0◦\ny\ny = cos θ\ny = cos(θ + 15◦)\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny = cos θ\nf(θ) = 1\n3 cos(θ −60◦)\nc)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ny = −2 cos θ\nd)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(30◦−θ)\ne)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ng(θ) = 1 + cos(θ −90◦)\nf)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(2θ + 60◦)\n2.\na) a = −1\nb) p = −180◦\nc) cos(θ −180◦) = −cos θ\nExercise 5 – 28: Revision\n1.\n1\n2\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1\n2 )\nAsymptotes: 90◦; 270◦\n2.\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: 90◦; 270◦\n3.\n1\n2\n3\n4\n5\n−1\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (116,6◦; 0); (296,6◦; 0)\ny-intercepts: (0◦; 2)\nAsymptotes: 90◦; 270◦\n4.\n1\n2\n−1\n−2\n−3\n−4\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1)\nAsymptotes: 90◦; 270◦\n500\n12.1.\nIntroduction\n\nExercise 5 – 29: Tangent functions of the form y = tan kθ\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦] Range: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 240◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −120◦; 120◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n4.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 270◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; 135◦\n501\nChapter 12.\nLinear programming\n\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−225◦; 0); (−45◦; 0); (135◦; 0);\n(315◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−330◦; 0); (−150◦; 0); (30◦; 0);\n(210◦; 0)\ny-intercepts: (0◦; −0,58)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−240◦; 0); (−60◦; 0); (120◦; 0);\n(300◦; 0)\ny-intercepts: (0◦; 1.73)\nAsymptotes: −330◦; −150◦; 30◦; 210◦\nExercise 5 – 31: The tangent function\n1.\na)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n−45◦\n−90◦\nθ\ny\nb)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\nθ\ny\nc)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nd)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\ny\n2. a = −1; k = 1\n2\n502\n12.1.\nIntroduction\n\nExercise 5 – 32: Mixed exercises\n1.\na) f(θ) = 3\n2 sin 2θ and g(θ) = −3\n2 tan θ\nb) f(θ) = −2 sin θ and\ng(θ) = 2 cos(θ + 360◦\nc) y = 3 tan θ\n2\nd) y = y = 2 cos θ + 2\n2.\na)\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\nθ\n0\ng\nf\ny\nb\nb\nb\nb\n(90◦; 2)\n(270◦; −2)\nb) 360◦\nc) 1\nd) At θ = 180◦\n3.\na) a = 2, b = −1 and c = 240◦\nb) 180◦\nc) θ = 60◦; 300◦\nd) y = −tan(θ −45◦)\n4.\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny1\ny2\ny\nb\nb\nExercise 5 – 33: End of chapter exercises\n2. a = −2; k = −1\n4.\na) y = x + 22 + 2\nb) y = x −12 + 5\n5. (−1; 0)\n6.\ny\nx\n0\n4\n−4\n4\n−4\ny =\n2\nx−3 −1\n7. y =\n1\n(x−1) + 2\n9.\na) a = −1\nb) f(−15) = 0,99997\nc) x = −1\nd) h(x) = −2(x−2) + 1\n10.\na) a = 256\nb) f(x) = 256\n\u0000 3\n4\n\u0001x\nc) f(13) = 6,08\n11.\na)\n1\n−1\n90◦\n180◦\n−90◦\n−180◦\nθ\n0\ny\nb)\n1\n−1\n90◦\n180◦\nθ\n0\ny\nd)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\n0\ny\ne)\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny\nf)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\n503\nChapter 12.\nLinear programming\n\n6\nTrigonometry\nExercise 6 – 1: Revision\n1.\na) True\nb) True\nc) False\nd) True\n2.\na) 50,2◦\nb) 40,5◦\nc) 26,6◦\nd) 109,8◦\ne) No solution\nf) 17,7◦\ng) 69,4◦\n3.\na) 17,3 cm\nb) 10 cm\nc) 64,8◦\n4.\na) 10 cm\nb) 5,2 cm and 19,3 cm\nc) 50 cm2\n5.\na) 2\nb) 0\nc) −1 1\n2\nd) 1\ne) 1\n6.\na) 60◦\nb)\n1\n2\nc) 1\n7. No\nExercise 6 – 2: Trigonometric identities\n1.\na) cos α\nb) tan2 θ\nc) cos2 θ\nd) 0\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1.\na)\n√\n3\n3\nb)\n1\n8\nc) 1\n2.\na)\n1−cos2 θ\ncos θ\nb) −1\n3.\na) 2t\nb) −1\nt\nExercise 6 – 4: Using reduction formula\n1.\na) −tan θ\nb) 1\nc) 1\n2. −cos β\n3.\na)\n1\n√\n3\nb) 2\nc) 2\nd) −3\n2\ne) −4\n√\n3\n5\n5.\na) −t\nb) 1 −t2\nc) ±\nt\n√\n1−t2\nExercise 6 – 5: Co-functions\n1.\na) cos θ\nb)\n3\n2\n2.\na) p\nb)\np\n1 −p2\nc) −\np\n√\n1−p2\nd) p\n504\n12.1.\nIntroduction\n\nExercise 6 – 6: Reduction formulae\n1.\na) sin2 θ\nb) cos2 θ\nc)\ni. 1\nii. tan2 θ\n2.\na) sin 17◦\nb) cos 33◦\nc) tan 68◦\nd) −cos 33◦\n3.\na)\n√\n3\nb)\n√\n3\n2\nc)\n1\n4\nd) 1\nExercise 6 – 7: Solving trigonometric equations\n1.\na) α = 60◦; 300◦\nb) α = 220,5◦; 319,5◦\nc) α = 79,2◦; 259,2◦\nd) α = 200,1◦; 339,9◦\ne) α = 36,9◦; 143,1◦\nf) α = 109,7◦; 289,7◦\n2.\na) θ = −323,1◦; −216,9◦; 36,9◦; 143,1◦\nb) θ = −221,4◦; −138,6◦; 138,6◦; 221,4◦\nc) θ = −278,5◦; −98,5◦; 81,5◦; 261,5◦\nd) θ = −90◦; 270◦\ne) θ = −293,6◦; −66,4◦; 66,4◦; 293,6◦\nExercise 6 – 8: General solution\n1.\na) θ = −128,36◦; −101,64◦; 51,64◦\nb) θ = −80,45◦; −9,54◦; 99,55◦; 170,46◦\nc) θ = −53,27◦; 126,73◦\nd) α = 0◦\ne) θ = −180◦; 0◦; 180◦\nf) θ = −180◦; 180◦\ng) θ = 84◦\nh) θ = −120◦; 120◦\ni) θ = −60◦; −30◦; 120◦; 150◦\n2.\na) θ = −20◦+ n . 360◦\nb) α = 30◦+ n . 120◦\nc) β = 10,25◦+ n . 45◦or\nβ = 55,25◦+ n . 45◦\nd) α = 70◦+ n . 360◦or\nα = 340◦+ n . 360◦\ne) θ = 140◦+ n . 240◦or\nθ = 220◦+ n . 240◦\nf) β = 15◦+ n . 180◦\nExercise 6 – 9: Solving trigonometric equations\n1.\na) θ = 45◦+ k . 180◦or\nθ = 135◦+ k . 180◦\nb) α = 50◦+ k . 360◦or\nα = 110◦+ k . 360◦\nc) θ = 60◦+ k . 720◦or\nθ = 660◦+ k . 720◦\nd) β = 146,6◦+ k . 180◦\ne) θ = 110,27◦+ k . 360◦or\nθ = 249,73◦+ k . 360◦\nf) α = 210◦+ k . 360◦or\nα = 330◦+ k . 360◦\ng) β = 23,3◦+ k . 120◦\nh) θ = 122◦+ k . 180◦\ni) α = 21◦+ k . 180◦or\nα = 39,5◦+ k . 90◦\nj) β = 22,5◦+ k . 90◦\n2. θ = 0◦, 180◦, 210◦, 330◦or 360◦\n3.\na) θ = 120◦+ k . 360◦or\nθ = 240◦+ k . 360◦\nb) θ = 0◦+ k . 180◦or\nθ = 146,3◦+ k . 180◦\nc) α = 36,9◦+ k . 360◦or\nα = 143,1◦+ k . 360◦or\nα = 216,9◦+ k . 360◦or\nα = 323,1◦+ k . 360◦\nd) β = 15◦+ k . 120◦or\nβ = 75◦+ k . 120◦\ne) α = 48,4◦+ k . 180◦\nf) θ = 63,4◦+ k . 180◦or\nθ = 116,6◦+ k . 180◦\ng) θ = 54,8◦+ k . 180◦or\nθ = 95,25◦+ k . 180◦\n4. β = −70,5◦or β = 109,5◦\n505\nChapter 12.\nLinear programming\n\nExercise 6 – 10: The area rule\n1.\na)\nP\nQ\nR\n30◦\n10\n7\nArea △PQR = 17,5 square units\nb)\nP\nQ\nR\n110◦\n9\n8\nArea △PQR = 33,8 square units\n2. Area △XY Z = 645,6 square units\n3. Area = 106,5 square units\n4.\nˆC = 72,2◦or ˆC = 107,8◦\nExercise 6 – 11: Sine rule\n1.\na)\nˆP = 92◦, q = 6,6, p = 7,4\nb)\nˆL = 87◦, l = 1,3, k = 0,89\nc)\nˆB = 76,8◦, b = 94,3, c = 91,3\nd)\nˆY = 84◦, y = 60, z = 38,8\n2.\nˆB = 32◦, AB = 23, BC = 39\n3. ST = 78,1 km\n4. m = 26,2\n5. BC = 3,2\nExercise 6 – 12: The cosine rule\n1.\na) a = 8,5, ˆC = 83,9◦, ˆB = 26,1◦\nb)\nˆR = 120◦, ˆS = 32,2◦, ˆT = 27,8◦\nc)\nˆ\nM = 27,7◦, ˆL = 40,5◦, ˆ\nK = 111,8◦\nd) h = 19,1, ˆJ = 18,2◦, ˆ\nK = 31,8◦\ne)\nˆD = 34◦, ˆE = 44,4◦, ˆF = 101,6◦\n2.\na) x = 4,4 km\nb) y = 63,5 cm\n3.\na)\nˆ\nK = 117,3◦\nb)\nˆQ = 78,5◦\nExercise 6 – 13: Area, sine and cosine rule\n1.\na) 7,78 km\nb) 6 km\n2. XZ = 1,73 km, XY = 0,87 km\n3.\na) 1053 km\nb) 4,42◦\n4. DC = x sin a sin(b+c)\nsin(a+c) sin b\n5.\nb) 438,5 km\n6. 9,38 m2\n7. DC = x sin α\nsin β\n506\n12.1.\nIntroduction\n\nExercise 6 – 14: End of chapter exercises\n1. sin2 A\n2. 1 1\n4\n3. cos α\n4. 3\n7.\na) −1\nb) θ = 135◦or θ = 315◦\n8.\na)\nb\nx\ny\n0\nθ\n(−12; −5)\nb) −5\n13 and 12\n13\nc) θ = 202,62◦\n9.\na) a = 1 and b = −\n√\n3\nb) −\n√\n3\n2\n10.\na) x = 50,9◦or x = 309,1◦\nb) x = 127,3◦or x = 307,3◦\nc) x = 26,6◦; 153,4◦206,6◦or 333,4◦\n11.\na) x = 55◦+ k . 360◦or\nx = 175◦+ k . 360◦\nb) x = 180◦+ k . 360◦\n12.\na) x = 28,6◦+ k . 180◦or\nx = 61,4◦+ k . 180◦\nb)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nb\nb\nb\nb\nθ\n0◦\ny\ny = sin 2α\nc) 28,6◦; 61,4◦; 208,6◦; 241,4◦\n13.\na) A ˆGN = α −β\nb)\nˆ\nA = 90◦−α\nd) H = 5 m\n14.\na) AC = 9,43 m\nb) AD = 6,2 m\nc) Area = 49,25 m2\nd) Area = 49,23 m2\n7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "2.1" }, { "title": "Completing the square", "content": "", "chapter_id": "2.2" }, { "title": "Quadratic formula", "content": "", "chapter_id": "2.3" }, { "title": "Substitution", "content": "", "chapter_id": "2.4" }, { "title": "Finding the equation", "content": "", "chapter_id": "2.5" }, { "title": "Nature of roots", "content": "", "chapter_id": "2.6" }, { "title": "Quadratic inequalities", "content": "", "chapter_id": "2.7" }, { "title": "Simultaneous equations", "content": "", "chapter_id": "2.8" }, { "title": "Word problems", "content": "", "chapter_id": "2.9" }, { "title": "Summary", "content": "3.3\nSummary\n99\n\n3\nNumber patterns\nIn earlier grades we learned about linear sequences, where the difference between\nconsecutive terms is constant. In this chapter, we will learn about quadratic sequences,\nwhere the difference between consecutive terms is not constant, but follows its own\npattern.\n3.1\nRevision\nEMBG2\nTerminology:\nSequence/pattern\nA sequence or pattern is an ordered set of numbers\nor variables.\nSuccessive/consecutive\nSuccessive or consecutive terms are terms that di-\nrectly follow one after another in a sequence.\nCommon difference\nThe common or constant difference (d) is the differ-\nence between any two consecutive terms in a linear\nsequence.\nGeneral term\nA mathematical expression that describes the se-\nquence and that generates any term in the pattern\nby substituting different values for n.\nConjecture\nA statement, consistent with known data, that has\nnot been proved true nor shown to be false.\nImportant: a series is not the same as a sequence or pattern. Different types of series\nare studied in Grade 12. In Grade 11 we study sequences only.\nSee video: 22FJ at www.everythingmaths.co.za\nDescribing patterns\nEMBG3\nTo describe terms in a pattern we use the following notation:\n• T1 is the first term of a sequence.\n• T4 is the fourth term of a sequence.\n• Tn is the general term and is often expressed as the nth term of a sequence.\nA sequence does not have to follow a pattern but when it does, we can write an\nequation for the general term. The general term can be used to calculate any term in\nthe sequence. For example, consider the following linear sequence: 1; 4; 7; 10; 13; . . .\nThe nth term is given by the equation Tn = 3n −2.\nYou can check this by substituting values for n:\nT1 = 3(1) −2 = 1\nT2 = 3(2) −2 = 4\nT3 = 3(3) −2 = 7\nT4 = 3(4) −2 = 10\nT5 = 3(5) −2 = 13\n86\n3.1.\nRevision\n\nIf we find the relationship between the position of a term and its value, we can describe\nthe pattern and find any term in the sequence.\nSee video: 22FK at www.everythingmaths.co.za\nLinear sequences\nEMBG4\nDEFINITION: Linear sequence\nA sequence of numbers in which there is a common difference (d) between any term\nand the term before it is called a linear sequence.\nImportant: d = T2 −T1, not T1 −T2.\nWorked example 1: Linear sequence\nQUESTION\nDetermine the common difference (d) and the general term for the following sequence:\n10; 7; 4; 1; . . .\nSOLUTION\nStep 1: Determine the common difference\nTo calculate the common difference, we find the difference between any term and the\nprevious term:\nd = Tn −Tn−1\nTherefore d = T2 −T1\n= 7 −10\n= −3\nor d = T3 −T2\n= 4 −7\n= −3\nor d = T4 −T3\n= 1 −4\n= −3\n10\n7\n4\n1\n−3\n−3\n−3\nStep 2: Determine the general term\nTo find the general term Tn, we must identify the relationship between:\n87", "chapter_id": "2.10" }, { "title": "Number patterns", "content": "Chapter 3.\nNumber patterns\n\n• the value of a number in the pattern and\n• the position of a number in the pattern\nposition\n1\n2\n3\n4\nvalue\n10\n7\n4\n1\nWe start with the value of the first term in the sequence. We need to write an expres-\nsion that includes the value of the common difference (d = −3) and the position of\nthe term (n = 1).\nT1 = 10\n= 10 + (0)(−3)\n= 10 + (1 −1)(−3)\nNow we write a similar expression for the second term.\nT2 = 7\n= 10 + (1)(−3)\n= 10 + (2 −1)(−3)\nWe notice a pattern forming that links the position of a number in the sequence to its\nvalue.\nTn = 10 + (n −1)(−3)\n= 10 −3n + 3\n= −3n + 13\nStep 3: Drawing a graph of the pattern\nWe can also represent this pattern graphically, as shown below.\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n0\n1\n2\n3\n4\n5\n6\nb\nb\nb\nb\nPattern number (n)\nTerm value Tn\nNotice that the position numbers (n) can be positive integers only.\nThis pattern can also be expressed in words: “each term in the sequence can be cal-\nculated by multiplying negative three and the position number, and then adding thir-\nteen.”\n88\n3.1.\nRevision\n\nSee video: 22FM at www.everythingmaths.co.za\nExercise 3 – 1: Linear sequences\n1. Write down the next three terms in each of the following sequences:\n45; 29; 13; −3; . . .\n2. The general term is given for each sequence below. Calculate the missing terms.\na) −4; −9; −14; . . . ; −24\nTn = 1 −5n\nb) 6; . . . ; 24; . . . ; 42\nTn = 9n −3\n3. Find the general formula for the following sequences and then find T10, T15 and\nT30:\na) 13; 16; 19; 22; . . .\nb) 18; 24; 30; 36; . . .\nc) −10; −15; −20; −25; . . .\n4. The seating in a classroom is arranged so that the first row has 20 desks, the\nsecond row has 22 desks, the third row has 24 desks and so on. Calculate how\nmany desks are in the ninth row.\n5.\na) Complete the following:\n13 + 31 = . . .\n24 + 42 = . . .\n38 + 83 = . . .\nb) Look at the numbers on the left-hand side, what do you notice about the\nunit digit and the tens-digit?\nc) Investigate the pattern by trying other examples of 2-digit numbers.\nd) Make a conjecture about the pattern that you notice.\ne) Prove this conjecture.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22FN\n2a. 22FP\n2b. 22FQ\n3a. 22FR\n3b. 22FS\n3c. 22FT\n4. 22FV\n5. 22FW\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n89\nChapter 3.\nNumber patterns\n\n3.2\nQuadratic sequences\nEMBG5\nInvestigation: Quadratic sequences\nb\nb\nb\nb\nb\nb\nb\nb\nb\n1. Study the dotted-tile pattern shown and answer the following questions.\na) Complete the fourth pattern in the diagram.\nb) Complete the table below:\npattern number\n1\n2\n3\n4\n5\n20\nn\ndotted tiles\n1\n3\n5\ndifference (d)\n−\n2\nc) What do you notice about the change in number of dotted tiles?\nd) Describe the pattern in words: “The number of dotted tiles...”.\ne) Write the general term: Tn = . . .\nf) Give the mathematical name for this kind of pattern.\ng) A pattern has 819 dotted tiles. Determine the value of n.\n2. Now study the number of blank tiles (tiles without dots) and answer the following\nquestions:\na) Complete the table below:\npattern number\n1\n2\n3\n4\n5\n10\nblank tiles\n3\n6\n11\nfirst difference\n−\n3\nsecond difference\n−\n−\nb) What do you notice about the change in the number of blank tiles?\nc) Describe the pattern in words: “The number of blank tiles...”.\nd) Write the general term: Tn = . . .\ne) Give the mathematical name for this kind of pattern.\nf) A pattern has 227 blank tiles. Determine the value of n.\ng) A pattern has 79 dotted tiles. Determine the number of blank tiles.\n90\n3.2.\nQuadratic sequences\n\nDEFINITION: Quadratic sequence\nA quadratic sequence is a sequence of numbers in which the second difference be-\ntween any two consecutive terms is constant.\nConsider the following example: 1; 2; 4; 7; 11; . . .\nThe first difference is calculated by finding the difference between consecutive terms:\n1\n2\n4\n7\n11\n+1\n+2\n+3\n+4\nThe second difference is obtained by taking the difference between consecutive first\ndifferences:\n1\n2\n3\n4\n+1\n+1\n+1\nWe notice that the second differences are all equal to 1. Any sequence that has a\ncommon second difference is a quadratic sequence.\nIt is important to note that the first differences of a quadratic sequence form a sequence.\nThis sequence has a constant difference between consecutive terms. In other words, a\nlinear sequence results from taking the first differences of a quadratic sequence.\nGeneral case\nIf the sequence is quadratic, the nth term is of the form Tn = an2 + bn + c.\nn = 1\nn = 2\nn = 3\nn = 4\nTn\na + b + c\n4a + 2b + c\n9a + 3b + c\n16a + 4b + c\n1st difference\n3a + b\n5a + b\n7a + b\n2nd difference\n2a\n2a\nIn each case, the common second difference is a 2a.\nExercise 3 – 2: Quadratic sequences\n1. Determine the second difference between the terms for the following se-\nquences:\na) 5; 20; 45; 80; . . .\nb) 6; 11; 18; 27; . . .\nc) 1; 4; 9; 16; . . .\nd) 3; 0; −5; −12; . . .\ne) 1; 3; 7; 13; . . .\nf) 0; −6; −16; −30; . . .\ng) −1; 2; 9; 20; . . .\nh) 1; −3; −9; −17; . . .\ni) 3a+1; 12a+1; 27a+1; 48a+1 . . .\nj) 2; 10; 24; 44; . . .\nk) t −2; 4t −1; 9t; 16t + 1; . . .\n91\nChapter 3.\nNumber patterns\n\n2. Complete the sequence by filling in the missing term:\na) 11; 21; 35; . . . ; 75\nb) 20; . . . ; 42; 56; 72\nc) . . . ; 37; 65; 101\nd) 3; . . . ; −13; −27; −45\ne) 24; 35; 48; . . . ; 80\nf) . . . ; 11; 26; 47\n3. Use the general term to generate the first four terms in each sequence:\na) Tn = n2 + 3n −1\nb) Tn = −n2 −5\nc) Tn = 3n2 −2n\nd) Tn = −2n2 + n + 1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22FX\n1b. 22FY\n1c. 22FZ\n1d. 22G2\n1e. 22G3\n1f. 22G4\n1g. 22G5\n1h. 22G6\n1i. 22G7\n1j. 22G8\n1k. 22G9\n2a. 22GB\n2b. 22GC\n2c. 22GD\n2d. 22GF\n2e. 22GG\n2f. 22GH\n3a. 22GJ\n3b. 22GK\n3c. 22GM\n3d. 22GN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nWorked example 2: Quadratic sequences\nQUESTION\nWrite down the next two terms and determine an equation for the nth term of the\nsequence 5; 12; 23; 38; . . .\nSOLUTION\nStep 1: Find the first differences between the terms\n5\n12\n23\n38\n+7\n+11\n+15\nStep 2: Find the second differences between the terms\n7\n11\n15\n+4\n+4\nSo there is a common second difference of 4. We can therefore conclude that this is a\nquadratic sequence of the form Tn = an2 + bn + c.\nContinuing the sequence, the next first differences will be:\n...15\n19\n23...\n+4\n+4\n92\n3.2.\nQuadratic sequences\n\nStep 3: Finding the next two terms in the sequence\nThe next two terms will be:\n...38\n57\n80...\n+19\n+23\nSo the sequence will be: 5; 12; 23; 38; 57; 80; . . .\nStep 4: Determine the general term for the sequence\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve a set of simultaneous equations to determine the values of a, b and c\nWe know that T1 = 5, T2 = 12 and T3 = 23\na + b + c = 5\n4a + 2b + c = 12\n9a + 3b + c = 23\nT2 −T1 = 4a + 2b + c −(a + b + c)\n12 −5 = 4a + 2b + c −a −b −c\n7 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n23 −12 = 9a + 3b + c −4a −2b −c\n11 = 5a + b\n. . . (2)\n(2) −(1) = 5a + b −(3a + b)\n11 −7 = 5a + b −3a −b\n4 = 2a\n∴a = 2\nUsing equation (1) :\n3(2) + b = 7\n∴b = 1\nAnd using\na + b + c = 5\n2 + 1 + c = 5\n∴c = 1\nStep 5: Write the general term for the sequence\nTn = 2n2 + n + 2\n93\nChapter 3.\nNumber patterns\n\nWorked example 3: Plotting a graph of terms in a sequence\nQUESTION\nConsider the following sequence:\n3; 6; 10; 15; 21; . . .\n1. Determine the general term (Tn) for the sequence.\n2. Is this a linear or a quadratic sequence?\n3. Plot a graph of Tn vs n.\nSOLUTION\nStep 1: Determine the first and second differences\nn = 1\nn = 2\nn = 3\nn = 4\nTn\n3\n6\n10\n15\n1st difference\n3\n4\n5\n2nd difference\n1\n1\nWe see that the first differences are not constant and form the sequence 3; 4; 5; . . . and\nthat there is a common second difference of 1. Therefore the sequence is quadratic\nand has a general term of the form Tn = an2 + bn + c.\nStep 2: Determine the general term Tn\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve this set of simultaneous equations to determine the values of a, b and c. We\nknow that T1 = 3, T2 = 6 and T3 = 10.\na + b + c = 3\n4a + 2b + c = 6\n9a + 3b + c = 10\nT2 −T1 = 4a + 2b + c −(a + b + c)\n6 −3 = 4a + 2b + c −a −b −c\n3 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n10 −6 = 9a + 3b + c −4a −2b −c\n4 = 5a + b\n. . . (2)\n94\n3.2.\nQuadratic sequences\n\n(2) −(1) = 5a + b −(3a + b)\n4 −3 = 5a + b −3a −b\n1 = 2a\n∴a = 1\n2\nUsing equation (1) :\n3\n\u00121\n2\n\u0013\n+ b = 3\n∴b = 3\n2\nAnd using\na + b + c = 3\n1\n2 + 3\n2 + c = 3\n∴c = 1\nTherefore the general term for the sequence is Tn = 1\n2n2 + 3\n2n + 1.\nStep 3: Plot a graph of Tn vs n\nUse the general term for the sequence, Tn = 1\n2n2 + 3\n2n + 1, to complete the table.\nn\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nTn\n3\n6\n10\n15\n21\n28\n36\n45\n55\n66\nUse the table to plot the graph:\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nTerm value (Tn)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nPosition number (n)\nT1\nT2\nT3\nT4\nT5\nT6\nT7\nT8\nT9\nT10\nIn this case it would not be accurate to join these points, since n indicates the position\nof a term in a sequence and can therefore only be a positive integer. We can, however,\nsee that the plot of the points lies in the shape of a parabola.\n95\nChapter 3.\nNumber patterns\n\nWorked example 4: Olympic Games soccer event\nQUESTION\nIn the first stage of the soccer event at the Olympic Games, there are teams from\nfour different countries in each group. Each country in a group must play every other\ncountry in the group once.\n1. How many matches will be played in each group in the first stage of the event?\n2. How many matches would be played if there are 5 teams in each group?\n3. How many matches would be played if there are 6 teams in each group?\n4. Determine the general formula of the sequence.\nSOLUTION\nStep 1: Determine the number of matches played if there are 4 teams in a group\nLet the teams from four different countries be A, B, C and D.\nteams in a group\nmatches played\nA\nAB, AC, AD\nB\nBC, BD\nC\nCD\nD\n4\n3 + 2 + 1 = 6\nAB means that team A plays team B and BA would be the same match as AB. So if\nthere are four different teams in a group, each group plays 6 matches.\nStep 2: Determine the number of matches played if there are 5 teams in a group\nLet the teams from five different countries be A, B, C, D and E.\nteams in a group\nmatches played\nA\nAB, AC, AD, AE\nB\nBC, BD, BE\nC\nCD, CE\nD\nDE\nE\n5\n4 + 3 + 2 + 1 = 10\nSo if there are five different teams in a group, each group plays 10 matches.\nStep 3: Determine the number of matches played if there are 6 teams in a group\nLet the teams from six different countries be A, B, C, D, E and F.\n96\n3.2.\nQuadratic sequences\n\nteams in a group\nmatches to be played\nA\nAB, AC, AD, AE, AF\nB\nBC, BD, BE, BF\nC\nCD, CE, CF\nD\nDE, DF\nE\nEF\nF\n5\n5 + 4 + 3 + 2 + 1 = 15\nSo if there are six different teams in a group, each group plays 15 matches.\nWe continue to increase the number of teams in a group and find that a group of 7\nteams plays 21 matches and a group of 8 teams plays 28 matches.\nStep 4: Consider the sequence\nWe examine the sequence to determine if it is linear or quadratic:\nn = 1\nn = 2\nn = 3\nn = 4\nn = 5\nTn\n6\n10\n15\n21\n28 . . .\nfirst difference\n4\n5\n6\n7\nsecond difference\n1\n1\n1\nWe see that the first differences are not constant and that there is a common second\ndifference of 1. Therefore the sequence is quadratic and has a general term of the form\nTn = an2 + bn + c.\nStep 5: Determine the general term Tn\nTo find the values of a, b and c for Tn = an2 + bn + c we look at the first 3 terms in the\nsequence:\nn = 1 : T1 = a + b + c\nn = 2 : T2 = 4a + 2b + c\nn = 3 : T3 = 9a + 3b + c\nWe solve a set of simultaneous equations to determine the values of a, b and c. We\nknow that T1 = 6, T2 = 10 and T3 = 15\na + b + c = 6\n4a + 2b + c = 10\n9a + 3b + c = 15\nT2 −T1 = 4a + 2b + c −(a + b + c)\n10 −6 = 4a + 2b + c −a −b −c\n4 = 3a + b\n. . . (1)\nT3 −T2 = 9a + 3b + c −(4a + 2b + c)\n15 −10 = 9a + 3b + c −4a −2b −c\n5 = 5a + b\n. . . (2)\n97\nChapter 3.\nNumber patterns\n\n(2) −(1) = 5a + b −(3a + b)\n5 −4 = 5a + b −3a −b\n1 = 2a\n∴a = 1\n2\nUsing equation (1) :\n3\n\u00121\n2\n\u0013\n+ b = 4\n∴b = 5\n2\nAnd using a + b + c = 6\n1\n2 + 5\n2 + c = 6\n∴c = 3\nTherefore the general term for the sequence is Tn = 1\n2n2 + 5\n2n + 3.\nExercise 3 – 3: Quadratic sequences\n1. Calculate the common second difference for each of the following quadratic\nsequences:\na) 3; 6; 10; 15; 21; ...\nb) 4; 9; 16; 25; 36; ...\nc) 7; 17; 31; 49; 71; ...\nd) 2; 10; 26; 50; 82; ...\ne) 31; 30; 27; 22; 15; ...\n2. Find the first five terms of the quadratic sequence defined by: Tn = 5n2 +3n+4.\n3. Given Tn = 4n2 + 5n + 10, find T9.\n4. Given Tn = 2n2, for which value of n does Tn = 32?\n5.\na) Write down the next two terms of the quadratic sequence: 16; 27; 42; 61; . . .\nb) Find the general formula for the quadratic sequence above.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22GP\n1b. 22GQ\n1c. 22GR\n1d. 22GS\n1e. 22GT\n2. 22GV\n3. 22GW\n4. 22GX\n5. 22GY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n98\n3.2.\nQuadratic sequences\n\n3.3\nSummary\nEMBG6\nSee presentation: 22GZ at www.everythingmaths.co.za\n• Tn is the general term of a sequence.\n• Successive or consecutive terms are terms that follow one after another in a\nsequence.\n• A linear sequence has a common difference (d) between any two successive\nterms.\nd = Tn −Tn−1\n• A quadratic sequence has a common second difference between any two suc-\ncessive terms.\n• The general term for a quadratic sequence is\nTn = an2 + bn + c\n• A general quadratic sequence:\nn = 1\nn = 2\nn = 3\nn = 4\nTn\na + b + c\n4a + 2b + c\n9a + 3b + c\n16a + 4b + c\n1st difference\n3a + b\n5a + b\n7a + b\n2nd difference\n2a\n2a\nExercise 3 – 4: End of chapter exercises\n1. Find the first five terms of the quadratic sequence defined by:\nTn = n2 + 2n + 1\n2. Determine whether each of the following sequences is:\n• a linear sequence,\n• a quadratic sequence,\n• or neither.\na) 6; 9; 14; 21; 30; ...\nb) 1; 7; 17; 31; 49; ...\nc) 8; 17; 32; 53; 80; ...\nd) 9; 26; 51; 84; 125; ...\ne) 2; 20; 50; 92; 146; ...\nf) 5; 19; 41; 71; 109; ...\ng) 2; 6; 10; 14; 18; ...\nh) 3; 9; 15; 21; 27; ...\ni) 1; 2,5; 5; 8,5; 13; ...\nj) 10; 24; 44; 70; 102; ...\nk) 21\n2; 6; 101\n2; 16; 221\n2; . . .\nl) 3p2; 6p2; 9p2; 12p2; 15p2; . . .\nm) 2k; 8k; 18k; 32k; 50k; . . .\n99\nChapter 3.\nNumber patterns\n\n3. Given the pattern: 16; x; 46; . . ., determine the value of x if the pattern is linear.\n4. Given Tn = 2n2, for which value of n does Tn = 242?\n5. Given Tn = 3n2, find T11.\n6. Given Tn = n2 + 4, for which value of n does Tn = 85?\n7. Given Tn = 4n2 + 3n −1, find T5.\n8. Given Tn = 3\n2n2, for which value of n does Tn = 96?\n9. For each of the following patterns, determine:\n• the next term in the pattern,\n• and the general term,\n• the tenth term in the pattern.\na) 3; 7; 11; 15; . . .\nb) 17; 12; 7; 2; . . .\nc)\n1\n2; 1; 11\n2; 2; . . .\nd) a; a + b; a + 2b; a + 3b; . . .\ne) 1; −1; −3; −5; . . .\n10. For each of the following sequences, find the equation for the general term and\nthen use the equation to find T100.\na) 4; 7; 12; 19; 28; ...\nb) 2; 8; 14; 20; 26; ...\nc) 7; 13; 23; 37; 55; ...\nd) 5; 14; 29; 50; 77; ...\nGiven: Tn = 3n −1\n11.\na) Write down the first five terms of the sequence.\nb) What do you notice about the difference between any two consecutive\nterms?\nc) Will this always be the case for a linear sequence?\nGiven the following sequence: −15; −11; −7; . . . ; 173\n12.\na) Determine the equation for the general term.\nb) Calculate how many terms there are in the sequence.\n13. Given 3; 7; 13; 21; 31; . . .\na) Thabang determines that the general term is Tn = 4n −1. Is he correct?\nExplain.\nb) Cristina determines that the general term is Tn = n2 + n + 1. Is she correct?\nExplain.\n100\n3.3.\nSummary\n\n14. Given the following pattern of blocks:\n2\n3\n4\na) Draw pattern 5.\nb) Complete the table below:\npattern number (n)\n2\n3\n4\n5\n10\n250\nn\nnumber of white blocks (w)\n4\n8\nc) Is this a linear or a quadratic sequence?\n15. Cubes of volume 1 cm3 are stacked on top of each other to form a tower:\n1\n2\n3\na) Complete the table for the height of the tower:\ntower number (n)\n1\n2\n3\n4\n10\nn\nheight of tower (h)\n2\nb) What type of sequence is this?\nc) Now consider the number of cubes in each tower and complete the table\nbelow:\ntower number (n)\n1\n2\n3\n4\nnumber of cubes (c)\n3\nd) What type of sequence is this?\ne) Determine the general term for this sequence.\nf) How many cubes are needed for tower number 21?\ng) How high will a tower of 496 cubes be?\n16. A quadratic sequence has a second term equal to 1, a third term equal to −6 and\na fourth term equal to −14.\na) Determine the second difference for this sequence.\nb) Hence, or otherwise, calculate the first term of the pattern.\n101\nChapter 3.\nNumber patterns\n\n17. There are 15 schools competing in the U16 girls hockey championship and every\nteam must play two matches — one home match and one away match.\na) Use the given information to complete the table:\nno. of schools\nno. of matches\n1\n0\n2\n3\n4\n5\nb) Calculate the second difference.\nc) Determine a general term for the sequence.\nd) How many matches will be played if there are 15 schools competing in the\nchampionship?\ne) If 600 matches must be played, how many schools are competing in the\nchampionship?\n18. The first term of a quadratic sequence is 4, the third term is 34 and the common\nsecond difference is 10. Determine the first six terms in the sequence.\n19. Challenge question:\nGiven that the general term for a quadratic sequences is Tn = an2 + bn + c, let\nd be the first difference and D be the second common difference.\na) Show that a = D\n2 .\nb) Show that b = d −3\n2D.\nc) Show that c = T1 −d + D.\nd) Hence, show that Tn = D\n2 n2 +\n\u0012\nd −3\n2D\n\u0013\nn + (T1 −d + D).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22H2\n2a. 22H3\n2b. 22H4\n2c. 22H5\n2d. 22H6\n2e. 22H7\n2f. 22H8\n2g. 22H9\n2h. 22HB\n2i. 22HC\n2j. 22HD\n2k. 22HF\n2l. 22HG\n2m. 22HH\n3. 22HJ\n4. 22HK\n5. 22HM\n6. 22HN\n7. 22HP\n8. 22HQ\n9a. 22HR\n9b. 22HS\n9c. 22HT\n9d. 22HV\n9e. 22HW\n10a. 22HX\n10b. 22HY\n10c. 22HZ\n10d. 22J2\n11. 22J3\n12. 22J4\n13. 22J5\n14. 22J6\n15. 22J7\n16. 22J8\n17. 22J9\n18. 22JB\n19. 22JC\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n102\n3.3.\nSummary\n\nCHAPTER\n4\nAnalytical geometry", "chapter_id": "18" }, { "title": "Revision", "content": "4.1\nRevision\n104\n4.2\nEquation of a line\n113\n4.3\nInclination of a line\n124\n4.4\nParallel lines\n132\n4.5\nPerpendicular lines\n136\n4.6\nSummary\n142\n\n4\nAnalytical geometry\nAnalytical geometry, also referred to as coordinate or Cartesian geometry, is the study\nof geometric properties and relationships between points, lines and angles in the Carte-\nsian plane. Geometrical shapes are defined using a coordinate system and algebraic\nprinciples. In this chapter we deal with the equation of a straight line, parallel and\nperpendicular lines and inclination of a line.\n4.1\nRevision\nEMBG7\nPoints A(x1; y1), B(x2; y2) and C(x2; y1) are shown in the diagram below:\nb\nb\nA(x1; y1)\nC(x2; y1)\nB(x2; y2)\nx\ny\n0\nTheorem of Pythagoras\nAB2 = AC2 + BC2\nDistance formula\nDistance between two points:\nAB =\np\n(x2 −x1)2 + (y2 −y1)2\nNotice that (x1 −x2)2 = (x2 −x1)2.\nSee video: 22JD at www.everythingmaths.co.za\nGradient\nGradient (m) describes the slope or steepness of the line joining two points. The\ngradient of a line is determined by the ratio of vertical change to horizontal change.\nmAB = y2 −y1\nx2 −x1\nor\nmAB = y1 −y2\nx1 −x2\nRemember to be consistent: m ̸= y1 −y2\nx2 −x1\n.\n104\n4.1.\nRevision\n\nHorizontal lines\nx\ny\n0\nm = 0\nVertical lines\nx\ny\n0\nm is undefined\nParallel lines\nθ\nθ\nx\ny\n0\nm1 = m2\nPerpendicular lines\nθ2\nθ1\nx\ny\n0\nm1 × m2 = −1\nMid-point of a line segment\nA(x1; y1)\nM(x; y)\nB(x2; y2)\nx\ny\n0\nThe coordinates of the mid-point M(x; y) of a line between any two points A(x1; y1)\nand B(x2; y2):\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nSee video: 22JF at www.everythingmaths.co.za\n105\nChapter 4.\nAnalytical geometry\n\nPoints on a straight line\nThe diagram shows points P(x1; y1), Q(x2; y2) and R(x; y) on a straight line.\nb\nb\nx\ny\nR(x; y)\n0\nb\nQ(x2; y2)\nP(x1; y1)\nWe know that mPR = mQR = mPQ.\nUsing mPR = mPQ, we obtain the following for any point (x; y) on a straight line\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 1: Revision\nQUESTION\nGiven the points P(−5; −4) and Q(0; 6):\n1. Determine the length of the line segment PQ.\n2. Determine the mid-point T(x; y) of the line segment PQ.\n3. Show that the line passing through R(1; −3\n4) and T(x; y) is perpendicular to the\nline PQ.\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n−2\n−4\n−6\n2\n−2\n−4\n−6\nb\nb\nb\nP(−5; −4)\nT(x; y)\nQ(0; 6)\nx\ny\n0\n106\n4.1.\nRevision\n\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q (x2; y2)\nx1 = −5;\ny1 = −4;\nx2 = 0;\ny2 = 6\nWrite down the distance formula\nPQ =\np\n(x2 −x1)2 + (y2 −y1)2\n=\np\n(0 −(−5))2 + (6 −(−4))2\n=\n√\n25 + 100\n=\n√\n125\n= 5\n√\n5\nThe length of the line segment PQ is 5\n√\n5 units.\nStep 3: Write down the mid-point formula and substitute the values\nT(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nx = x1 + x2\n2\n= −5 + 0\n2\n= −5\n2\ny = y1 + y2\n2\n= −4 + 6\n2\n= 2\n2\n= 1\nThe mid-point of PQ is T(−5\n2; 1).\nStep 4: Determine the gradients of PQ and RT\nm = y2 −y1\nx2 −x1\nmPQ = 6 −(−4)\n0 −(−5)\n= 10\n5\n= 2\n107\nChapter 4.\nAnalytical geometry\n\nmRT = −3\n4 −1\n1 −(−5\n2)\n= −7\n4\n7\n2\n= −7\n4 × 2\n7\n= −1\n2\nCalculate the product of the two gradients:\nmRT × mPQ = −1\n2 × 2\n= −1\nTherefore PQ is perpendicular to RT.\nQuadrilaterals\n• A quadrilateral is a closed shape consisting of four straight line segments.\n• A parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n– Both pairs of opposite sides are equal in length.\n– Both pairs of opposite angles are equal.\n– The diagonals bisect each other.\n• A rectangle is a parallelogram that has all four angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other.\n– The diagonals are equal in length.\n108\n4.1.\nRevision\n\n• A rhombus is a parallelogram that has all four sides equal in length.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n×\n×\n××\n•\n•\n•\n•\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals of a rhombus bisect both pairs of opposite angles.\n• A square is a rhombus that has all four interior angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n•\n•\n••\n••\n••\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals are equal in length.\n– The diagonals bisect both pairs of interior opposite angles (that is, all angles\nare 45◦).\n• A trapezium is a quadrilateral with one pair of opposite sides parallel.\n• A kite is a quadrilateral with two pairs of adjacent sides equal.\nA\nB\nC\nD\nb\nb\n××\n– One pair of opposite angles are equal (the angles are between unequal\nsides).\n– The diagonal between equal sides bisects the other diagonal.\n– The diagonal between equal sides bisects the interior angles.\n– The diagonals intersect at 90◦.\n109\nChapter 4.\nAnalytical geometry\n\nWorked example 2: Quadrilaterals\nQUESTION\nPoints A (−1; 0), B (0; 3), C (8; 11) and D (x; y) are points on the Cartesian plane.\nDetermine D (x; y) if ABCD is a parallelogram.\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n−1\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\nb\nb\nx\ny\nC(8; 11)\n0\nD(x; y)\nA(−1; 0)\nb\nB(0; 3)\nM\nThe mid-point of AC will be the same as the mid-point of BD. We first find the\nmid-point of AC and then use it to determine the coordinates of point D.\nStep 2: Assign values to (x1; y1) and (x2; y2)\nLet the mid-point of AC be M(x; y)\nx1 = −1;\ny1 = 0;\nx2 = 8;\ny2 = 11\nStep 3: Write down the mid-point formula\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nStep 4: Substitute the values and calculate the coordinates of M\nM(x; y) =\n\u0012−1 + 8\n2\n; 0 + 11\n2\n\u0013\n=\n\u00127\n2; 11\n2\n\u0013\n110\n4.1.\nRevision\n\nStep 5: Use the coordinates of M to determine D\nM is also the mid-point of BD so we use M\n\u0000 7\n2; 11\n2\n\u0001\nand B (0; 3) to find D (x; y)\nStep 6: Substitute values and determine x and y\nM =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n∴\n\u00127\n2; 11\n2\n\u0013\n=\n\u00120 + x\n2\n; 3 + y\n2\n\u0013\n7\n2 = 0 + x\n2\n7 = 0 + x\n∴x = 7\n11\n2 = 3 + y\n2\n11 = 3 + y\n∴y = 8\nStep 7: Alternative method: inspection\nSince we are given that ABCD is a parallelogram, we can use the properties of a\nparallelogram and the given points to determine the coordinates of D.\nFrom the sketch we expect that point D will lie below C.\nConsider the given points A, B and C:\n• Opposite sides of a parallelogram are parallel, therefore BC must be parallel to\nAD and their gradients must be equal.\n• The vertical change from B to C is 8 units up.\n• Therefore the vertical change from A to D is also 8 units up (y = 0 + 8 = 8).\n• The horizontal change from B to C is 8 units to the right.\n• Therefore the horizontal change from A to D is also 8 units to the right (x =\n−1 + 8 = 7).\nor\n• Opposite sides of a parallelogram are parallel, therefore AB must be parallel to\nDC and their gradients must be equal.\n• The vertical change from A to B is 3 units up.\n111\nChapter 4.\nAnalytical geometry\n\n• Therefore the vertical change from C to D is 3 units down (y = 11 −3 = 8).\n• The horizontal change from A to B is 1 unit to the right.\n• Therefore the horizontal change from C to D is 1 unit to the left (x = 8 −1 = 7).\nStep 8: Write the final answer\nThe coordinates of D are (7; 8).\nExercise 4 – 1: Revision\n1. Determine the length of the line segment between the following points:\na) P(−3; 5) and Q(−1; −5)\nb) R(0,75; 3) and S(0,75; −4)\nc) T(2x; y −2) and U(3x + 1; y −2)\n2. Given Q(4; 1), T(p; 3) and length QT =\n√\n8 units, determine the value of p.\n3. Determine the gradient of the line AB if:\na) A(−5; 3) and B(−7; 4)\nb) A(3; −2) and B(1; −8)\n4. Prove that the line PQ, with P(0; 3) and Q(5; 5), is parallel to the line 5y + 5 =\n2x.\n5. Given the points A(−1; −1), B(2; 5), C(−1; −5\n2) and D(x; −4) and AB ⊥CD,\ndetermine the value of x.\n6. Calculate the coordinates of the mid-point P(x; y) of the line segment between\nthe points:\na) M(3; 5) and N(−1; −1)\nb) A(−3; −4) and B(2; 3)\n7. The line joining A(−2; 4) and B(x; y) has the mid-point C(1; 3). Determine the\nvalues of x and y.\n8. Given\nquadrilateral\nABCD\nwith\nvertices\nA(0; 3), B(4; 3), C(5; −1)\nand\nD(1; −1).\na) Determine the equation of the line AD and the line BC.\nb) Show that AD ∥BC.\nc) Calculate the lengths of AD and BC.\nd) Determine the equation of the diagonal BD.\ne) What type of quadrilateral is ABCD?\n112\n4.1.\nRevision\n\n9. MPQN is a parallelogram with points M(−5; 3), P(−1; 5) and Q(4; 5). Draw a\nsketch and determine the coordinates of N(x; y).\n10. PQRS is a quadrilateral with points P(−3; 1), Q(1; 3), R(6; 1) and S(2; −1) in\nthe Cartesian plane.\na) Determine the lengths of PQ and SR.\nb) Determine the mid-point of PR.\nc) Show that PQ ∥SR.\nd) Determine the equations of the line PS and the line SR.\ne) Is PS ⊥SR? Explain your answer.\nf) What type of quadrilateral is PQRS?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22JG\n1b. 22JH\n1c. 22JJ\n2. 22JK\n3a. 22JM\n3b. 22JN\n4. 22JP\n5. 22JQ\n6a. 22JR\n6b. 22JS\n7. 22JT\n8. 22JV\n9. 22JW\n10. 22JX\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.2\nEquation of a line\nEMBG8\nWe can derive different forms of the straight line equation. The different forms are\nused depending on the information provided in the problem:\n• The two-point form of the straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• The gradient–point form of the straight line equation: y −y1 = m(x −x1)\n• The gradient–intercept form of the straight line equation: y = mx + c\nThe two-point form of the straight line equation\nEMBG9\nb\nb\n(x1; y1)\n(x2; y2)\nx\ny\n0\n113\nChapter 4.\nAnalytical geometry\n\nGiven any two points (x1; y1) and (x2; y2), we can determine the equation of the line\npassing through the two points using the equation:\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 3: The two-point form of the straight line equation\nQUESTION\nFind the equation of the straight line passing through P (−1; −5) and Q (5; 4).\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nP(−1; −5)\nQ(5; 4)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q(x2; y2)\nx1 = −1;\ny1 = −5;\nx2 = 5;\ny2 = 4\nStep 3: Write down the two-point form of the straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\n114\n4.2.\nEquation of a line\n\nStep 4: Substitute the values and make y the subject of the equation\ny −(−5)\nx −(−1) = 4 −(−5)\n5 −(−1)\ny + 5\nx + 1 = 9\n6\ny + 5 = 3\n2(x + 1)\ny + 5 = 3\n2x + 3\n2\ny = 3\n2x −7\n2\nStep 5: Write the final answer\ny = 3\n2x −31\n2\nExercise 4 – 2: The two-point form of the straight line equation\nDetermine the equation of the straight line passing through the points:\n1. (3; 7) and (−6; 1)\n2. (1; −11\n4 ) and (2\n3; −7\n4)\n3. (−2; 1) and (3; 6)\n4. (2; 3) and (3; 5)\n5. (1; −5) and (−7; −5)\n6. (−4; 0) and (1; 15\n4 )\n7. (s; t) and (t; s)\n8. (−2; −8) and (1; 7)\n9. (2p; q) and (0; −q)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22JY\n2. 22JZ\n3. 22K2\n4. 22K3\n5. 22K4\n6. 22K5\n7. 22K6\n8. 22K7\n9. 22K8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n115\nChapter 4.\nAnalytical geometry\n\nThe gradient–point form of the straight line equation\nEMBGB\nWe derive the gradient–point form of the straight line equation using the definition of\ngradient and the two-point form of a straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nSubstitute m = y2 −y1\nx2 −x1\non the right-hand side of the equation\ny −y1\nx −x1\n= m\nMultiply both sides of the equation by (x −x1)\ny −y1 = m(x −x1)\nTo use this equation, we need to know the gradient of the line and the coordinates of\none point on the line.\nSee video: 22K9 at www.everythingmaths.co.za\nWorked example 4: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −1\n3 and passing through\nthe point (−1; 1).\nSOLUTION\nStep 1: Draw a sketch\nWe notice that m < 0, therefore the graph decreases as x increases.\n1\n2\n3\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\n(−1; 1)\nx\ny\n0\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\n116\n4.2.\nEquation of a line\n\nSubstitute the value of the gradient\ny −y1 = −1\n3(x −x1)\nSubstitute the coordinates of the given point\ny −1 = −1\n3(x −(−1))\ny −1 = −1\n3(x + 1)\ny = −1\n3x −1\n3 + 1\n= −1\n3x + 2\n3\nStep 3: Write the final answer\nThe equation of the straight line is y = −1\n3x + 2\n3.\nIf we are given two points on a straight line, we can also use the gradient–point form\nto determine the equation of a straight line. We first calculate the gradient using the\ntwo given points and then substitute either of the two points into the gradient–point\nform of the equation.\nWorked example 5: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line passing through (−3; 2) and (5; 8).\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n8\n2\n4\n6\n−2\n−4\nb\nb\n(−3; 2)\n(5; 8)\nx\ny\n0\n117\nChapter 4.\nAnalytical geometry\n\nStep 2: Assign variables to the coordinates of the given points\nx1 = −3;\ny1 = 2;\nx2 = 5;\ny2 = 8\nStep 3: Calculate the gradient using the two given points\nm = y2 −y1\nx2 −x1\n=\n8 −2\n5 −(−3)\n= 6\n8\n= 3\n4\nStep 4: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the value of the gradient\ny −y1 = 3\n4(x −x1)\nSubstitute the coordinates of a given point\ny −y1 = 3\n4(x −x1)\ny −2 = 3\n4(x −(−3))\ny −2 = 3\n4(x + 3)\ny = 3\n4x + 9\n4 + 2\n= 3\n4x + 17\n4\nStep 5: Write the final answer\nThe equation of the straight line is y = 3\n4x + 41\n4.\nSee video: 22KB at www.everythingmaths.co.za\n118\n4.2.\nEquation of a line\n\nExercise 4 – 3: Gradient–point form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (−1; 10\n3 ) and with m = 2\n3.\n2. with m = −1 and passing through the point (−2; 0).\n3. passing through the point (3; −1) and with m = −1\n3.\n4. parallel to the x-axis and passing through the point (0; 11).\n5. passing through the point (1; 5) and with m = −2.\n6. perpendicular to the x-axis and passing through the point (−3\n2; 0).\n7. with m = −0,8 and passing through the point (10; −7).\n8. with undefined gradient and passing through the point (4; 0).\n9. with m = 3a and passing through the point (−2; −6a + b).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KC\n2. 22KD\n3. 22KF\n4. 22KG\n5. 22KH\n6. 22KJ\n7. 22KK\n8. 22KM\n9. 22KN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe gradient–intercept form of a straight line equation EMBGC\nUsing the gradient–point form, we can also derive the gradient–intercept form of the\nstraight line equation.\nStarting with the equation\ny −y1 = m(x −x1)\nExpand the brackets and make y the subject of the formula\ny −y1 = mx −mx1\ny = mx −mx1 + y1\ny = mx + (y1 −mx1)\nWe define constant c such that c = y1 −mx1 so that we get the equation\ny = mx + c\nThis is also called the standard form of the straight line equation.\n119\nChapter 4.\nAnalytical geometry\n\nNotice that when x = 0, we have\ny = m(0) + c\n= c\nTherefore c is the y-intercept of the straight line.\nWorked example 6: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −2 and passing through\nthe point (−1; 7).\nSOLUTION\nStep 1: Slope of the line\nWe notice that m < 0, therefore the graph decreases as x increases.\n2\n4\n6\n8\n−2\n2\n4\n−2\n−4\nb\n(−1; 7)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\nSubstitute the value of the gradient\ny = −2x + c\n120\n4.2.\nEquation of a line\n\nSubstitute the coordinates of the given point and find c\ny = −2x + c\n7 = −2(−1) + c\n7 −2 = c\n∴c = 5\nThis gives the y-intercept (0; 5).\nStep 3: Write the final answer\nThe equation of the straight line is y = −2x + 5.\nIf we are given two points on a straight line, we can also use the gradient–intercept\nform to determine the equation of a straight line. We solve for the two unknowns m\nand c using simultaneous equations — using the methods of substitution or elimina-\ntion.\nWorked example 7: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line passing through the points (−2; −7) and\n(3; 8).\nSOLUTION\nStep 1: Draw a sketch\n4\n8\n−4\n−8\n2\n4\n−2\n−4\nb\nb\n(−2; −7)\n(3; 8)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\n121\nChapter 4.\nAnalytical geometry\n\nStep 3: Substitute the coordinates of the given points\n−7 = m(−2) + c\n−7 = −2m + c\n. . . (1)\n8 = m(3) + c\n8 = 3m + c\n. . . (2)\nWe have two equations with two unknowns; we can therefore solve using simultane-\nous equations.\nStep 4: Make the coefficient of one of the variables the same in both equations\nWe notice that the coefficient of c in both equations is 1, therefore we can subtract\none equation from the other to eliminate c:\n−7 = −2m + c\n−(8 = 3m + c)\n−15 = −5m\n∴3 = m\nSubstitute m = 3 into either of the two equations and determine c:\n−7 = −2m + c\n−7 = −2(3) + c\n∴c = −1\nor\n8 = 3m + c\n8 = 3(3) + c\n∴c = −1\nStep 5: Write the final answer\nThe equation of the straight line is y = 3x −1.\n122\n4.2.\nEquation of a line\n\nExercise 4 – 4: The gradient–intercept form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (1\n2; 4) and\nwith m = 2.\n2. passing through the points (1\n2; −2)\nand (2; 4).\n3. passing through the points (2; −3)\nand (−1; 0).\n4. passing through the point (2; −6\n7)\nand with m = −3\n7.\n5. which cuts the y-axis at y = −1\n5 and\nwith m = 1\n2.\n6.\nb\nb\n(−1; −4)\n(2; 2)\nx\ny\n0\n7.\nb −3\n2\nx\ny\n0\n8.\nb\n(−2; −2)\n4\nx\ny\n0\n9.\nb (−2; 10)\nx\ny\n0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KP\n2. 22KQ\n3. 22KR\n4. 22KS\n5. 22KT\n6. 22KV\n7. 22KW\n8. 22KX\n9. 22KY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n123\nChapter 4.\nAnalytical geometry\n\n4.3\nInclination of a line\nEMBGD\n1\n2\n3\n1\n2\n3\nθ\n∆y\n∆x\nx\ny\nThe diagram shows that a straight line makes an angle θ with the positive x-axis. This\nis called the angle of inclination of a straight line.\nWe notice that if the gradient changes, then the value of θ also changes, therefore the\nangle of inclination of a line is related to its gradient. We know that gradient is the\nratio of a change in the y-direction to a change in the x-direction:\nm = ∆y\n∆x\nFrom trigonometry we know that the tangent function is defined as the ratio:\ntan θ = opposite side\nadjacent side\nAnd from the diagram we see that\ntan θ = ∆y\n∆x\n∴m = tan θ\nfor 0◦≤θ < 180◦\nTherefore the gradient of a straight line is equal to the tangent of the angle formed\nbetween the line and the positive direction of the x-axis.\nVertical lines\n• θ = 90◦\n• Gradient is undefined since there is no change in the x-values (∆x = 0).\n• Therefore tan θ is also undefined (the graph of tan θ has an asymptote at θ =\n90◦).\n124\n4.3.\nInclination of a line\n\nHorizontal lines\n• θ = 0◦\n• Gradient is equal to 0 since there is no change in the y-values (∆y = 0).\n• Therefore tan θ is also equal to 0 (the graph of tan θ passes through the origin\n(0◦; 0).\nLines with negative gradients\nIf a straight line has a negative gradient (m < 0, tan θ < 0), then the angle formed\nbetween the line and the positive direction of the x-axis is obtuse.\nθ\nx\ny\n0\nFrom the CAST diagram in trigonometry, we know that the tangent function is negative\nin the second and fourth quadrant. If we are calculating the angle of inclination for a\nline with a negative gradient, we must add 180◦to change the negative angle in the\nfourth quadrant to an obtuse angle in the second quadrant:\nIf we are given a straight line with gradient m = −0,7, then we can determine the\nangle of inclination using a calculator:\ntan θ = m\n= −0,7\n∴θ = tan−1(−0,7)\n= −35,0◦\nThis negative angle lies in the fourth quadrant. We must add 180◦to get an obtuse\nangle in the second quadrant:\nθ = −35,0◦+ 180◦\n= 145◦\n125\nChapter 4.\nAnalytical geometry\n\nAnd we can always use our calculator to check that the obtuse angle θ = 145◦gives a\ngradient of m = −0,7.\n35◦\n180◦−35◦= 145◦\nx\ny\n0\nExercise 4 – 5: Angle of inclination\n1. Determine the gradient (correct to 1 decimal place) of each of the following\nstraight lines, given that the angle of inclination is equal to:\na) 60◦\nb) 135◦\nc) 0◦\nd) 54◦\ne) 90◦\nf) 45◦\ng) 140◦\nh) 180◦\ni) 75◦\n2. Determine the angle of inclination (correct to 1 decimal place) for each of the\nfollowing:\na) a line with m = 3\n4\nb) 2y −x = 6\nc) the line passes through the points (−4; −1) and (2; 5)\nd) y = 4\ne) x = 3y + 1\n2\nf) x = −0,25\ng) the line passes through the points (2; 5) and (2\n3; 1)\nh) a line with gradient equal to 0,577\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22KZ\n1b. 22M2\n1c. 22M3\n1d. 22M4\n1e. 22M5\n1f. 22M6\n1g. 22M7\n1h. 22M8\n1i. 22M9\n2a. 22MB\n2b. 22MC\n2c. 22MD\n2d. 22MF\n2e. 22MG\n2f. 22MH\n2g. 22MJ\n2h. 22MK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n126\n4.3.\nInclination of a line\n\nWorked example 8: Inclination of a straight line\nQUESTION\nDetermine the angle of inclination (correct to 1 decimal place) of the straight line\npassing through the points (2; 1) and (−3; −9).\nSOLUTION\nStep 1: Draw a sketch\n2\n−2\n−4\n−6\n−8\n1\n2\n−1\n−2\n−3\nb\nb\n(2; 1)\n(−3; −9)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nx1 = 2;\ny1 = 1;\nx2 = −3;\ny2 = −9\nStep 3: Determine the gradient of the line\nm = y2 −y1\nx2 −x1\n= −9 −1\n−3 −2\n= −10\n−5\n∴m = 2\nStep 4: Use the gradient to determine the angle of inclination of the line\ntan θ = m\n= 2\n∴θ = tan−1 2\n= 63,4◦\nImportant: make sure your calculator is in DEG (degrees) mode.\nStep 5: Write the final answer\nThe angle of inclination of the straight line is 63,4◦.\n127\nChapter 4.\nAnalytical geometry\n\nWorked example 9: Inclination of a straight line\nQUESTION\nDetermine the equation of the straight line passing through the point (3; 1) and with\nan angle of inclination of 135◦.\nSOLUTION\nStep 1: Use the angle of inclination to determine the gradient of the line\nm = tan θ\n= tan 135◦\n∴m = −1\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = −1\ny −y1 = −(x −x1)\nSubstitute the given point (3; 1)\ny −1 = −(x −3)\ny = −x + 3 + 1\n= −x + 4\nStep 3: Write the final answer\nThe equation of the straight line is y = −x + 4.\nWorked example 10: Inclination of a straight line\nQUESTION\nDetermine the acute angle (correct to 1 decimal place) between the line passing\nthrough the points M(−1; 13\n4) and N(4; 3) and the straight line y = −3\n2x + 4.\nSOLUTION\nStep 1: Draw a sketch\nDraw the line through points M(−1; 13\n4) and N(4; 3) and the line y = −3\n2x + 4 on a\nsuitable system of axes. Label α and β, the angles of inclination of the two lines. Label\nθ, the acute angle between the two straight lines.\n128\n4.3.\nInclination of a line\n\n2\n4\n6\n−2\n2\n4\n−2\n−4\n−6\n−8\nb\nb\nβ\nˆB1\nα\nθ\nx\ny\n0\nM(−1; 7\n4)\nN(4; 3)\nNotice that α and θ are acute angles and β is an obtuse angle.\nˆB1 = 180◦−β\n(∠on str. line)\nand θ = α + ˆB1\n(ext. ∠of △= sum int. opp)\n∴θ = α + (180◦−β)\n= 180◦+ α −β\nStep 2: Use the gradient to determine the angle of inclination β\nFrom the equation y = −3\n2x + 4 we see that m < 0, therefore β is an obtuse angle\nsuch that 90◦< β < 180◦.\ntan β = m\n= −3\n2\ntan−1\n\u0012\n−3\n2\n\u0013\n= −56,3◦\nThis negative angle lies in the fourth quadrant. We know that the angle of inclination\nβ is an obtuse angle that lies in the second quadrant, therefore\nβ = −56,3◦+ 180◦\n= 123,7◦\nStep 3: Determine the gradient and angle of inclination of the line through M and\nN\n129\nChapter 4.\nAnalytical geometry\n\nDetermine the gradient\nm = y2 −y1\nx2 −x1\n=\n3 −7\n4\n4 −(−1)\n=\n5\n4\n5\n= 1\n4\nDetermine the angle of inclination\ntan α = m\n= 1\n4\n∴α = tan−1\n\u00121\n4\n\u0013\n= 14,0◦\nStep 4: Write the final answer\nθ = 180◦+ α −β\n= 180◦+ 14,0◦−123,7◦\n= 70,3◦\nThe acute angle between the two straight lines is 70,3◦.\nExercise 4 – 6: Inclination of a straight line\n1. Determine the angle of inclination for each of the following:\na) a line with m = 4\n5\nb) x + y + 1 = 0\nc) a line with m = 5,69\nd) the line that passes through (1; 1) and (−2; 7)\ne) 3 −2y = 9x\nf) the line that passes through (−1; −6) and (−1\n2; −11\n2 )\ng) 5 = 10y −15x\n130\n4.3.\nInclination of a line\n\nh)\nb\nx\ny\n(2; 3)\n−1\n0\ni)\nb\nx\ny\n(6; 0)\n2\n0\nj)\nb\nx\ny\n(−3; 3)\n−3\n0\n2. Determine the acute angle between the line passing through the points A(−2; 1\n5)\nand B(0; 1) and the line passing through the points C(1; 0) and D(−2; 6).\n3. Determine the angle between the line y + x = 3 and the line x = y + 1\n2.\n4. Find the angle between the line y = 2x and the line passing through the points\n(−1; 7\n3) and (0; 2).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22MM\n1b. 22MN\n1c. 22MP\n1d. 22MQ\n1e. 22MR\n1f. 22MS\n1g. 22MT\n1h. 22MV\n1i. 22MW\n1j. 22MX\n2. 22MY\n3. 22MZ\n4. 22N2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n131\nChapter 4.\nAnalytical geometry\n\n4.4\nParallel lines\nEMBGF\nInvestigation: Parallel lines\n1. Draw a sketch of the line passing through the points P(−1; 0) and Q(1; 4) and\nthe line passing through the points R(1; 2) and S(2; 4).\n2. Label and measure α and β, the angles of inclination of straight lines PQ and\nRS respectively.\n3. Describe the relationship between α and β.\n4. “α and β are alternate angles, therefore PQ ∥RS.” Is this a true statement? If\nnot, provide a correct statement.\n5. Use your calculator to determine tan α and tan β.\n6. Complete the sentence: . . . . . . lines have . . . . . . angles of inclination.\n7. Determine the equations of the straight lines PQ and RS.\n8. What do you notice about mPQ and mRS?\n9. Complete the sentence: . . . . . . lines have . . . . . . gradients.\nAnother method of determining the equation of a straight line is to be given a point on\nthe unknown line, (x1; y1), and the equation of a line which is parallel to the unknown\nline.\nLet the equation of the unknown line be y = m1x + c1 and the equation of the given\nline be y = m2x + c2.\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are parallel then\nm1 = m2\n132\n4.4.\nParallel lines\n\nImportant: when determining the gradient of a line using the coefficient of x, make\nsure the given equation is written in the gradient–intercept (standard) form. y = mx+c\nSubstitute the value of m2 and the given point (x1; y1), into the gradient–intercept form\nof a straight line equation\ny −y1 = m(x −x1)\nand determine the equation of the unknown line.\nWorked example 11: Parallel lines\nQUESTION\nDetermine the equation of the line that passes through the point (−1; 1) and is parallel\nto the line y −2x + 1 = 0.\nSOLUTION\nStep 1: Write the equation in gradient–intercept form\nWe write the given equation in gradient–intercept form and determine the value of m.\ny = 2x −1\nWe know that the two lines are parallel, therefore m1 = m2 = 2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = 2\ny −y1 = 2(x −x1)\nSubstitute the given point (−1; 1)\ny −1 = 2(x −(−1))\ny −1 = 2x + 2\ny = 2x + 2 + 1\n= 2x + 3\n133\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\n(−1; 1)\ny = 2x −1\ny = 2x + 3\nx\ny\n0\nA sketch was not required, but it is always helpful and can be used to check answers.\nStep 3: Write the final answer\nThe equation of the straight line is y = 2x + 3.\nWorked example 12: Parallel lines\nQUESTION\nLine AB passes through the point A(0; 3) and has an angle of inclination of 153,4◦.\nDetermine the equation of the line CD which passes through the point C(2; −3) and\nis parallel to AB.\nSOLUTION\nStep 1: Use the given angle of inclination to determine the gradient\nmAB = tan θ\n= tan 153,4◦\n= −0,5\nStep 2: Parallel lines have equal gradients\nSince we are given AB ∥CD,\nmCD = mAB = −0,5\n134\n4.4.\nParallel lines\n\nStep 3: Write down the gradient–point form of a straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient mCD = −0,5.\ny −y1 = −1\n2(x −x1)\nSubstitute the given point (2; −3).\ny −(−3) = −1\n2(x −2)\ny + 3 = −1\n2x + 1\ny = −1\n2x −2\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\nb C(2; −3)\ny = −1\n2x −2\ny = −1\n2x + 3\nx\ny\n0\nA sketch was not required, but it is always useful.\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n2x −2.\nSee video: 22N3 at www.everythingmaths.co.za\n135\nChapter 4.\nAnalytical geometry\n\nExercise 4 – 7: Parallel lines\n1. Determine whether or not the following two lines are parallel:\na) y + 2x = 1 and −2x + 3 = y\nb)\ny\n3 + x + 5 = 0 and 2y + 6x = 1\nc) y = 2x −7 and the line passing through (1; −2) and (1\n2; −1)\nd) y + 1 = x and x + y = 3\ne) The line passing through points (−2; −1) and (−4; −3) and the line −y +\nx −4 = 0\nf) y −1 = 1\n3x and the line passing through points (−2; 4) and (1; 5)\n2. Determine the equation of the straight line that passes through the point (1; −5)\nand is parallel to the line y + 2x −1 = 0.\n3. Determine the equation of the straight line that passes through the point (−2; −6)\nand is parallel to the line 2y + 1 = 6x.\n4. Determine the equation of the straight line that passes through the point (−2; −2)\nand is parallel to the line with angle of inclination θ = 56,31◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is parallel to the line with angle of inclination θ = 145◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22N4\n1b. 22N5\n1c. 22N6\n1d. 22N7\n1e. 22N8\n1f. 22N9\n2. 22NB\n3. 22NC\n4. 22ND\n5. 22NF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.5\nPerpendicular lines\nEMBGG\nInvestigation: Perpendicular lines\n1. Draw a sketch of the line passing through the points A(−2; −3) and B(2; 5) and\nthe line passing through the points C(−1; 1\n2) and D(4; −2).\n2. Label and measure α and β, the angles of inclination of straight lines AB and\nCD respectively.\n3. Label and measure θ, the angle between the lines AB and CD.\n4. Describe the relationship between the lines AB and CD.\n5. “θ is a reflex angle, therefore AB ⊥CD.” Is this a true statement? If not, provide\na correct statement.\n136\n4.5.\nPerpendicular lines\n\n6. Determine the equation of the straight line AB and the line CD.\n7. Use your calculator to determine tan α × tan β.\n8. Determine mAB × mCD.\n9. What do you notice about these products?\n10. Complete the sentence: if two lines are . . . . . . to each other, then the product of\ntheir . . . . . . is equal . . . . . .\n11. Complete the sentence: if the gradient of a straight line is equal to the negative\n. . . . . . of the gradient of another straight line, then the two lines are . . . . . .\nDeriving the formula: m1 × m2 = −1\nb\nb\nA(4; 3)\nB(−3; 4)\nθ\n90◦+ θ\nO\ny\nx\nConsider the point A(4; 3) on the Cartesian plane with an angle of inclination A ˆOX =\nθ. Rotate through an angle of 90◦and place point B at (−3; 4) so that we have the\nangle of inclination B ˆOX = 90◦+ θ.\nWe determine the gradient of OA:\nmOA = y2 −y1\nx2 −x1\n= 3 −0\n4 −0\n= 3\n4\nAnd determine the gradient of OB:\nmOB = y2 −y1\nx2 −x1\n= 4 −0\n−3 −0\n= 4\n−3\n137\nChapter 4.\nAnalytical geometry\n\nBy rotating through an angle of 90◦we know that OB ⊥OA:\nmOA × mOB = 3\n4 × 4\n−3\n= −1\nWe can also write that\nmOA = −\n1\nmOB\nb\nb\nA(x; y)\nB(−y; x)\nθ\n90◦+ θ\nO\ny\nx\nIf we have the general point A(x; y) with an angle of inclination A ˆOX = θ and point\nB(−y; x) such that B ˆOX = 90◦+ θ, then we know that\nmOA = y\nx\nmOB = −x\ny\n∴mOA × mOB = y\nx × −x\ny\n= −1\nAnother method of determining the equation of a straight line is to be given a point on\nthe line, (x1; y1), and the equation of a line which is perpendicular to the unknown\nline. Let the equation of the unknown line be y = m1x + c1 and the equation of the\ngiven line be y = m2x + c2.\nθ2\nθ1\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are perpendicular then\nm1 × m2 = −1\nNote: this rule does not apply to vertical or horizontal lines.\n138\n4.5.\nPerpendicular lines\n\nWhen determining the gradient of a line using the coefficient of x, make sure the\ngiven equation is written in the gradient–intercept (standard) form y = mx + c. Then\nwe know that\nm1 = −1\nm2\nSubstitute the value of m1 and the given point (x1; y1), into the gradient–intercept form\nof the straight line equation y −y1 = m(x −x1) and determine the equation of the\nunknown line.\nWorked example 13: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point T(2; 2) and per-\npendicular to the line 3y + 2x −6 = 0.\nSOLUTION\nStep 1: Write the equation in standard form\nLet the gradient of the unknown line be m1 and the given gradient be m2. We write\nthe given equation in gradient–intercept form and determine the value of m2.\n3y + 2x −6 = 0\n3y = −2x + 6\ny = −2\n3x + 2\n∴m2 = −2\n3\nWe know that the two lines are perpendicular, therefore m1 × m2 = −1. Therefore\nm1 = 3\n2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m1 = 3\n2.\ny −y1 = 3\n2(x −x1)\nSubstitute the given point T(2; 2).\ny −2 = 3\n2(x −2)\ny −2 = 3\n2x −3\ny = 3\n2x −1\n139\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\nb T(2; 2)\ny = −2\n3x + 2\ny = 3\n2x −1\nx\ny\n0\nA sketch was not required, but it is useful for checking the answer.\nStep 3: Write the final answer\nThe equation of the straight line is y = 3\n2x −1.\nWorked example 14: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point (2; 1\n3) and per-\npendicular to the line with an angle of inclination of 71,57◦.\nSOLUTION\nStep 1: Use the given angle of inclination to determine gradient\nLet the gradient of the unknown line be m1 and let the given gradient be m2.\nm2 = tan θ\n= tan 71,57◦\n= 3,0\nStep 2: Determine the unknown gradient\nSince we are given that the two lines are perpendicular,\nm1 × m2 = −1\n∴m1 = −1\n3\n140\n4.5.\nPerpendicular lines\n\nStep 3: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient m1 = −1\n3.\ny −y1 = −1\n3(x −x1)\nSubstitute the given point (2; 1\n3).\ny −\n\u00121\n3\n\u0013\n= −1\n3(x −2)\ny −1\n3 = −1\n3x + 2\n3\ny = −1\n3x + 1\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n3x + 1.\nSee video: 22NG at www.everythingmaths.co.za\nExercise 4 – 8: Perpendicular lines\n1. Calculate whether or not the following two lines are perpendicular:\na) y −1 = 4x and 4y + x + 2 = 0\nb) 10x = 5y −1 and 5y −x −10 = 0\nc) x = y −5 and the line passing through (−1; 5\n4) and (3; −11\n4 )\nd) y = 2 and x = 1\ne)\ny\n3 = x and 3y + x = 9\nf) 1 −2x = y and the line passing through (2; −1) and (−1; 5)\ng) y = x + 2 and 2y + 1 = 2x\n2. Determine the equation of the straight line that passes through the point (−2; −4)\nand is perpendicular to the line y + 2x = 1.\n3. Determine the equation of the straight line that passes through the point (2; −7)\nand is perpendicular to the line 5y −x = 0.\n141\nChapter 4.\nAnalytical geometry\n\n4. Determine the equation of the straight line that passes through the point (3; −1)\nand is perpendicular to the line with angle of inclination θ = 135◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is perpendicular to the line y = 4\n3.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NH\n1b. 22NJ\n1c. 22NK\n1d. 22NM\n1e. 22NN\n1f. 22NP\n1g. 22NQ\n2. 22NR\n3. 22NS\n4. 22NT\n5. 22NV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.6\nSummary\nEMBGH\nSee presentation: 22NW at www.everythingmaths.co.za\n• Distance between two points: d =\np\n(x2 −x1)2 + (y2 −y1)2\n• Gradient of a line between two points: m = y2 −y1\nx2 −x1\n• Mid-point of a line: M(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n• Parallel lines: m1 = m2\n• Perpendicular lines: m1 × m2 = −1\n• General form of a straight line equation: ax + by + c = 0\n• Two-point form of a straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• Gradient–point form of a straight line equation: y −y1 = m(x −x1)\n• Gradient–intercept form of a straight line equation (standard form): y = mx + c\n• Angle of inclination of a straight line: θ, the angle formed between the line and\nthe positive x-axis; m = tan θ\n142\n4.6.\nSummary\n\nExercise 4 – 9: End of chapter exercises\n1. Determine the equation of the line:\na) through points (−1; 3) and (1; 4)\nb) through points (7; −3) and (0; 4)\nc) parallel to y = 1\n2x + 3 and passing through (−2; 3)\nd) perpendicular to y = −1\n2x + 3 and passing through (−1; 2)\ne) perpendicular to 3y + x = 6 and passing through the origin\n2. Determine the angle of inclination of the following lines:\na) y = 2x −3\nb) y = 1\n3x −7\nc) 4y = 3x + 8\nd) y = −2\n3x + 3\ne) 3y + x −3 = 0\n3. P(2; 3), Q(−4; 0) and R(5; −3) are the vertices of △PQR in the Cartesian plane.\nPR intersects the x-axis at S. Determine the following:\na) the equation of the line PR\nb) the coordinates of point S\nc) the angle of inclination of PR (correct to two decimal places)\nd) the gradient of line PQ\ne) Q ˆPR\nf) the equation of the line perpendicular to PQ and passing through the origin\ng) the mid-point M of QR\nh) the equation of the line parallel to PR and passing through point M\n4. Points A(−3; 5), B(−7; −4) and C(2; 0) are given.\na) Plot the points on the Cartesian plane.\nb) Determine the coordinates of D if ABCD is a parallelogram.\nc) Prove that ABCD is a rhombus.\n5.\nb\nb\nb\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\nx\ny\n0\nM\nN\nP\n143\nChapter 4.\nAnalytical geometry\n\nConsider the sketch above, with the following lines shown:\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\na) Determine the coordinates of the point N.\nb) Determine the coordinates of the point P.\nc) Determine the equation of the vertical line MN.\nd) Determine the length of the vertical line MN.\ne) Find M ˆNP.\nf) Determine the equation of the line parallel to NP and passing through the\npoint M.\n6. The following points are given: A(−2; 3), B(2; 4), C(3; 0).\na) Plot the points on the Cartesian plane.\nb) Prove that △ABC is a right-angled isosceles triangle.\nc) Determine the equation of the line AB.\nd) Determine the coordinates of D if ABCD is a square.\ne) Determine the coordinates of E, the mid-point of BC.\n7. Given points S(2; 5), T(−3; −4) and V (4; −2).\na) Determine the equation of the line ST.\nb) Determine the size of T ˆSV .\n8. Consider triangle FGH with vertices F(−1; 3), G(2; 1) and H(4; 4).\na) Sketch △FGH on the Cartesian plane.\nb) Show that △FGH is an isosceles triangle.\nc) Determine the equation of the line PQ, perpendicular bisector of FH.\nd) Does G lie on the line PQ?\ne) Determine the equation of the line parallel to GH and passing through\npoint F.\n9. Given the points A(−1; 5), B(5; −3) and C(0; −6). M is the mid-point of AB\nand N is the mid-point of AC.\na) Draw a sketch on the Cartesian plane.\nb) Show that the coordinates of M and N are (2; 1) and (−1\n2; −1\n2) respectively.\nc) Use analytical geometry methods to prove the mid-point theorem. (Prove\nthat NM ∥CB and NM = 1\n2CB.)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NX\n1b. 22NY\n1c. 22NZ\n1d. 22P2\n1e. 22P3\n2a. 22P4\n2b. 22P5\n2c. 22P6\n2d. 22P7\n2e. 22P8\n3. 22P9\n4. 22PB\n5. 22PC\n6. 22PD\n7. 22PF\n8. 22PG\n9. 22PH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n144\n4.6.\nSummary\n\nCHAPTER\n5\nFunctions\n5.1\nQuadratic functions\n146\n5.2\nAverage gradient\n164\n5.3\nHyperbolic functions\n170\n5.4\nExponential functions\n184\n5.5\nThe sine function\n197\n5.6\nThe cosine function\n209\n5.7\nThe tangent function\n222\n5.8\nSummary\n235\n\n5\nFunctions\nA function describes a specific relationship between two variables; where an indepen-\ndent (input) variable has exactly one dependent (output) variable. Every element in the\ndomain maps to only one element in the range. Functions can be one-to-one relations\nor many-to-one relations. A many-to-one relation associates two or more values of the\nindependent variable with a single value of the dependent variable. Functions allow\nus to visualise relationships in the form of graphs, which are much easier to read and\ninterpret than lists of numbers.\n5.1\nQuadratic functions\nEMBGJ\nRevision\nEMBGK\nFunctions of the form y = ax2 + q\nFunctions of the general form y = ax2 + q are called parabolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = ax2 + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\nThe turning point of f(x) is\nabove the x-axis.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\nThe turning point of f(x) is be-\nlow the x-axis.\n– q is also the y-intercept of the\nparabola.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\n• The effect of a on shape\n– For a > 0; the graph of f(x) is a “smile” and has a minimum turning point\n(0; q). As the value of a becomes larger, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n– For a < 0; the graph of f(x) is a “frown” and has a maximum turning point\n(0; q). As the value of a becomes smaller, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n146\n5.1.\nQuadratic functions\n\nExercise 5 – 1: Revision\n1. On separate axes, accurately draw each of the following functions.\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = x2\nb) y2 = 1\n2x2\nc) y3 = −x2 −1\nd) y4 = −2x2 + 4\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nyint = 0\nvalue of a\na = 1\neffect of a\nstandard\nparabola\nturning point\n(0; 0)\naxis of symmetry\nx = 0\n(y-axis)\ndomain\n{x : x ∈R}\nrange\n{y : y ≥0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22PJ\n1b. 22PK\n1c. 22PM\n1d. 22PN\n2. 22PP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22PQ at www.everythingmaths.co.za\n147\nChapter 5.\nFunctions\n\nFunctions of the form y = a(x + p)2 + q\nEMBGM\nWe now consider parabolic functions of the form y = a(x + p)2 + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a parabolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = x2\nb) y2 = (x −2)2\nc) y3 = (x −1)2\nd) y4 = (x + 1)2\ne) y5 = (x + 2)2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = x2 + 2\nb) y2 = (x −2)2 −1\nc) y3 = (x −1)2 + 1\nd) y4 = (x + 1)2 + 1\ne) y5 = (x + 2)2 −1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of q\n3. Consider the three functions given below and answer the questions that follow:\n• y1 = (x −2)2 + 1\n• y2 = 2(x −2)2 + 1\n• y3 = −1\n2(x −2)2 + 1\na) What is the value of a for y2?\nb) Does y1 have a minimum or maximum turning point?\n148\n5.1.\nQuadratic functions\n\nc) What are the coordinates of the turning point of y2?\nd) Compare the graphs of y1 and y2. Discuss the similarities and differences.\ne) What is the value of a for y3?\nf) Will the graph of y3 be narrower or wider than the graph of y1?\ng) Determine the coordinates of the turning point of y3.\nh) Compare the graphs of y1 and y3. Describe any differences.\nSee video: 22PR at www.everythingmaths.co.za\nThe effect of the parameters on y = a(x + p)2 + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects whether the turning point is to the left of the y-axis (p > 0)\nor to the right of the y-axis (p < 0). The axis of symmetry is the line x = −p.\nThe effect of q is a vertical shift. The value of q affects whether the turning point of the\ngraph is above the x-axis (q > 0) or below the x-axis (q < 0).\nThe value of a affects the shape of the graph. If a < 0, the graph is a “frown” and has\na maximum turning point. If a > 0 then the graph is a “smile” and has a minimum\nturning point. When a = 0, the graph is a horizontal line y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\nSee simulation: 22PS at www.everythingmaths.co.za\n149\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form f(x) = y = a(x + p)2 + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative. If a > 0\nwe have:\n(x + p)2 ≥0\n(perfect square is always positive)\n∴a(x + p)2 ≥0\n(a is positive)\n∴a(x + p)2 + q ≥q\n∴f(x) ≥q\nThe range is therefore {y : y ≥q, y ∈R} if a > 0. Similarly, if a < 0, the range is\n{y : y ≤q, y ∈R}.\nWorked example 1: Domain and range\nQUESTION\nState the domain and range for g(x) = −2(x −1)2 + 3.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n(x −1)2 ≥0\n−2(x −1)2 ≤0\n−2(x −1)2 + 3 ≤3\ng(x) ≤3\nTherefore the range is {g(x) : g(x) ≤3} or in interval notation (−∞; 3].\nNotice in the example above that it helps to have the function in the form y = a(x +\np)2 + q.\nWe use the method of completing the square to write a quadratic function of the\ngeneral form y = ax2 + bx + c in the form y = a(x + p)2 + q (see Chapter 2).\n150\n5.1.\nQuadratic functions\n\nExercise 5 – 2: Domain and range\nGive the domain and range for each of the following functions:\n1. f(x) = (x −4)2 −1\n2. g(x) = −(x −5)2 + 4\n3. h(x) = x2 −6x + 9\n4. j(x) = −2(x + 1)2\n5. k(x) = −x2 + 2x −3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PT\n2. 22PV\n3. 22PW\n4. 22PX\n5. 22PY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nEvery point on the y-axis has an x-coordinate of 0, therefore to calculate the y-intercept\nwe let x = 0.\nFor example, the y-intercept of g(x) = (x −1)2 + 5 is determined by setting x = 0:\ng(x) = (x −1)2 + 5\ng(0) = (0 −1)2 + 5\n= 6\nThis gives the point (0; 6).\nThe x-intercept:\nEvery point on the x-axis has a y-coordinate of 0, therefore to calculate the x-intercept\nwe let y = 0.\nFor example, the x-intercept of g(x) = (x −1)2 + 5 is determined by setting y = 0:\ng(x) = (x −1)2 + 5\n0 = (x −1)2 + 5\n−5 = (x −1)2\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n151\nChapter 5.\nFunctions\n\nExercise 5 – 3: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = (x + 4)2 −1\n2. g(x) = 16 −8x + x2\n3. h(x) = −x2 + 4x −3\n4. j(x) = 4(x −3)2 −1\n5. k(x) = 4(x −3)2 + 1\n6. l(x) = 2x2 −3x −4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PZ\n2. 22Q2\n3. 22Q3\n4. 22Q4\n5. 22Q5\n6. 22Q6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTurning point\nThe turning point of the function f(x) = a(x+p)2 +q is determined by examining the\nrange of the function:\n• If a > 0, f(x) has a minimum turning point and the range is [q; ∞):\nThe minimum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\n• If a < 0, f(x) has a maximum turning point and the range is (−∞; q]:\nThe maximum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\nTherefore the turning point of the quadratic function f(x) = a(x + p)2 + q is (−p; q).\nAlternative form for quadratic equations:\nWe can also write the quadratic equation in the form\ny = a(x −p)2 + q\nThe effect of p is still a horizontal shift, however notice that:\n• For p > 0, the graph is shifted to the right by p units.\n• For p < 0, the graph is shifted to the left by p units.\nThe turning point is (p; q) and the axis of symmetry is the line x = p.\n152\n5.1.\nQuadratic functions\n\nWorked example 2: Turning point\nQUESTION\nDetermine the turning point of g(x) = 3x2 −6x −1.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q\nWe use the method of completing the square:\ng(x) = 3x2 −6x −1\n= 3(x2 −2x) −1\n= 3\n\u0000(x −1)2 −1\n\u0001\n−1\n= 3(x −1)2 −3 −1\n= 3(x −1)2 −4\nStep 2: Determine turning point (−p; q)\nFrom the equation g(x) = 3(x −1)2 −4 we know that the turning point for g(x) is\n(1; −4).\nWorked example 3: Turning point\nQUESTION\n1. Show that the x-value for the turning point of h(x) = ax2 + bx + c is given by\nx = −b\n2a.\n2. Hence, determine the turning point of k(x) = 2 −10x + 5x2.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q and show that p =\nb\n2a\nWe use the method of completing the square:\nh(x) = ax2 + bx + c\n= a\n\u0012\nx2 + b\nax + c\na\n\u0013\nTake half the coefficient of the x term and square it; then add and subtract it from the\n153\nChapter 5.\nFunctions\n\nexpression.\nh(x) = a\n \nx2 + b\nax +\n\u0012 b\n2a\n\u00132\n−\n\u0012 b\n2a\n\u00132\n+ c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2\n4a2 + c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a2\n!\n= a\n\u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a\nFrom the above we have that the turning point is at x = −p = −b\n2a and y = q =\n−b2−4ac\n4a\n.\nStep 2: Determine the turning point of k(x)\nWrite the equation in the general form y = ax2 + bx + c.\nk(x) = 5x2 −10x + 2\nTherefore a = 5; b = −10; c = 2.\nUse the results obtained above to determine x = −b\n2a:\nx = −\n\u0012−10\n2(5)\n\u0013\n= 1\nSubstitute x = 1 to obtain the corresponding y-value :\ny = 5x2 −10x + 2\n= 5(1)2 −10(1) + 2\n= 5 −10 + 2\n= −3\nThe turning point of k(x) is (1; −3).\nExercise 5 – 4: Turning points\nDetermine the turning point of each of the following:\n1. y = x2 −6x + 8\n2. y = −x2 + 4x −3\n3. y = 1\n2(x + 2)2 −1\n4. y = 2x2 + 2x + 1\n154\n5.1.\nQuadratic functions\n\n5. y = 18 + 6x −3x2\n6. y = −2[(x + 1)2 + 3]\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22Q7\n2. 22Q8\n3. 22Q9\n4. 22QB\n5. 22QC\n6. 22QD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxis of symmetry\nThe axis of symmetry for f(x) = a(x + p)2 + q is the vertical line x = −p. The axis of\nsymmetry passes through the turning point (−p; q) and is parallel to the y-axis.\ny\nx\n0\nb\nx = −p\nf(x) = a(x + p)2 + q\nExercise 5 – 5: Axis of symmetry\n1. Determine the axis of symmetry of each of the following:\na) y = 2x2 −5x −18\nb) y = 3(x −2)2 + 1\nc) y = 4x −x2\n2. Write down the equation of a parabola where the y-axis is the axis of symmetry.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QF\n1b. 22QG\n1c. 22QH\n2. 22QJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n155\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) = a(x + p)2 + q\nIn order to sketch graphs of the form f(x) = a(x + p)2 + q, we need to determine five\ncharacteristics:\n• sign of a\n• turning point\n• y-intercept\n• x-intercept(s) (if they exist)\n• domain and range\nSee video: 22QK at www.everythingmaths.co.za\nWorked example 4: Sketching a parabola\nQUESTION\nSketch the graph of y = −1\n2(x + 1)2 −3.\nMark the intercepts, turning point and the axis of symmetry. State the domain and\nrange of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = a(x + p)2 + q\nWe notice that a < 0, therefore the graph is a “frown” and has a maximum turning\npoint.\nStep 2: Determine the turning point (−p; q)\nFrom the equation we know that the turning point is (−1; −3).\nStep 3: Determine the axis of symmetry x = −p\nFrom the equation we know that the axis of symmetry is x = −1.\nStep 4: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = −1\n2 ((0) + 1)2 −3\n= −1\n2 −3\n= −31\n2\nThis gives the point (0; −31\n2).\n156\n5.1.\nQuadratic functions\n\nStep 5: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = −1\n2 (x + 1)2 −3\n3 = −1\n2 (x + 1)2\n−6 = (x + 1)2\nwhich has no real solutions. Therefore, there are no x-intercepts and the graph lies\nbelow the x-axis.\nStep 6: Plot the points and sketch the graph\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\nb\ny\nx\n0\n(0; −3 1\n2)\n(−1; −3)\nStep 7: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≤−3, y ∈R}\nSee video: 22QM at www.everythingmaths.co.za\nWorked example 5: Sketching a parabola\nQUESTION\nSketch the graph of y = 1\n2x2 −4x + 7\n2.\nDetermine the intercepts, turning point and the axis of symmetry. Give the domain\nand range of the function.\nSOLUTION\n157\nChapter 5.\nFunctions\n\nStep 1: Examine the equation of the form y = ax2 + bx + c\nWe notice that a > 0, therefore the graph is a “smile” and has a minimum turning\npoint.\nStep 2: Determine the turning point and the axis of symmetry\nCheck that the equation is in standard form and identify the coefficients.\na = 1\n2;\nb = −4;\nc = 7\n2\nCalculate the x-value of the turning point using\nx = −b\n2a\n= −\n \n−4\n2\n\u0000 1\n2\n\u0001\n!\n= 4\nTherefore the axis of symmetry is x = 4.\nSubstitute x = 4 into the original equation to obtain the corresponding y-value.\ny = 1\n2x2 −4x + 7\n2\n= 1\n2(4)2 −4(4) + 7\n2\n= 8 −16 + 7\n2\n= −41\n2\nThis gives the point\n\u00004; −41\n2\n\u0001\n.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = 1\n2(0)2 −4(0) + 7\n2\n= 7\n2\nThis gives the point\n\u00000; 7\n2\n\u0001\n.\nStep 4: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = 1\n2x2 −4x + 7\n2\n= x2 −8x + 7\n= (x −1)(x −7)\nTherefore x = 1 or x = 7. This gives the points (1; 0) and (7; 0).\nStep 5: Plot the points and sketch the graph\n158\n5.1.\nQuadratic functions\n\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\nb\nb\ny\nx\n0\n(4; −41\n2)\n(0; 3 1\n2)\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≥−41\n2, y ∈R}\nSee video: 22QN at www.everythingmaths.co.za\nInvestigation: Shifting the equation of a parabola\nCarl and Eric are doing their Mathematics homework and decide to check each others\nanswers.\nHomework question:\nIf the parabola y = 3x2 + 1 is shifted 2 units to the right, determine the equation of the\nnew parabola.\n• Carl’s answer:\nA shift to the right means moving in the positive x direction, therefore x is re-\nplaced with x + 2 and the new equation is y = 3(x + 2)2 + 1.\n• Eric’s answer:\nWe replace x with x −2, therefore the new equation is y = 3(x −2)2 + 1.\nWork together in pairs. Discuss the two different answers and decide which one is\ncorrect. Use calculations and sketches to help explain your reasoning.\n159\nChapter 5.\nFunctions\n\nWriting an equation of a shifted parabola\nThe parabola is shifted horizontally:\n• If the parabola is shifted m units to the right, x is replaced by (x −m).\n• If the parabola is shifted m units to the left, x is replaced by (x + m).\nThe parabola is shifted vertically:\n• If the parabola is shifted n units down, y is replaced by (y + n).\n• If the parabola is shifted n units up, y is replaced by (y −n).\nWorked example 6: Shifting a parabola\nQUESTION\nGiven y = x2 −2x −3.\n1. If the parabola is shifted 1 unit to the right, determine the new equation of the\nparabola.\n2. If the parabola is shifted 3 units down, determine the new equation of the\nparabola.\nSOLUTION\nStep 1: Determine the new equation of the shifted parabola\n1. The parabola is shifted 1 unit to the right, so x must be replaced by (x −1).\ny = x2 −2x −3\n= (x −1)2 −2(x −1) −3\n= x2 −2x + 1 −2x + 2 −3\n= x2 −4x\nBe careful not to make a common error: replacing x with x + 1 for a shift to the\nright.\n2. The parabola is shifted 3 units down, so y must be replaced by (y + 3).\ny + 3 = x2 −2x −3\ny = x2 −2x −3 −3\n= x2 −2x −6\n160\n5.1.\nQuadratic functions\n\nExercise 5 – 6: Sketching parabolas\n1. Sketch graphs of the following functions and determine:\n• intercepts\n• turning point\n• axes of symmetry\n• domain and range\na) y = −x2 + 4x + 5\nb) y = 2(x + 1)2\nc) y = 3x2 −2(x + 2)\nd) y = 3(x −2)2 + 1\n2. Draw the following graphs on the same system of axes:\nf(x) = −x2 + 7\ng(x) = −(x −2)2 + 7\nh(x) = (x −2)2 −7\n3. Draw a sketch of each of the following graphs:\na) y = ax2 + bx + c if a > 0, b > 0, c < 0.\nb) y = ax2 + bx + c if a < 0, b = 0, c > 0.\nc) y = ax2 + bx + c if a < 0, b < 0, b2 −4ac < 0.\nd) y = (x + p)2 + q if p < 0, q < 0 and the x-intercepts have different signs.\ne) y = a(x + p)2 + q if a < 0, p < 0, q > 0 and one root is zero.\nf) y = a(x + p)2 + q if a > 0, p = 0, b2 −4ac > 0.\n4. Determine the new equation (in the form y = ax2 + bx + c) if:\na) y = 2x2 + 4x + 2 is shifted 3 units to the left.\nb) y = −(x + 1)2 is shifted 1 unit up.\nc) y = 3(x −1)2 + 2\n\u0000x −1\n2\n\u0001\nis shifted 2 units to the right.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QP\n1b. 22QQ\n1c. 22QR\n1d. 22QS\n2. 22QT\n3a. 22QV\n3b. 22QW\n3c. 22QX\n3d. 22QY\n3e. 22QZ\n3f. 22R2\n4a. 22R3\n4b. 22R4\n4c. 22R5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n161\nChapter 5.\nFunctions\n\nFinding the equation of a parabola from the graph\nIf the intercepts are given, use y = a(x −x1)(x −x2).\nExample:\ny\nx\n0\n(0; 2)\n(4; 0)\n(−1; 0)\nx-intercepts: (−1; 0) and (4; 0)\ny = a(x −x1)(x −x2)\n= a(x + 1)(x −4)\n= ax2 −3ax −4a\ny-intercept: (0; 2)\n−4a = 2\na = −1\n2\nEquation of the parabola:\ny = ax2 −3ax −4a\n= −1\n2x2 −3\n\u0012\n−1\n2\n\u0013\nx −4\n\u0012\n−1\n2\n\u0013\n= −1\n2x2 + 3\n2x + 2\nIf the x-intercepts and another point are given, use y = a(x −x1)(x −x2).\nExample:\nb\ny\nx\n0\n(−1; 12)\n(1; 0)\n(5; 0)\nx-intercepts: (1; 0) and (5; 0)\ny = a(x −x1)(x −x2)\n= a(x −1)(x −5)\n= ax2 −6ax + 5a\nSubstitute the point: (−1; 12)\n12 = a(−1)2 −6a(−1) + 5a\n12 = a + 6a + 5a\n12 = 12a\n1 = a\nEquation of the parabola:\ny = ax2 −6ax + 5a\n= x2 −6x + 5\nIf the turning point and another point are given, use y = a(x + p)2 + q.\nExample:\nb\nb\ny\nx\n0\n(1; 5)\n(−3; 1)\nTurning point: (−3; 1)\ny = a(x + p)2 + q\n= a(x + 3)2 + 1\n= ax2 + 6ax + 9a + 1\nSubstitute the point: (1; 5)\n5 = a(1)2 + 6a(1) + 9a + 1\n4 = 16a\n1\n4 = a\nEquation of the parabola:\ny = 1\n4(x + 3)2 + 1\n162\n5.1.\nQuadratic functions\n\nExercise 5 – 7: Finding the equation\nDetermine the equations of the following graphs. Write your answers in the form\ny = a(x + p)2 + q.\n1.\nb\nb\ny\nx\n0\n3\n(−1; 6)\n2.\nb\ny\nx\n0\n(−1; 3)\n5\n3.\nb\nb\ny\nx\n0\n(1; 6)\n−2\n4.\nb\nb\nb\ny\nx\n0\n(1; 6)\n(3; 4)\n4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22R6\n2. 22R7\n3. 22R8\n4. 22R9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n163\nChapter 5.\nFunctions\n\n5.2\nAverage gradient\nEMBGN\nWe notice that the gradient of a curve changes at every point on the curve, therefore\nwe need to work with the average gradient. The average gradient between any two\npoints on a curve is the gradient of the straight line passing through the two points.\ny\nx\n0\nb\nb\nA(−3; 7)\nC(−1; −1)\nFor the diagram above, the gradient of the line AC is\nGradient = yA −yC\nxA −xC\n= 7 −(−1)\n−3 −(−1)\n= 8\n−2\n= −4\nThis is the average gradient of the curve between the points A and C.\nWhat happens to the gradient if we fix the position of one point and move the second\npoint closer to the fixed point?\nSee video: 22RB at www.everythingmaths.co.za\nInvestigation: Gradient at a single point on a curve\nThe curve shown here is defined by y = −2x2 −5.\nPoint B is fixed at (0; −5) and the position of point\nA varies.\nComplete the table below by calculating the y-\ncoordinates of point A for the given x-coordinates\nand then calculating the average gradient between\npoints A and B.\ny\nx\n0\nb\nb\nA\nB(0; −5)\n164\n5.2.\nAverage gradient\n\nxA\nyA\nAverage gradient\n−2\n−1,5\n−1\n−0,5\n0\n0,5\n1\n1,5\n2\n1. What happens to the average gradient as A moves towards B?\n2. What happens to the average gradient as A moves away from B?\n3. What is the average gradient when A overlaps with B?\nIn the example above, the gradient of the straight line that passes through points A and\nC changes as A moves closer to C. At the point where A and C overlap, the straight\nline only passes through one point on the curve. This line is known as a tangent to the\ncurve.\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb A\nC\ny\nx\n0\nb\nb\nA\nC\nWe therefore introduce the idea of the gradient at a single point on a curve. The\ngradient at a point on a curve is the gradient of the tangent to the curve at the given\npoint.\n165\nChapter 5.\nFunctions\n\nWorked example 7: Average gradient\nQUESTION\ny\nx\n0\nb\nb\nP(a; g(a))\nQ(a + h; g(a + h))\ng(x) = x2\n1. Find the average gradient between two points P (a; g(a)) and Q (a + h; g(a + h))\non a curve g(x) = x2.\n2. Determine the average gradient between P (2; g(2)) and Q (5; g(5)).\n3. Explain what happens to the average gradient if Q moves closer to P.\nSOLUTION\nStep 1: Assign labels to the x-values for the given points\nx1 = a\nx2 = a + h\nStep 2: Determine the corresponding y-coordinates\nUsing the function g(x) = x2, we can determine:\ny1 = g(a)\n= a2\ny2 = g(a + h)\n= (a + h)2\n= a2 + 2ah + h2\n166\n5.2.\nAverage gradient\n\nStep 3: Calculate the average gradient\ny2 −y1\nx2 −x1\n=\n\u0000a2 + 2ah + h2\u0001\n−\n\u0000a2\u0001\n(a + h) −(a)\n= a2 + 2ah + h2 −a2\na + h −a\n= 2ah + h2\nh\n= h(2a + h)\nh\n= 2a + h\nThe average gradient between P (a; g(a)) and Q (a + h; g(a + h)) on the curve g(x) =\nx2 is 2a + h.\nStep 4: Calculate the average gradient between P (2; g(2)) and Q (5; g(5))\nThe x-coordinate of P is a and the x-coordinate of Q is a+h therefore if we know that\na = 2 and a + h = 5, then h = 3.\nThe average gradient is therefore 2a + h = 2 (2) + (3) = 7\nStep 5: When Q moves closer to P\nWhen point Q moves closer to point P, h gets smaller.\nWhen the point Q overlaps with the point P, h = 0 and the gradient is given by 2a.\nWe can write the equation for average gradient in another form. Given a curve f(x)\nwith two points P and Q with P (a; f(a)) and Q (a + h; f(a + h)). The average gradi-\nent between P and Q is:\nAverage gradient = yQ −yP\nxQ −xP\n= f(a + h) −f(a)\n(a + h) −(a)\n= f(a + h) −f(a)\nh\nThis result is important for calculating the gradient at a point on a curve and will be\nexplored in greater detail in Grade 12.\n167\nChapter 5.\nFunctions\n\nWorked example 8: Average gradient\nQUESTION\nGiven f(x) = −2x2.\n1. Draw a sketch of the function and determine the average gradient between the\npoints A, where x = 1, and B, where x = 3.\n2. Determine the gradient of the curve at point A.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that a < 0, therefore the graph is a “frown” and has a\nmaximum turning point. We also see that when x = 0, y = 0, therefore the graph\npasses through the origin.\nStep 2: Draw a rough sketch\ny = −2x2\nx\ny\n0\nb\nb\nA\nB\nStep 3: Calculate the average gradient between A and B\nAverage gradient = f(3) −f(1)\n3 −1\n= −2(3)2 −(−2(1)2)\n2\n= −18 + 2\n2\n= −16\n2\n= −8\n168\n5.2.\nAverage gradient\n\nStep 4: Calculate the average gradient for f(x)\nAverage gradient = f(a + h) −f(a)\n(a + h) −a\n= −2(a + h)2 −(−2a2)\nh\n= −2a2 −4ah −2h2 + 2a2\nh\n= −4ah −2h2\nh\n= h(−4a −2h)\nh\n= −4a −2h\nAt point A, h = 0 and a = 1. Therefore\nAverage gradient = −4a −2h\n= −4(1) −2(0)\n= −4\nExercise 5 – 8:\n1.\na) Determine the average gradient of the curve f(x) = x (x + 3) between\nx = 5 and x = 3.\nb) Hence, state what you can deduce about the function f between x = 5 and\nx = 3.\n2. A (1; 3) is a point on f(x) = 3x2.\na) Draw a sketch of f(x) and label point A.\nb) Determine the gradient of the curve at point A.\nc) Determine the equation of the tangent line at A.\n3. Given: g(x) = −x2 + 1.\na) Draw a sketch of g(x).\nb) Determine the average gradient of the curve between x = −2 and x = 1.\nc) Determine the gradient of g at x = 2.\nd) Determine the gradient of g at x = 0.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RC\n2. 22RD\n3. 22RF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n169\nChapter 5.\nFunctions\n\n5.3\nHyperbolic functions\nEMBGP\nRevision\nEMBGQ\nFunctions of the form y = a\nx + q\nFunctions of the general form y = a\nx + q are called hyperbolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = a\nx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted vertically upwards by q units.\n– For q < 0, f(x) is shifted vertically downwards by q units.\n– The horizontal asymptote is the line y = q.\n– The vertical asymptote is the y-axis, the line x = 0.\n• The effect of a on shape and quad-\nrants\n– For a > 0, f(x) lies in the first\nand third quadrants.\n– For a > 1, f(x) will be further\naway from both axes than y =\n1\nx.\n– For 0 < a < 1, as a tends to\n0, f(x) moves closer to the axes\nthan y = 1\nx.\n– For a < 0, f(x) lies in the sec-\nond and fourth quadrants.\n– For a < −1, f(x) will be further\naway from both axes than y =\n−1\nx.\n– For −1 < a < 0, as a tends to\n0, f(x) moves closer to the axes\nthan y = −1\nx.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\nExercise 5 – 9: Revision\n1. Consider the following hyperbolic functions:\n• y1 = 1\nx\n• y2 = −4\nx\n• y3 = 4\nx −2\n• y4 = −4\nx + 1\n170\n5.3.\nHyperbolic functions\n\nComplete the table to summarise the properties of the hyperbolic function:\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nlies in I and III quad\nasymptotes\ny-axis, x = 0\nx-axis, y = 0\naxes of symmetry\ny = x\ny = −x\ndomain\n{x : x ∈R, x ̸= 0}\nrange\n{y : y ∈R, y ̸= 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22RH at www.everythingmaths.co.za\nFunctions of the form y =\na\nx+p + q\nEMBGR\nWe now consider hyperbolic functions of the form y =\na\nx+p + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a hyperbolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 1\nx\nb) y2 =\n1\nx−2\nc) y3 =\n1\nx−1\nd) y4 =\n1\nx+1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nintercept(s)\nasymptotes\naxes of symmetry\ndomain\nrange\neffect of p\n171\nChapter 5.\nFunctions\n\n2. Complete the following sentences for functions of the form y =\na\nx+p + q:\na) A change in p causes a . . . . . . shift.\nb) If the value of p increases, the graph and the vertical asymptote . . . . . .\nc) If the value of q changes, then the . . . . . . asymptote of the hyperbola will\nshift.\nd) If the value of p decreases, the graph and the vertical asymptote . . . . . .\nThe effect of the parameters on y =\na\nx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects the vertical asymptote, the line x = −p.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position on the Cartesian plane.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n172\n5.3.\nHyperbolic functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y =\na\nx+p + q:\nDomain and range\nThe domain is {x : x ∈R, x ̸= −p}. If x = −p, the dominator is equal to zero and the\nfunction is undefined.\nWe see that\ny =\na\nx + p + q\ncan be re-written as:\ny −q =\na\nx + p\nIf x ̸= −p then:\n(y −q) (x + p) = a\nx + p =\na\ny −q\nThe range is therefore {y : y ∈R, y ̸= q}.\nThese restrictions on the domain and range determine the vertical asymptote x = −p\nand the horizontal asymptote y = q.\nWorked example 9: Domain and range\nQUESTION\nDetermine the domain and range for g(x) =\n2\nx+1 + 2.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R, x ̸= −1} since g(x) is undefined for x = −1.\nStep 2: Determine the range\nLet g(x) = y:\ny =\n2\nx + 1 + 2\ny −2 =\n2\nx + 1\n(y −2)(x + 1) = 2\nx + 1 =\n2\ny −2\nTherefore the range is {g(x) : g(x) ∈R, g(x) ̸= 2}.\n173\nChapter 5.\nFunctions\n\nExercise 5 – 10: Domain and range\nDetermine the domain and range for each of the following functions:\n1. y = 1\nx + 1\n2. g(x) =\n8\nx−8 + 4\n3. y = −\n4\nx+1 −3\n4. x =\n2\n3−y + 5\n5. (y −2)(x + 2) = 3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RJ\n2. 22RK\n3. 22RM\n4. 22RN\n5. 22RP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n2\nx+1+2\nis determined by setting x = 0:\ng(x) =\n2\nx + 1 + 2\ng(0) =\n2\n0 + 1 + 2\n= 2 + 2\n= 4\nThis gives the point (0; 4).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n2\nx+1+2\nis determined by setting y = 0:\ng(x) =\n2\nx + 1 + 2\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\n174\n5.3.\nHyperbolic functions\n\nExercise 5 – 11: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) =\n1\nx+4 −2\n2. g(x) = −5\nx + 2\n3. j(x) =\n2\nx−1 + 3\n4. h(x) =\n3\n6−x + 1\n5. k(x) =\n5\nx+2 −1\n2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RQ\n2. 22RR\n3. 22RS\n4. 22RT\n5. 22RV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptotes\nThere are two asymptotes for functions of the form y =\na\nx+p + q. The asymptotes\nindicate the values of x for which the function does not exist. In other words, the\nvalues that are excluded from the domain and the range. The horizontal asymptote is\nthe line y = q and the vertical asymptote is the line x = −p.\nExercise 5 – 12: Asymptotes\nDetermine the asymptotes for each of the following functions:\n1. y =\n1\nx+4 −2\n2. y = −5\nx\n3. y =\n3\n2−x + 1\n4. y = 1\nx −8\n5. y = −\n2\nx−2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RW\n2. 22RX\n3. 22RY\n4. 22RZ\n5. 22S2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxes of symmetry\nThere are two lines about which a hyperbola is symmetrical.\nFor the standard hyperbola y = 1\nx, we see that if we replace x ⇒y and y ⇒x, we get\ny = 1\nx. Similarly, if we replace x ⇒−y and y ⇒−x, the function remains the same.\nTherefore the function is symmetrical about the lines y = x and y = −x.\nFor the shifted hyperbola y =\na\nx+p + q, the axes of symmetry intersect at the point\n(−p; q).\n175\nChapter 5.\nFunctions\n\nTo determine the axes of symmetry we define the two straight lines y1 = m1x + c1 and\ny2 = m2x + c2. For the standard and shifted hyperbolic function, the gradient of one\nof the lines of symmetry is 1 and the gradient of the other line of symmetry is −1. The\naxes of symmetry are perpendicular to each other and the product of their gradients\nequals −1. Therefore we let y1 = x+c1 and y2 = −x+c2. We then substitute (−p; q),\nthe point of intersection of the axes of symmetry, into both equations to determine the\nvalues of c1 and c2.\nWorked example 10: Axes of symmetry\nQUESTION\nDetermine the axes of symmetry for y =\n2\nx+1 −2.\nSOLUTION\nStep 1: Determine the point of intersection (−p; q)\nFrom the equation we see that p = 1 and q = −2. So the axes of symmetry will\nintersect at (−1; −2).\nStep 2: Define two straight line equations\ny1 = x + c1\ny2 = −x + c2\nStep 3: Solve for c1 and c2\nUse (−1; −2) to solve for c1:\ny1 = x + c1\n−2 = −1 + c1\n−1 = c1\nUse (−1; −2) to solve for c2:\ny2 = −x + c2\n−2 = −(−1) + c2\n−3 = c2\nStep 4: Write the final answer\nThe axes of symmetry for y =\n2\nx+1 −2 are the lines\ny1 = x −1\ny2 = −x −3\n176\n5.3.\nHyperbolic functions\n\n1\n2\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny1 = x −1\ny2 = −x −3\nExercise 5 – 13: Axes of symmetry\n1. Complete the following for f(x) and g(x):\n• Sketch the graph.\n• Determine (−p; q).\n• Find the axes of symmetry.\nCompare f(x) and g(x) and also their axes of symmetry. What do you notice?\na) f(x) = 2\nx\ng(x) = 2\nx + 1\nb) f(x) = −3\nx\ng(x) = −\n3\nx+1\nc) f(x) = 5\nx\ng(x) =\n5\nx−1 −1\n2. A hyperbola of the form k(x) =\na\nx+p + q passes through the point (4; 3). If the\naxes of symmetry intersect at (−1; 2), determine the equation of k(x).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S3\n1b. 22S4\n1c. 22S5\n2. 22S6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n177\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) =\na\nx+p + q\nIn order to sketch graphs of functions of the form, f(x) =\na\nx+p +q, we need to calculate\nfive characteristics:\n• quadrants\n• asymptotes\n• y-intercept\n• x-intercept\n• domain and range\nWorked example 11: Sketching a hyperbola\nQUESTION\nSketch the graph of y =\n2\nx+1 + 2. Determine the intercepts, asymptotes and axes of\nsymmetry. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Determine the asymptotes\nFrom the equation we know that p = 1 and q = 2.\nTherefore the horizontal asymptote is the line y = 2 and the vertical asymptote is the\nline x = −1.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n2\n0 + 1 + 2\n= 4\nThis gives the point (0; 4).\nStep 4: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n178\n5.3.\nHyperbolic functions\n\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\nStep 5: Determine the axes of symmetry\nUsing (−1; 2) to solve for c1:\ny1 = x + c1\n2 = −1 + c1\n3 = c1\ny2 = −x + c2\n2 = −(−1) + c2\n1 = c2\nTherefore the axes of symmetry are y = x + 3 and y = −x + 1.\nStep 6: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny = x + 3\ny = −x + 1\nStep 7: State the domain and range\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 2}\n179\nChapter 5.\nFunctions\n\nWorked example 12: Sketching a hyperbola\nQUESTION\nUse horizontal and vertical shifts to sketch the graph of f(x) =\n1\nx−2 + 3.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Sketch the standard hyperbola y = 1\nx\nStart with a sketch of the standard hyperbola g(x) = 1\nx. The vertical asymptote is x = 0\nand the horizontal asymptote is y = 0.\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 3: Determine the vertical shift\nFrom the equation we see that q = 3, which means g(x) must shifted 3 units up. The\nhorizontal asymptote is also shifted 3 units up to y = 3 .\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 4: Determine the horizontal shift\nFrom the equation we see that p = −2, which means g(x) must shifted 2 units to the\nright. The vertical asymptote is also shifted 2 units to the right.\n180\n5.3.\nHyperbolic functions\n\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n5\n6\n−1\n−2\ny\nx\n0\nStep 5: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n1\n0 −2 + 3\n= 21\n2\nThis gives the point (0; 21\n2).\nStep 6: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n1\nx −2 + 3\n−3 =\n1\nx −2\n−3(x −2) = 1\n−3x + 6 = 1\n−3x = −5\nx = 5\n3\nThis gives the point (5\n3; 0).\nStep 7: Determine the domain and range\nDomain: {x : x ∈R, x ̸= 2}\nRange: {y : y ∈R, y ̸= 3}\n181\nChapter 5.\nFunctions\n\nWorked example 13: Finding the equation of a hyperbola from the graph\nQUESTION\nUse the graph below to determine the values of a, p and q for y =\na\nx+p + q.\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\nSOLUTION\nStep 1: Examine the graph and deduce the sign of a\nWe notice that the graph lies in the second and fourth quadrants, therefore a < 0.\nStep 2: Determine the asymptotes\nFrom the graph we see that the vertical asymptote is x = −1, therefore p = 1. The\nhorizontal asymptote is y = 3, and therefore q = 3.\ny =\na\nx + 1 + 3\nStep 3: Determine the value of a\nTo determine the value of a we substitute a point on the graph, namely (0; 0):\ny =\na\nx + 1 + 3\n0 =\na\n0 + 1 + 3\n∴−3 = a\nStep 4: Write the final answer\ny = −\n3\nx + 1 + 3\n182\n5.3.\nHyperbolic functions\n\nExercise 5 – 14: Sketching graphs\n1. Draw the graphs of the following functions and indicate:\n• asymptotes\n• intercepts, where applicable\n• axes of symmetry\n• domain and range\na) y = 1\nx + 2\nb) y =\n1\nx+4 −2\nc) y = −\n1\nx+1 + 3\nd) y = −\n5\nx−2 1\n2 −2\ne) y =\n8\nx−8 + 4\n2. Given the graph of the hyperbola of the form y =\n1\nx+p + q, determine the values\nof p and q.\ny\nx\n−2\n−1\n3. Given a sketch of the function of the form y =\na\nx+p + q, determine the values of\na, p and q.\ny\nx\n2\n2\n4.\na) Draw the graph of f(x) = −3\nx, x > 0.\nb) Determine the average gradient of the graph between x = 1 and x = 3.\nc) Is the gradient at (1\n2; −6) less than or greater than the average gradient be-\ntween x = 1 and x = 3? Illustrate this on your graph.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S7\n1b. 22S8\n1c. 22S9\n1d. 22SB\n1e. 22SC\n2. 22SD\n3. 22SF\n4. 22SG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n183\nChapter 5.\nFunctions\n\n5.4\nExponential functions\nEMBGS\nRevision\nEMBGT\nFunctions of the form y = abx + q\nFunctions of the general form y = abx + q, for b > 0, are called exponential functions,\nwhere a, b and q are constants.\nThe effects of a, b and q on f(x) = abx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\n– The horizontal asymptote is the\nline y = q.\n• The effect of a on shape\n– For a > 0, f(x) is increasing.\n– For a < 0, f(x) is decreasing.\nThe graph is reflected about the\nhorizontal asymptote.\n• The effect of b on direction\nAssuming a > 0:\n– If b > 1, f(x) is an increasing\nfunction.\n– If 0 < b < 1, f(x) is a decreas-\ning function.\n– If b ≤0, f(x) is not defined.\nb > 1\na < 0\na > 0\nq > 0\nq < 0\n0 < b < 1\na < 0\na > 0\nq > 0\nq < 0\nExercise 5 – 15: Revision\n1. On separate axes, accurately draw each of the following functions:\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = 3x\nb) y2 = −2 × 3x\nc) y3 = 2 × 3x + 1\nd) y4 = 3x −2\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\n184\n5.4.\nExponential functions\n\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nincreasing\nasymptote\nx-axis, y = 0\ndomain\n{x : x ∈R}\nrange\n{y : y ∈R, y > 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22SH\n1b. 22SJ\n1c. 22SK\n1d. 22SM\n2. 22SN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = ab(x+p) + q\nEMBGV\nWe now consider exponential functions of the form y = ab(x+p) + q and the effects of\nparameter p.\nSee video: 22SP at www.everythingmaths.co.za\nInvestigation: The effects of a, p and q on an exponential graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 2x\nb) y2 = 2(x−2)\nc) y3 = 2(x−1)\nd) y4 = 2(x+1)\ne) y5 = 2(x+2)\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptote\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = 2(x−1) + 2\n185\nChapter 5.\nFunctions\n\nb) y2 = 3 × 2(x−1) + 2\nc) y3 = 1\n2 × 2(x−1) + 2\nd) y4 = 0 × 2(x−1) + 2\ne) y5 = −3 × 2(x−1) + 2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptotes\ndomain\nrange\neffect of a\nThe effect of the parameters on y = abx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position relative to the horizontal\nasymptote.\n• For a > 0, the graph lies above the horizontal asymptote, y = q.\n• For a < 0, the graph lies below the horizontal asymptote, y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n186\n5.4.\nExponential functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y = ab(x+p) + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative.\nIf a > 0 we have:\nb(x+p) > 0\nab(x+p) > 0\nab(x+p) + q > q\nf(x) > q\nThe range is therefore {y : y > q, y ∈R}.\nSimilarly, if a < 0, the range is {y : y < q, y ∈R}.\nWorked example 14: Domain and range\nQUESTION\nState the domain and range for g(x) = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n3(x+1) > 0\n5 × 3(x+1) > 0\n5 × 3(x+1) −1 > −1\n∴g(x) > −1\nTherefore the range is {g(x) : g(x) > −1} or in interval notation (−1; ∞).\n187\nChapter 5.\nFunctions\n\nExercise 5 – 16: Domain and range\nGive the domain and range for each of the following functions:\n1. y =\n\u0000 3\n2\n\u0001(x+3)\n2. f(x) = −5(x−2) + 1\n3. y + 3 = 2(x+1)\n4. y = n + 3(x−m)\n5.\ny\n2 = 3(x−1) −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SQ\n2. 22SR\n3. 22SS\n4. 22ST\n5. 22SV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting x = 0:\ng(0) = 3 × 2(0+1) + 2\n= 3 × 2 + 2\n= 8\nThis gives the point (0; 8).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting y = 0:\n0 = 3 × 2(x+1) + 2\n−2 = 3 × 2(x+1)\n−2\n3 = 2(x+1)\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n188\n5.4.\nExponential functions\n\nExercise 5 – 17: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = 2(x+1) −8\n2. y = 2 × 3(x−1) −18\n3. y + 5(x+2) = 5\n4. y = 1\n2\n\u0000 3\n2\n\u0001(x+3) −0,75\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SW\n2. 22SX\n3. 22SY\n4. 22SZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptote\nExponential functions of the form y = ab(x+p) + q have a horizontal asymptote, the\nline y = q.\nWorked example 15: Asymptote\nQUESTION\nDetermine the asymptote for y = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the asymptote\nThe asymptote of g(x) can be calculated as:\n3(x+1) ̸= 0\n5 × 3(x+1) ̸= 0\n5 × 3(x+1) −1 ̸= −1\n∴y ̸= −1\nTherefore the asymptote is the line y = −1.\n189\nChapter 5.\nFunctions\n\nExercise 5 – 18: Asymptote\nGive the asymptote for each of the following functions:\n1. y = −5(x+1)\n2. y = 3(x−2) + 1\n3.\n\u0010\n3y\n2\n\u0011\n= 5(x+3) −1\n4. y = 7(x+1) −2\n5.\ny\n2 + 1 = 3(x+2)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T2\n2. 22T3\n3. 22T4\n4. 22T5\n5. 22T6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching graphs of the form f(x) = ab(x+p) + q\nIn order to sketch graphs of functions of the form, f(x) = ab(x+p) + q, we need to\ndetermine five characteristics:\n• shape\n• y-intercept\n• x-intercept\n• asymptote\n• domain and range\nWorked example 16: Sketching an exponential graph\nQUESTION\nSketch the graph of 2y = 10 × 2(x+1) −5.\nMark the intercept(s) and asymptote. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nWe notice that a > 0 and b > 1, therefore the function is increasing.\n190\n5.4.\nExponential functions\n\nStep 2: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\n2y = 10 × 2(0+1) −5\n= 10 × 2 −5\n= 15\n∴y = 71\n2\nThis gives the point (0; 71\n2).\nStep 3: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 = 10 × 2(x+1) −5\n5 = 10 × 2(x+1)\n1\n2 = 2(x+1)\n2−1 = 2(x+1)\n∴−1 = x + 1\n(same base)\n−2 = x\nThis gives the point (−2; 0).\nStep 4: Determine the asymptote\nThe horizontal asymptote is the line y = −5\n2.\nStep 5: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\n−3\n1\n2\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y > −5\n2, y ∈R}\n191\nChapter 5.\nFunctions\n\nWorked example 17: Finding the equation of an exponential function from a\ngraph\nQUESTION\nUse the given graph of y = −2 × 3(x+p) + q to determine the values of p and q.\n1\n2\n3\n4\n5\n6\n7\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nFrom the graph we see that the function is decreasing. We also note that a = −2 and\nb = 3. We need to solve for p and q.\nStep 2: Use the asymptote to determine q\nThe horizontal asymptote y = 6 is given, therefore we know that q = 6.\ny = −2 × 3(x+p) + 6\nStep 3: Use the x-intercept to determine p\nSubstitute (2; 0) into the equation and solve for p:\ny = −2 × 3(x+p) + 6\n0 = −2 × 3(2+p) + 6\n−6 = −2 × 3(2+p)\n3 = 3(2+p)\n∴1 = 2 + p\n(same base)\n∴p = −1\nStep 4: Write the final answer\ny = −2 × 3(x−1) + 6\n192\n5.4.\nExponential functions\n\nExercise 5 – 19: Mixed exercises\n1. Given the graph of the hyperbola of the form h(x) = k\nx, x < 0, which passes\nthough the point A(−1\n2; −6).\nb\ny\nx\n0\nA(−1\n2; −6)\na) Show that k = 3.\nb) Write down the equation for the new function formed if h(x):\ni. is shifted 3 units vertically upwards\nii. is shifted to the right by 3 units\niii. is reflected about the y-axis\niv. is shifted so that the asymptotes are x = 0 and y = −1\n4\nv. is shifted upwards to pass through the point (−1; 1)\nvi. is shifted to the left by 2 units and 1 unit vertically downwards (for\nx < 0)\n2. Given the graphs of f(x) = a(x + p)2 and g(x) = a\nx.\nThe axis of symmetry for f(x) is x = −1 and f(x) and g(x) intersect at point M.\nThe line y = 2 also passes through M.\nb\ny\nx\n0\nM\n−1\n2\nf\ng\n193\nChapter 5.\nFunctions\n\nDetermine:\na) the coordinates of M\nb) the equation of g(x)\nc) the equation of f(x)\nd) the values for which f(x) < g(x)\ne) the range of f(x)\n3. On the same system of axes, sketch:\na) the graphs of k(x) = 2(x + 1\n2)2 −41\n2 and h(x) = 2(x+ 1\n2 ). Determine all\nintercepts, turning point(s) and asymptotes.\nb) the reflection of h(x) about the x-axis. Label this function as j(x).\n4. Sketch the graphs of y = ax2 + bx + c for:\na) a < 0, b > 0, b2 < 4ac\nb) a > 0, b > 0, one root = 0\n5. On separate systems of axes, sketch the graphs:\ny =\n2\nx−2\ny = 2\nx −2\ny = −2(x−2)\n6. For the diagrams shown below, determine:\n• the equations of the functions; f(x) = a(x + p)2 + q, g(x) = ax2 + q,\nh(x) = a\nx, x < 0 and k(x) = bx + q\n• the axes of symmetry of each function\n• the domain and range of each function\na)\nb\ny\nx\n0\n(2; 3)\nf\n194\n5.4.\nExponential functions\n\nb)\nb\ny\nx\n0\n(−2; −1)\ng\nh\n−2\nc)\ny\nx\n0\nk\ny = 2x + 1\n2\n7. Given the graph of the function Q(x) = ax.\nb\nb\nb\ny\nx\n0\nQ = ax\n(−2; p)\n1\n(1; 1\n3)\na) Show that a = 1\n3.\nb) Find the value of p if the point (−2; p) is on Q.\nc) Calculate the average gradient of the curve between x = −2 and x = 1.\nd) Determine the equation of the new function formed if Q is shifted 2 units\nvertically downwards and 2 units to the left.\n8. Find the equation for each of the functions shown below:\na) f(x) = 2x + q\ng(x) = mx + c\n195\nChapter 5.\nFunctions\n\nb\ny\nx\n0\nf\ng\n−1\n2\n−2\nb) h(x) =\nk\nx+p + q\nb\nb\ny\nx\n0\n1\n−2\n−1\n2\nh\n9. Given: the graph of k(x) = −x2 + 3x + 10 with turning point at D. The graph of\nthe straight line h(x) = mx + c passing through points B and C is also shown.\nb\nb\nb\ny\nx\n0\nB\nA\nE\nF\nD\nC\nk\nh\nDetermine:\na) the lengths AO, OB, OC and DE\nb) the equation of DE\nc) the equation of h(x)\nd) the x-values for which k(x) < 0\n196\n5.4.\nExponential functions\n\ne) the x-values for which k(x) ≥h(x)\nf) the length of DF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T7\n2. 22T8\n3. 22T9\n4. 22TB\n5. 22TC\n6a. 22TD\n6b. 22TF\n6c. 22TG\n7. 22TH\n8a. 22TJ\n8b. 22TK\n9. 22TM\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIMPORTANT: Trigonometric functions are examined in PAPER 2.\n5.5\nThe sine function\nEMBGW\nRevision\nEMBGX\nFunctions of the form y = sin θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• Period of one complete wave is 360◦.\n• Amplitude is the maximum height of the wave above and below the x-axis and\nis always positive. Amplitude = 1.\n• Domain: [0◦; 360◦]\nFor y = sin θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Maximum turning point: (90◦; 1)\n• Minimum turning point: (270◦; −1)\n197\nChapter 5.\nFunctions\n\nFunctions of the form y = a sin θ + q\nThe effects of a and q on f(θ) = a sin θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 20: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦.\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function also determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n198\n5.5.\nThe sine function\n\n1. y1 = sin θ\n2. y2 = −2 sin θ\n3. y3 = sin θ + 1\n4. y4 = 1\n2 sin θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TN\n2. 22TP\n3. 22TQ\n4. 22TR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin kθ\nEMBGY\nWe now consider cosine functions of the form y = sin kθ and the effects of k.\nInvestigation: The effects of k on a sine graph\n1. Complete the following table for y1 = sin θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−270◦\n−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n2. Use the table of values to plot the graph of y1 = sin θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = sin(−θ)\nb) y3 = sin 2θ\nc) y4 = sin θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n199\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = sin θ and y2 = sin(−θ)?\n6. Is sin(−θ) = −sin θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = sin kθ?\nThe effect of the parameter on y = sin kθ\nThe value of k affects the period of the sine function. If k is negative, then the graph is\nreflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the sine function decreases.\nFor 0 < k < 1, the period of the sine function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\nsin(−θ) = −sin θ\nCalculating the period:\nTo determine the period of y = sin kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k (this means that k is always considered to be\npositive).\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n200\n5.5.\nThe sine function\n\nWorked example 18: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = sin θ\nb) y2 = sin 3θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin kθ\nNotice that k > 1 for y2 = sin 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\nsin θ\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\nsin 3θ\n2\n1\n0,38\n−0,71\n−0,92\n0\n0,92\n0,71\n−0,38\n−1\nStep 3: Sketch the sine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin 3\n2θ\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin 3θ\n2\nperiod\n360◦\n240◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(90◦; 1)\n(−180◦; 1) and (60◦; 1)\nminimum turning points\n(−90◦; −1)\n(−60◦; −1) and (180◦; 1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\n201\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = sin kθ\n= sin 0◦\n= 0\nThis gives the point (0◦; 0).\nExercise 5 – 21: Sine functions of the form y = sin kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦and for each graph deter-\nmine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = sin 3θ\nb) g(θ) = sin θ\n3\nc) h(θ) = sin(−2θ)\nd) k(θ) = sin 3θ\n4\n2. For each graph of the form f(θ) = sin kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n202\n5.5.\nThe sine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22TS\n1b. 22TT\n1c. 22TV\n1d. 22TW\n2a. 22TX\n2b. 22TY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin(θ + p)\nEMBGZ\nInvestigation: The effects of p on a sine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = sin θ\nb) y2 = sin(θ −90◦)\nc) y3 = sin(θ −60◦)\nd) y4 = sin(θ + 90◦)\ne) y5 = sin(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of p\n203\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = sin(θ + p)\nThe effect of p on the sine function is a horizontal shift, also called a phase shift; the\nentire graph slides to the left or to the right.\n• For p > 0, the graph of the sine function shifts to the left by p.\n• For p < 0, the graph of the sine function shifts to the right by p.\np > 0\np < 0\nWorked example 19: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = sin θ\nb) y2 = sin(θ −30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin(θ + p)\nNotice that for y1 = sin θ we have p = 0 (no phase shift) and for y2 = sin(θ −30◦),\np < 0 therefore the graph shifts to the right by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n0\n1\n0\n−1\n0\n1\n0\n−1\n0\nsin(θ −30◦)\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n204\n5.5.\nThe sine function\n\nStep 3: Sketch the sine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = sin θ\ny2 = sin(θ −30◦)\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin(θ −30◦)\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−270◦; 1) and (90◦; 1)\n(−240◦; 1) and\n(120◦; 1)\nminimum turning points\n(−90◦; −1) and\n(270◦; −1)\n(−60◦; −1) and\n(300◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; −1\n2)\nx-intercept(s)\n(−360◦; 0), (−180◦; 0),\n(0◦; 0), (180◦; 0) and\n(360◦; 0)\n(−330◦; 0), (−150◦; 0),\n(30◦; 0) and (210◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\n205\nChapter 5.\nFunctions\n\nExercise 5 – 22: Sine functions of the form y = sin(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = sin(θ + 30◦)\n2. g(θ) = sin(θ −45◦)\n3. h(θ) = sin(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TZ\n2. 22V2\n3. 22V3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching sine graphs\nEMBH2\nWorked example 20: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(45◦−θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin(θ + p).\nf(θ) = sin(45◦−θ)\n= sin(−θ + 45◦)\n= sin (−(θ −45◦))\n= −sin(θ −45◦)\nTo draw a graph of the above function, we know that the standard sine graph, y = sin θ,\n206\n5.5.\nThe sine function\n\nmust:\n• be reflected about the x-axis\n• be shifted to the right by 45◦\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n0,71\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = −sin(θ −45◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 360◦\nAmplitude: 1\nDomain: [−360◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (315◦; 1)\nMinimum turning point: (135◦; −1)\ny-intercepts: (0◦; 0,71)\nx-intercept: (45◦; 0) and (225◦; 0)\nWorked example 21: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(3θ + 60◦) for 0◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin k(θ + p).\nf(θ) = sin(3θ + 60◦)\n= sin 3(θ + 20◦)\n207\nChapter 5.\nFunctions\n\nTo draw a graph of the above equation, the standard sine graph, y = sin θ, must be\nchanged in the following ways:\n• decrease the period by a factor of 3;\n• shift to the left by 20◦.\nStep 2: Complete a table of values\nθ\n0◦\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nf(θ)\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nθ\nf(θ)\nf(θ) = sin 3(θ + 20◦)\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 180◦]\nRange: [−1; 1]\nMaximum turning point: (10◦; 1) and (130◦; 1)\nMinimum turning point: (70◦; −1)\ny-intercept: (0◦; 0,87)\nx-intercepts: (40◦; 0), (100◦; 0) and (160◦; 0)\nExercise 5 – 23: The sine function\n1. Sketch the following graphs on separate axes:\na) y = 2 sin θ\n2 for −360◦≤θ ≤360◦\nb) f(θ) = 1\n2 sin(θ −45◦) for −90◦≤θ ≤90◦\nc) y = sin(θ + 90◦) + 1 for 0◦≤θ ≤360◦\nd) y = sin(−3θ\n2 ) for −180◦≤θ ≤180◦\ne) y = sin(30◦−θ) for −360◦≤θ ≤360◦\n2. Given the graph of the function y = a sin(θ + p), determine the values of a and\np.\n208\n5.5.\nThe sine function\n\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nθ\nf(θ)\nCan you describe this graph in terms of cos θ?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22V4\n1b. 22V5\n1c. 22V6\n1d. 22V7\n1e. 22V8\n2. 22V9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n5.6\nThe cosine function\nEMBH3\nRevision\nEMBH4\nFunctions of the form y = cos θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• The period is 360◦and the amplitude is 1.\n• Domain: [0◦; 360◦]\nFor y = cos θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (90◦; 0), (270◦; 0)\n• y-intercept: (0◦; 1)\n• Maximum turning points: (0◦; 1), (360◦; 1)\n• Minimum turning point: (180◦; −1)\n209\nChapter 5.\nFunctions\n\nFunctions of the form y = a cos θ + q\nCosine functions of the general form y = a cos θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a cos θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 24: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function in the previous problem determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n210\n5.6.\nThe cosine function\n\n1. y1 = cos θ\n2. y2 = −3 cos θ\n3. y3 = cos θ + 2\n4. y4 = 1\n2 cos θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VB\n2. 22VC\n3. 22VD\n4. 22VF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos(kθ)\nEMBH5\nWe now consider cosine functions of the form y = cos kθ and the effects of k.\nInvestigation: The effects of k on a cosine graph\n1. Complete the following table for y1 = cos θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ncos θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ncos θ\n2. Use the table of values to plot the graph of y1 = cos θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = cos(−θ)\nb) y3 = cos 3θ\nc) y4 = cos 3θ\n4\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n211\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = cos θ and y2 = cos(−θ)?\n6. Is cos(−θ) = −cos θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = cos kθ?\nThe effect of the parameter k on y = cos kθ\nThe value of k affects the period of the cosine function.\n• For k > 0:\nFor k > 1, the period of the cosine function decreases.\nFor 0 < k < 1, the period of the cosine function increases.\n• For k < 0:\nFor −1 < k < 0, the period increases.\nFor k < −1, the period decreases.\nNegative angles:\ncos(−θ) = cos θ\nNotice that for negative values of θ, the graph is not reflected about the x-axis.\nCalculating the period:\nTo determine the period of y = cos kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k.\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n212\n5.6.\nThe cosine function\n\nWorked example 22: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = cos θ\nb) y2 = cos θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos kθ\nNotice that for y2 = cos θ\n2, k < 1 therefore the period of the graph increases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ncos θ\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\n−0,71\n−1\ncos θ\n2\n0\n0,38\n0,71\n0,92\n1\n0,92\n0,71\n0,38\n0\nStep 3: Sketch the cosine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = cos θ\n2\ny2 = cos θ\nStep 4: Complete the table\ny1 = cos θ\ny2 = cos θ\n2\nperiod\n360◦\n720◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[0; 1]\nmaximum turning points\n(0◦; 1)\n(0◦; 1)\nminimum turning points\n(−180◦; −1) and (180◦; −1)\nnone\ny-intercept(s)\n(0◦; 1)\n(0◦; 1)\nx-intercept(s)\n(−90◦; 0) and (90◦; 0)\n(−180◦; 0) and (180◦; 0)\n213\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos kθ\n= cos 0◦\n= 1\nThis gives the point (0◦; 1).\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = cos 2θ\nb) g(θ) = cos θ\n3\nc) h(θ) = cos(−2θ)\nd) k(θ) = cos 3θ\n4\n2. For each graph of the form f(θ) = cos kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n214\n5.6.\nThe cosine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nbA(135◦; 0)\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VG\n1b. 22VH\n1c. 22VJ\n1d. 22VK\n2a. 22VM\n2b. 22VN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos (θ + p)\nEMBH6\nWe now consider cosine functions of the form y = cos(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a cosine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = cos θ\nb) y2 = cos(θ −90◦)\nc) y3 = cos(θ −60◦)\nd) y4 = cos(θ + 90◦)\ne) y5 = cos(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning points\nminimum turning points\ny-intercept(s)\nx-intercept(s)\neffect of p\n215\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = cos(θ + p)\nThe effect of p on the cosine function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the cosine function shifts to the left by p degrees.\n• For p < 0, the graph of the cosine function shifts to the right by p degrees.\np > 0\np < 0\nWorked example 23: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = cos θ\nb) y2 = cos(θ + 30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos(θ + p)\nNotice that for y1 = cos θ we have p = 0 (no phase shift) and for y2 = cos(θ + 30◦),\np < 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\ncos θ\n1\n0\n−1\n0\n1\n0\n−1\n0\n1\ncos(θ +30◦)\n0,87\n−0,5\n−0,87\n0,5\n0,87\n−0,5\n−0,87\n0,5\n0,87\n216\n5.6.\nThe cosine function\n\nStep 3: Sketch the cosine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = cos θ\ny2 = cos(θ + 30◦)\nStep 4: Complete the table\ny1\ny2\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−360◦; 1), (0◦; 1) and\n(360◦; 1)\n(−30◦; 1) and (330◦; 1)\nminimum turning points\n(−180◦; −1) and\n(180◦; −1)\n(−210◦; −1) and\n(150◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,87)\nx-intercept(s)\n(−270◦; 0), (−90◦; 0),\n(90◦; 0) and (270◦; 0)\n(−300◦; 0), (−120◦; 0),\n(60◦; 0) and (240◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos(θ + p)\n= cos(0◦+ p)\n= cos p\nThis gives the point (0◦; cos p).\n217\nChapter 5.\nFunctions\n\nExercise 5 – 26: Cosine functions of the form y = cos(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = cos(θ + 45◦)\n2. g(θ) = cos(θ −30◦)\n3. h(θ) = cos(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VP\n2. 22VQ\n3. 22VR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching cosine graphs\nEMBH7\nWorked example 24: Sketching a cosine graph\nQUESTION\nSketch the graph of f(θ) = cos(180◦−3θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = cos k(θ + p).\nf(θ) = cos(180◦−3θ)\n= cos(−3θ + 180◦)\n= cos (−3(θ −60◦))\n= cos 3(θ −60◦)\nTo draw a graph of the above function, the standard cosine graph, y = cos θ, must be\nchanged in the following ways:\n218\n5.6.\nThe cosine function\n\n• decrease the period by a factor of 3\n• shift to the right by 60◦.\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n−1\n0,71\n0\n−0,71\n1\n−0,71\n0\n0,71\n−1\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = cos 3(θ −60◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (60◦; 1), (180◦; 1) and (300◦; 1)\nMinimum turning point: (0◦; −1), (120◦; −1), (240◦; −1) and (360◦; −1)\ny-intercepts: (0◦; −1)\nx-intercept: (30◦; 0), (90◦; 0), (150◦; 0), (210◦; 0), (270◦; 0) and (330◦; 0)\nWorked example 25: Finding the equation of a cosine graph\nQUESTION\nGiven the graph of y = a cos(kθ+p), determine the values of a, k, p and the minimum\nturning point.\nθ\ny\ny = a cos(θ + p)\nb\n(45◦; 2)\n−45◦\n315◦\n219\nChapter 5.\nFunctions\n\nSOLUTION\nStep 1: Determine the value of k\nFrom the sketch we see that the period of the graph is 360◦, therefore k = 1.\ny = a cos(θ + p)\nStep 2: Determine the value of a\nFrom the sketch we see that the maximum turning point is (45◦; 2), so we know that\nthe amplitude of the graph is 2 and therefore a = 2.\ny = 2 cos(θ + p)\nStep 3: Determine the value of p\nCompare the given graph with the standard cosine function y = cos θ and notice the\ndifference in the maximum turning points. We see that the given function has been\nshifted to the right by 45◦, therefore p = 45◦.\ny = 2 cos(θ −45◦)\nStep 4: Determine the minimum turning point\nAt the minimum turning point, y = −2:\ny = 2 cos(θ −45◦)\n−2 = 2 cos(θ −45◦)\n−1 = cos(θ −45◦)\ncos−1(−1) = θ −45◦\n180◦= θ −45◦\n225◦= θ\nThis gives the point (225◦; −2).\n220\n5.6.\nThe cosine function\n\nExercise 5 – 27: The cosine function\n1. Sketch the following graphs on separate axes:\na) y = cos(θ + 15◦) for −180◦≤θ ≤180◦\nb) f(θ) = 1\n3 cos(θ −60◦) for −90◦≤θ ≤90◦\nc) y = −2 cos θ for 0◦≤θ ≤360◦\nd) y = cos(30◦−θ) for −360◦≤θ ≤360◦\ne) g(θ) = 1 + cos(θ −90◦) for 0◦≤θ ≤360◦\nf) y = cos(2θ + 60◦) for −360◦≤θ ≤360◦\n2. Two girls are given the following graph:\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\ny\nb\na) Audrey decides that the equation for the graph is a cosine function of the\nform y = a cos θ. Determine the value of a.\nb) Megan thinks that the equation for the graph is a cosine function of the\nform y = cos(θ + p). Determine the value of p.\nc) What can they conclude?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VS\n1b. 22VT\n1c. 22VV\n1d. 22VW\n1e. 22VX\n1f. 22VY\n2. 22VZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n221\nChapter 5.\nFunctions\n\n5.7\nThe tangent function\nEMBH8\nRevision\nEMBH9\nFunctions of the form y = tan θ for 0◦≤θ ≤360◦\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nf(θ)\nθ\nThe dashed vertical lines are called the asymptotes. The asymptotes are at the values\nof θ where tan θ is not defined.\n• Period: 180◦\n• Domain: {θ : 0◦≤θ ≤360◦, θ ̸= 90◦; 270◦}\n• Range: {f(θ) : f(θ) ∈R}\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Asymptotes: the lines θ = 90◦and θ = 270◦\nFunctions of the form y = a tan θ + q\nTangent functions of the general form y = a tan θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a tan θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted vertically upwards by q units.\n– For q < 0, f(θ) is shifted vertically downwards by q units.\n• The effect of a on shape\n– For a > 1, branches of f(θ) are steeper.\n– For 0 < a < 1, branches of f(θ) are less steep and curve more.\n222\n5.7.\nThe tangent function\n\n– For a < 0, there is a reflection about the x-axis.\n– For −1 < a < 0, there is a reflection about the x-axis and the branches of\nthe graph are less steep.\n– For a < −1, there is a reflection about the x-axis and the branches of the\ngraph are steeper.\na < 0\na > 0\nq > 0\nb\n0\nb\n0\nq = 0\nb\n0\nb\n0\nq < 0\nb\n0\nb\n0\nExercise 5 – 28: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function determine the following:\n• Period\n• Domain and range\n223\nChapter 5.\nFunctions\n\n• x- and y-intercepts\n• Asymptotes\n1. y1 = tan θ −1\n2\n2. y2 = −3 tan θ\n3. y3 = tan θ + 2\n4. y4 = 2 tan θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W2\n2. 22W3\n3. 22W4\n4. 22W5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = tan(kθ)\nEMBHB\nInvestigation: The effects of k on a tangent graph\n1. Complete the following table for y1 = tan θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ntan θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ntan θ\n2. Use the table of values to plot the graph of y1 = tan θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = tan(−θ)\nb) y3 = tan 3θ\nc) y4 = tan θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of k\n224\n5.7.\nThe tangent function\n\n5. What do you notice about y1 = tan θ and y2 = tan(−θ)?\n6. Is tan(−θ) = −tan θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = tan kθ?\nThe effect of the parameter on y = tan kθ\nThe value of k affects the period of the tangent function. If k is negative, then the\ngraph is reflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the tangent function decreases.\nFor 0 < k < 1, the period of the tangent function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\ntan(−θ) = −tan θ\nCalculating the period:\nTo determine the period of y = tan kθ we use,\nPeriod = 180◦\n|k|\nwhere |k| is the absolute value of k.\nk > 0\nk < 0\n225\nChapter 5.\nFunctions\n\nWorked example 26: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan 3θ\n2\n2. For each function determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan kθ\nNotice that k > 1 for y2 = tan 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan 3θ\n2\nUNDEF\n−0,41\n1\n−2,41\n0\n2,41\n−1\n0,41\nUNDEF\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan 3θ\n2\nperiod\n180◦\n120◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦< θ < 180◦, θ ̸=\n−60◦; 60◦}\nrange\n{f(θ) : f(θ) ∈R}\n{f(θ) : f(θ) ∈R}\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −180◦; −60◦and 180◦\n226\n5.7.\nThe tangent function\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan kθ:\nDomain and range\nThe domain of one branch is {θ : −90◦\nk\n< θ < 90◦\nk , θ ∈R} because f(θ) is undefined\nfor θ = −90◦\nk and θ = 90◦\nk .\nThe range is {f(θ) : f(θ) ∈R} or (−∞; ∞).\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0 and solving for f(θ).\ny = tan kθ\n= tan 0◦\n= 0\nThis gives the point (0◦; 0).\nAsymptotes\nThese are the values of kθ for which tan kθ is undefined.\nExercise 5 – 29: Tangent functions of the form y = tan kθ\nSketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan 2θ\n2. g(θ) = tan 3θ\n4\n3. h(θ) = tan(−2θ)\n4. k(θ) = tan 2θ\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W6\n2. 22W7\n3. 22W8\n4. 22W9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n227\nChapter 5.\nFunctions\n\nFunctions of the form y = tan (θ + p)\nEMBHC\nWe now consider tangent functions of the form y = tan(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a tangent graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = tan θ\nb) y2 = tan(θ −60◦)\nc) y3 = tan(θ −90◦)\nd) y4 = tan(θ + 60◦)\ne) y5 = tan(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of p\nThe effect of the parameter on y = tan(θ + p)\nThe effect of p on the tangent function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the tangent function shifts to the left by p.\n• For p < 0, the graph of the tangent function shifts to the right by p.\np > 0\np < 0\n228\n5.7.\nThe tangent function\n\nWorked example 27: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan(θ + 30◦)\nFor each function determine the following:\n2.\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan(θ + p)\nNotice that for y1 = tan θ we have p = 0◦(no phase shift) and for y2 = tan(θ + 30◦),\np > 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−180◦−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan(θ+30◦)\n0,58\n3,73\n−1,73\n−0,27\n0,58\n3,73\n−1,73\n−0,27\n0,58\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan(θ + 30◦)\nperiod\n180◦\n180◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦≤θ ≤\n180◦, θ ̸= −120◦; 60◦}\nrange\n(−∞; ∞)\n(−∞; ∞)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,58)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and (180◦; 0)\n(−30◦; 0) and (150◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −120◦and θ = 60◦\n229\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan(θ + p):\nDomain and range\nThe domain of one branch is {θ : θ ∈(−90◦−p; 90◦−p)} because the function is\nundefined for θ = −90◦−p and θ = 90◦−p.\nThe range is {f(θ) : f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = tan(θ + p)\n= tan(0◦+ p)\n= tan p\nThis gives the point (0◦; tan p).\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan(θ + 45◦)\n2. g(θ) = tan(θ −30◦)\n3. h(θ) = tan(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WB\n2. 22WC\n3. 22WD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n230\n5.7.\nThe tangent function\n\nSketching tangent graphs\nEMBHD\nWorked example 28: Sketching a tangent graph\nQUESTION\nSketch the graph of f(θ) = tan 1\n2(θ −30◦) for −180◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that 0 < k < 1, therefore the branches of the graph will be\nless steep than the standard tangent graph y = tan θ. We also notice that p < 0 so the\ngraph will be shifted to the right on the x-axis.\nStep 2: Determine the period\nThe period for f(θ) = tan 1\n2(θ −30◦) is:\nPeriod = 180◦\n|k|\n= 180◦\n1\n2\n= 360◦\nStep 3: Determine the asymptotes\nThe standard tangent graph, y = tan θ, for −180◦≤θ ≤180◦is undefined at θ = −90◦\nand θ = 90◦. Therefore we can determine the asymptotes of f(θ) = tan 1\n2(θ −30◦):\n•\n−90◦\n0,5 + 30◦= −150◦\n•\n90◦\n0,5 + 30◦= 210◦\nThe asymptote at θ = 210◦lies outside the required interval.\n231\nChapter 5.\nFunctions\n\nStep 4: Plot the points and join with a smooth curve\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 360◦\nDomain: {θ : −180◦≤θ ≤180◦, θ ̸= −150◦}\nRange: (−∞; ∞)\ny-intercepts: (0◦; −0,27)\nx-intercept: (30◦; 0)\nAsymptotes: θ = −150◦\nExercise 5 – 31: The tangent function\n1. Sketch the following graphs on separate axes:\na) y = tan θ −1 for −90◦≤θ ≤90◦\nb) f(θ) = −tan 2θ for 0◦≤θ ≤90◦\nc) y = 1\n2 tan(θ + 45◦) for 0◦≤θ ≤360◦\nd) y = tan(30◦−θ) for −180◦≤θ ≤180◦\n2. Given the graph of y = a tan kθ, determine the values of a and k.\nθ\nf(θ)\nb\nb\n(90◦; −1)\n360◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WF\n1b. 22WG\n1c. 22WH\n1d. 22WJ\n2. 22WK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n232\n5.7.\nThe tangent function\n\nExercise 5 – 32: Mixed exercises\n1. Determine the equation for each of the following:\na) f(θ) = a sin kθ and g(θ) = a tan θ\nθ\ny\nb\nb\nf\n(45◦; −3\n2 )\ng\n(180◦; 0)\n(135◦; −1 1\n2 )\nb) f(θ) = a sin kθ and g(θ) = a cos(θ + p)\nθ\n0\ny\nb\n(−90◦; 2)\n−180◦\n180◦\nf and g\nc) y = a tan kθ\nθ\n0\ny\nb\n(90◦; 3)\n360◦\n180◦\nd) y = a cos θ + q\nθ\n0\ny\n4\n360◦\n180◦\n233\nChapter 5.\nFunctions\n\n2. Given the functions f(θ) = 2 sin θ and g(θ) = cos θ + 1:\na) Sketch the graphs of both functions on the same system of axes, for 0◦≤\nθ ≤360◦. Indicate the turning points and intercepts on the diagram.\nb) What is the period of f?\nc) What is the amplitude of g?\nd) Use your sketch to determine how many solutions there are for the equation\n2 sin θ −cos θ = 1. Give one of the solutions.\ne) Indicate on your sketch where on the graph the solution to 2 sin θ = −1 is\nfound.\n3. The sketch shows the two functions f(θ) = a cos θ and g(θ) = tan θ for 0◦≤θ ≤\n360◦. Points P(135◦; b) and Q(c; −1) lie on g(θ) and f(θ) respectively.\nθ\n0\ny\nb\nb\n360◦\n180◦\nP\nQ\ng\nf\n−1\n2\n−2\na) Determine the values of a, b and c.\nb) What is the period of g?\nc) Solve the equation cos θ = 1\n2 graphically and show your answer(s) on the\ndiagram.\nd) Determine the equation of the new graph if g is reflected about the x-axis\nand shifted to the right by 45◦.\n4. Sketch the graphs of y1 = −1\n2 sin(θ + 30◦) and y2 = cos(θ −60◦), on the same\nsystem of axes for 0◦≤θ ≤360◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WM\n1b. 22WN\n1c. 22WP\n1d. 22WQ\n2. 22WR\n3. 22WS\n4. 22WT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n234\n5.7.\nThe tangent function\n\n5.8\nSummary\nEMBHF\nSee presentation: 22WV at www.everythingmaths.co.za\n1. Parabolic functions:\nStandard form: y = ax2 + bx + c\n• y-intercept: (0; c)\n• x-intercept: x = −b±\n√\nb2−4ac\n2a\n• Turning point:\n\u0010\n−b\n2a; −b2\n4a + c\n\u0011\n• Axis of symmetry: x = −b\n2a\nCompleted square form: y = a(x + p)2 + q\n• Turning point: (−p; q)\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n2. Average gradient:\n• Average gradient = y2−y1\nx2−x1\n3. Hyperbolic functions:\nStandard form: y = k\nx\n• k > 0: first and third quadrant\n• k < 0: second and fourth quadrant\nShifted form: y =\nk\nx+p + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: x = −p and y = q\n4. Exponential functions:\nStandard form: y = abx\n• a > 0: above x-axis\n• a < 0: below x-axis\n• b > 1: increasing function if a > 0; decreasing function if a < 0\n• 0 < b < 1: decreasing function if a > 0; increasing function if a < 0\nShifted form: y = ab(x+p) + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n235\nChapter 5.\nFunctions\n\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: y = q\n5. Sine functions:\nShifted form: y = a sin(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• sin(−θ) = −sin θ\n6. Cosine functions:\nShifted form: y = a cos(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• cos(−θ) = cos θ\n7. Tangent functions:\nShifted form: y = a tan(kθ + p) + q\n• Period = 180◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• tan(−θ) = −tan θ\n• Asymptotes: 90◦−p\nk\n± 180◦n\nk\n, n ∈Z\n236\n5.8.\nSummary\n\nExercise 5 – 33: End of chapter exercises\n1. Show that if a\n<\n0,\nthen the range of f(x)\n=\na(x + p)2 + q is\n{f(x) : f(x) ∈(−∞, q]}.\n2. If (2; 7) is the turning point of f(x) = −2x2 −4ax + k, find the values of the\nconstants a and k.\n3. The following graph is represented by the equation f(x) = ax2 + bx. The coor-\ndinates of the turning point are (3; 9). Show that a = −1 and b = 6.\nb (3; 9)\nx\n0\ny\n4. Given: f(x) = x2 −2x + 3. Give the equation of the new graph originating if:\na) the graph of f is moved three units to the left.\nb) the x-axis is moved down three units.\n5. A parabola with turning point (−1; −4) is shifted vertically by 4 units upwards.\nWhat are the coordinates of the turning point of the shifted parabola?\n6. Plot the graph of the hyperbola defined by y = 2\nx for −4 ≤x ≤4. Suppose\nthe hyperbola is shifted 3 units to the right and 1 unit down. What is the new\nequation then?\n7. Based on the graph of y =\nk\n(x+p) + q, determine the equation of the graph with\nasymptotes y = 2 and x = 1 and passing through the point (2; 3).\ny\nx\n0\n2\n1\nb (2; 3)\n237\nChapter 5.\nFunctions\n\n8. The columns in the table below give the y-values for the following functions:\ny = ax, y = ax+1 and y = ax + 1. Match each function to the correct column.\nx\nA\nB\nC\n−2\n7,25\n6,25\n2,5\n−1\n3,5\n2,5\n1\n0\n2\n1\n0,4\n1\n1,4\n0,4\n0,16\n2\n1,16\n0,16\n0,064\n9. The graph of f(x) = 1 + a . 2x (a is a constant) passes through the origin.\na) Determine the value of a.\nb) Determine the value of f(−15) correct to five decimal places.\nc) Determine the value of x, if P (x; 0,5) lies on the graph of f.\nd) If the graph of f is shifted 2 units to the right to give the function h, write\ndown the equation of h.\n10. The graph of f(x) = a . bx (a ̸= 0) has the point P (2; 144) on f.\na) If b = 0,75, calculate the value of a.\nb) Hence write down the equation of f.\nc) Determine, correct to two decimal places, the value of f(13).\nd) Describe the transformation of the curve of f to h if h(x) = f(−x).\n11. Using your knowledge of the effects of p and k draw a rough sketch of the fol-\nlowing graphs without a table of values.\na) y = sin 3θ for −180◦≤θ ≤180◦\nb) y = −cos 2θ for 0◦≤θ ≤180◦\nc) y = tan 1\n2θ for 0◦≤θ ≤360◦\nd) y = sin(θ −45◦) for −360◦≤θ ≤360◦\ne) y = cos(θ + 45◦) for 0◦≤θ ≤360◦\nf) y = tan(θ −45◦) for 0◦≤θ ≤360◦\ng) y = 2 sin 2θ for −180◦≤θ ≤180◦\nh) y = sin(θ + 30◦) + 1 for −360◦≤θ ≤0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WW\n2. 22WX\n3. 22WY\n4. 22WZ\n5. 22X2\n6. 22X3\n7. 22X4\n8. 22X5\n9. 22X6\n10. 22X7\n11a. 22X8\n11b. 22X9\n11c. 22XB\n11d. 22XC\n11e. 22XD\n11f. 22XF\n11g. 22XG\n11h. 22XH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n238\n5.8.\nSummary\n\nCHAPTER\n6\nTrigonometry\n6.1\nRevision\n240\n6.2\nTrigonometric identities\n247\n6.3\nReduction formula\n253\n6.4\nTrigonometric equations\n266\n6.5\nArea, sine, and cosine rules\n280\n6.6\nSummary\n301\n\n6\nTrigonometry\n6.1\nRevision\nEMBHG\nTrigonometric ratios\nb\nb\nb\ny\nx\nP(x; y)\nQ(−x; y)\nO\nα\nβ\nr\nr\nWe plot the points P(x; y) and Q(−x; y) in the Cartesian plane and measure the angles\nfrom the positive x-axis to the terminal arms (OP and OQ).\nP(x; y) lies in the first quadrant with P ˆOX = α and Q(−x; y) lies in the second\nquadrant with Q ˆOX = β.\nUsing the theorem of Pythagoras we have that\nOP 2 = x2 + y2\nAnd OQ2 = (−x)2 + y2\n= x2 + y2\n∴OP = OQ\nLet OP = OQ = r.\nTrigonometric ratios\nsin α = y\nr\ncos α = x\nr\ntan α = y\nx\nIn the second quadrant we notice that −x < 0\nsin β = y\nr\ncos β = −x\nr\ntan β = −y\nx\n240\n6.1.\nRevision\n\nSimilarily, in the third and fourth quadrants the sign of the trigonometric ratios depends\non the signs of x and y:\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nSpecial angles\n30◦\n60◦\n1\n√\n3\n2\n45◦\n45◦\n1\n1\n√\n2\nθ\n0◦\n30◦\n45◦\n60◦\n90◦\ncos θ\n1\n√\n3\n2\n1\n√\n2\n1\n2\n0\nsin θ\n0\n1\n2\n1\n√\n2\n√\n3\n2\n1\ntan θ\n0\n1\n√\n3\n1\n√\n3\nundef\nSee video: 22XJ at www.everythingmaths.co.za\n241\nChapter 6.\nTrigonometry\n\nSolving equations\nWorked example 1: Solving equations\nQUESTION\nDetermine the values of a and b in the right-angled triangle TUW (correct to one\ndecimal place):\nT\nW\nU\n47◦\nb\n30\na\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of a\nsin θ = opposite side\nhypotenuse\nsin 47◦= 30\na\na =\n30\nsin 47◦\n∴a = 41,0\nStep 3: Determine the value of b\nAlways try to use the information that is given for calculations and not answers that you\nhave worked out in case you have made an error. For example, avoid using a = 41,0\nto determine the value of b.\ntan θ = opposite side\nadjacent side\ntan 47◦= 30\nb\nb =\n30\ntan 47◦\n∴b = 28,0\nStep 4: Write the final answer\na = 41,0 units and b = 28,0 units.\n242\n6.1.\nRevision\n\nFinding an angle\nWorked example 2: Finding an angle\nQUESTION\nCalculate the value of θ in the right-angled triangle MNP (correct to one decimal\nplace):\nP\nN\nM\n41\nθ\n24\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of θ\ntan θ = opposite side\nadjacent side\ntan θ = 41\n24\n∴θ = tan−1\n\u001241\n24\n\u0013\nθ = 59,7◦\n243\nChapter 6.\nTrigonometry\n\nWorked example 3: Finding an angle\nQUESTION\nGiven 2 sin θ\n2 = cos 43◦, for θ ∈[0◦; 90◦], determine the value of θ (correct to one\ndecimal place).\nSOLUTION\nStep 1: Simplify the equation\nAvoiding rounding off in calculations until you have determined the final answer. In\nthe calculation below, the dots indicate that the number has not been rounded so that\nthe answer is as accurate as possible.\n2 sin θ\n2 = cos 43◦\nsin θ\n2 = cos 43◦\n2\nθ\n2 = sin−1(0,365 . . .)\nθ = 2(21,449 . . .)\n∴θ = 42,9◦\nTwo-dimensional problems\nWorked example 4: Flying a kite\nQUESTION\nThelma flies a kite on a 22 m piece of string and the height of the kite above the\nground is 20,4 m. Determine the angle of inclination of the string (correct to one\ndecimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the opposite and adjacent sides and the hy-\npotenuse\nLet the angle of inclination of the string be θ.\n244\n6.1.\nRevision\n\nKite\nThelma\n20,4\nθ\n22\nStep 2: Use an appropriate trigonometric ratio to find θ\nsin θ = opposite side\nhypotenuse\n= 20,4\n22\nθ = sin−1(0,927 . . .)\n∴θ = 68,0◦\nExercise 6 – 1: Revision\n1. If p = 49◦and q = 32◦, use a calculator to determine whether the following\nstatements are true of false:\na) sin p + 3 sin p = 4 sin p\nb) sin q\ncos q = tan q\nc) cos(p −q) = cos p −cos q\nd) sin(2p) = 2 sin p cos p\n2. Determine the following angles (correct to one decimal place):\na) cos α = 0,64\nb) sin θ + 2 = 2,65\nc) 1\n2 cos 2β = 0,3\nd) tan θ\n3 = sin 48◦\ne) cos 3p = 1,03\nf) 2 sin 3β + 1 = 2,6\ng) sin θ\ncos θ = 42\n3\n3. In △ABC, A ˆCB = 30◦, AC = 20 cm and BC = 22 cm. The perpendicular\nline from A intersects BC at T.\n245\nChapter 6.\nTrigonometry\n\nDetermine:\nA\nC\nB\nT\n20 cm\n22 cm\n30◦\na) the length TC\nb) the length AT\nc) the angle B ˆAT\n4. A rhombus has a perimeter of 40 cm and one of the internal angles is 30◦.\na) Determine the length of the sides.\nb) Determine the lengths of the diagonals.\nc) Calculate the area of the rhombus.\n5. Simplify the following without using a calculator:\na) 2 sin 45◦× 2 cos 45◦\nb) cos2 30◦−sin2 60◦\nc) sin 60◦cos 30◦−cos 60◦sin 30◦−tan 45◦\nd) 4 sin 60◦cos 30◦−2 tan 45◦+ tan 60◦−2 sin 60◦\ne) sin 60◦×\n√\n2 tan 45◦+ 1 −sin 30◦\n6. Given the diagram below.\nb\nx\ny\n0\nB(2; 2\n√\n3)\nβ\nDetermine the following without using a calculator:\na) β\nb) cos β\nc) cos2 β + sin2 β\n7. The 10 m ladder of a fire truck leans against the wall of a burning building at an\nangle of 60◦. The height of an open window is 9 m from the ground. Will the\nladder reach the window?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22XK\n2a. 22XM\n2b. 22XN\n2c. 22XP\n2d. 22XQ\n2e. 22XR\n2f. 22XS\n2g. 22XT\n3. 22XV\n4. 22XW\n5a. 22XX\n5b. 22XY\n5c. 22XZ\n5d. 22Y2\n5e. 22Y3\n6. 22Y4\n7. 22Y5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n246\n6.1.\nRevision\n\n6.2\nTrigonometric identities\nEMBHH\nAn identity is a mathematical statement that equates one quantity with another. Trigono-\nmetric identities allow us to simplify a given expression so that it contains sine and co-\nsine ratios only. This enables us to solve equations and also to prove other identities.\nQuotient identity\nInvestigation: Quotient identity\n1. Complete the table without using a calculator, leaving your answer in surd form\nwhere applicable:\nθ = 45◦\nθ\n3\n5\nx\ny\n(3; 2)\nθ\nb\nsin θ\ncos θ\nsin θ\ncos θ\ntan θ\n2. Examine the last two rows of the table and make a conjecture.\n3. Are there any values of θ for which your conjecture would not be true? Explain\nyour answer.\nWe know that tan θ is defined as:\ntan θ = opposite side\nadjacent side\nUsing the diagram below and the theorem of Pythagoras, we can write the tangent\nfunction in terms of x, y and r:\nx\ny\n(x; y)\nθ\nb\nO\n247\nChapter 6.\nTrigonometry\n\ntan θ = y\nx\n= y\nx × r\nr\n= y\nr × r\nx\n= y\nr ÷ x\nr\n= sin θ ÷ cos θ\n= sin θ\ncos θ\nThis is the quotient identity:\ntan θ = sin θ\ncos θ\nNotice that tan θ is undefined if cos θ = 0, therefore θ ̸= k × 90◦, where k is an odd\ninteger.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\ntan θ\nSquare identity\nInvestigation: Square identity\n1. Use a calculator to complete the following table:\nsin2 80◦+ cos2 80◦=\ncos2 23◦+ sin2 23◦=\nsin 50◦+ cos 50◦=\nsin2 67◦−cos2 67◦=\nsin2 67◦+ cos2 67◦=\n2. What do you notice? Make a conjecture.\n3. Draw a sketch and prove your conjecture in general terms, using x, y and r.\n248\n6.2.\nTrigonometric identities\n\nx\ny\n(x; y)\nα\nb\nO\nr\nUsing the theorem of Pythagoras, we can write the sine and cosine functions in terms\nof x, y and r:\nsin2 θ + cos2 θ =\n\u0010y\nr\n\u00112\n+\n\u0010x\nr\n\u00112\n= y2\nr2 + x2\nr2\n= y2 + x2\nr2\n= r2\nr2\n= 1\nThis is the square identity:\nsin2 θ + cos2 θ = 1\nOther forms of the square identity\nComplete the following:\n1. sin2 θ = 1 −. . . . . .\n2. cos θ = ±√. . . . . .\n3. sin2 θ = (1 + . . . . . .)(1 −. . . . . .)\n4. cos2 θ −1 = . . . . . .\nHere are some useful tips for proving identities:\n• Change all trigonometric ratios to sine and cosine.\n• Choose one side of the equation to simplify and show that it is equal to the other\nside.\n• Usually it is better to choose the more complicated side to simplify.\n• Sometimes we need to simplify both sides of the equation to show that they are\nequal.\n• A square root sign often indicates that we need to use the square identity.\n• We can also add to the expression to make simplifying easier:\n– replace 1 with sin2 θ + cos2 θ.\n– multiply by 1 in the form of a suitable fraction, for example 1 + sin θ\n1 + sin θ.\n249\nChapter 6.\nTrigonometry\n\nSee video: 22Y6 at www.everythingmaths.co.za\nWorked example 5: Trigonometric identities\nQUESTION\nSimplify the following:\n1. tan2 θ × cos2 θ\n2.\n1\ncos2 θ −tan2 θ\nSOLUTION\nStep 1: Write the expression in terms of sine and cosine only\nWe use the square and quotient identities to write the given expression in terms of sine\nand cosine and then simplify as far as possible.\n1.\ntan2 θ × cos2 θ =\n\u0012 sin θ\ncos θ\n\u00132\n× cos2 θ\n= sin2 θ\ncos2 θ × cos2 θ\n= sin2 θ\n2.\n1\ncos2 θ −tan2 θ =\n1\ncos2 θ −\n\u0012 sin θ\ncos θ\n\u00132\n=\n1\ncos2 θ −sin2 θ\ncos2 θ\n= 1 −sin2 θ\ncos2 θ\n= cos2 θ\ncos2 θ\n= 1\n250\n6.2.\nTrigonometric identities\n\nWorked example 6: Trigonometric identities\nQUESTION\nProve: 1 −sin α\ncos α\n=\ncos α\n1 + sin α\nSOLUTION\nStep 1: Note restrictions\nWhen working with fractions, we must be careful that the denominator does not equal\n0. Therefore cos θ ̸= 0 for the fraction on the left-hand side and sin θ + 1 ̸= 0 for the\nfraction on the right-hand side.\nStep 2: Simplify the left-hand side\nThis is not an equation that needs to be solved. We are required to show that one side\nof the equation is equal to the other. We can choose either of the two sides to simplify.\nLHS = 1 −sin α\ncos α\n= 1 −sin α\ncos α\n× 1 + sin α\n1 + sin α\nNotice that we have not changed the equation — this is the same as multiplying by 1\nsince the numerator and the denominator are the same.\nStep 3: Determine the lowest common denominator and simplify\nLHS =\n1 −sin2 α\ncos α(1 + sin α)\n=\ncos2 α\ncos α(1 + sin α)\n=\ncos α\n1 + sin α\n= RHS\n251\nChapter 6.\nTrigonometry\n\nExercise 6 – 2: Trigonometric identities\n1. Reduce the following to one trigonometric ratio:\na) sin α\ntan α\nb) cos2 θ tan2 θ + tan2 θ sin2 θ\nc) 1 −sin θ cos θ tan θ\nd)\n\u00121 −cos2 β\ncos2 β\n\u0013\n−tan2 β\n2. Prove the following identities and state restrictions where appropriate:\na) 1 + sin θ\ncos θ\n=\ncos θ\n1 −sin θ\nb) sin2α + (cos α −tan α) (cos α + tan α) = 1 −tan2α\nc)\n1\ncos θ −cos θtan2θ\n1\n= cos θ\nd)\n2 sin θ cos θ\nsin θ + cos θ = sin θ + cos θ −\n1\nsin θ + cos θ\ne)\n\u0012cos β\nsin β + tan β\n\u0013\ncos β =\n1\nsin β\nf)\n1\n1 + sin θ +\n1\n1 −sin θ = d\n2 tan θ\nsin θ cos θ\ng) (1 + tan2 α) cos α\n(1 −tan α)\n=\n1\ncos α −sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Y7\n1b. 22Y8\n1c. 22Y9\n1d. 22YB\n2a. 22YC\n2b. 22YD\n2c. 22YF\n2d. 22YG\n2e. 22YH\n2f. 22YJ\n2g. 22YK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n252\n6.2.\nTrigonometric identities\n\n6.3\nReduction formula\nEMBHJ\nAny trigonometric function whose argument is 90◦± θ; 180◦± θ and 360◦± θ can be\nwritten simply in terms of θ.\nDeriving reduction formulae\nEMBHK\nInvestigation: Reduction formulae for function values of 180◦± θ\n1. Function values of 180◦−θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the y-axis, determine the coordi-\nnates of P ′.\nb) Write down values for sin θ, cos θ and tan θ.\nc) Use the coordinates for P ′ to determine sin(180◦−θ), cos(180◦−θ),\ntan(180◦−θ).\nd) From your results determine a relationship between the trigonometric func-\ntion values of (180◦−θ) and θ.\n253\nChapter 6.\nTrigonometry\n\n2. Function values of 180◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦+ θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the origin (the two points are sym-\nmetrical about both the x-axis and the y-axis), determine the coordinates of\nP ′.\nb) Use the coordinates for P ′ to determine sin(180◦+ θ), cos(180◦+ θ) and\ntan(180◦+ θ).\nc) From your results determine a relationship between the trigonometric func-\ntion values of (180◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(180◦−θ) = . . . . . .\nb) cos(180◦−θ) = . . . . . .\nc) tan(180◦−θ) = . . . . . .\nd) sin(180◦+ θ) = . . . . . .\ne) cos(180◦+ θ) = . . . . . .\nf) tan(180◦+ θ) = . . . . . .\n254\n6.3.\nReduction formula\n\nWorked example 7: Reduction formulae for function values of 180◦± θ\nQUESTION\nWrite the following as a single trigonometric ratio:\nsin 163◦\ncos 197◦+ tan 17◦+ cos(180◦−θ) × tan(180◦+ θ)\nSOLUTION\nStep 1: Use reduction formulae to write the trigonometric function values in terms\nof acute angles and θ\n= sin(180◦−17◦)\ncos(180◦+ 17◦) + tan 17◦+ (−cos θ) × tan θ\nStep 2: Simplify\n=\nsin 17◦\n−cos 17◦+ tan 17◦−cos θ × sin θ\ncos θ\n= −tan 17◦+ tan 17◦−sin θ\n= −sin θ\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1. Determine the value of the following expressions without using a calculator:\na) tan 150◦sin 30◦−cos 210◦\nb) (1 + cos 120◦)(1 −sin2 240◦)\nc) cos2 140◦+ sin2 220◦\n2. Write the following in terms of a single trigonometric ratio:\na) tan(180◦−θ) × sin(180◦+ θ)\nb) tan(180◦+ θ) cos(180◦−θ)\nsin(180◦−θ)\n255\nChapter 6.\nTrigonometry\n\n3. If t = tan 40◦, express the following in terms of t:\na) tan 140◦+ 3 tan 220◦\nb) cos 220◦\nsin 140◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YM\n1b. 22YN\n1c. 22YP\n2a. 22YQ\n2b. 22YR\n3. 22YS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of (360◦± θ) and (−θ)\n1. Function values of (360◦−θ) and (−θ)\nIn the Cartesian plane we measure angles from the positive x-axis to the terminal\narm, which means that an anti-clockwise rotation gives a positive angle. We can\ntherefore measure negative angles by rotating in a clockwise direction.\nFor an acute angle θ, we know that −θ will lie in the fourth quadrant.\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n360◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the x-axis (y = 0), determine the\ncoordinates of P ′.\nb) Use the coordinates of P ′ to determine sin(360◦−θ), cos(360◦−θ) and\ntan(360◦−θ).\nc) Use the coordinates of P ′ to determine sin(−θ), cos(−θ) and tan(−θ).\nd) From your results determine a relationship between the function values of\n(360◦−θ) and −θ.\n256\n6.3.\nReduction formula\n\ne) Complete the following reduction formulae:\ni. sin(360◦−θ) = . . . . . .\nii. cos(360◦−θ) = . . . . . .\niii. tan(360◦−θ) = . . . . . .\niv. sin(−θ) = . . . . . .\nv. cos(−θ) = . . . . . .\nvi. tan(−θ) = . . . . . .\n2. Function values of 360◦+ θ\nWe can also have an angle that is larger than 360◦. The angle completes a\nrevolution of 360◦and then continues to give an angle of θ.\nComplete the following reduction formulae:\na) sin(360◦+ θ) = . . . . . .\nb) cos(360◦+ θ) = . . . . . .\nc) tan(360◦+ θ) = . . . . . .\nFrom working with functions, we know that the graph of y = sin θ has a period of\n360◦. Therefore, one complete wave of a sine graph is the same as one complete\nrevolution for sin θ in the Cartesian plane.\n0\n1\n−1\n90◦\n180◦\n270◦\n360◦\n1st\n2nd\n3rd\n4th\npositive\npositive\nnegative\nnegative\n0◦/360◦\n90◦\n180◦\n270◦\n2nd\npos.\nneg.\n3rd\nneg.\n4th\n1st\npos.\nWe can also have multiple revolutions. The periodicity of the trigonometric graphs\nshows this clearly. A complete sine or cosine curve is completed in 360◦.\ny = cos θ\ny = sin θ\nθ\ny\n257\nChapter 6.\nTrigonometry\n\nIf k is any integer, then\nsin(k . 360◦+ θ) = sin θ\ncos(k . 360◦+ θ) = cos θ\ntan(k . 360◦+ θ) = tan θ\nWorked example 8: Reduction formulae for function values of 360◦± θ\nQUESTION\nIf f = tan 67◦, express the following in terms of f\nsin 293◦\ncos 427◦+ tan(−67◦) + tan 1147◦\nSOLUTION\nStep 1: Using reduction formula\n= sin(360◦−67◦)\ncos(360◦+ 67◦) −tan(67◦) + tan (3(360◦) + 67◦)\n= −sin 67◦\ncos 67◦−tan 67◦+ tan 67◦\n= −tan 67◦\n= −f\nWorked example 9: Using reduction formula\nQUESTION\nEvaluate without using a calculator:\ntan2 210◦−(1 + cos 120◦) sin2 405◦\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and special angles\n258\n6.3.\nReduction formula\n\n= tan2(180◦+ 30◦) −(1 + cos(180◦−60◦)) sin2(360◦+ 45◦)\n= tan2 30◦−(1 + (−cos 60◦)) sin2 45◦\n=\n\u0012 1\n√\n3\n\u00132\n−\n\u0012\n1 −1\n2\n\u0013 \u0012 1\n√\n2\n\u00132\n= 1\n3 −\n\u00121\n2\n\u0013 \u00121\n2\n\u0013\n= 1\n3 −1\n4\n= 1\n12\nExercise 6 – 4: Using reduction formula\n1. Simplify the following:\na) tan(180◦−θ) sin(360◦+ θ)\ncos(180◦+ θ) tan(360◦−θ)\nb) cos2(360◦+ θ) + cos(180◦+ θ) tan(360◦−θ) sin(360◦+ θ)\nc)\nsin(360◦+ α) tan(180◦+ α)\ncos(360◦−α) tan2(360◦+ α)\n2. Write the following in terms of cos β:\ncos(360◦−β) cos(−β) −1\nsin(360◦+ β) tan(360◦−β)\n3. Simplify the following without using a calculator:\na)\ncos 300◦tan 150◦\nsin 225◦cos(−45◦)\nb) 3 tan 405◦+ 2 tan 330◦cos 750◦\nc) cos 315◦cos 405◦+ sin 45◦sin 135◦\nsin 750◦\nd) tan 150◦cos 390◦−2 sin 510◦\ne) 2 sin 120◦+ 3 cos 765◦−2 sin 240◦−3 cos 45◦\n5 sin 300◦+ 3 tan 225◦−6 cos 60◦\n4. Given 90◦< α < 180◦, use a sketch to help explain why:\na) sin(−α) = −sin α\nb) cos(−α) = −cos α\n259\nChapter 6.\nTrigonometry\n\n5. If t = sin 43◦, express the following in terms of t:\na) sin 317◦\nb) cos2 403◦\nc) tan(−43◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YT\n1b. 22YV\n1c. 22YW\n2. 22YX\n3a. 22YY\n3b. 22YZ\n3c. 22Z2\n3d. 22Z3\n3e. 22Z4\n4. 22Z5\n5. 22Z6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of 90◦± θ\nIn any right-angled triangle, the two acute angles are complements of each other, ˆA +\nˆC = 90◦\nA\nB\nC\nb\na\nc\nComplete the following:\nIn △ABC\nsin ˆC = c\nb = cos . . .\ncos ˆC = a\nb = sin . . .\nComplementary angles are positive acute angles that add up to 90◦. For example 20◦\nand 70◦are complementary angles.\n260\n6.3.\nReduction formula\n\nIn the figure P(\n√\n3; 1) and P ′ lie on a circle with radius 2. OP makes an angle of\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\nP ′\n2\n2\n90◦−θ\n1. Function values of 90◦−θ\na) If points P and P ′ are symmetrical about the line y = x, determine the\ncoordinates of P ′.\nb) Use the coordinates for P ′ to determine sin(90◦−θ) and cos(90◦−θ).\nc) From your results determine a relationship between the function values of\n(90◦−θ) and θ.\n2. Function values of 90◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nθ\nb\nP\nO\nx\ny\nθ\nb\nP ′\n90◦+ θ\n2\n2\n261\nChapter 6.\nTrigonometry\n\na) If point P is rotated through 90◦to get point P ′, determine the coordinates\nof P ′.\nb) Use the coordinates for P ′ to determine sin(90◦+ θ) and cos(90◦+ θ).\nc) From your results determine a relationship between the function values of\n(90◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(90◦−θ) = . . . . . .\nb) cos(90◦−θ) = . . . . . .\nc) sin(90◦+ θ) = . . . . . .\nd) cos(90◦+ θ) = . . . . . .\nSine and cosine are known as co-functions. Two functions are called co-functions if\nf (A) = g (B) whenever A + B = 90◦(that is, A and B are complementary angles).\nThe function value of an angle is equal to the co-function of its complement.\nThus for sine and cosine we have\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nThe sine and cosine graphs illustrate this clearly: the two graphs are identical except\nthat they have a 90◦phase difference.\nθ\ny\ny = cos θ\ny = sin θ\n262\n6.3.\nReduction formula\n\nWorked example 10: Using the co-function rule\nQUESTION\nWrite each of the following in terms of sin 40◦:\n1. cos 50◦\n2. sin 320◦\n3. cos 230◦\n4. cos 130◦\nSOLUTION\n1. cos 50◦= sin(90◦−50◦) = sin 40◦\n2. sin 320◦= sin(360◦−40◦) = −sin 40◦\n3. cos 230◦= cos(180◦+ 50◦) = −cos 50◦= −cos(90◦−40◦) = −sin 40◦\n4. cos 130◦= cos(90◦+ 40◦) = −sin 40◦\nFunction values of θ −90◦\nWe can write sin(θ −90◦) as\nsin(θ −90◦) = sin [−(90◦−θ)]\n= −sin(90◦−θ)\n= −cos θ\nsimilarly, we can show that cos (θ −90◦) = sin θ\nTherefore, sin (θ −90◦) = −cos θ and cos (θ −90◦) = sin θ.\nWorked example 11: Co-functions\nQUESTION\nExpress the following in terms of t if t = sin θ:\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and co-functions\n263\nChapter 6.\nTrigonometry\n\nUse the CAST diagram to check in which quadrants the trigonometric ratios are positive\nand negative.\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\n=cos[−(90◦−θ)] cos[2(360◦) + θ] tan[−(360◦−θ)]\nsin2(360◦+ θ) cos(90◦+ θ)\n=sin θ cos θ tan θ\nsin2 θ(−sin θ)\n= −cos θ\n\u0000 sin θ\ncos θ\n\u0001\nsin2 θ\n= −\n1\nsin θ\n= −1\nt\nExercise 6 – 5: Co-functions\n1. Simplify the following:\na) cos(90◦+ θ) sin(θ + 90◦)\nsin(−θ)\nb) 2 sin(90◦−x) + sin(90◦+ x)\nsin(90◦−x) + cos(180◦+ x)\n2. Given cos 36◦= p, express the following in terms on p:\na) sin 54◦\nb) sin 36◦\nc) tan 126◦\nd) cos 324◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Z7\n1b. 22Z8\n2. 22Z9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n264\n6.3.\nReduction formula\n\nReduction formulae and co-functions:\n1. The reduction formulae hold for any angle θ. For convenience, we assume θ is\nan acute angle (0◦< θ < 90◦).\n2. When determining function values of (180◦±θ), (360◦±θ) and (−θ) the function\ndoes not change.\n3. When determining function values of (90◦±θ) and (θ±90◦) the function changes\nto its co-function.\nsecond quadrant (180◦−θ) or (90◦+ θ)\nfirst quadrant (θ) or (90◦−θ)\nsin(180◦−θ) = + sin θ\nall trig functions are positive\ncos(180◦−θ) = −cos θ\nsin(360◦+ θ) = sin θ\ntan(180◦−θ) = −tan θ\ncos(360◦+ θ) = cos θ\nsin(90◦+ θ) = + cos θ\ntan(360◦+ θ) = tan θ\ncos(90◦+ θ) = −sin θ\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nthird quadrant (180◦+ θ)\nfourth quadrant (360◦−θ)\nsin(180◦+ θ) = −sin θ\nsin(360◦−θ) = −sin θ\ncos(180◦+ θ) = −cos θ\ncos(360◦−θ) = + cos θ\ntan(180◦+ θ) = + tan θ\ntan(360◦−θ) = −tan θ\nExercise 6 – 6: Reduction formulae\n1. Write A and B as a single trigonometric ratio:\na) A = sin(360◦−θ) cos(180◦−θ) tan(360◦+ θ)\nb) B = cos(360◦+ θ) cos(−θ) sin(−θ)\ncos(90◦+ θ)\nc) Hence, determine:\ni. A + B = . . .\nii.\nA\nB = . . .\n2. Write the following as a function of an acute angle:\na) sin 163◦\nb) cos 327◦\nc) tan 248◦\nd) cos(−213◦)\n3. Determine the value of the following, without using a calculator:\na) sin(−30◦)\ntan(150◦) + cos 330◦\nb) tan 300◦cos 120◦\nc) (1 −cos 30◦)(1 −cos 210◦)\nd) cos 780◦−(sin 315◦)(cos 405◦)\n4. Prove that the following identity is true and state any restrictions:\nsin(180◦+ α) tan(360◦+ α) cos α\ncos(90◦−α)\n= sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22ZB\n2a. 22ZC\n2b. 22ZD\n2c. 22ZF\n2d. 22ZG\n3a. 22ZH\n3b. 22ZJ\n3c. 22ZK\n3d. 22ZM\n4. 22ZN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n265\nChapter 6.\nTrigonometry\n\n6.4\nTrigonometric equations\nEMBHM\nSolving trigonometric equations requires that we find the value of the angles that satisfy\nthe equation. If a specific interval for the solution is given, then we need only find the\nvalue of the angles within the given interval that satisfy the equation. If no interval is\ngiven, then we need to find the general solution. The periodic nature of trigonometric\nfunctions means that there are many values that satisfy a given equation, as shown in\nthe diagram below.\n1\n−1\n90◦180◦270◦360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nθ\n0\ny\ny = 0,5\ny = sin θ\nWorked example 12: Solving trigonometric equations\nQUESTION\nSolve for θ (correct to one decimal place), given tan θ = 5 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to solve for θ\ntan θ = 5\n∴θ = tan−1 5\n= 78,7◦\nThis value of θ is an acute angle which lies in the first quadrant and is called the\nreference angle.\nStep 2: Use the CAST diagram to determine in which quadrants tan θ is positive\nThe CAST diagram indicates that tan θ is positive in the first and third quadrants, there-\nfore we must determine the value of θ such that 180◦< θ < 270◦.\nUsing reduction formulae, we know that tan(180◦+ θ) = tan θ\nθ = 180◦+ 78,7◦\n∴θ = 258,7◦\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 78,7◦or θ = 258,7◦.\n266\n6.4.\nTrigonometric equations\n\nWorked example 13: Solving trigonometric equations\nQUESTION\nSolve for α (correct to one decimal place), given cos α = −0,7 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we do not include the negative sign. The reference\nangle must be an acute angle in the first quadrant, where all the trigonometric functions\nare positive.\nref ∠= cos−1 0,7\n= 45,6◦\nStep 2: Use the CAST diagram to determine in which quadrants cos α is negative\nThe CAST diagram indicates that cos α is negative in the second and third quadrants,\ntherefore we must determine the value of α such that 90◦< α < 270◦.\nUsing reduction formulae, we know that cos(180◦−α) = −cos α and cos(180◦+α) =\n−cos α\nIn the second quadrant:\nα = 180◦−45,6◦\n= 134,4◦\nIn the third quadrant:\nα = 180◦+ 45,6◦\n= 225,6◦\nNote: the reference angle (45,6◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nα = 134,4◦or α = 225,6◦.\n267\nChapter 6.\nTrigonometry\n\nWorked example 14: Solving trigonometric equations\nQUESTION\nSolve for β (correct to one decimal place), given sin β = −0,5 and β ∈[−360◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we use a positive value.\nref ∠= sin−1 0,5\n= 30◦\nStep 2: Use the CAST diagram to determine in which quadrants sin β is negative\nThe CAST diagram indicates that sin β is negative in the third and fourth quadrants.\nWe also need to find the values of β such that −360◦≤β ≤360◦.\nUsing reduction formulae, we know that sin(180◦+β) = −sin β and sin(360◦−β) =\n−sin β\nIn the third quadrant:\nβ = 180◦+ 30◦\n= 210◦\nor β = −180◦+ 30◦\n= −150◦\nIn the fourth quadrant:\nβ = 360◦−30◦\n= 330◦\nor β = 0◦−30◦\n= −30◦\nNotice: the reference angle (30◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nβ = −150◦, −30◦, 210◦or 330◦.\n268\n6.4.\nTrigonometric equations\n\nExercise 6 – 7: Solving trigonometric equations\n1. Determine the values of α for α ∈[0◦; 360◦] if:\na) 4 cos α = 2\nb) sin α + 3,65 = 3\nc) tan α = 51\n4\nd) cos α + 0,939 = 0\ne) 5 sin α = 3\nf)\n1\n2 tan α = −1,4\n2. Determine the values of θ for θ ∈[−360◦; 360◦] if:\na) sin θ = 0,6\nb) cos θ + 3\n4 = 0\nc) 3 tan θ = 20\nd) sin θ = cos 180◦\ne) 2 cos θ = 4\n5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22ZP\n1b. 22ZQ\n1c. 22ZR\n1d. 22ZS\n1e. 22ZT\n1f. 22ZV\n2a. 22ZW\n2b. 22ZX\n2c. 22ZY\n2d. 22ZZ\n2e. 2322\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe general solution\nEMBHN\nIn the previous worked example, the solution was restricted to a certain interval. How-\never, the periodicity of the trigonometric functions means that there are an infinite\nnumber of positive and negative angles that satisfy an equation. If we do not restrict\nthe solution, then we need to determine the general solution to the equation. We know\nthat the sine and cosine functions have a period of 360◦and the tangent function has\na period of 180◦.\nMethod for finding the general solution:\n1. Determine the reference angle (use a positive value).\n2. Use the CAST diagram to determine where the function is positive or negative\n(depending on the given equation).\n3. Find the angles in the interval [0◦; 360◦] that satisfy the equation and add multi-\nples of the period to each answer.\n4. Check answers using a calculator.\n269\nChapter 6.\nTrigonometry\n\nWorked example 15: Finding the general solution\nQUESTION\nDetermine the general solution for sin θ = 0,3 (correct to one decimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nsin θ = 0,3\n∴ref ∠= sin−1 0,3\n= 17,5◦\nStep 2: Use CAST diagram to determine in which quadrants sin θ is positive\nThe CAST diagram indicates that sin θ is positive in the first and second quadrants.\nUsing reduction formulae, we know that sin(180◦−θ) = sin θ.\nIn the first quadrant:\nθ = 17,5◦\n∴θ = 17,5◦+ k . 360◦\nIn the second quadrant:\nθ = 180◦−17,5◦\n∴θ = 162,5◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 4:\nθ = 17,5◦+ 4(360)◦\n∴θ = 1457,5◦\nAnd sin 1457,5◦= 0,3007 . . .\nThis solution is correct.\nSimilarly, if we let k = −2:\nθ = 162,5◦−2(360)◦\n∴θ = −557,5◦\nAnd sin(−557,5◦) = 0,3007 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 17,5◦+ k . 360◦or θ = 162,5◦+ k . 360◦.\n270\n6.4.\nTrigonometric equations\n\nWorked example 16: Finding the general solution\nQUESTION\nDetermine the general solution for cos 2θ = −0,6427 (give answers correct to one\ndecimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nref ∠= sin−1 0,6427\n= 50,0◦\nStep 2: Use CAST diagram to determine in which quadrants cos θ is negative\nThe CAST diagram shows that cos θ is negative in the second and third quadrants.\nTherefore we use the reduction formulae cos(180◦−θ) = −cos θ and cos(180◦+θ) =\n−cos θ.\nIn the second quadrant:\n2θ = 180◦−50◦+ k . 360◦\n= 130◦+ k . 360◦\n∴θ = 65◦+ k . 180◦\nIn the third quadrant:\n2θ = 180◦+ 50◦+ k . 360◦\n= 230◦+ k . 360◦\n∴θ = 115◦+ k . 180◦\nwhere k ∈Z.\nRemember: also divide the period (360◦) by the coefficient of θ.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 2:\nθ = 65◦+ 2(180◦)\n∴θ = 425◦\nAnd cos 2(425)◦= −0,6427 . . .\nThis solution is correct.\n271\nChapter 6.\nTrigonometry\n\nSimilarly, if we let k = −5:\nθ = 115◦−5(180◦)\n∴θ = −785◦\nAnd cos 2(−785◦) = −0,6427 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 65◦+ k . 180◦or θ = 115◦+ k . 180◦.\nWorked example 17: Finding the general solution\nQUESTION\nDetermine the general solution for tan(2α −10◦) = 2,5 such that −180◦≤α ≤180◦\n(give answers correct to one decimal place).\nSOLUTION\nStep 1: Make a substitution\nTo solve this equation, it can be useful to make a substitution: let x = 2α −10◦.\ntan(x) = 2,5\nStep 2: Use a calculator to find the reference angle\ntan x = 2,5\n∴ref ∠= tan−1 2,5\n= 68,2◦\nStep 3: Use CAST diagram to determine in which quadrants the tangent function is\npositive\nWe see that tan x is positive in the first and third quadrants, so we use the reduction\nformula tan(180◦+ x) = tan x. It is also important to remember that the period of the\ntangent function is 180◦.\n272\n6.4.\nTrigonometric equations\n\nIn the first quadrant:\nx = 68,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 68,2◦+ k . 180◦\n2α = 78,2◦+ k . 180◦\n∴α = 39,1◦+ k . 90◦\nIn the third quadrant:\nx = 180◦+ 68,2◦+ k . 180◦\n= 248,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 248,2◦+ k . 180◦\n2α = 258,2◦+ k . 180◦\n∴α = 129,1◦+ k . 90◦\nwhere k ∈Z.\nRemember: to divide the period (180◦) by the coefficient of α.\nStep 4: Find the answers within the given interval\nSubstitute suitable values of k to determine the values of α that lie within the interval\n(−180◦≤α ≤180◦).\nI: α = 39,1◦+ k . 90◦\nIII: α = 129,1◦+ k . 90◦\nk = 0\n39,1◦\n129,1◦\nk = 1\n129,1◦\n219,1◦\n(outside)\nk = 2\n219,1◦\n(outside)\nk = −1\n−50,9◦\n39,1◦\nk = −2\n−140,9◦\n−50,9◦\nk = −3\n−230,9◦\n(outside)\n−140,9◦\nk = −4\n−230,9◦\n(outside)\nNotice how some of the values repeat. This is because of the periodic nature of the\ntangent function. Therefore we need only determine the solution:\nα = 39,1◦+ k . 90◦\nfor k ∈Z.\nStep 5: Write the final answer\nα = −140,9◦; −50,9◦; 39,1◦or 129,1◦.\n273\nChapter 6.\nTrigonometry\n\nWorked example 18: Finding the general solution using co-functions\nQUESTION\nDetermine the general solution for sin(θ −20◦) = cos 2θ.\nSOLUTION\nStep 1: Use co-functions to simplify the equation\nsin(θ −20◦) = cos 2θ\n= sin(90◦−2θ)\n∴θ −20◦= 90◦−2θ + k . 360◦,\nk ∈Z\n3θ = 110◦+ k . 360◦\n∴θ = 36,7◦+ k . 120◦\nStep 2: Use the CAST diagram to determine the correct quadrants\nSince the original equation equates a sine and cosine function, we need to work in the\nquadrant where both functions are positive or in the quadrant where both functions\nare negative so that the equation holds true. We therefore determine the solution using\nthe first and third quadrants.\nIn the first quadrant: θ = 36,7◦+ k . 120◦.\nIn the third quadrant:\n3θ = 180◦+ 110◦+ k . 360◦\n= 290◦+ k . 360◦\n∴θ = 96,6◦+ k . 120◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 36,7◦+ k . 120◦or θ = 96,6◦+ k . 120◦\n274\n6.4.\nTrigonometric equations\n\nExercise 6 – 8: General solution\n1.\n• Find the general solution for each equation.\n• Hence, find all the solutions in the interval [−180◦; 180◦].\na) cos(θ + 25◦) = 0,231\nb) sin 2α = −0,327\nc) 2 tan β = −2,68\nd) cos α = 1\ne) 4 sin θ = 0\nf) cos θ = −1\ng) tan θ\n2 = 0,9\nh) 4 cos θ + 3 = 1\ni) sin 2θ = −\n√\n3\n2\n2. Find the general solution for each equation.\na) cos(θ + 20◦) = 0\nb) sin 3α = −1\nc) tan 4β = 0,866\nd) cos(α −25◦) = 0,707\ne) 2 sin 3θ\n2 = −1\nf) 5 tan(β + 15◦) =\n5\n√\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2323\n1b. 2324\n1c. 2325\n1d. 2326\n1e. 2327\n1f. 2328\n1g. 2329\n1h. 232B\n1i. 232C\n2a. 232D\n2b. 232F\n2c. 232G\n2d. 232H\n2e. 232J\n2f. 232K\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSolving quadratic trigonometric equations\nWe can use our knowledge of algebraic equations to solve quadratic trigonometric\nequations.\nWorked example 19: Quadratic trigonometric equations\nQUESTION\nFind the general solution of 4 sin2 θ = 3.\nSOLUTION\nStep 1: Simplify the equation and determine the reference angle\n4 sin2 θ = 3\nsin2 θ = 3\n4\n∴sin θ = ±\nr\n3\n4\n= ±\n√\n3\n2\n∴ref ∠= 60◦\n275\nChapter 6.\nTrigonometry\n\nStep 2: Determine in which quadrants the sine function is positive and negative\nThe CAST diagram shows that sin θ is positive in the first and second quadrants and\nnegative in the third and fourth quadrants.\nPositive in the first and second quadrants:\nθ = 60◦+ k . 360◦\nor θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nNegative in the third and fourth quadrants:\nθ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor θ = 360◦−60◦+ k . 360◦\n= 300◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 60◦+ k . 360◦or 120◦+ k . 360◦or 240◦+ k . 360◦or 300◦+ k . 360◦\nWorked example 20: Quadratic trigonometric equations\nQUESTION\nFind θ if 2 cos2 θ −cos θ −1 = 0 for θ ∈[−180◦; 180◦].\nSOLUTION\nStep 1: Factorise the equation\n2 cos2 θ −cos θ −1 = 0\n(2 cos θ + 1)(cos θ −1) = 0\n∴2 cos θ + 1 = 0 or cos θ −1 = 0\n276\n6.4.\nTrigonometric equations\n\nStep 2: Simplify the equations and solve for θ\n2 cos θ + 1 = 0\n2 cos θ = −1\ncos θ = −1\n2\n∴ref ∠= 60◦\nII quadrant: θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nIII quadrant: θ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor\ncos θ −1 = 0\ncos θ = 1\n∴ref ∠= 0◦\nII and IV quadrants: θ = k . 360◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of θ that lie within the the given interval θ ∈[−180◦; 180◦] by\nsubstituting suitable values of k.\nIf k = −1,\nθ = 240◦+ k . 360◦\n= 240◦−(360◦)\n= −120◦\nIf k = 0,\nθ = 120◦+ k . 360◦\n= 120◦+ 0(360◦)\n= 120◦\nIf k = 1,\nθ = k . 360◦\n= 0(360◦)\n= 0◦\n277\nChapter 6.\nTrigonometry\n\nStep 4: Alternative method: substitution\nWe can simplify the given equation by letting y = cos θ and then factorising as:\n2y2 −y −1 = 0\n(2y + 1)(y −1) = 0\n∴y = −1\n2 or y = 1\nWe substitute y = cos θ back into these two equations and solve for θ.\nStep 5: Write the final answer\nθ = −120◦; 0◦; 120◦\nWorked example 21: Quadratic trigonometric equations\nQUESTION\nFind α if 2 sin2 α −sin α cos α = 0 for α ∈[0◦; 360◦].\nSOLUTION\nStep 1: Factorise the equation by taking out a common factor\n2 sin2 α −sin α cos α = 0\nsin α(2 sin α −cos α) = 0\n∴sin α = 0 or 2 sin α −cos α = 0\nStep 2: Simplify the equations and solve for α\nsin α = 0\n∴ref ∠= 0◦\n∴α = 0◦+ k . 360◦\nor α = 180◦+ k . 360◦\nand since 360◦= 2 × 180◦\nwe therefore have α = k . 180◦\n278\n6.4.\nTrigonometric equations\n\nor\n2 sin α −cos α = 0\n2 sin α = cos α\nTo simplify further, we divide both sides of the equation by cos α.\n2 sin α\ncos α = cos α\ncos α\n(cos α ̸= 0)\n2 tan α = 1\ntan α = 1\n2\n∴ref ∠= 26,6◦\n∴α = 26,6◦+ k . 180◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of α that lie within the the given interval α ∈[0◦; 360◦] by\nsubstituting suitable values of k.\nIf k = 0:\nα = 0◦\nor α = 26,6◦\nIf k = 1:\nα = 180◦\nor α = 26,6◦+ 180◦\n= 206,6◦\nIf k = 2:\nα = 360◦\nStep 4: Write the final answer\nα = 0◦; 26,6◦; 180◦; 206,6◦; 360◦\n279\nChapter 6.\nTrigonometry\n\nExercise 6 – 9: Solving trigonometric equations\n1. Find the general solution for each of the following equations:\na) cos 2θ = 0\nb) sin(α + 10◦) =\n√\n3\n2\nc) 2 cos θ\n2 −\n√\n3 = 0\nd)\n1\n2 tan(β −30◦) = −1\ne) 5 cos θ = tan 300◦\nf) 3 sin α = −1,5\ng) sin 2β = cos(β + 20◦)\nh) 0,5 tan θ + 2,5 = 1,7\ni) sin(3α −10◦) = sin(α + 32◦)\nj) sin 2β = cos 2β\n2. Find θ if sin2 θ + 1\n2 sin θ = 0 for θ ∈[0◦; 360◦].\n3. Determine the general solution for each of the following:\na) 2 cos2 θ −3 cos θ = 2\nb) 3 tan2 θ + 2 tan θ = 0\nc) cos2 α = 0,64\nd) sin(4β + 35◦) = cos(10◦−β)\ne) sin(α + 15◦) = 2 cos(α + 15◦)\nf) sin2 θ −4 cos2 θ = 0\ng) cos(2θ + 30◦)\n2\n+ 0,38 = 0\n4. Find β if 1\n3 tan β = cos 200◦for β ∈[−180◦; 180◦].\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 232M\n1b. 232N\n1c. 232P\n1d. 232Q\n1e. 232R\n1f. 232S\n1g. 232T\n1h. 232V\n1i. 232W\n1j. 232X\n2. 232Y\n3a. 232Z\n3b. 2332\n3c. 2333\n3d. 2334\n3e. 2335\n3f. 2336\n3g. 2337\n4. 2338\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n6.5\nArea, sine, and cosine rules\nEMBHP\nThere are three identities relating to the trigonometric functions that make working\nwith triangles easier:\n1. the area rule\n2. the sine rule\n3. the cosine rule\n280\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nEMBHQ\nInvestigation: The area rule\n1. Consider △ABC:\nB\nA\nC\n10\n54◦\n7\nComplete the following:\na) Area △ABC = 1\n2 × . . . × AC\nb) sin ˆB = . . . and AC = . . . × . . .\nc) Therefore area △ABC = . . . × . . . × . . . × . . .\n2. Consider △A′B′C′:\nB′\nA′\nC′\n10\n54◦\n7\nComplete the following:\na) How is △A′B′C′ different from △ABC?\nb) Calculate area △A′B′C′.\n3. Use your results to write a general formula for determining the area of △PQR:\nQ\nP\nR\nr\np\nq\n281\nChapter 6.\nTrigonometry\n\nFor any △ABC with AB = c, BC = a and AC = b, we can construct a perpendicular\nheight (h) from vertex A to the line BC:\nB\nA\nC\nc\na\nb\nh\nIn △ABC:\nsin ˆB = h\nc\n∴h = c sin ˆB\nAnd we know that\nArea △ABC = 1\n2 × a × h\n= 1\n2 × a × c sin ˆB\n∴Area △ABC = 1\n2ac sin ˆB\nAlternatively, we could write that\nsin ˆC = h\nb\n∴h = b sin ˆC\nAnd then we would have that\nArea △ABC = 1\n2 × a × h\n= 1\n2ab sin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nThe area rule\nIn any △ABC:\nArea △ABC = 1\n2bc sin ˆA\n= 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n282\n6.5.\nArea, sine, and cosine rules\n\nWorked example 22: The area rule\nQUESTION\nFind the area of △ABC (correct to two decimal places):\nA\n7\nB\nC\n50◦\nSOLUTION\nStep 1: Use the given information to determine unknown angles and sides\nAB = AC = 7\n(given)\n∴ˆB = ˆC = 50◦\n(∠s opp. equal sides)\nAnd ˆA = 180◦−50◦−50◦\n(∠s sum of △ABC)\n∴ˆA = 80◦\nStep 2: Use the area rule to calculate the area of △ABC\nNotice that we do not know the length of side a and must therefore choose the form\nof the area rule that does not include this side of the triangle.\nIn △ABC:\nArea = 1\n2bc sin ˆA\n= 1\n2(7)(7) sin 80◦\n= 24,13\nStep 3: Write the final answer\nArea of △ABC = 24,13 square units.\n283\nChapter 6.\nTrigonometry\n\nWorked example 23: The area rule\nQUESTION\nShow that the area of △DEF = 1\n2df sin ˆE.\nD\nF\nE\nH\ne\nd\nf\nh\n1\n2\nSOLUTION\nStep 1: Construct a perpendicular height h\nDraw DH such that DH ⊥EF and let DH = h, D ˆEF = ˆE1 and D ˆEH = ˆE2.\nIn △DHE:\nsin ˆE2 = h\nf\nh = f sin(180◦−ˆE1)\n(∠s on str. line)\n= f sin ˆE1\nStep 2: Use the area rule to calculate the area of △DEF\nIn △DEF:\nArea = 1\n2d × h\n= 1\n2df sin ˆE1\n284\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nIn any △PQR:\nP\nP\nQ\nQ\nR\nR\nq\nq\nr\nr\np\np\nArea △PQR = 1\n2qr sin ˆP\n= 1\n2pr sin ˆQ\n= 1\n2pq sin ˆR\nThe area rule states that the area of any triangle is equal to half the product of the\nlengths of the two sides of the triangle multiplied by the sine of the angle included by\nthe two sides.\nExercise 6 – 10: The area rule\n1. Draw a sketch and calculate the area of △PQR given:\na) ˆQ = 30◦; r = 10 and p = 7\nb) ˆR = 110◦; p = 8 and q = 9\n2. Find the area of △XY Z given XZ = 52 cm, XY = 29 cm and ˆX = 58,9◦.\n3. Determine the area of a parallelogram in which two adjacent sides are 10 cm\nand 13 cm and the angle between them is 55◦.\n4. If the area of △ABC is 5000 m2 with a = 150 m and b = 70 m, what are the two\npossible sizes of ˆC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2339\n1b. 233B\n2. 233C\n3. 233D\n4. 233F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n285\nChapter 6.\nTrigonometry\n\nThe sine rule\nEMBHR\nSo far we have only applied the trigonometric ratios to right-angled triangles. We now\nexpand the application of the trigonometric ratios to triangles that do not have a right\nangle:\nInvestigation: The sine rule\nIn △ABC, AC = 15, BC = 11 and ˆA = 48◦. Find ˆB.\nA\nC\nB\nb = 15\nF\na = 11\n48◦\n1. Method 1: using the sine ratio\na) Draw a sketch of △ABC.\nb) Construct CF ⊥AB.\nc) In △CBF:\nCF\n. . . = sin ˆB\n∴CF = . . . × sin ˆB\nd) In △CAF:\nCF\n15 = . . .\n∴CF = 15 × . . .\ne) Therefore we have that:\nCF = 15 × . . .\nand CF = . . . × sin ˆB\n∴15 × . . . = . . . × sin ˆB\n∴sin ˆB = . . . . . . . . .\n∴ˆB = . . .\n2. Method 2: using the area rule\n286\n6.5.\nArea, sine, and cosine rules\n\na) In △ABC:\nArea △ABC = 1\n2AB × AC × . . .\n= 1\n2AB × . . . × . . .\nb) And we also know that\nArea △ABC = 1\n2AB × . . . × sin ˆB\nc) We can equate these two equations and solve for ˆB:\n1\n2AB × . . . × sin ˆB = 1\n2AB × . . . × . . .\n∴. . . × sin ˆB = . . . × . . .\n∴sin ˆB = . . . × . . .\n∴ˆB = . . .\n3. Use your results to write a general formula for the sine rule given △PQR:\nP\nQ\nR\nq\nr\np\nFor any triangle ABC with AB = c, BC = a and AC = b, we can construct a perpen-\ndicular height (h) at F:\nA\nC\nB\nb\nF\na\nh\nc\nMethod 1: using the sine ratio\nIn △ABF:\nsin ˆB = h\nc\n∴h = c sin ˆB\n287\nChapter 6.\nTrigonometry\n\nIn △ACF:\nsin ˆC = h\nb\n∴h = b sin ˆC\nWe can equate the two equations\nc sin ˆB = b sin ˆC\n∴sin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that:\nsin ˆA\na\n= sin ˆC\nc\nor\na\nsin ˆA\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nMethod 2: using the area rule\nIn △ABC:\nArea △ABC = 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n∴1\n2ac sin ˆB = 1\n2ab sin ˆC\nc sin ˆB = b sin ˆC\nsin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\n288\n6.5.\nArea, sine, and cosine rules\n\nThe sine rule\nIn any △ABC:\nA\nC\nB\nb\na\nc\nsin ˆA\na\n= sin ˆB\nb\n= sin ˆC\nc\na\nsin ˆA\n=\nb\nsin ˆB\n=\nc\nsin ˆC\nSee video: 233G at www.everythingmaths.co.za\nWorked example 24: The sine rule\nQUESTION\nGiven △TRS with S ˆTR = 55◦, TR = 30 and R ˆST = 40◦, determine RS, ST and\nT ˆRS.\nSOLUTION\nStep 1: Draw a sketch\nLet RS = t, ST = r and TR = s.\nR\nS\nT\n55◦\n30\n40◦\nStep 2: Find T ˆRS using angles in a triangle\nT ˆRS + R ˆST + S ˆTR = 180◦\n(∠s sum of △TRS)\n∴T ˆRS = 180◦−40◦−55◦\n= 85◦\n289\nChapter 6.\nTrigonometry\n\nStep 3: Determine t and r using the sine rule\nt\nsin ˆT\n=\ns\nsin ˆS\nt\nsin 55◦=\n30\nsin 40◦\n∴t =\n30\nsin 40◦× sin 55◦\n= 38,2\nr\nsin ˆR\n=\ns\nsin ˆS\nr\nsin 85◦=\n30\nsin 40◦\n∴r =\n30\nsin 40◦× sin 85◦\n= 46,5\nWorked example 25: The sine rule\nQUESTION\nProve the sine rule for △MNP with MS ⊥NP.\nM\nP\nN\nS\nn\nm\np\nh\n1\n2\nSOLUTION\nStep 1: Use the sine ratio to express the angles in the triangle in terms of the length\nof the sides\nIn △MSN:\nsin ˆN2 = h\np\n∴h = p sin ˆN2\nand ˆN2 = 180◦−ˆN1\n∠s on str. line\n∴h = p sin(180◦−ˆN1)\n= p sin ˆN1\n290\n6.5.\nArea, sine, and cosine rules\n\nIn △MSP:\nsin ˆP = h\nn\n∴h = n sin ˆP\nStep 2: Equate the two equations to derive the sine rule\np sin ˆN1 = n sin ˆP\n∴sin ˆN1\nn\n= sin ˆP\np\nor\nn\nsin ˆN1\n=\np\nsin ˆP\nThe ambiguous case\nIf two sides and an interior angle of a triangle are given, and the side opposite the given\nangle is the shorter of the two sides, then we can draw two different triangles (△NMP\nand △NMP ′), both having the given dimensions. We call this the ambiguous case\nbecause there are two ways of interpreting the given information and it is not certain\nwhich is the required solution.\nM\nP ′\np\nN\nP\nn\nn\nWorked example 26: The ambiguous case\nQUESTION\nIn △ABC, AB = 82, BC = 65 and ˆA = 50◦. Draw △ABC and find ˆC (correct to\none decimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the ambiguous case\nWe notice that for the given dimensions of △ABC, the side BC opposite ˆA is shorter\nthan AB. This means that we can draw two different triangles with the given dimen-\nsions.\nA\nB\nC\n50◦\n65\n82\nA′\nC′\nB′\n65\n82\n50◦\n291\nChapter 6.\nTrigonometry\n\nStep 2: Solve for unknown angle using the sine rule\nIn △ABC:\nsin ˆA\nBC = sin ˆC\nAB\nsin 50◦\n65\n= sin ˆC\n82\n∴sin 50◦\n65\n× 82 = sin ˆC\n∴ˆC = 75,1◦\nIn △A′B′C′:\nWe know that sin(180 −ˆC) = sin ˆC, which means we can also have the solution\nˆC′ = 180◦−75,1◦\n= 104,9◦\nBoth solutions are correct.\nWorked example 27: Lighthouses\nQUESTION\nThere is a coastline with two lighthouses, one on either side of a beach. The two\nlighthouses are 0,67 km apart and one is exactly due east of the other. The lighthouses\ntell how close a boat is by taking bearings to the boat (a bearing is an angle measured\nclockwise from north). These bearings are shown on the diagram below.\nCalculate how far the boat is from each lighthouse.\nˆA = 127◦\nˆB = 255◦\nC\nSOLUTION\nWe see that the two lighthouses and the boat form a triangle. Since we know the\ndistance between the lighthouses and we have two angles we can use trigonometry\n292\n6.5.\nArea, sine, and cosine rules\n\nto find the remaining two sides of the triangle, the distance of the boat from the two\nlighthouses.\nb\nA\nb B\nb\nC\n15◦\n37◦\n128◦\n0,67 km\nWe need to determine the lengths of the two sides AC and BC. We can use the sine\nrule to find the missing lengths.\nBC\nsin ˆA\n= AB\nsin ˆC\nBC = AB . sin ˆA\nsin ˆC\n= (0,67 km) sin 37◦\nsin 128◦\n= 0,51 km\nAC\nsin ˆB\n= AB\nsin ˆC\nAC = AB . sin ˆB\nsin ˆC\n= (0,67 km) sin 15◦\nsin 128◦\n= 0,22 km\nExercise 6 – 11: Sine rule\n1. Find all the unknown sides and angles of the following triangles:\na) △PQR in which ˆQ = 64◦; ˆR = 24◦and r = 3\nb) △KLM in which ˆK = 43◦; ˆ\nM = 50◦and m = 1\nc) △ABC in which ˆA = 32,7◦; ˆC = 70,5◦and a = 52,3\nd) △XY Z in which ˆX = 56◦; ˆZ = 40◦and x = 50\n2. In △ABC, ˆA = 116◦; ˆC = 32◦and AC = 23 m. Find the lengths of the sides\nAB and BC.\n3. In △RST, ˆR = 19◦; ˆS = 30◦and RT = 120 km. Find the length of the side\nST.\n4. In △KMS, ˆK = 20◦; ˆ\nM = 100◦and s = 23 cm. Find the length of the side m.\n293\nChapter 6.\nTrigonometry\n\n5. In △ABD, ˆB = 90◦, AB = 10 cm and A ˆDB = 40◦. In △BCD, ˆC = 106◦and\nC ˆDB = 15◦. Determine BC.\nA\nB\nD\nC\n10\n106◦\n15◦\n40◦\n6. In △ABC, ˆA = 33◦, AC = 21 mm and AB = 17 mm. Can you determine BC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233H\n1b. 233J\n1c. 233K\n1d. 233M\n2. 233N\n3. 233P\n4. 233Q\n5. 233R\n6. 233S\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe cosine rule\nEMBHS\nInvestigation: The cosine rule\nIf a triangle is given with two sides and the included angle known, then we can not\nsolve for the remaining unknown sides and angles using the sine rule. We therefore\ninvestigate the cosine rule:\nIn △ABC, AB = 21, AC = 17 and ˆA = 33◦. Find ˆB.\nA\nC\nB\nH\n21\nc\n17\n33◦\n1. Determine CB:\na) Construct CH ⊥AB.\nb) Let AH = c and therefore HB = . . .\n294\n6.5.\nArea, sine, and cosine rules\n\nc) Applying the theorem of Pythagoras in the right-angled triangles:\nIn△CHB:\nCB2 = BH2 + CH2\n= (. . .)2 + CH2\n= 212 −(2)(21)c + c2 + CH2 . . . . . . (1)\nIn △CHA:\nCA2 = c2 + CH2\n172 = c2 + CH2 . . . . . . (2)\nSubstitute equation (2) into equation (1):\nCB2 = 212 −(2)(21)c + 172\nNow c is the only remaining unknown. In △CHA:\nc\n17 = cos 33◦\n∴c = 17 cos 33◦\nTherefore we have that\nCB2 = 212 −(2)(21)c + 172\n= 212 −(2)(21)(17 cos 33◦) + 172\n= 212 + 172 −(2)(21)(17) cos 33◦\n= 131,189 . . .\n∴CB = 11,5\n2. Use your results to write a general formula for the cosine rule given △PQR:\nP\nQ\nR\nq\nr\np\nThe cosine rule relates the length of a side of a triangle to the angle opposite it and the\nlengths of the other two sides.\n295\nChapter 6.\nTrigonometry\n\nConsider △ABC with CD ⊥AB:\nb\nD\nb\nA\nb\nB\nbC\nh\na\nb\nc\nc −d\nd\nIn △DCB: a2 = (c −d)2 + h2 from the theorem of Pythagoras.\nIn △ACD: b2 = d2 + h2 from the theorem of Pythagoras.\nSince h2 is common to both equations we can write:\na2 = (c −d)2 + h2\n∴h2 = a2 −(c −d)2\nAnd b2 = d2 + h2\n∴h2 = b2 −d2\n∴b2 −d2 = a2 −(c −d)2\na2 = b2 + (c2 −2cd + d2) −d2\n= b2 + c2 −2cd\nIn order to eliminate d we look at △ACD, where we have: cos ˆA = d\nb. So, d = b cos ˆA.\nSubstituting back we get: a2 = b2 + c2 −2bc cos ˆA.\nThe cosine rule\nIn any △ABC:\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\na2 = b2 + c2 −2bc cos ˆA\nb2 = a2 + c2 −2ac cos ˆB\nc2 = a2 + b2 −2ab cos ˆC\nSee video: 233T at www.everythingmaths.co.za\n296\n6.5.\nArea, sine, and cosine rules\n\nWorked example 28: The cosine rule\nQUESTION\nDetermine the length of QR.\nP\nR\nQ\n13 cm\n4 cm\n70◦\nSOLUTION\nStep 1: Use the cosine rule to solve for the unknown side\nQR2 = PR2 + QP 2 −2(PR)(QP) cos ˆP\n= 42 + 132 −2(4)(13) cos 70◦\n= 149,42 . . .\n∴QR = 12,2\nStep 2: Write the final answer\nQR = 12,2 cm\nWorked example 29: The cosine rule\nQUESTION\nDetermine ˆA.\n5\n7\n8\nA\nB\nC\nSOLUTION\nApplying the cosine rule:\na2 = b2 + c2 −2bc cos ˆA\n∴cos ˆA = b2 + c2 −a2\n2bc\n= 82 + 52 −72\n2 . 8 . 5\n= 0,5\n∴ˆA = 60◦\n297\nChapter 6.\nTrigonometry\n\nIt is very important:\n• not to round off before the final answer as this will affect accuracy;\n• to take the square root;\n• to remember to give units where applicable.\nHow to determine which rule to use:\n1. Area rule:\n• if no perpendicular height is given\n2. Sine rule:\n• if no right angle is given\n• if two sides and an angle are given (not the included angle)\n• if two angles and a side are given\n3. Cosine rule:\n• if no right angle is given\n• if two sides and the included angle are given\n• if three sides are given\nExercise 6 – 12: The cosine rule\n1. Solve the following triangles (that is, find all unknown sides and angles):\na) △ABC in which ˆA = 70◦; b = 4 and c = 9\nb) △RST in which RS = 14; ST = 26 and RT = 16\nc) △KLM in which KL = 5; LM = 10 and KM = 7\nd) △JHK in which ˆH = 130◦; JH = 13 and HK = 8\ne) △DEF in which d = 4; e = 5 and f = 7\n2. Find the length of the third side of the △XY Z where:\na) ˆX = 71,4◦; y = 3,42 km and z = 4,03 km\nb) x = 103,2 cm; ˆY = 20,8◦and z = 44,59 cm\n3. Determine the largest angle in:\na) △JHK in which JH = 6; HK = 4 and JK = 3\nb) △PQR where p = 50; q = 70 and r = 60\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233V\n1b. 233W\n1c. 233X\n1d. 233Y\n1e. 233Z\n2a. 2342\n2b. 2343\n3a. 2344\n3b. 2345\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n298\n6.5.\nArea, sine, and cosine rules\n\nSee video: 2346 at www.everythingmaths.co.za\nExercise 6 – 13: Area, sine and cosine rule\n1. Q is a ship at a point 10 km due south of another ship P. R is a lighthouse on\nthe coast such that ˆP = ˆQ = 50◦.\n10 km\nP\nQ\nR\n50◦\n50◦\nDetermine:\na) the distance QR\nb) the shortest distance from the lighthouse to the line joining the two ships\n(PQ).\n2. WXY Z is a trapezium, WX ∥Y Z with WX = 3 m; Y Z = 1,5 m; ˆZ = 120◦\nand ˆW = 30◦.\nDetermine the distances XZ and XY .\n1,5 m\n3 m\n30◦\n120◦\nW\nX\nY\nZ\n3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has\nbeen flying 3◦off course. At this point the plane is 500 km from Johannesburg.\nThe direct distance between Cape Town and Johannesburg airports is 1552 km.\nDetermine, to the nearest km:\na) The distance the plane has to travel to get to Cape Town and hence the\nextra distance that the plane has had to travel due to the pilot’s error.\nb) The correction, to one hundredth of a degree, to the plane’s heading (or\ndirection).\n4. ABCD is a trapezium (meaning that AB ∥CD). AB = x; B ˆAD = a; B ˆCD = b\nand B ˆDC = c.\nFind an expression for the length of CD in terms of x, a, b and c.\nA\nB\nC\nD\na\nb\nc\nx\n299\nChapter 6.\nTrigonometry\n\n5. A surveyor is trying to determine the distance between points X and Z. However\nthe distance cannot be determined directly as a ridge lies between the two points.\nFrom a point Y which is equidistant from X and Z, he measures the angle X ˆY Z.\nY\nX\nZ\nx\nθ\na) If XY = x and X ˆY Z = θ, show that XZ = x\np\n2(1 −cos θ).\nb) Calculate XZ (to the nearest kilometre) if x = 240 km and θ = 132◦.\n6. Find the area of WXY Z (to two decimal places):\nW\nX\nY\nZ\n120◦\n3\n4\n3,5\n7. Find the area of the shaded triangle in terms of x, α, β, θ and φ:\nA\nB\nC\nD\nE\nx\nα\nβ\nθ\nφ\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2347\n2. 2348\n3. 2349\n4. 234B\n5. 234C\n6. 234D\n7. 234F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n300\n6.5.\nArea, sine, and cosine rules\n\n6.6\nSummary\nEMBHT\nSee presentation: 234G at www.everythingmaths.co.za\nsquare identity\nquotient identity\ncos2 θ + sin2 θ = 1\ntan θ = sin θ\ncos θ\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\nnegative angles\nperiodicity identities\nco-function identities\nsin(−θ) = −sin θ\nsin(θ ± 360◦) = sin θ\nsin(90◦−θ) = cos θ\ncos(−θ) = cos θ\ncos(θ ± 360◦) = cos θ\ncos(90◦−θ) = sin θ\nsine rule\narea rule\ncosine rule\nsin A\na\n= sin B\nb\n= sin C\nc\narea △ABC = 1\n2bc sin A\na2 = b2 + c2 −2bc cos A\na\nsin A =\nb\nsin B =\nc\nsin C\narea △ABC = 1\n2ac sin B\nb2 = a2 + c2 −2ac cos B\narea △ABC = 1\n2ab sin C\nc2 = a2 + b2 −2ab cos C\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nGeneral solution:\n301\nChapter 6.\nTrigonometry\n\n1.\nIf sin θ = x\nθ = sin−1 x + k . 360◦\nor θ =\n\u0000180◦−sin−1 x\n\u0001\n+ k . 360◦\n2.\nIf cos θ = x\nθ = cos−1 x + k . 360◦\nor θ =\n\u0000360◦−cos−1 x\n\u0001\n+ k . 360◦\n3.\nIf tan θ = x\nθ = tan−1 x + k . 180◦\nfor k ∈Z.\nHow to determine which rule to use:\n1. Area rule:\n• no perpendicular height is given\n2. Sine rule:\n• no right angle is given\n• two sides and an angle are given (not the included angle)\n• two angles and a side are given\n3. Cosine rule:\n• no right angle is given\n• two sides and the included angle angle are given\n• three sides are given\nExercise 6 – 14: End of chapter exercises\n1. Write the following as a single trigonometric ratio:\ncos(90◦−A) sin 20◦\nsin(180◦−A) cos 70◦+ cos(180◦+ A) sin(90◦+ A)\n2. Determine the value of the following expression without using a calculator:\nsin 240◦cos 210◦−tan2 225◦cos 300◦cos 180◦\n302\n6.6.\nSummary\n\n3. Simplify:\nsin(180◦+ θ) sin(θ + 360◦)\nsin(−θ) tan(θ −360◦)\n4. Without the use of a calculator, evaluate:\n3 sin 55◦sin2 325◦\ncos(−145◦)\n−3 cos 395◦sin 125◦\n5. Prove the following identities:\na)\n1\n(cos x −1)(cos x + 1) =\n−1\ntan2 x cos2 x\nb) (1 −tan α) cos α = sin(90 + α) + cos(90 + α)\n6.\na) Prove: tan y +\n1\ntan y =\n1\ncos2 y tan y\nb) For which values of y ∈[0◦; 360◦] is the identity above undefined?\n7.\na) Simplify: sin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\nb) Hence, solve the equation\nsin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\n= tan θ\nfor θ ∈[0◦; 360◦].\n8. Given 12 tan θ = 5 and θ > 90◦.\na) Draw a sketch.\nb) Determine without using a calculator sin θ and cos(180◦+ θ).\nc) Use a calculator to find θ (correct to two decimal places).\n9.\nθ\nP(a; b)\n2\nx\ny\nO\nb\nIn the figure, P is a point on the Cartesian plane such that OP = 2 units and\nθ = 300◦. Without the use of a calculator, determine:\na) the values of a and b\nb) the value of sin(180◦−θ)\n10. Solve for x with x ∈[−180◦; 180◦] (correct to one decimal place):\na) 2 sin x\n2 = 0,86\n303\nChapter 6.\nTrigonometry\n\nb) tan(x + 10◦) = cos 202,6◦\nc) cos2 x −4 sin2 x = 0\n11. Find the general solution for the following equations:\na)\n1\n2 sin(x −25◦) = 0,25\nb) sin2 x + 2 cos x = −2\n12. Given the equation: sin 2α = 0,84\na) Find the general solution of the equation.\nb) Illustrate how this equation could be solved graphically for α ∈[0◦; 360◦].\nc) Write down the solutions for sin 2α = 0,84 for α ∈[0◦; 360◦].\n13.\nA\nT\nG\nN\nH\nn\nα\nβ\nA is the highest point of a vertical tower AT. At point N on the tower, n metres\nfrom the top of the tower, a bird has made its nest. The angle of inclination from\nG to point A is α and the angle of inclination from G to point N is β.\na) Express A ˆGN in terms of α and β.\nb) Express ˆA in terms of α and/or β.\nc) Show that the height of the nest from the ground (H) can be determined by\nthe formula\nH = n cos α sin β\nsin(α −β)\nd) Calculate the height of the nest H if n = 10 m, α = 68◦and β = 40◦(give\nyour answer correct to the nearest metre).\n304\n6.6.\nSummary\n\n14.\nA\nD\nB\nC\n11\n8\n5\nMr. Collins wants to pave his trapezium-shaped backyard, ABCD. AB ∥DC\nand ˆB = 90◦. DC = 11 m, AB = 8 m and BC = 5 m.\na) Calculate the length of the diagonal AC.\nb) Calculate the length of the side AD.\nc) Calculate the area of the patio using geometry.\nd) Calculate the area of the patio using trigonometry.\n15.\nA\nC\nB\n2t\nF\nt\nn\nn\n2n\nα\nIn △ABC, AC = 2A, AF = BF, A ˆFB = α and FC = 2AF. Prove that\ncos α = 1\n4.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 234H\n2. 234J\n3. 234K\n4. 234M\n5a. 234N\n5b. 234P\n6. 234Q\n7. 234R\n8. 234S\n9. 234T\n10a. 234V\n10b. 234W\n10c. 234X\n11a. 234Y\n11b. 234Z\n12. 2352\n13. 2353\n14. 2354\n15. 2355\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n305\nChapter 6.\nTrigonometry\n\n\nCHAPTER\n7\nMeasurement\n7.1\nArea of a polygon\n308\n7.2\nRight prisms and cylinders\n311\n7.3\nRight pyramids, right cones and spheres\n318\n7.4\nMultiplying a dimension by a constant factor\n322\n7.5\nSummary\n326\n\n7\nMeasurement\nThis chapter is a revision of perimeters and areas of two dimensional objects and\nvolumes of three dimensional objects. We also examine different combinations of\ngeometric objects and calculate areas and volumes in a variety of real-life contexts.\nSee video: 2356 at www.everythingmaths.co.za\n7.1\nArea of a polygon\nEMBHV\nSquare\ns\ns\nArea = s2\nRectangle\nh\nb\nArea = b × h\nTriangle\nh\nb\nArea = 1\n2b × h\nSee video: 2357 at www.everythingmaths.co.za\nTrapezium\nh\nb\na\nArea = 1\n2 (a + b) × h\nParallelogram\nh\nb\nArea = b × h\nCircle\nb r\nArea = πr2\n(Circumference = 2πr)\nSee video: 2358 at www.everythingmaths.co.za\n308\n7.1.\nArea of a polygon\n\nWorked example 1: Finding the area of a polygon\nQUESTION\nABCD is a parallelogram with DC = 15 cm, h = 8 cm and BF = 9 cm.\nA\nB\nC\nD\nH\n9 cm\n15 cm\nh\nF\nCalculate:\n1. the area of ABCD\n2. the perimeter of ABCD\nSOLUTION\nStep 1: Determine the area\nThe area of a parallelogram ABCD = base × height:\nArea = 15 × 8\n= 120 cm2\nStep 2: Determine the perimeter\nThe perimeter of a parallelogram ABCD = 2DC + 2BC.\nTo find the length of BC, we use AF ⊥BC and the theorem of Pythagoras.\nIn △ABF:\nAF 2 = AB2 −BF 2\n= 152 −92\n= 144\n∴AF = 12 cm\nAreaABCD = BC × AF\n120 = BC × 12\n∴BC = 10 cm\n∴PerimeterABCD = 2(15) + 2(10)\n= 50 cm\n309\nChapter 7.\nMeasurement\n\nExercise 7 – 1: Area of a polygon\n1. Vuyo and Banele are having a competition to see who can build the best kite\nusing balsa wood (a lightweight wood) and paper. Vuyo decides to make his kite\nwith one diagonal 1 m long and the other diagonal 60 cm long. The intersection\nof the two diagonals cuts the longer diagonal in the ratio 1 : 3.\nBanele also uses diagonals of length 60 cm and 1 m, but he designs his kite to\nbe rhombus-shaped.\na) Draw a sketch of Vuyo’s kite and write down all the known measurements.\nb) Determine how much balsa wood Vuyo will need to build the outside frame\nof the kite (give answer correct to the nearest cm).\nc) Calculate how much paper he will need to cover the frame of the kite.\nd) Draw a sketch of Banele’s kite and write down all the known measure-\nments.\ne) Determine how much wood and paper Banele will need for his kite.\nf) Compare the two designs and comment on the similarities and differences.\nWhich do you think is the better design? Motivate your answer.\n2. O is the centre of the bigger semi-circle with a radius of 10 units. Two smaller\nsemi-circles are inscribed into the bigger one, as shown on the diagram. Calcu-\nlate the following (in terms of π):\nO\nb\nb\na) The area of the shaded figure.\nb) The perimeter enclosing the shaded area.\n3. Karen’s engineering textbook is 30 cm long and 20 cm wide. She notices that\nthe dimensions of her desk are in the same proportion as the dimensions of her\ntextbook.\na) If the desk is 90 cm wide, calculate the area of the top of the desk.\nb) Karen uses some cardboard to cover each corner of her desk with an isosce-\nles triangle, as shown in the diagram:\n150 mm\n150 mm\ndesk\nCalculate the new perimeter and area of the visible part of the top of her\ndesk.\n310\n7.1.\nArea of a polygon\n\nc) Use this new area to calculate the dimensions of a square desk with the\nsame desk top area.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2359\n2. 235B\n3. 235C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.2\nRight prisms and cylinders\nEMBHW\nA right prism is a geometric solid that has a polygon as its base and vertical sides\nperpendicular to the base. The base and top surface are the same shape and size. It is\ncalled a “right” prism because the angles between the base and sides are right angles.\nA triangular prism has a triangle as its base, a rectangular prism has a rectangle as its\nbase, and a cube is a rectangular prism with all its sides of equal length. A cylinder is\nanother type of right prism which has a circle as its base. Examples of right prisms are\ngiven below: a rectangular prism, a cube, a triangular prism and a cylinder.\nSurface area of prisms and cylinders\nEMBHX\nSurface area is the total area of the exposed or outer surfaces of a prism. This is easier\nto understand if we imagine the prism to be a cardboard box that we can unfold. A\nsolid that is unfolded like this is called a net. When a prism is unfolded into a net, we\ncan clearly see each of its faces. In order to calculate the surface area of the prism, we\ncan then simply calculate the area of each face, and add them all together.\nFor example, when a triangular prism is unfolded into a net, we can see that it has\ntwo faces that are triangles and three faces that are rectangles. To calculate the surface\narea of the prism, we find the area of each triangle and each rectangle, and add them\ntogether.\nIn the case of a cylinder the top and bottom faces are circles and the curved surface\nflattens into a rectangle with a length that is equal to the circumference of the circular\nbase. To calculate the surface area we therefore find the area of the two circles and the\nrectangle and add them together.\n311\nChapter 7.\nMeasurement\n\nBelow are examples of right prisms that have been unfolded into nets. A rectangular\nprism unfolded into a net is made up of six rectangles.\nA cube unfolded into a net is made up of six identical squares.\nA triangular prism unfolded into a net is made up of two triangles and three rectangles.\nThe sum of the lengths of the rectangles is equal to the perimeter of the triangles.\nA cylinder unfolded into a net is made up of two identical circles and a rectangle with\nlength equal to the circumference of the circles.\n312\n7.2.\nRight prisms and cylinders\n\nWorked example 2: Calculating surface area\nQUESTION\nA box of chocolates has the following dimensions:\nlength = 25 cm\nwidth = 20 cm\nheight = 4 cm\n25 cm\n20 cm\n4 cm\nAnd a cylindrical tin of biscuits has the following dimensions:\ndiameter = 20 cm\nheight = 20 cm\nb\n20 cm\n20 cm\n1. Calculate the area of the wrapping paper needed to cover the entire box (assume\nno overlapping at the corners).\n2. Determine if this same sheet of wrapping paper would be enough to cover the\ntin of biscuits.\nSOLUTION\nStep 1: Determine the area of the rectangular box\nSurface area = 2 × (25 × 20) + 2 × (20 × 4) + 2 × (25 × 4)\n= 1360 cm2\n313\nChapter 7.\nMeasurement\n\nStep 2: Determine the area of the cylindrical tin\nThe radius of the cylinder = 20\n2 = 10 cm.\nSurface area = 2 × π(10)2 + 2π(10)(20)\n= 1885 cm2\nStep 3: Write the final answer\nNo, the area of the sheet of wrapping paper used to cover the box is not big enough to\ncover the tin.\nExercise 7 – 2: Calculating surface area\n1. A popular chocolate container is an equilateral right triangular prism with sides\nof 34 mm. The box is 170 mm long. Calculate the surface area of the box (to the\nnearest square centimetre).\n34 mm\n34 mm\n34 mm\n170 mm\n2. Gordon buys a cylindrical water tank to catch rain water off his roof. He discov-\ners a full 2 ℓtin of green paint in his garage and decides to paint the tank (not the\nbase). If he uses 250 ml to cover 1 m2, will he have enough green paint to cover\nthe tank with one layer of paint?\nDimensions of the tank:\ndiameter = 1,1 m\nheight = 1,4 m\nb\n1,1 m\n1,4 m\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235D\n2. 235F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n314\n7.2.\nRight prisms and cylinders\n\nVolume of prisms and cylinders\nEMBHY\nVolume, sometimes also called capacity, is the three dimensional space occupied by\nan object, or the contents of an object. It is measured in cubic units.\nThe volume of a right prism is simply calculated by multiplying the area of the base of\na solid by the height of the solid.\nRectangular\nprism\nl\nb\nh\nVolume = area of base × height\n= area of rectangle × height\n= l × b × h\nTriangular\nprism\nH\nb\nh\nVolume = area of base × height\n= area of triangle × height\n=\n\u00121\n2b × h\n\u0013\n× H\nCylinder\nh\nr\nVolume = area of base × height\n= area of circle × height\n= πr2 × h\nSee video: 235G at www.everythingmaths.co.za\n315\nChapter 7.\nMeasurement\n\nWorked example 3: Calculating volume\nQUESTION\nA rectangular glass vase with dimensions 28 cm × 18 cm × 8 cm is used for flower\narrangements. A florist uses a platic cylindrical jug to pour water into the glass vase.\nThe jug has a diameter of 142 mm and a height of 28 cm.\n28 cm\n18 cm\n8 cm\n142 mm\n28 cm\njug\nvase\n1. Will the plastic jug hold 5 ℓof water?\n2. Will a full jug of water be enough to fill the glass vase?\nSOLUTION\nStep 1: Determine the volume of the plastic jug\nThe diameter of the jug is 142 mm, therefore the radius =\n142\n2×10 = 7,1 cm.\nVolume of a cylinder = area of the base × height\nVolume of the jug = πr2 × h\n= π × (7,1)2 × 28\n= 4434 cm3\nAnd 1000 cm3 = 1 ℓ\n∴Volume of the jug = 4434\n1000\n= 4,434 ℓ\nNo, the capacity of the jug is not enough to hold 5 ℓof water.\n316\n7.2.\nRight prisms and cylinders\n\nStep 2: Determine the volume of the glass vase\nVolume of a rectangular prism = area of the base × height\nVolume of the vase = l × b × h\n= 28 × 18 × 8\n= 4032 cm3\n∴Volume of the vase = 4032\n1000\n= 4,032 ℓ\nYes, the volume of the jug is greater than the volume of the vase.\nExercise 7 – 3: Calculating volume\n1. The roof of Phumza’s house is the shape of a right-angled trapezium. A cylindri-\ncal water tank is positioned next to the house so that the rain on the roof runs\ninto the tank. The diameter of the tank is 140 cm and the height is 2,2 m.\n10 m\n8 m\n7,5 m\n2,2 m\n140 cm\na) Determine the area of the roof.\nb) Determine how many litres of water the tank can hold.\n2. The length of a side of a hexagonal sweet tin is 8 cm and its height is equal to\nhalf of the side length.\nA\nB\nC\nD\nE\nF\n8 cm\nh\na) Show that the interior angles are equal to 120◦.\n317\nChapter 7.\nMeasurement\n\nb) Determine the length of the line AE.\nc) Calculate the volume of the tin.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235H\n2. 235J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.3\nRight pyramids, right cones and spheres\nEMBHZ\nA pyramid is a geometric solid that has a polygon as its base and sides that converge\nat a point called the apex. In other words the sides are not perpendicular to the base.\nb\nThe triangular pyramid and square pyramid take their names from the shape of their\nbase. We call a pyramid a “right pyramid” if the line between the apex and the centre\nof the base is perpendicular to the base. Cones are similar to pyramids except that\ntheir bases are circles instead of polygons. Spheres are solids that are perfectly round\nand look the same from any direction.\nSurface area of pyramids, cones and spheres\nEMBJ2\nSquare\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n= b2 + 4\n\u0000 1\n2bhs\n\u0001\n= b (b + 2hs)\n318\n7.3.\nRight pyramids, right cones and spheres\n\nTriangular\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n=\n\u0000 1\n2b × hb\n\u0001\n+ 3\n\u0000 1\n2b × hs\n\u0001\n= 1\n2b (hb + 3hs)\nRight cone\nh\nr\nH\nSurface area = area of base +\narea of walls\n= πr2 + 1\n2 × 2πrh\n= πr (r + h)\nSphere\nb\nr\nSurface area = 4πr2\nVolume of pyramids, cones and spheres\nEMBJ3\nSquare\npyramid\nb\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × b2 × H\nTriangular\npyramid\nb\nh\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × 1\n2bh × H\n319\nChapter 7.\nMeasurement\n\nRight cone\nr\nH\nVolume = 1\n3 × area of base ×\nheight of cone\n= 1\n3 × πr2 × H\nSphere\nb\nr\nVolume = 4\n3πr3\nSee video: 235K at www.everythingmaths.co.za\nWorked example 4: Finding surface area and volume\nQUESTION\nThe Southern African Large Telescope (SALT) is housed in a cylindrical building with\na domed roof in the shape of a hemisphere. The height of the building wall is 17 m\nand the diameter is 26 m.\n17 m\n26 m\n1. Calculate the total surface area of the building.\n2. Calculate the total volume of the building.\n320\n7.3.\nRight pyramids, right cones and spheres\n\nSOLUTION\nStep 1: Calculate the total surface area\nTotal surface area = area of the dome + area of the cylinder\nSurface area =\n\u00141\n2(4πr2)\n\u0015\n+ [2πr × h]\n= 1\n2(4π)(13)2 + 2π(13)(17)\n= 2450 m2\nStep 2: Calculate the total volume\nTotal volume = volume of the dome + volume of the cylinder\nVolume =\n\u00141\n2 ×\n\u00124\n3πr3\n\u0013\u0015\n+\n\u0002\nπr2h\n\u0003\n= 2\n3π(13)3 + π(11)2(13)\n= 9543 m3\nExercise 7 – 4: Finding surface area and volume\n1. An ice-cream cone has a diameter of 52,4 mm and a total height of 146 mm.\n52,4 mm\n146 mm\na) Calculate the surface area of the ice-cream and the cone.\nb) Calculate the total volume of the ice-cream and the cone.\nc) How many ice-cream cones can be made from a 5 ℓtub of ice-cream (as-\nsume the cone is completely filled with ice-cream)?\n321\nChapter 7.\nMeasurement\n\nd) Consider the net of the cone given below. R is the length from the tip of\nthe cone to its perimeter, P.\nP\nR\nb\nM\ni. Determine the value of R.\nii. Calculate the length of arc P.\niii. Determine the length of arc M.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.4\nMultiplying a dimension by a constant factor\nEMBJ4\nWhen one or more of the dimensions of a prism or cylinder is multiplied by a constant,\nthe surface area and volume will change. The new surface area and volume can be\ncalculated by using the formulae from the preceding section.\nIt is important to see a relationship between the change in dimensions and the resulting\nchange in surface area and volume. These relationships make it simpler to calculate\nthe new volume or surface area of an object when its dimensions are scaled up or\ndown.\nConsider a rectangular prism of dimensions l, b and h. Below we multiply one, two\nand three of its dimensions by a constant factor of 5 and calculate the new volume and\nsurface area.\n322\n7.4.\nMultiplying a dimension by a constant factor\n\nDimensions\nVolume\nSurface\nOriginal dimensions\nl\nb\nh\nV = l × b × h\n= lbh\nA\n= 2 [(l × h) + (l × b) + (b × h)]\n= 2 (lh + lb + bh)\nMultiply one\ndimension by 5\nl\nb\n5h\nV1 = l × b × 5h\n= 5 (lbh)\n= 5V\nA1\n= 2 [(l × 5h) + (l × b) + (b × 5h)]\n= 2 (5lh + lb + 5bh)\nMultiply two\ndimensions by 5\n5l\nb\n5h\nV = 5l × b × 5h\n= 5 . 5(lbh)\n= 52V\nA2\n= 2 [(5l × 5h) + (5l × b) + (b × 5h)]\n= 2 × 5(5lh + lb + bh)\nMultiply all three\ndimensions by 5\n5l\n5b\n5h\nV = 5l × 5b × 5h\n= 53(lbh)\n= 53V\nA3\n= 2 [(5l × 5h) + (5l × 5b) + (5b × 5h)]\n= 2 × (52lh + 52lb + 52bh)\n= 52 × 2(lh + lb + bh)\n= 52A\nMultiply all three\ndimensions by k\nkl\nkb\nkh\nV = kl × kb × kh\n= k3(lbh)\n= k3V\nAk\n= 2 [(kl × kh) + (kl × kb) + (kb × kh)]\n= 2 × (k2lh + k2lb + k2bh)\n= k2 × 2(lh + lb + bh)\n= k2A\n323\nChapter 7.\nMeasurement\n\nWorked example 5: The effects of k\nQUESTION\nThe Nash family wants to build a television room onto their house. The dad draws up\nthe plans for the new square room of length k metres. The mum looks at the plans and\ndecides that the area of the room needs to be doubled. To achieve this:\n• the mum suggests doubling the length of the sides of the room\n• the dad recommends adding 2 m to the length of the sides\n• the daughter suggests multiplying the length of the sides by a factor of\n√\n2\n• the son suggests doubling only the width of the room\nWho’s suggestion will double the area of the square room? Show all calculations.\nSOLUTION\nStep 1: Draw a sketch\nk\nk\n2k\n2k\nk + 2\nk + 2\n√\n2k\n√\n2k\n2k\nk\nArea O\nArea M\nArea D\nArea d\nArea s\nStep 2: Calculate and compare\nFirst calculate the area of the square room in the original plan:\nArea O = length × length\n= k2\nTherefore, double the area of the room would be 2k2.\n324\n7.4.\nMultiplying a dimension by a constant factor\n\nConsider the mum’s suggestion of doubling the length of the sides of the room:\nArea M = length × length\n= 2k × 2k\n= 4k2\nThis area would be 4 times the original area.\nThe dad suggests adding 2 m to the length of the sides of the room:\nArea D = length × length\n= (k + 2) × (k + 2)\n= k2 + 4k + 2\n̸= 2k2\nThis is not double the original area.\nThe daughter suggests multiplying the length of the sides by a factor of\n√\n2:\nArea d = length × length\n=\n√\n2k ×\n√\n2k\n= 2k2\nThe daughter’s suggestion would double the area of the room. Practically, the length\nof the room could be multiplied by\n√\n2 ≈1,41 which would given an area of 1,96 m2.\nThe son suggests doubling only the width of the room:\nArea s = length × length\n= 2k × k\n= 2k2\nThe son’s suggestion would double the area of the room, however the room would no\nlonger be a square.\nStep 3: Write the final answer\nThe daughter’s suggestion of multiplying the length of the sides of the room by a factor\nof\n√\n2 would keep the shape of the room a square and would double the area of the\nroom.\nExercise 7 – 5: The effects of k\n1. Complete the following sentences:\na) If one dimension of a cube is multiplied by a factor 1\n2, the volume of the\ncube . . .\nb) If two dimensions of a cube are multiplied by a factor 7, the volume of the\ncube . . .\n325\nChapter 7.\nMeasurement\n\nc) If three dimensions of a cube are multiplied by a factor 3, then:\ni. each side of the cube will . . .\nii. the outer surface area of the cube will . . .\niii. the volume of the cube will . . .\nd) If each side of a cube is halved, then:\ni. the outer surface area of the cube will . . .\nii. the volume of the cube will . . .\n2. The municipality intends building a swimming pool of volume W 3 cubic metres.\nHowever, they realise that it will be very expensive to fill the pool with water, so\nthey decide to make the pool smaller.\na) The length and breadth of the pool are reduced by a factor of\n7\n10. Express\nthe new volume in terms of W.\nb) The dimensions of the pool are reduced so that the volume of the pool\ndecreases by a factor of 0,8. Determine the new dimensions of the pool in\nterms of W (remember that the pool must be a cube).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235N\n2. 235P\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.5\nSummary\nEMBJ5\nSee presentation: 235Q at www.everythingmaths.co.za\n1. Area is the two dimensional space inside the boundary of a flat object.\n2. Area formulae:\n• square: s2\n• rectangle: b × h\n• triangle: 1\n2b × h\n• trapezium: 1\n2 (a + b) × h\n• parallelogram: b × h\n• circle: πr2\n3. Surface area is the total area of the exposed or outer surfaces of a prism.\n4. A net is the unfolded “plan” of a solid.\n5. Volume is the three dimensional space occupied by an object, or the contents\nof an object.\n• Volume of a rectangular prism: l × b × h\n326\n7.5.\nSummary\n\n• Volume of a triangular prism:\n\u0000 1\n2b × h\n\u0001\n× H\n• Volume of a square prism or cube: s3\n• Volume of a cylinder: πr2 × h\n6. A pyramid is a geometric solid that has a polygon as its base and sides that\nconverge at a point called the apex. The sides are not perpendicular to the base.\n7. Surface area formulae:\n• square pyramid: b (b + 2h)\n• triangular pyramid: 1\n2b (hb + 3hs)\n• right cone: πr (r + hs)\n• sphere: 4πr2\n8. Volume formulae:\n• square pyramid: 1\n3 × b2 × H\n• triangular pyramid: 1\n3 × 1\n2bh × H\n• right cone: 1\n3 × πr2 × H\n• sphere: 4\n3πr3\nExercise 7 – 6: End of chapter exercises\n1.\na) Describe this figure in terms of a prism.\nb) Draw a net of this figure.\n2. Which of the following is a net of a cube?\na)\nb)\nc)\nd)\ne)\n327\nChapter 7.\nMeasurement\n\n3. Name and draw the following figures:\na) A prism with the least number of sides.\nb) A pyramid with the least number of vertices.\nc) A right prism with a kite base.\n4.\na)\ni. Determine how much paper is needed to make a box of width 16 cm,\nheight 3 cm and length 20 cm (assume no overlapping at corners).\nii. Give a mathematical name for the shape of the box.\niii. Calculate the volume of the box.\nb) Determine how much paper is needed to make a cube with a capacity of\n1 ℓ.\nc) Compare the box and the cube. Which has the greater volume and which\nrequires the most paper to make?\n5. ABCD is a rhombus with sides of length 3\n2x millimetres. The diagonals intersect\nat O and length DO = x millimetres. Express the area of ABCD in terms of x.\nO\nB\nD\nx\nC\nA\n3\n2x\n6. The diagram shows a rectangular pyramid with a base of length 80 cm and\nbreadth 60 cm. The vertical height of the pyramid is 45 cm.\n60 cm\n80 cm\n45 cm\nb\nh\nH\na) Calculate the volume of the pyramid.\nb) Calculate H and h.\nc) Calculate the surface area of the pyramid.\n7. A group of children are playing soccer in a field. The soccer ball has a capacity\nof 5000 cc (cubic centimetres). A drain pipe in the corner of the field has a\ndiameter of 20 cm. Is it possible for the children to lose their ball down the pipe?\nShow your calculations.\n328\n7.5.\nSummary\n\n8. A litre of washing powder goes into a standard cubic container at the factory.\na) Determine the length of the sides of the container.\nb) Determine the dimensions of the cubic container required to hold double\nthe volume of washing powder.\n9. A cube has sides of length k units.\na) Describe the effect on the volume of the cube if the height is tripled.\nb) If all three dimensions of the cube are tripled, determine the effect on the\nouter surface area.\nc) If all three dimensions of the cube are tripled, determine the effect on the\nvolume.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235R\n2. 235S\n3a. 235T\n3b. 235V\n3c. 235W\n4. 235X\n5. 235Y\n6. 235Z\n7. 2362\n8. 2363\n9. 2364\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n329\nChapter 7.\nMeasurement\n\n\nCHAPTER\n8\nEuclidean geometry\n8.1\nRevision\n332\n8.2\nCircle geometry\n333\n8.3\nSummary\n363\n\n8\nEuclidean geometry\n8.1\nRevision\nEMBJ6\nParallelogram\nEMBJ7\nA parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nSummary of the properties of a parallelogram:\n• Both pairs of opposite sides are parallel.\n• Both pairs of opposite sides are equal in length.\n• Both pairs of opposite angles are equal.\n• Both diagonals bisect each other.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\nThe mid-point theorem\nEMBJ8\nThe line joining the mid-points of two sides of a triangle is parallel to the third side\nand equal to half the length of the third side.\nA\nB\nC\nD\nE\nGiven: AD = DB and AE = EC, we can conclude that DE ∥BC and DE = 1\n2BC.\n332\n8.1.\nRevision\n\n8.2\nCircle geometry\nEMBJ9\nTerminology\nThe following terms are regularly used when referring to circles:\n• Arc — a portion of the circumference of a circle.\n• Chord — a straight line joining the ends of an arc.\n• Circumference — the perimeter or boundary line of a circle.\n• Radius (r) — any straight line from the centre of the circle to a point on the\ncircumference.\n• Diameter — a special chord that passes through the centre of the circle. A di-\nameter is a straight line segment from one point on the circumference to another\npoint on the circumference that passes through the centre of the circle.\n• Segment — part of the circle that is cut off by a chord. A chord divides a circle\ninto two segments.\n• Tangent — a straight line that makes contact with a circle at only one point on\nthe circumference.\nb\nb\nA\nB\nO\nP\na\nr\nc\nchord\ntangent\ndiameter\nradius\nsegment\nSee video: 2365 at www.everythingmaths.co.za\nAxioms\nAn axiom is an established or accepted principle. For this section, the following are\naccepted as axioms.\n333\nChapter 8.\nEuclidean geometry\n\n1. The theorem of Pythagoras states that the square of the hypotenuse of a right-\nangled triangle is equal to the sum of the squares of the other two sides.\n(AC)2 = (AB)2 + (BC)2\nC\nB\nA\n(AC)2\n(AB)2\n(BC)2\n2. A tangent is perpendicular to the radius (OT ⊥ST), drawn at the point of contact\nwith the circle.\nT\nS\nb\nO\nTheorems\nEMBJB\nA theorem is a hypothesis (proposition) that can be shown to be true by accepted\nmathematical operations and arguments. A proof is the process of showing a theorem\nto be correct.\nThe converse of a theorem is the reverse of the hypothesis and the conclusion. For\nexample, given the theorem “if A, then B”, the converse is “if B, then A”.\n334\n8.2.\nCircle geometry\n\nTheorem: Perpendicular line from circle centre bisects chord\nSTATEMENT\nIf a line is drawn from the centre of a circle perpendicular to a chord, then it bisects\nthe chord.\n(Reason: ⊥from centre bisects chord)\nGiven:\nCircle with centre O and line OP perpendicular to chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = PB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA2 = OP 2 + AP 2\n(Pythagoras)\nOB2 = OP 2 + BP 2\n(Pythagoras)\nand\nOA = OB\n(equal radii)\n∴AP 2 = BP 2\n∴AP = BP\nTherefore OP bisects AB.\nAlternative proof:\nIn △OPA and in △OPB,\nO ˆPA = O ˆPB\n(given OP ⊥AB)\nOA = OB\n(equal radii)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(RHS)\n∴AP = PB\nTherefore OP bisects AB.\n335\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Line from circle centre to mid-point of\nchord is perpendicular\nSTATEMENT\nIf a line is drawn from the centre of a circle to the mid-point of a chord, then the line\nis perpendicular to the chord.\n(Reason: line from centre to mid-point ⊥)\nGiven:\nCircle with centre O and line OP to mid-point P on chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nOP ⊥AB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA = OB\n(equal radii)\nAP = PB\n(given)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(SSS)\n∴O ˆPA = O ˆPB\nand O ˆPA + O ˆPB = 180◦\n(∠on str. line)\n∴O ˆPA = O ˆPB = 90◦\nTherefore OP ⊥AB.\nSee video: 2366 at www.everythingmaths.co.za\n336\n8.2.\nCircle geometry\n\nTheorem: Perpendicular bisector of chord passes through circle centre\nSTATEMENT\nIf the perpendicular bisector of a chord is drawn, then the line will pass through the\ncentre of the circle.\n(Reason: ⊥bisector through centre)\nGiven:\nCircle with mid-point P on chord AB.\nLine QP is drawn such that Q ˆPA = Q ˆPB = 90◦.\nLine RP is drawn such that R ˆPA = R ˆPB = 90◦.\nb\nb\nA\nB\nQ\nP\nR\nRequired to prove:\nCircle centre O lies on the line PR\nPROOF\nDraw lines QA and QB.\nDraw lines RA and RB.\nIn △QPA and in △QPB,\nAP = PB\n(given)\nQP = QP\n(common side)\nQ ˆPA = Q ˆPB = 90◦\n(given)\n∴△QPA ≡△QPB\n(SAS)\n∴QA = QB\nSimilarly it can be shown that in △RPA and in △RPB, RA = RB.\nWe conclude that all the points that are equidistant from A and B will lie on the\nline PR extended. Therefore the centre O, which is equidistant to all points on the\ncircumference, must also lie on the line PR.\n337\nChapter 8.\nEuclidean geometry\n\nWorked example 1: Perpendicular line from circle centre bisects chord\nQUESTION\nGiven OQ ⊥PR and PR = 8 units, determine the value of x.\nO\nx\n5\nP\nQ\nR\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nPQ = QR = 4\n(⊥from centre bisects chord)\nStep 2: Solve for x\nIn △OQP:\nPQ = 4\n(⊥from centre bisects chord)\nOP 2 = OQ2 + QP 2\n(Pythagoras)\n52 = x2 + 42\n∴x2 = 25 −16\nx2 = 9\nx = 3\nStep 3: Write the final answer\nx = 3 units.\n338\n8.2.\nCircle geometry\n\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. In the circle with centre O, OQ ⊥PR,\nOQ = 4 units and PR = 10. Determine\nx.\nO\n4\nP\nQ\nR\nx\n2. In the circle with centre O and radius\n= 10 units, OQ ⊥PR and PR = 8. De-\ntermine x.\nO\nx\n10\nP\nQ\nR\n3. In the circle with centre O, OQ ⊥PR,\nPR = 12 units and SQ = 2 units. Deter-\nmine x.\nO\nx\nP\nQ\nR\nS\n4. In the circle with centre O, OT ⊥SQ,\nOT ⊥PR, OP = 10 units, ST = 5 units\nand PU = 8 units. Determine TU.\nO\nV\n8\nP\nR\nU\n10\n5\nT\nQ\nS\n5. In the circle with centre O, OT ⊥QP,\nOS ⊥PR, OT = 5 units, PQ = 24 units\nand PR = 25 units. Determine OS = x.\nO\nx\nP\nS\n5\nT\nQ\nR\n25\n24\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2367\n2. 2368\n3. 2369\n4. 236B\n5. 236C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n339\nChapter 8.\nEuclidean geometry\n\nInvestigation: Angles subtended by an arc at the centre and the circumference of\na circle\n1. Measure angles x and y in each of the following graphs:\nb\nx1\ny1\nb\nx2\ny2\nb\nx3\ny3\n2. Complete the table:\nx\ny\n3. Use your results to make a conjecture about the relationship between angles\nsubtended by an arc at the centre of a circle and angles at the circumference of\na circle.\n4. Now draw three of your own similar diagrams and measure the angles to check\nyour conjecture.\n340\n8.2.\nCircle geometry\n\nTheorem: Angle at the centre of a circle is twice the size of the angle at the cir-\ncumference\nSTATEMENT\nIf an arc subtends an angle at the centre of a circle and at the circumference, then the\nangle at the centre is twice the size of the angle at the circumference.\n(Reason: ∠at centre = 2∠at circum.)\nGiven:\nCircle with centre O, arc AB subtending A ˆOB at the centre of the circle, and A ˆPB at\nthe circumference.\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nRequired to prove:\nA ˆOB = 2A ˆPB\nPROOF\nDraw PO extended to Q and let A ˆOQ = ˆO1 and B ˆOQ = ˆO2.\nˆO1 = A ˆPO + P ˆAO\n(ext. ∠△= sum int. opp. ∠s)\nand A ˆPO = P ˆAO\n(equal radii, isosceles △APO)\n∴ˆO1 = A ˆPO + A ˆPO\nˆO1 = 2A ˆPO\nSimilarly, we can also show that ˆO2 = 2B ˆPO.\nFor the first two diagrams shown above we have that:\nA ˆOB = ˆO1 + ˆO2\n= 2A ˆPO + 2B ˆPO\n= 2(A ˆPO + B ˆPO)\n∴A ˆOB = 2(A ˆPB)\nAnd for the last diagram:\nA ˆOB = ˆO2 −ˆO1\n= 2B ˆPO −2A ˆPO\n= 2(B ˆPO −A ˆPO)\n∴A ˆOB = 2(A ˆPB)\n341\nChapter 8.\nEuclidean geometry\n\nWorked example 2: Angle at the centre of circle is twice angle at circumference\nQUESTION\nGiven HK, the diameter of the circle passing through centre O.\nb\nJ\nH\nK\nO\na\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nStep 2: Solve for a\nIn △HJK:\nH ˆOK = 180◦\n(∠on str. line)\n= 2a\n(∠at centre = 2∠at circum.)\n∴2a = 180◦\na = 180◦\n2\n= 90◦\nStep 3: Conclusion\nThe diameter of a circle subtends a right angle at the circumference (angles in a semi-\ncircle).\n342\n8.2.\nCircle geometry\n\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\nGiven O is the centre of the circle, determine the unknown angle in each of the fol-\nlowing diagrams:\n1.\nb\nJ\nH\nK\nO\nb\n45◦\n2.\nbO\nJ\nK\nH\n45◦\nc\n3.\nb\nO\nK\nJ\n100◦\nH\nd\n4.\nb\nO\nH\nJ\ne\nK\n35◦\n5.\nb\nO\nJ\nK\nH\n120◦\nf\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236D\n2. 236F\n3. 236G\n4. 236H\n5. 236J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n343\nChapter 8.\nEuclidean geometry\n\nInvestigation: Subtended angles in the same segment of a circle\n1. Measure angles a, b, c, d and e in the diagram below:\na\ne\nd\nb\nc\nP\nQ\n2. Choose any two points on the circumference of the circle and label them A and\nB.\n3. Draw AP and BP, and measure A ˆPB.\n4. Draw AQ and BQ, and measure A ˆQB.\n5. What do you observe? Make a conjecture about these types of angles.\nTheorem: Subtended angles in the same segment of a circle are equal\nSTATEMENT\nIf the angles subtended by a chord of the circle are on the same side of the chord, then\nthe angles are equal.\n(Reason: ∠s in same seg.)\nGiven:\nCircle with centre O, and points P and Q on the circumference of the circle. Arc AB\nsubtends A ˆPB and A ˆQB in the same segment of the circle.\n344\n8.2.\nCircle geometry\n\nbO\nA\nB\nP\nQ\nRequired to prove:\nA ˆPB = A ˆQB\nPROOF\nA ˆOB = 2A ˆPB\n(∠at centre = 2∠at circum.)\nA ˆOB = 2A ˆQB\n(∠at centre = 2∠at circum.)\n∴2A ˆPB = 2A ˆQB\nA ˆPB = A ˆQB\nEqual arcs subtend equal angles\nFrom the theorem above we can deduce that if angles at the circumference of a circle\nare subtended by arcs of equal length, then the angles are equal. In the figure below,\nnotice that if we were to move the two chords with equal length closer to each other,\nuntil they overlap, we would have the same situation as with the theorem above. This\nshows that the angles subtended by arcs of equal length are also equal.\nb\nb\n345\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Concyclic points\nSTATEMENT\nIf a line segment subtends equal angles at two other points on the same side of the line\nsegment, then these four points are concyclic (lie on a circle).\nGiven:\nLine segment AB subtending equal angles at points P and Q on the same side of the\nline segment AB.\nA\nB\nR\nQ\nP\nRequired to prove:\nA, B, P and Q lie on a circle.\nPROOF\nProof by contradiction:\nPoints on the circumference of a circle: we know that there are only two possible\noptions regarding a given point — it either lies on circumference or it does not.\nWe will assume that point P does not lie on the circumference.\nWe draw a circle that cuts AP at R and passes through A, B and Q.\nA ˆQB = A ˆRB\n(∠s in same seg.)\nbut A ˆQB = A ˆPB\n(given)\n∴A ˆRB = A ˆPB\nbut A ˆRB = A ˆPB + R ˆBP\n(ext. ∠△= sum int. opp.)\n∴R ˆBP = 0◦\nTherefore the assumption that the circle does not pass through P must be false.\nWe can conclude that A, B, Q and P lie on a circle (A, B, Q and P are concyclic).\n346\n8.2.\nCircle geometry\n\nWorked example 3: Concyclic points\nQUESTION\nGiven FH ∥EI and E ˆIF = 15◦, determine the value of b.\nE\nF\nG\nH\nI\n15◦\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nH ˆFI = 15◦\n(alt. ∠, FH ∥EI)\nand b = H ˆFI\n(∠s in same seg.)\n∴b = 15◦\nExercise 8 – 3: Subtended angles in the same segment\n1. Find the values of the unknown angles.\na)\nA\nB\nC\nD\n21◦\na\nb)\nJ\nK\nL\nM\n24◦\nc\n102◦\nd\nc)\nN\nO\nP\nQ\n17◦\nd\n347\nChapter 8.\nEuclidean geometry\n\n2.\nR\nS\nT\nU\nV\n45◦\n35◦\n15◦\ne\na) Given T ˆV S = S ˆV R, deter-\nmine the value of e.\nb) Is TV a diameter of the cir-\ncle? Explain your answer.\n3.\nb\nW\nX\nY\nZ\nO\n35◦\nf\nT\n1\n2\nGiven circle with centre O, WT =\nTY and X ˆWT = 35◦. Determine\nf.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236K\n1b. 236M\n1c. 236N\n2. 236P\n3. 236Q\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nCyclic quadrilaterals\nCyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference\nof a circle (concyclic).\nInvestigation: Cyclic quadrilaterals\nConsider the diagrams given below:\nCircle 1\nCircle 2\nCircle 3\nA\nB\nC\nD\nA\nB\nC\nD\nA\nB\nC\nD\n348\n8.2.\nCircle geometry\n\n1. Complete the following:\nABCD is a cyclic quadrilateral because . . . . . .\n2. Complete the table:\nCircle 1\nCircle 2\nCircle 3\nˆA =\nˆB =\nˆC =\nˆD =\nˆA + ˆC =\nˆB + ˆD =\n3. Use your results to make a conjecture about the relationship between angles of\ncyclic quadrilaterals.\nTheorem: Opposite angles of a cyclic quadrilateral\nSTATEMENT\nThe opposite angles of a cyclic quadrilateral are supplementary.\n(Reason: opp. ∠s cyclic quad.)\nGiven:\nCircle with centre O with points A, B, P and Q on the circumference such that ABPQ\nis a cyclic quadrilateral.\nbO\nA\nB\nP\nQ\n1\n2\nRequired to prove:\nA ˆBP + A ˆQP = 180◦and Q ˆAB + Q ˆPB = 180◦\n349\nChapter 8.\nEuclidean geometry\n\nPROOF\nDraw AO and OP. Label ˆO1 and ˆO2.\nˆO1 = 2A ˆBP\n(∠at centre = 2∠at circum.)\nˆO2 = 2A ˆQP\n(∠at centre = 2∠at circum.)\nand ˆO1 + ˆO2 = 360◦\n(∠s around a point)\n∴2A ˆBP + 2A ˆQP = 360◦\nA ˆBP + A ˆQP = 180◦\nSimilarly, we can show that Q ˆAB + Q ˆPB = 180◦.\nConverse: interior opposite angles of a quadrilateral\nIf the interior opposite angles of a quadrilateral are supplementary, then the quadrilat-\neral is cyclic.\nExterior angle of a cyclic quadrilateral\nIf a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle.\nb\nb\nWorked example 4: Opposite angles of a cyclic quadrilateral\nQUESTION\nGiven the circle with centre O and cyclic quadrilateral PQRS. SQ is drawn and\nS ˆPQ = 34◦. Determine the values of a, b and c.\nbO\nP\nQ\nR\nS\na\nb\nc\n34◦\n350\n8.2.\nCircle geometry\n\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nS ˆPQ + c = 180◦\n(opp. ∠s cyclic quad supp.)\n∴c = 180◦−34◦\n= 146◦\na = 90◦\n(∠in semi circle)\nIn △PSQ:\na + b + 34◦= 180◦\n(∠sum of △)\n∴b = 180◦−90◦−34◦\n= 56◦\nMethods for proving a quadrilateral is cyclic\nThere are three ways to prove that a quadrilateral is a cyclic quadrilateral:\nMethod of proof\nReason\nR\nQ\nS\nP\nIf ˆP + ˆR = 180◦or ˆS +\nˆQ = 180◦, then PQRS is\na cyclic quad.\nopp.\nint.\nangles\nsuppl.\nR\nQ\nS\nP\nIf ˆP = ˆQ or ˆS = ˆR, then\nPQRS is a cyclic quad.\nangles in the same\nseg.\nR\nQ\nS\nP\nT\nIf T ˆQR = ˆS, then PQRS\nis a cyclic quad.\next.\nangle equal to\nint. opp. angle\n351\nChapter 8.\nEuclidean geometry\n\nWorked example 5: Proving a quadrilateral is a cyclic quadrilateral\nQUESTION\nProve that ABDE is a cyclic quadrilateral.\nbO\nE\nC\nD\nA\nB\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Prove that ABDE is a cyclic quadrilateral\nD ˆBC = 90◦\n(∠in semi circle)\nand ˆE = 90◦\n(given)\n∴D ˆBC = ˆE\n∴ABDE is a cyclic quadrilateral\n(ext. ∠equals int. opp. ∠)\nExercise 8 – 4: Cyclic quadrilaterals\n1. Find the values of the unknown angles.\na)\nX\nY\nZ\nW\na\nb\n106◦\n87◦\nb)\nH\nI\nJ\nK\nL\n114◦\na\nc)\nU\nV\nW\nX\n57◦\na\n86◦\n352\n8.2.\nCircle geometry\n\n2. Prove that ABCD is a cyclic quadrilateral:\na) D\nC\n72◦\nB\nA\n32◦\nM\n40◦\nb) D\nC\n70◦\nB\nA\n35◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236R\n1b. 236S\n1c. 236T\n2a. 236V\n2b. 236W\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTangent line to a circle\nA tangent is a line that touches the circumference of a circle at only one place. The\nradius of a circle is perpendicular to the tangent at the point of contact.\nb\nO\n353\nChapter 8.\nEuclidean geometry\n\nTheorem: Two tangents drawn from the same point outside a circle\nSTATEMENT\nIf two tangents are drawn from the same point outside a circle, then they are equal in\nlength.\n(Reason: tangents from same point equal)\nGiven:\nCircle with centre O and tangents PA and PB, where A and B are the respective\npoints of contact for the two lines.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = BP\nPROOF\nIn △AOP and △BOP,\nO ˆAP = O ˆBP = 90◦\n(tangent ⊥radius)\nAO = BO\n(equal radii)\nOP = OP\n(common side)\n∴△AOP ≡△BOP\n(RHS)\n∴AP = BP\n354\n8.2.\nCircle geometry\n\nWorked example 6: Tangents from the same point outside a circle\nQUESTION\nIn the diagram below AE = 5 cm, AC = 8 cm and CE = 9 cm. Determine the values\nof a, b and c.\nA\nB\nC\nD\nE\nF\nAE = 5 cm\nAC = 8 cm\nCE = 9 cm\na\nb\nc\nb\nb\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for a, b and c\nAB = AF = a\n(tangents from A)\nEF = ED = c\n(tangents from E)\nCB = CD = b\n(tangents from C)\n∴AE = a + c = 5\nand AC = a + b = 8\nand CE = b + c = 9\nStep 3: Solve for the unknown variables using simultaneous equations\na + c = 5\n. . . (1)\na + b = 8\n. . . (2)\nb + c = 9\n. . . (3)\nSubtract equation (1) from equation (2) and then substitute into equation (3):\n(2) −(1)\nb −c = 8 −5\n= 3\n∴b = c + 3\nSubstitute into (3)\nc + 3 + c = 9\n2c = 6\nc = 3\n∴a = 2\nand b = 6\n355\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 5: Tangents to a circle\nFind the values of the unknown lengths.\n1.\nb\nG\nH\nI\nJ\nd\n5 cm\n8 cm\n2.\nb\nK\nL\nM\nN\nO\nP\ne\nLN = 7,5 cm\n2 cm\n6 cm\n3.\nb\nb\nR\nQ\nS\nf\n3 cm\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236X\n2. 236Y\n3. 236Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Tangent-chord theorem\nConsider the diagrams given below:\nDiagram 1\nDiagram 2\nDiagram 3\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\n1. Measure the following angles with a protractor and complete the table:\nDiagram 1\nDiagram 2\nDiagram 3\nA ˆBC =\nˆD =\nˆE =\n2. Use your results to complete the following: the angle between a tangent to a\ncircle and a chord is . . . . . . to the angle in the alternate segment.\n356\n8.2.\nCircle geometry\n\nTheorem: Tangent-chord theorem\nSTATEMENT\nThe angle between a tangent to a circle and a chord drawn at the point of contact, is\nequal to the angle which the chord subtends in the alternate segment.\n(Reason: tan. chord theorem)\nGiven:\nCircle with centre O and tangent SR touching the circle at B. Chord AB subtends ˆP1\nand ˆQ1.\nb\nO\nA\nB\nP\n1\n1\nQ\nT\n1\nS\nR\nRequired to prove:\n1. A ˆBR = A ˆPB\n2. A ˆBS = A ˆQB\nPROOF\nDraw diameter BT and join T to A.\nLet A ˆTB = T1.\nA ˆBS + A ˆBT = 90◦\n(tangent ⊥radius)\nB ˆAT = 90◦\n(∠in semi circle)\n∴A ˆBT + T1 = 90◦\n(∠sum of △BAT)\n∴A ˆBS = T1\nbut Q1 = T1\n(∠s in same segment)\n∴Q1 = A ˆBS\nA ˆBS + A ˆBR = 180◦\n(∠s on str. line)\nˆQ1 + ˆP1 = 180◦\n(opp. ∠s cyclic quad. supp.)\n∴A ˆBS + A ˆBR = Q1 + P1\nand A ˆBS = Q1\n∴A ˆBR = P1\n357\nChapter 8.\nEuclidean geometry\n\nWorked example 7: Tangent-chord theorem\nQUESTION\nDetermine the values of h and s.\nP\nO\nQ\nS\nR\nh + 20◦s\n4h\n4h −70◦\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for h\nO ˆQS = S ˆRQ\n(tangent chord theorem)\nh + 20◦= 4h −70◦\n90◦= 3h\n∴h = 30◦\nStep 3: Solve for s\nP ˆQR = Q ˆSR\n(tangent chord theorem)\ns = 4h\n= 4(30◦)\n= 120◦\n358\n8.2.\nCircle geometry\n\nExercise 8 – 6: Tangent-chord theorem\n1. Find the values of the unknown letters, stating reasons.\nQ\nR\nS\nO\nP\na\nb\n33◦\na)\nO\nP\nQ\nR\nS\nc\nd\n72◦\nb)\nO\nP\nQ\nR\nS\ng\nf\n38◦\n47◦\nc)\nR\nP\nO\nQ\nl\n1\n1\n66◦\nd)\nO\nP\nQ\nR\nS\ni\nj\nk\n39◦\n101◦\ne)\nO\nR\nQ\nS\nT\nm\nn\no\n34◦\nf)\nO\n•\nP\nR\nQ\nS\nT\np\nq\nr\n52◦\ng)\n359\nChapter 8.\nEuclidean geometry\n\n2. O is the centre of the circle and SPT is a tangent, with OP ⊥ST. Determine\na, b and c, giving reasons.\nO•\nS\nT\nP\nM\nN\na\nb\nc\n64◦\n3.\nP\nL\nA\nB\nC\n1 2\n3\n1\n2\nD\nGiven AB = AC, AP ∥BC and ˆA2 = ˆB2. Prove:\na) PAL is a tangent to the circle ABC.\nb) AB is a tangent to the circle ADP.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2372\n1b. 2373\n1c. 2374\n1d. 2375\n1e. 2376\n1f. 2377\n1g. 2378\n2. 2379\n3. 237B\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nConverse: tangent-chord theorem\nIf a line drawn through the end point of a chord forms an angle equal to the angle\nsubtended by the chord in the alternate segment, then the line is a tangent to the\ncircle.\n(Reason: ∠between line and chord = ∠in alt. seg. )\n360\n8.2.\nCircle geometry\n\nWorked example 8: Applying the theorems\nQUESTION\nA\nD\nB\nC\nO\nE\nF\nBD is a tangent to the circle with centre O, with BO ⊥AD.\nProve that:\n1. CFOE is a cyclic quadrilateral\n2. FB = BC\n3. ∠A ˆOC = 2B ˆFC\n4. Will DC be a tangent to the circle passing through C, F, O and E? Motivate your\nanswer.\nSOLUTION\nStep 1: Prove CFOE is a cyclic quadrilateral by showing opposite angles are supple-\nmentary\nBO ⊥OD\n(given)\n∴F ˆOE = 90◦\nF ˆCE = 90◦\n(∠in semi circle)\n∴CFOE is a cyclic quad.\n(opp. ∠s suppl.)\nStep 2: Prove BFC is an isosceles triangle\nTo show that FB = BC we first prove △BFC is an isosceles triangle by showing that\nB ˆFC = B ˆCF.\nB ˆCF = C ˆEO\n(tangent-chord)\nC ˆEO = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴B ˆFC = B ˆCF\n∴FB = BC\n(△BFC isosceles)\n361\nChapter 8.\nEuclidean geometry\n\nStep 3: Prove A ˆOC = 2B ˆFC\nA ˆOC = 2A ˆEC\n(∠at centre = 2∠at circum.)\nand A ˆEC = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴A ˆOC = 2B ˆFC\nStep 4: Determine if DC is a tangent to the circle through C, F, O and E\nProof by contradiction.\nLet us assume that DC is a tangent to the circle passing through the points C, F, O\nand E:\n∴D ˆCE = C ˆOE\n(tangent-chord)\nAnd using the circle with centre O and tangent BD we have that:\nD ˆCE = C ˆAE\n(tangent-chord)\nbut C ˆAE = 1\n2C ˆOE\n(∠at centre = 2∠at circum.)\n∴D ˆCE ̸= C ˆOE\nTherefore our assumption is not correct and we can conclude that DC is not a tangent\nto the circle passing through the points C, F, O and E.\nWorked example 9: Applying the theorems\nQUESTION\nA\nB\nC\nD\nE\nF\nG\nH\nFD is drawn parallel to the tangent CB\n362\n8.2.\nCircle geometry\n\nProve that:\n1. FADE is a cyclic quadrilateral\n2. F ˆEA = ˆB\nSOLUTION\nStep 1: Prove FADE is a cyclic quadrilateral using angles in the same segment\nF ˆDC = D ˆCB\n(alt. ∠s FD ∥CB)\nand D ˆCB = C ˆAE\n(tangent-chord)\n∴F ˆDC = C ˆAE\n∴FADE is a cyclic quad.\n(∠s in same seg.)\nStep 2: Prove F ˆEA = ˆB\nF ˆDA = ˆB\n(corresp. ∠s FD ∥CB)\nand F ˆEA = F ˆDA\n(∠s same seg. cyclic quad. FADE)\n∴F ˆEA = ˆB\n8.3\nSummary\nEMBJC\nSee presentation: 237C at www.everythingmaths.co.za\n• Arc An arc is a portion of the circumference of a circle.\n• Chord - a straight line joining the ends of an arc.\n• Circumference - perimeter or boundary line of a circle.\n• Radius (r) - any straight line from the centre of the circle to a point on the cir-\ncumference.\n• Diameter - a special chord that passes through the centre of the circle. A diame-\nter is the length of a straight line segment from one point on the circumference to\nanother point on the circumference, that passes through the centre of the circle.\n• Segment A segment is a part of the circle that is cut off by a chord. A chord\ndivides a circle into two segments.\n• Tangent - a straight line that makes contact with a circle at only one point on the\ncircumference.\n• A tangent line is perpendicular to the radius, drawn at the point of contact with\nthe circle.\n363\nChapter 8.\nEuclidean geometry\n\nb O\nM\nA\nB\n• If O is the centre and OM ⊥AB, then AM =\nMB.\n• If O is the centre and AM\n= MB, then\nA ˆ\nMO = B ˆ\nMO = 90◦.\n• If AM = MB and OM ⊥AB, then ⇒MO\npasses through centre O.\nb\n2x\nx\n2y\ny\nx\nIf an arc subtends an angle at the centre of a cir-\ncle and at the circumference, then the angle at the\ncentre is twice the size of the angle at the circum-\nference.\nb\nb\nAngles at the circumference subtended by the same\narc (or arcs of equal length) are equal.\nA\nB\nC\nD\n1\n2\nE\nThe four sides of a cyclic quadrilateral ABCD are\nchords of the circle with centre O.\n• ˆA + ˆC = 180◦(opp. ∠s supp.)\n• ˆB + ˆD = 180◦(opp. ∠s supp.)\n• E ˆBC = ˆD (ext. ∠cyclic quad.)\n• ˆA1 = ˆA2 = ˆC (vert. opp. ∠, ext. ∠cyclic\nquad.)\nA\nB\nC\nD\nProving a quadrilateral is cyclic: If ˆA + ˆC = 180◦or\nˆB+ ˆD = 180◦, then ABCD is a cyclic quadrilateral.\n364\n8.3.\nSummary\n\nA\nB\nC\nD\n1\n1\nIf ˆA1 = ˆC or ˆD1 = ˆB, then ABCD is a cyclic\nquadrilateral.\nA\nB\nC\nD\nIf ˆA = ˆB or ˆC = ˆD, then ABCD is a cyclic quadri-\nlateral.\nb\nA\nB\nO\nT\nIf AT and BT are tangents to circle O, then\n• OA ⊥AT (tangent ⊥radius)\n• OB ⊥BT (tangent ⊥radius)\n• TA = TB (tangents from same point equal)\nA\nB\nT\nD\nC\nx\ny\nx\ny\n• If DC is a tangent, then D ˆTA = T ˆBA and\nC ˆTB = T ˆAB\n• If D ˆTA = T ˆBA or C ˆTB = T ˆAB, then DC is\na tangent touching at T\n365\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 7: End of chapter exercises\n1.\nO\n•\nA\nB\nC\nD\nE\nF\n×\n×\nx\nAOC is a diameter of the circle with centre O. F is the mid-point of chord EC.\nB ˆOC = C ˆOD and ˆB = x. Express the following angles in terms of x, stating\nreasons:\na) ˆA\nb) C ˆOD\nc) ˆD\n2.\nM•\nD\nE\nF\nG\n1 2\n1\n2\n1\n2\n1 2\nD, E, F and G are points on circle with centre M.\nˆF1 = 7◦and ˆD2 = 51◦.\nDetermine the sizes of the following angles, stating reasons:\na)\nˆ\nM1\nb) ˆD1\nc) ˆF2\nd) ˆG\ne) ˆE1\n366\n8.3.\nSummary\n\n3.\nM\n•\nO•\nD\nA\nB\nC\n1 2\nO is a point on the circle with centre M. O is also the centre of a second circle.\nDA cuts the smaller circle at C and ˆD1 = x. Express the following angles in\nterms of x, stating reasons:\na) ˆD2\nb) O ˆAB\nc) O ˆBA\nd) A ˆOB\ne) ˆC\n4.\nO•\nA\nB\nC\nM\nO is the centre of the circle with radius 5 cm and chord BC = 8 cm. Calculate\nthe lengths of:\na) OM\nb) AM\nc) AB\n5.\nO•\nA\nB\nC\n70◦\nx\nAO ∥CB in circle with centre O. A ˆOB = 70◦and O ˆAC = x. Calculate the\nvalue of x, giving reasons.\n367\nChapter 8.\nEuclidean geometry\n\n6.\nO\n•\nP\nQ\nR\nS\nT\nx\nPQ is a diameter of the circle with centre O. SQ bisects P ˆQR and P ˆQS = x.\na) Write down two other angles that are also equal to x.\nb) Calculate P ˆOS in terms of x, giving reasons.\nc) Prove that OS is a perpendicular bisector of PR.\n7.\nO•\nA\nB\nC\nD\n35◦\nB ˆOD is a diameter of the circle with centre O. AB = AD and O ˆCD = 35◦.\nCalculate the value of the following angles, giving reasons:\na) O ˆDC\nb) C ˆOD\nc) C ˆBD\nd) B ˆAD\ne) A ˆDB\n8.\nO\n•\nR\nP\nT\nQ\nx\ny\nQP in the circle with centre O is protracted to T so that PR = PT. Express y in\nterms of x.\n368\n8.3.\nSummary\n\n9.\nO•\nA\nB\nC\nD\nE\nP\nF\nO is the centre of the circle with diameter AB. CD ⊥AB at P and chord DE\ncuts AB at F. Prove that:\na) C ˆBP = D ˆPB\nb) C ˆED = 2C ˆBA\nc) A ˆBD = 1\n2C ˆOA\n10.\nO\n•\nP\nQ\nR\nx\nS\nIn the circle with centre O, OR ⊥QP, PQ = 30 mm and RS = 9 mm. Deter-\nmine the length of OQ.\n11.\nM •\nP\nQ\nR\nS\nT\nP, Q, R and S are points on the circle with centre M. PS and QR are extended\nand meet at T. PQ = PR and P ˆQR = 70◦.\na) Determine, stating reasons, three more angles equal to 70◦.\nb) If Q ˆPS = 80◦, calculate S ˆRT, S ˆTR and P ˆQS.\nc) Explain why PQ is a tangent to the circle QST at point Q.\nd) Determine P ˆ\nMQ.\n369\nChapter 8.\nEuclidean geometry\n\n12.\nO\n•\nA\nP\nQ\nC\nB\nPOQ is a diameter of the circle with centre O. QP is protruded to A and AC is\na tangent to the circle. BA ⊥AQ and BCQ is a straight line. Prove:\na) P ˆCQ = B ˆAP\nb) BAPC is a cyclic quadrilateral\nc) AB = AC\n13.\nO•\nT\nC\nA\nB\nx\nTA and TB are tangents to the circle with centre O. C is a point on the circum-\nference and A ˆTB = x. Express the following in terms of x, giving reasons:\na) A ˆBT\nb) O ˆBA\nc) ˆC\n14.\nO•\nA\nB\nC\nE\nD\nAOB is a diameter of the circle\nAECB with centre O. OE ∥BC\nand cuts AC at D.\na) Prove AD = DC\nb) Show that A ˆBC is bisected\nby EB\nc) If O ˆEB = x, express B ˆAC\nin terms of x\nd) Calculate the radius of the\ncircle if AC = 10 cm and\nDE = 1 cm\n370\n8.3.\nSummary\n\n15.\nV\nQ\nS\nR\nP\nT\nW\nx\ny\nPQ and RS are chords of the circle and PQ ∥RS. The tangent to the circle at\nQ meets RS protruded at T. The tangent at S meets QT at V . QS and PR are\ndrawn.\nLet T ˆQS = x and Q ˆRP = y. Prove that:\na) T ˆV S = 2Q ˆRS\nb) QV SW is a cyclic quadrilateral\nc) Q ˆPS + ˆT = P ˆRT\nd) W is the centre of the circle\n16.\nF\nD\nB\nC\nE\nA\nK\nT\n1\n2\n1\n2\n1\n2\n3\n4\nThe two circles shown intersect at points F and D. BFT is a tangent to the\nsmaller circle at F. Straight line AFE is drawn such that DF = EF. CDE is a\nstraight line and chord AC and BF cut at K. Prove that:\na) BT ∥CE\nb) BCEF is a parallelogram\nc) AC = BF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237D\n2. 237F\n3. 237G\n4. 237H\n5. 237J\n6. 237K\n7. 237M\n8. 237N\n9. 237P\n10. 237Q\n11. 237R\n12. 237S\n13. 237T\n14. 237V\n15. 237W\n16. 237X\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n371\nChapter 8.\nEuclidean geometry\n\n\nCHAPTER\n9\nFinance, growth and decay\n9.1\nRevision\n374\n9.2\nSimple and compound depreciation\n377\n9.3\nTimelines\n388\n9.4\nNominal and effective interest rates\n394\n9.5\nSummary\n398\n\n9\nFinance, growth and decay\n9.1\nRevision\nEMBJD\nSimple interest is the interest calculated only on the initial amount invested, the prin-\ncipal amount. Compound interest is the interest earned on the principal amount and\non its accumulated interest. This means that interest is being earned on interest. The\naccumulated amount is the final amount; the sum of the principal amount and the\namount of interest earned.\nFormula for simple interest:\nA = P(1 + in)\nFormula for compound interest:\nA = P(1 + i)n\nwhere\nA = accumulated amount\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nWorked example 1: Simple and compound interest\nQUESTION\nSam wants to invest R 3450 for 5 years. Wise Bank offers a savings account which pays\nsimple interest at a rate of 12,5% per annum, and Grand Bank offers a savings account\npaying compound interest at a rate of 10,4% per annum. Which bank account would\ngive Sam the greatest accumulated balance at the end of the 5 year period?\nSOLUTION\nStep 1: Calculation using the simple interest formula\nWrite down the known variables and the simple interest formula\nP = 3450\ni = 0,125\nn = 5\nA = P(1 + in)\nSubstitute the values to determine the accumulated amount for the Wise Bank savings\n374\n9.1.\nRevision\n\naccount.\nA = 3450(1 + 0,125 × 5)\n= R 5606,25\nStep 2: Calculation using the compound interest formula\nWrite down the known variables and the compound interest formula.\nP = 3450\ni = 0,104\nn = 5\nA = P(1 + i)n\nSubstitute the values to determine the accumulated amount for the Grand Bank savings\naccount.\nA = 3450(1 + 0,104)5\n= R 5658,02\nStep 3: Write the final answer\nThe Grand Bank savings account would give Sam the highest accumulated balance at\nthe end of the 5 year period.\nWorked example 2: Finding i\nQUESTION\nBongani decides to put R 30 000 in an investment account. What compound interest\nrate must the investment account achieve for Bongani to double his money in 6 years?\nGive your answer correct to one decimal place.\nSOLUTION\nStep 1: Write down the known variables and the compound interest formula\nA = 60 000\nP = 30 000\nn = 6\nA = P(1 + i)n\n375\nChapter 9.\nFinance, growth and decay\n\nStep 2: Substitute the values and solve for i\n60 000 = 30 000(1 + i)6\n60 000\n30 000 = (1 + i)6\n2 = (1 + i)6\n6√\n2 = 1 + i\n6√\n2 −1 = i\n∴i = 0,122 . . .\nStep 3: Write the final answer and comment\nWe round up to a rate of 12,3% p.a. to make sure that Bongani doubles his invest-\nment.\nExercise 9 – 1: Revision\n1. Determine the value of an investment of R 10 000 at 12,1% p.a. simple interest\nfor 3 years.\n2. Calculate the value of R 8000 invested at 8,6% p.a. compound interest for 4\nyears.\n3. Calculate how much interest John will earn if he invests R 2000 for 4 years at:\na) 6,7% p.a. simple interest\nb) 5,4% p.a. compound interest\n4. The value of an investment grows from R 2200 to R 3850 in 8 years. Determine\nthe simple interest rate at which it was invested.\n5. James had R 12 000 and invested it for 5 years. If the value of his investment is\nR 15 600, what compound interest rate did it earn?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237Y\n2. 237Z\n3. 2382\n4. 2383\n5. 2384\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n376\n9.1.\nRevision\n\n9.2\nSimple and compound depreciation\nEMBJF\nAs soon as a new car leaves the dealership, its value decreases and it is considered\n“second-hand”.\nVehicles, equipment, machinery and other similar assets, all lose\nvalue over time as a result of usage and age. This loss in value is called deprecia-\ntion. Assets that have a relatively long useful lifetime, such as machines, trucks, farm-\ning equipment etc., depreciate slower than assets like office equipment, computers,\nfurniture etc. which need to be replaced more often and therefore depreciate more\nquickly.\nDepreciation is used to calculate the value of a company’s assets, which determines\nhow much tax a company must pay. Companies can take depreciation into account as\nan expense, and thereby reduce their taxable income. A lower taxable income means\nthat the company will pay less income tax to SARS (South African Revenue Service).\nWe can calculate two different kinds of depreciation: simple decay and compound\ndecay. Decay is also a term used to describe a reduction or decline in value. Simple\ndecay is also called straight-line depreciation and compound decay can also be re-\nferred to as reducing-balance depreciation. In the straight-line method the value of the\nasset is reduced by a constant amount each year, which is calculated on the principal\namount. In reducing-balance depreciation we calculate the depreciation on the re-\nduced value of the asset. This means that the value of an asset decreases by a different\namount each year.\nInvestigation: Simple and compound depreciation\n1. Mr. Sontange buys an Opel Fiesta for R 72 000. He expects that the value of the\ncar will depreciate by R 6000 every year. He draws up a table to calculate the\ndepreciated value of his Opel Fiesta.\nComplete Mr. Sontange’s table of values for the 7 year period:\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 6000\nR 66 000\n2\nR 66 000\nR 6000\n3\n4\n5\n6\n7\n2. His son, David, does not agree that the value of the car will reduce by the same\namount each year. David thinks that the car will depreciate by 10% every year.\nComplete David’s table of values:\n377\nChapter 9.\nFinance, growth and decay\n\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 7200\nR 64 800\n2\nR 64 800\nR 6480\n3\n4\n5\n6\n7\n3. Compare and discuss the results of the two different tables.\n4. Consider the graph below, which represents Mr. Sontange’s table of values:\n10 000\n20 000\n30 000\n40 000\n50 000\n60 000\n70 000\n80 000\n1\n2\n3\n4\n5\n6\n7\n8\n0\nTime (years)\nValue (Rands)\na) Draw a similar graph using David’s table of values.\nb) Interpret the two graphs and discuss the differences between them.\nc) Explain how the graphs can be used to determine the total depreciation in\neach case.\nd)\ni. Draw two new graphs by plotting the maximum value of each bar.\nii. Join the points with a line to show the general trend.\niii. Is it mathematically correct to join these points? Explain your answer.\n378\n9.2.\nSimple and compound depreciation\n\nSimple depreciation\nEMBJG\nWorked example 3: Straight-line depreciation\nQUESTION\nA new smartphone costs R 6000 and depreciates at 22% p.a. on a straight-line basis.\nDetermine the value of the smartphone at the end of each year over a 4 year period.\nSOLUTION\nStep 1: Calculate depreciation amount\nDepreciation = 6000 × 22\n100\n= 1320\nTherefore the smartphone depreciates by R 1320 every year.\nStep 2: Complete a table of values\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 6000\nR 1320\nR 4680\n2\nR 4680\nR 1320\nR 3360\n3\nR 3360\nR 1320\nR 2040\n4\nR 2040\nR 1320\nR 720\nWe notice that\nTotal depreciation = P × i × n\nwhere\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nTherefore the depreciated value of the asset (also called the book value) can be calcu-\nlated as:\nA = P(1 −in)\nNote the similarity to the simple interest formula A = P(1 + in). Interest increases the\nvalue of the principal amount, whereas with simple decay, depreciation reduces the\nvalue of the principal amount.\nImportant: to get an accurate answer do all calculations in one step on your calculator.\nDo not round off answers in your calculations until the final answer. In the worked\nexamples in this chapter, we use dots to show that the answer has not been rounded\noff. We always round the final answer to two decimal places (cents).\n379\nChapter 9.\nFinance, growth and decay\n\nWorked example 4: Straight-line depreciation method\nQUESTION\nA car is valued at R 240 000. If it depreciates at 15% p.a. using straight-line deprecia-\ntion, calculate the value of the car after 5 years.\nSOLUTION\nStep 1: Write down the known variables and the simple decay formula\nP = 240 000\ni = 0,15\nn = 5\nA = P(1 −in)\nStep 2: Substitute the values and solve for A\nA = 240 000(1 −0,15 × 5)\n= 240 000(0,25)\n= 60 000\nStep 3: Write the final answer\nAt the end of 5 years, the car is worth R 60 000.\nWorked example 5: Simple decay\nQUESTION\nA small business buys a photocopier for R 12 000. For the tax return the owner depre-\nciates this asset over 3 years using a straight-line depreciation method. What amount\nwill he fill in on his tax form at the end of each year?\nSOLUTION\nStep 1: Write down the known variables\nThe owner of the business wants the photocopier to have a book value of R 0 after 3\nyears.\nA = 0\nP = 12 000\nn = 3\n380\n9.2.\nSimple and compound depreciation\n\nTherefore we can calculate the annual depreciation as\nDepreciation = P\nn\n= 12 000\n3\n= R 4000\nStep 2: Determine the book value at the end of each year\nBook value end of first year = 12 000 −4000\n= R 8000\nBook value end of second year = 8000 −4000\n= R 4000\nBook value end of third year = 4000 −4000\n= R 0\nExercise 9 – 2: Simple decay\n1. A business buys a truck for R 560 000. Over a period of 10 years the value of\nthe truck depreciates to R 0 using the straight-line method. What is the value of\nthe truck after 8 years?\n2. Harry wants to buy his grandpa’s donkey for R 800. His grandpa is quite pleased\nwith the offer, seeing that it only depreciated at a rate of 3% per year using the\nstraight-line method. Grandpa bought the donkey 5 years ago. What did grandpa\npay for the donkey then?\n3. Seven years ago, Rocco’s drum kit cost him R 12 500. It has now been valued at\nR 2300. What rate of simple depreciation does this represent?\n4. Fiona buys a DStv satellite dish for R 3000. Due to weathering, its value depre-\nciates simply at 15% per annum. After how long will the satellite dish have a\nbook value of zero?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2385\n2. 2386\n3. 2387\n4. 2388\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n381\nChapter 9.\nFinance, growth and decay\n\nCompound depreciation\nEMBJH\nWorked example 6: Reducing-balance depreciation\nQUESTION\nA second-hand farm tractor worth R 60 000 has a limited useful life of 5 years and\ndepreciates at 20% p.a. on a reducing-balance basis. Determine the value of the\ntractor at the end of each year over the 5 year period.\nSOLUTION\nStep 1: Write down the known variables\nP = 60 000\ni = 0,2\nn = 5\nWhen we calculate depreciation using the reducing-balance method:\n1. the depreciation amount changes for each year.\n2. the depreciation amount gets smaller each year.\n3. the book value at the end of a year becomes the principal amount for the next\nyear.\n4. the asset will always have some value (the book value will never equal zero).\nStep 2: Complete a table of values\nYear\nBook value\nDepreciation\nValue at end of\nyear\n1\nR 60 000\n60 000 × 0,2 = 12 000\nR 48 000\n2\nR 48 000\n48 000 × 0,2 = 9600\nR 38 400\n3\nR 38 400\n38 400 × 0,2 = 7680\nR 30 720\n4\nR 30 720\n30 720 × 0,2 = 6144\nR 24 576\n5\nR 24 576\n24 576 × 0,2 = 4915,20\nR 19 660,80\n382\n9.2.\nSimple and compound depreciation\n\nNotice in the example above that we could also write the book value at the end of\neach year as:\nBook value end of first year\n= 60 000(1 −0,2)\nBook value end of second year = 48 000(1 −0,2) = 60 000(1 −0,2)2\nBook value end of third year\n= 38 400(1 −0,2) = 60 000(1 −0,2)3\nBook value end of fourth year = 30 720(1 −0,2) = 60 000(1 −0,2)4\nBook value end of fifth year\n= 24 576(1 −0,2) = 60 000(1 −0,2)5\nUsing the formula for simple decay and the observed pattern in the calculation above,\nwe obtain the following formula for compound decay:\nA = P(1 −i)n\nwhere\nA = book value or depreciated value\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nAgain, notice the similarity to the compound interest formula A = P(1 + i)n.\nWorked example 7: Reducing-balance depreciation\nQUESTION\nThe number of pelicans at the Berg river mouth is decreasing at a compound rate of\n12% p.a. If there are currently 3200 pelicans in the wetlands of the Berg river mouth,\nwhat will the population be in 5 years?\nSOLUTION\nStep 1: Write down the known variables and the compound decay formula\nP = 3200\ni = 0,12\nn = 5\nA = P(1 −i)n\nStep 2: Substitute the values and solve for A\nA = 3200(1 −0,12)5\n= 3200(0,88)5\n= 1688,7421 . . .\nStep 3: Write the final answer\nIn 5 years, the pelican population will be approximately 1689.\n383\nChapter 9.\nFinance, growth and decay\n\nWorked example 8: Compound decay\nQUESTION\n1. A school buys a minibus for R 950 000, which depreciates at 13,5% per annum.\nDetermine the value of the minibus after 3 years if the depreciation is calculated:\na) on a straight-line basis.\nb) on a reducing-balance basis.\n2. Which is the better option?\nSOLUTION\nStep 1: Write down known variables\nP = 950 000\ni = 0,135\nn = 3\nStep 2: Use the simple decay formula and solve for A\nA = 950 000(1 −3 × 0,135)\n= 950 000(0,865)\n= 565 250\n∴A = R 565 250\nStep 3: Use the compound decay formula and solve for A\nA = 950 000(1 −0,135)3\n= 950 000(0,865)3\n= 614 853,89\n∴A = R 614 853,89\nStep 4: Interpret the answers\nAfter a period of 3 years, the value of the minibus calculated on the straight-line\nmethod is less than the value of the minibus calculated on the reducing-balance\nmethod. The value of the minibus depreciated less on the reducing-balance basis\nbecause the amount of depreciation is calculated on a smaller amount every year,\nwhereas the straight-line method is based on the full value of the minibus every year.\n384\n9.2.\nSimple and compound depreciation\n\nWorked example 9: Compound depreciation\nQUESTION\nFarmer Jack bought a tractor and it has depreciated by 20% p.a. on a reducing-balance\nbasis. If the current value of the tractor is R 52 429, calculate how much Farmer Jack\npaid for his tractor if he bought it 7 years ago.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 52 429\ni = 0,2\nn = 7\nA = P(1 −i)n\nStep 2: Substitute the values and solve for P\n52 429 = P(1 −0,2)7\n= P(0,8)7\n∴P = 52 429\n(0,8)7\n= 250 000,95 . . .\nStep 3: Write the final answer\n7 years ago, Farmer Jack paid R 250 000 for his tractor.\nExercise 9 – 3: Compound depreciation\n1. Jwayelani buys a truck for R 89 000 and depreciates it by 9% p.a. using the\ncompound depreciation method. What is the value of the truck after 14 years?\n2. The number of cormorants at the Amanzimtoti river mouth is decreasing at a\ncompound rate of 8% p.a. If there are now 10 000 cormorants, how many will\nthere be in 18 years’ time?\n3. On January 1, 2008 the value of my Kia Sorento is R 320 000. Each year after\nthat, the car’s value will decrease 20% of the previous year’s value. What is the\nvalue of the car on January 1, 2012?\n385\nChapter 9.\nFinance, growth and decay\n\n4. The population of Bonduel decreases at a reducing-balance rate of 9,5% per\nannum as people migrate to the cities. Calculate the decrease in population over\na period of 5 years if the initial population was 2 178 000.\n5. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses\n3% of its water each day. How much does it weigh after a month of 31 days?\n6. Richard bought a car 15 years ago and it depreciated by 17% p.a. on a com-\npound depreciation basis. How much did he pay for the car if it is now worth\nR 5256?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2389\n2. 238B\n3. 238C\n4. 238D\n5. 238F\n6. 238G\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFinding i\nEMBJJ\nWorked example 10: Finding i for simple decay\nQUESTION\nAfter 4 years, the value of a computer is halved. Assuming simple decay, at what\nannual rate did it depreciate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and simple decay formula\nLet the value of the computer be x, therefore:\nA = x\n2\nP = x\nn = 4\nA = P(1 −in)\nStep 2: Substitute the values and solve for i\n386\n9.2.\nSimple and compound depreciation\n\nx\n2 = x(1 −3i)\n1\n2 = 1 −3i\n∴3i = 1 −1\n2\n∴i = 0,1667\nStep 3: Write the final answer\nThe computer depreciated at a rate of 16,67% p.a.\nWorked example 11: Finding i for compound decay\nQUESTION\nCristina bought a fridge at the beginning of 2009 for R 8999 and sold it at the end\nof 2011 for R 4500. At what rate did the value of her fridge depreciate assuming a\nreducing-balance method? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 4500\nP = 8999\nn = 3\nA = P(1 −i)n\nStep 2: Substitute the values and solve for i\n4500 = 8999(1 −i)3\n4500\n8999 = (1 −i)3\n3\nr\n4500\n8999 = 1 −i\n∴i = 1 −\n3\nr\n4500\n8999\n= 0,206\nStep 3: Write the final answer\nCristina’s fridge depreciated at a rate of 20,6% p.a.\n387\nChapter 9.\nFinance, growth and decay\n\nExercise 9 – 4: Finding i\n1. A machine costs R 45 000 and has a scrap value of R 9000 after 10 years. Deter-\nmine the annual rate of depreciation if it is calculated on the reducing balance\nmethod.\n2. After 15 years, an aeroplane is worth 1\n6 of its original value. At what annual rate\nwas depreciation compounded?\n3. Mr. Mabula buys furniture for R 20 000. After 6 years he sells the furniture for\nR 9300. Calculate the annual compound rate of depreciation of the furniture.\n4. Ayanda bought a new car 7 years ago for double what it is worth today. At what\nyearly compound rate did her car depreciate?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238H\n2. 238J\n3. 238K\n4. 238M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.3\nTimelines\nEMBJK\nInterest can be compounded more than once a year. For example, an investment can\nbe compounded monthly or quarterly. Below is a table of compounding terms and\ntheir corresponding numeric value (p). When amounts are compounded more than\nonce per annum, we multiply the number of years by p and we also divide the interest\nrate by p.\nTerm\np\nyearly / annually\n1\nhalf-yearly / bi-annually\n2\nquarterly\n4\nmonthly\n12\nweekly\n52\ndaily\n365\nWorked example 12: Timelines\nQUESTION\nR 5500 is invested for a period of 4 years in a savings account. For the first year, the\ninvestment grows at a simple interest rate of 11% p.a. and then at a rate of 12,5%\np.a. compounded quarterly for the rest of the period. Determine the value of the\ninvestment at the end of the 4 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\n388\n9.3.\nTimelines\n\nT0\nT1\nT2\nT3\nT4\n11% p.a. simple interest\n12,5% p.a. compounded quarterly\nR 5500\nIn the timeline above, the intervals are given in years. For example, T0 is the start of\nthe investment, T1 is the end of the first year and T4 is the end of the fourth year.\nStep 2: Use the simple interest formula to calculate A at T1\nA = P(1 + in)\n= 5500(1 + 0,11)\n= R 6105\nStep 3: Use the compound interest formula to calculate A at T4\nThe investment is compounded quarterly, therefore:\nn = 3 × 4\n= 12\nand i = 0,125\n4\nAlso notice that the accumulated amount at the end of the first year becomes the\nprincipal amount at the beginning of the second year.\nA = P(1 + i)n\n= 6105\n\u0012\n1 + 0,125\n4\n\u001312\n= R 8831,88\nStep 4: Write the final answer\nThe value of the investment at the end of the 4 years is R 8831,88.\n389\nChapter 9.\nFinance, growth and decay\n\nWorked example 13: Timelines\nQUESTION\nR 150 000 is deposited in an investment account for a period of 6 years at an interest\nrate of 12% p.a. compounded half-yearly for the first 4 years and then 8,5% p.a.\ncompounded yearly for the rest of the period. A deposit of R 8000 is made into the\naccount after the first year and then another deposit of R 2000 is made 5 years after\nthe initial investment. Calculate the value of the investment at the end of the 6 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n8,5% p.a. compounded yearly\nR 15 000\nT5\nT6\n12% p.a. compounded half-yearly\n+R 8000\n+R 2000\nRemember to show when the additional deposits of R 8000 and R 2000 where made\ninto the account. It is very important to note that the interest rate changes at T4.\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nBetween T0 and T4:\nWe notice that interest for the first 4 years is compounded half-yearly, therefore:\nn1 = 4 × 2\n= 8\nand i1 = 0,12\n2\nBetween T4 and T6:\nn2 = 2\nand i2 = 0,085\nTherefore the total growth of the initial deposit over the 6 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\nStep 3: The deposit at T1\nBetween T1 and T4:\n390\n9.3.\nTimelines\n\nInterest on this deposit is compounded half-yearly for 3 years, therefore:\nn3 = 3 × 2\n= 6\nand i3 = 0,12\n2\nBetween T4 and T6:\nn4 = 2\nand i4 = 0,085\nTherefore the total growth of the deposit over the 5 years is:\nA = P(1 + i3)n3(1 + i4)n4\n= 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2\nStep 4: The deposit at T5\nAccumulate interest for only 1 year:\nA = P(1 + i)n\n= 2000(1 + 0,085)1\nStep 5: Determine the total calculation\nTo get as accurate an answer as possible, we do the the calculation on the calculator\nin one step. Using the memory and answer recall function on the calculator, we avoid\nrounding off until we get the final answer.\nA = 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\n+ 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2 + 2000(1 + 0,085)1\n= R 296 977,00\nStep 6: Write the final answer\nThe value of the investment at the end of the 6 years is R 296 977,00.\n391\nChapter 9.\nFinance, growth and decay\n\nWorked example 14: Timelines\nQUESTION\nR 60 000 is invested in an account which offers interest at 7% p.a.\ncompounded\nquarterly for the first 18 months. Thereafter the interest rate changes to 5% p.a. com-\npounded monthly. Three years after the initial investment, R 5000 is withdrawn from\nthe account. How much will be in the account at the end of 5 years?\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n5% p.a. compounded monthly\nR 60 000\nT5\n7% p.a. compounded quarterly\n−R 5000\nRemember to show when the withdrawal of R 5000 was taken out of the account. It is\nalso important to note that the interest rate changes after 18 months (T1 1\n2 ).\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nInterest for the first 1,5 years is compounded quarterly, therefore:\nn1 = 1,5 × 4\n= 6\nand i1 = 0,07\n4\nInterest for the remaining 3,5 years is compounded monthly, therefore:\nn2 = 3,5 × 12\n= 42\nand i2 = 0,05\n12\nTherefore the total growth of the initial deposit over the 5 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n392\n9.3.\nTimelines\n\nStep 3: The withdrawal at T3\nWe calculate the interest that the R 5000 would have earned if it had remained in the\naccount:\nn = 2 × 12\n= 24\nand i = 0,05\n12\nTherefore we have that:\nA = P(1 + i)n\n= 5000\n\u0012\n1 + 0,05\n12\n\u001324\nStep 4: Determine the total calculation\nWe subtract the withdrawal and the interest it would have earned from the accumu-\nlated amount at the end of the 5 years:\nA = 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n−5000\n\u0012\n1 + 0,05\n12\n\u001324\n= R 73 762,19\nStep 5: Write the final answer\nThe value of the investment at the end of the 5 years is R 73 762,19.\nExercise 9 – 5: Timelines\n1. After a 20-year period Josh’s lump sum investment matures to an amount of\nR 313 550. How much did he invest if his money earned interest at a rate of\n13,65% p.a. compounded half yearly for the first 10 years, 8,4% p.a. com-\npounded quarterly for the next five years and 7,2% p.a. compounded monthly\nfor the remaining period?\n2. Sindisiwe wants to buy a motorcycle. The cost of the motorcycle is R 55 000.\nIn 1998 Sindisiwe opened an account at Sutherland Bank with R 16 000. Then\nin 2003 she added R 2000 more into the account. In 2007 Sindisiwe made\nanother change: she took R 3500 from the account. If the account pays 6% p.a.\ncompounded half-yearly, will Sindisiwe have enough money in the account at\nthe end of 2012 to buy the motorcycle?\n3. A loan has to be returned in two equal semi-annual instalments. If the rate of\ninterest is 16% per annum, compounded semi-annually and each instalment is\nR 1458, find the sum borrowed.\n393\nChapter 9.\nFinance, growth and decay\n\n4. A man named Phillip invests R 10 000 into an account at North Bank at an\ninterest rate of 7,5% p.a. compounded monthly. After 5 years the bank changes\nthe interest rate to 8% p.a. compounded quarterly. How much money will\nPhillip have in his account 9 years after the original deposit?\n5. R 75 000 is invested in an account which offers interest at 11% p.a.\ncom-\npounded monthly for the first 24 months.\nThen the interest rate changes to\n7,7% p.a. compounded half-yearly. If R 9000 is withdrawn from the account\nafter one year and then a deposit of R 3000 is made three years after the initial\ninvestment, how much will be in the account at the end of 6 years?\n6. Christopher wants to buy a computer, but right now he doesn’t have enough\nmoney. A friend told Christopher that in 5 years the computer will cost R 9150.\nHe decides to start saving money today at Durban United Bank. Christopher\ndeposits R 5000 into a savings account with an interest rate of 7,95% p.a. com-\npounded monthly.\nThen after 18 months the bank changes the interest rate\nto 6,95% p.a. compounded weekly. After another 6 months, the interest rate\nchanges again to 7,92% p.a.\ncompounded two times per year.\nHow much\nmoney will Christopher have in the account after 5 years, and will he then have\nenough money to buy the computer?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238N\n2. 238P\n3. 238Q\n4. 238R\n5. 238S\n6. 238T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.4\nNominal and effective interest rates\nEMBJM\nWe have seen that although interest is quoted as a percentage per annum it can be\ncompounded more than once a year. We therefore need a way of comparing interest\nrates. For example, is an annual interest rate of 8% compounded quarterly higher or\nlower than an interest rate of 8% p.a. compounded yearly?\nInvestigation: Nominal and effective interest rates\n1. Calculate the accumulated amount at the end of one year if R 1000 is invested\nat 8% p.a. compound interest:\nA = P(1 + i)n\n= . . . . . .\n2. Calculate the value of R 1000 if it is invested for one year at 8% p.a. com-\npounded:\n394\n9.4.\nNominal and effective interest rates\n\nFrequency\nCalculation\nAccumulated\namount\nInterest\namount\nhalf-yearly\nA = 1000\n\u0010\n1 + 0,08\n2\n\u00111×2\nR 1081,60\nR 81,60\nquarterly\nmonthly\nweekly\ndaily\n3. Use your results from the table above to calculate the effective rate that the\ninvestment of R 1000 earns in one year:\nFrequency\nAccumulated\namount\nCalculation\nEffective\ninterest\nrate\nhalf-yearly\nR 1081,60\n1081,60 = 1000(1 + i)\n1081,60\n1000\n= 1 + i\n1081,60\n1000\n−1 = i\n∴i = 0,0816\ni = 8,16%\nquarterly\nmonthly\nweekly\ndaily\n4. If you wanted to borrow R 10 000 from the bank, would it be better to pay it\nback at an interest rate of 22% p.a. compounded quarterly or 22% compounded\nmonthly? Show your calculations.\nAn interest rate compounded more than once a year is called the nominal interest rate.\nIn the investigation above, we determined that the nominal interest rate of 8% p.a.\ncompounded half-yearly is actually an effective rate of 8,16% p.a.\nGiven a nominal interest rate i(m) compounded at a frequency of m times per year\nand the effective interest rate i, the accumulated amount calculated using both interest\nrates will be equal so we can write:\nP(1 + i) = P\n \n1 + i(m)\nm\n!m\n∴1 + i =\n \n1 + i(m)\nm\n!m\n395\nChapter 9.\nFinance, growth and decay\n\nWorked example 15: Nominal and effective interest rates\nQUESTION\nInterest on a credit card is quoted as 23% p.a. compounded monthly. What is the\neffective annual interest rate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down the known variables\nInterest is being added monthly, therefore:\nm = 12\ni(12) = 0,23\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i\n1 + i =\n\u0012\n1 + 0,23\n12\n\u001312\n∴i = 1 −\n\u0012\n1 + 0,23\n12\n\u001312\n= 25,59%\nStep 3: Write the final answer\nThe effective interest rate is 25,59% per annum.\nWorked example 16: Nominal and effective interest rates\nQUESTION\nDetermine the nominal interest rate compounded quarterly if the effective interest rate\nis 9% per annum (correct to two decimal places).\nSOLUTION\nStep 1: Write down the known variables\n396\n9.4.\nNominal and effective interest rates\n\nInterest is being added quarterly, therefore:\nm = 4\ni = 0,09\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i(m)\n1 + 0,09 =\n \n1 + i(4)\n4\n!4\n4p\n1,09 = 1 + i(4)\n4\n4p\n1,09 −1 = i(4)\n4\n4\n\u0010\n4p\n1,09 −1\n\u0011\n= i(4)\n∴i(4) = 8,71%\nStep 3: Write the final answer\nThe nominal interest rate is 8,71% p.a. compounded quarterly.\nExercise 9 – 6: Nominal and effect interest rates\n1. Determine the effective annual interest rate if the nominal interest rate is:\na) 12% p.a. compounded quarterly.\nb) 14,5% p.a. compounded weekly.\nc) 20% p.a. compounded daily.\n2. Consider the following:\n• 16,8% p.a. compounded annually.\n• 16,4% p.a. compounded monthly.\n• 16,5% p.a. compounded quarterly.\na) Determine the effective annual interest rate of each of the nominal rates\nlisted above.\nb) Which is the best interest rate for an investment?\nc) Which is the best interest rate for a loan?\n397\nChapter 9.\nFinance, growth and decay\n\n3. Calculate the effective annual interest rate equivalent to a nominal interest rate\nof 8,75% p.a. compounded monthly.\n4. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every\nfour months on her investment of R 85 000.\nCalculate the effective rate per\nannum.\n5. Determine which of the following would be the better agreement for paying back\na student loan:\na) 9,1% p.a. compounded quarterly.\nb) 9% p.a. compounded monthly.\nc) 9,3% p.a. compounded half-yearly.\n6. Miranda invests R 8000 for 5 years for her son’s study fund. Determine how\nmuch money she will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 6% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 238V\n1b. 238W\n1c. 238X\n2. 238Y\n3. 238Z\n4. 2392\n5. 2393\n6. 2394\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.5\nSummary\nEMBJN\nSee presentation: 2395 at www.everythingmaths.co.za\n• Simple interest: A = P(1 + in)\n• Compound interest: A = P(1 + i)n\n• Simple depreciation: A = P(1 −in)\n• Compound depreciation: A = P(1 −i)n\n• Nominal and effective annual interest rates: 1 + i =\n\u0010\n1 + i(m)\nm\n\u0011m\n398\n9.5.\nSummary\n\nExercise 9 – 7: End of chapter exercises\n1. Thabang buys a Mercedes worth R 385 000 in 2007. What will the value of the\nMercedes be at the end of 2013 if:\na) the car depreciates at 6% p.a. straight-line depreciation.\nb) the car depreciates at 6% p.a. reducing-balance depreciation.\n2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R 8900.\nThe interest rate is quoted as 11% per annum based on simple interest. Calculate\nthe required monthly payment for this contract.\n3. A computer is purchased for R 16 000. It depreciates at 15% per annum.\na) Determine the book value of the computer after 3 years if depreciation is\ncalculated according to the straight-line method.\nb) Find the rate according to the reducing-balance method that would yield,\nafter 3 years, the same book value as calculated in the previous question.\n4. Maggie invests R 12 500 for 5 years at 12% per annum compounded monthly\nfor the first 2 years and 14% per annum compounded semi-annually for the next\n3 years. How much will Maggie receive in total after 5 years?\n5. Tintin invests R 120 000. He is quoted a nominal interest rate of 7,2% per an-\nnum compounded monthly.\na) Calculate the effective rate per annum (correct to two decimal places).\nb) Use the effective rate to calculate the value of Tintin’s investment if he\ninvested the money for 3 years.\nc) Suppose Tintin invests his money for a total period of 4 years, but after 18\nmonths makes a withdrawal of R 20 000, how much will he receive at the\nend of the 4 years?\n6. Ntombi opens accounts at a number of clothing stores and spends freely. She\ngets herself into terrible debt and she cannot pay off her accounts. She owes\nFashion World R 5000 and the shop agrees to let her pay the bill at a nominal\ninterest rate of 24% compounded monthly.\na) How much money will she owe Fashion World after two years?\nb) What is the effective rate of interest that Fashion World is charging her?\n7. John invests R 30 000 in the bank for a period of 18 months. Calculate how\nmuch money he will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 8% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\ndaily\n399\nChapter 9.\nFinance, growth and decay\n\n8. Convert an effective annual interest rate of 11,6% p.a. to a nominal interest rate\ncompounded:\na) half-yearly\nb) quarterly\nc) monthly\n9. Joseph must sell his plot on the West Coast and he needs to get R 300 000 on the\nsale of the land. If the estate agent charges him 7% commission on the selling\nprice, what must the buyer pay for the plot?\n10. Mrs. Brown retired and received a lump sum of R 200 000. She deposited the\nmoney in a fixed deposit savings account for 6 years. At the end of the 6 years\nthe value of the investment was R 265 000. If the interest on her investment was\ncompounded monthly, determine:\na) the nominal interest rate per annum\nb) the effective annual interest rate\n11. R 145 000 is invested in an account which offers interest at 9% p.a.\ncom-\npounded half-yearly for the first 2 years. Then the interest rate changes to 4%\np.a. compounded quarterly. Four years after the initial investment, R 20 000 is\nwithdrawn. 6 years after the initial investment, a deposit of R 15 000 is made.\nDetermine the balance of the account at the end of 8 years.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2396\n2. 2397\n3. 2398\n4. 2399\n5. 239B\n6. 239C\n7. 239D\n8. 239F\n9. 239G\n10. 239H\n11. 239J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n400\n9.5.\nSummary\n\nCHAPTER\n10\nProbability\n10.1\nRevision\n402\n10.2\nDependent and independent events\n411\n10.3\nMore Venn diagrams\n419\n10.4\nTree diagrams\n426\n10.5\nContingency tables\n431\n10.6\nSummary\n435\n\n10\nProbability\n10.1\nRevision\nEMBJP\nTerminology\nEMBJQ\nOutcome: a single observation of an uncertain or random process (called an experi-\nment). For example, when you accidentally drop a book, it might fall on its cover, on\nits back or on its side. Each of these options is a possible outcome.\nSample space of an experiment: the set of all possible outcomes of the experiment. For\nexample, the sample space when you roll a single 6-sided die is the set {1; 2; 3; 4; 5; 6}.\nFor a given experiment, there is exactly one sample space. The sample space is de-\nnoted by the letter S.\nEvent: a set of outcomes of an experiment. For example, during radioactive decay of\n1 gramme of uranium-234, one possible event is that the number of alpha-particles\nemitted during 1 microsecond is between 225 and 235.\nProbability of an event: a real number between 0 and 1 that describes how likely it\nis that the event will occur. A probability of 0 means the outcome of the experiment\nwill never be in the event set. A probability of 1 means the outcome of the experiment\nwill always be in the event set. When all possible outcomes of an experiment have\nequal chance of occurring, the probability of an event is the number of outcomes in\nthe event set as a fraction of the number of outcomes in the sample space.\nRelative frequency of an event: the number of times that the event occurs during\nexperimental trials, divided by the total number of trials conducted. For example, if\nwe flip a coin 10 times and it landed on heads 3 times, then the relative frequency of\nthe heads event is 3\n10 = 0,3.\nUnion of events: the set of all outcomes that occur in at least one of the events. For\n2 events called A and B, we write the union as “A or B”. Another way of writing the\nunion is using set notation: A ∪B.\nIntersection of events: the set of all outcomes that occur in all of the events. For 2\nevents called A and B, we write the intersection as “A and B”. Another way of writing\nthe intersection is using set notation: A ∩B.\nMutually exclusive events: events with no outcomes in common, that is (A and B) =\n∅. Mutually exclusive events can never occur simultaneously. For example the event\nthat a number is even and the event that the same number is odd are mutually exclu-\nsive, since a number can never be both even and odd.\nComplementary events: two mutually exclusive events that together contain all the\noutcomes in the sample space. For an event called A, we write the complement as\n“not A”. Another way of writing the complement is as A′.\nSee video: 239K at www.everythingmaths.co.za\n402\n10.1.\nRevision\n\nIdentities\nEMBJR\nThe addition rule (also called the sum rule) for any 2 events, A and B is\nP(A or B) = P(A) + P(B) −P(A and B)\nThis rule relates the probabilities of 2 events with the probabilities of their union and\nintersection.\nThe addition rule for 2 mutually exclusive events is\nP(A or B) = P(A) + P(B)\nThis rule is a special case of the previous rule. Because the events are mutually exclu-\nsive, P(A and B) = 0.\nThe complementary rule is\nP(not A) = 1 −P(A)\nThis rule is a special case of the previous rule. Since A and (not A) are mutually\nexclusive, P(A or (not A)) = 1.\nSee video: 239M at www.everythingmaths.co.za\nWorked example 1: Events\nQUESTION\nYou take all the hearts from a deck of cards. You then select a random card from the set\nof hearts. What is the sample space? What is the probability of each of the following\nevents?\n1. The card is the ace of hearts.\n2. The card has a prime number on it.\n3. The card has a letter of the alphabet on it.\nSOLUTION\nStep 1: Write down the sample space\nSince we are considering only one suit from the deck of cards (the hearts), we need to\nwrite down only the letters and numbers on the cards. Therefore the sample space is\nS = {A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K}\nStep 2: Write down the event sets\n• ace of hearts: {A}\n• prime number: {2; 3; 5; 7}\n• letter of alphabet: {A; J; Q; K}\n403\nChapter 10.\nProbability\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are 13 elements in the\nsample space. So the probability of each event is\n• ace of hearts:\n1\n13\n• prime number:\n4\n13\n• letter of alphabet:\n4\n13\nWorked example 2: Events\nQUESTION\nYou roll two 6-sided dice. Let E be the event that the total number of dots on the dice\nis 10. Let F be the event that at least one die is a 3.\n1. Write down the event sets for E and F.\n2. Determine the probabilities for E and F.\n3. Are E and F mutually exclusive? Why or why not?\nSOLUTION\nStep 1: Write down the sample space\nThe sample space of a single 6-sided die is just {1; 2; 3; 4; 5; 6}. To get the sample\nspace of two 6-sided dice, we have to take every possible pair of numbers from 1 to 6.\nS =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(1; 1)\n(1; 2)\n(1; 3)\n(1; 4)\n(1; 5)\n(1; 6)\n(2; 1)\n(2; 2)\n(2; 3)\n(2; 4)\n(2; 5)\n(2; 6)\n(3; 1)\n(3; 2)\n(3; 3)\n(3; 4)\n(3; 5)\n(3; 6)\n(4; 1)\n(4; 2)\n(4; 3)\n(4; 4)\n(4; 5)\n(4; 6)\n(5; 1)\n(5; 2)\n(5; 3)\n(5; 4)\n(5; 5)\n(5; 6)\n(6; 1)\n(6; 2)\n(6; 3)\n(6; 4)\n(6; 5)\n(6; 6)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nStep 2: Write down the events\nFor E the dice have to add to 10.\nE = {(4; 6); (5; 5); (6; 4)}\nFor F at least one die has to be 3.\nF = {(1; 3); (3; 1); (2; 3); (3; 2); (3; 3); (4; 3); (3; 4); (5; 3); (3; 5); (6; 3); (3; 6)}\n404\n10.1.\nRevision\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are\n• 6 × 6 = 36 outcomes in the sample space, S;\n• 3 outcomes in event E; and\n• 11 outcomes in event F.\nTherefore\nP(E) = 3\n36 = 1\n12\nand\nP(F) = 11\n36\nStep 4: Are they mutually exclusive\nTo test whether two events are mutually exclusive, we have to test whether their in-\ntersection is empty. Since E has no outcomes that contain a 3 on one of the dice,\nthe intersection of E and F is empty: (E and F) = ∅. This means that the events are\nmutually exclusive.\nSee video: 239N at www.everythingmaths.co.za\nExercise 10 – 1: Revision\n1. A bag contains r red balls, b blue balls and y yellow balls. What is the probability\nthat a ball drawn from the bag at random is yellow?\n2. A packet has yellow and pink sweets. The probability of taking out a pink sweet\nis 7\n12. What is the probability of taking out a yellow sweet?\n3. You flip a coin 4 times. What is the probability that you get 2 heads and 2 tails?\nWrite down the sample space and the event set to determine the probability of\nthis event.\n4. In a class of 37 children, 15 children walk to school, 20 children have pets at\nhome and 12 children who have a pet at home also walk to school. How many\nchildren walk to school and do not have a pet at home?\n5. You roll two 6-sided dice and are interested in the following two events:\n• A: the sum of the dice equals 8\n• B: at least one of the dice shows a 1\nShow that these events are mutually exclusive.\n405\nChapter 10.\nProbability\n\n6. You ask a friend to think of a number from 1 to 100. You then ask her the\nfollowing questions:\n• Is the number even?\n• Is the number divisible by 7?\nHow many possible numbers are less than 80 if she answered “yes” to both\nquestions?\n7. In a group of 42 pupils, all but 3 had a packet of chips or a Fanta or both. If 23\nhad a packet of chips and 7 of these also had a Fanta, what is the probability that\none pupil chosen at random has:\na) both chips and Fanta\nb) only Fanta\n8. Tamara has 18 loose socks in a drawer. Eight of these are orange and two are\npink. Calculate the probability that the first sock taken out at random is:\na) orange\nb) not orange\nc) pink\nd) not pink\ne) orange or pink\nf) neither orange nor pink\n9. A box contains coloured blocks. The number of blocks of each colour is given\nin the following table.\nColour\nPurple\nOrange\nWhite\nPink\nNumber of blocks\n24\n32\n41\n19\nA block is selected randomly. What is the probability that the block will be:\na) purple\nb) purple or white\nc) pink and orange\nd) not orange?\n10. The surface of a soccer ball is made up of 32 faces. 12 faces are regular pen-\ntagons, each with a surface area of about 37 cm2. The other 20 faces are regular\nhexagons, each with a surface area of about 56 cm2.\nYou roll the soccer ball. What is the probability that it stops with a pentagon\ntouching the ground?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 239P\n2. 239Q\n3. 239R\n4. 239S\n5. 239T\n6. 239V\n7. 239W\n8. 239X\n9. 239Y\n10. 239Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n406\n10.1.\nRevision\n\nVenn diagrams\nEMBJS\nA Venn diagram is used to show how events are related to one another.\nA Venn\ndiagram can be very helpful when doing calculations with probabilities. In a Venn\ndiagram each event is represented by a shape, often a circle or a rectangle. The region\ninside the shape represents the outcomes included in the event and the region outside\nthe shape represents the outcomes that are not in the event.\nS\nA\nB\nA and B\nA Venn diagram representing a sample space, S, as a square; and two events, A and\nB, as circles. The intersection of the two circles contains outcomes that are in both A\nand B.\nVenn diagrams can be used in slightly different ways and it is important to notice the\ndifferences between them. The following 3 examples show how a Venn diagram is\nused to represent\n• the outcomes included in each event;\n• the number of outcomes in each event; and\n• the probability of each event.\nWorked example 3: Venn diagrams with outcomes\nQUESTION\nChoose a number between 1 and 20. Draw a Venn diagram to answer the following\nquestions.\n1. What is the probability that the number is a multiple of 3?\n2. What is the probability that the number is a multiple of 5?\n3. What is the probability that the number is a multiple of 3 or 5?\n4. What is the probability that the number is a multiple of 3 and 5?\nSOLUTION\nStep 1: Draw a Venn diagram\nThe Venn diagram should show the sample space of all numbers from 1 to 20. It should\nalso show an event set that contains all the multiples of 3, let A = {3; 6; 9; 12; 15; 18},\n407\nChapter 10.\nProbability\n\nand another event set that contains all the multiples of 5, let B = {5; 10; 15; 20}. Note\nthat there is one shared outcome between these two events, namely 15.\n3\n18\n15\n12\n9\n6\n10\n20\n5\n1\n2\n4\n7\n8\n11\n13\n14\n16\n17\n19\nStep 2: Compute probabilities\nThe probability of an event is the number of outcomes in the event set divided by the\nnumber of outcomes in the sample space. There are 20 outcomes in the sample space.\n1. Since there are 6 outcomes in the multiples of 3 event set, the probability of a\nmultiple of 3 is P(A) = 6\n20 = 3\n10.\n2. Since there are 4 outcomes in the multiples of 5 event set, the probability of a\nmultiple of 5 is P(B) = 4\n20 = 1\n5.\n3. The event that the number is a multiple of 3 or 5 is the union of the above two\nevent sets. There are 9 elements in the union of the event sets, so the probability\nis 9\n20.\n4. The event that the number is a multiple of 3 and 5 is the intersection of the\ntwo event sets. There is 1 element in the intersection of the event sets, so the\nprobability is 1\n20.\nWorked example 4: Venn diagrams with counts\nQUESTION\nIn a group of 50 learners, 35 take Mathematics and 30 take History, while 12 take\nneither of the two subjects. Draw a Venn diagram representing this information. If a\nlearner is chosen at random from this group, what is the probability that he takes both\nMathematics and History?\nSOLUTION\nStep 1: Draw outline of Venn diagram\nThere are 2 events in this question, namely\n• M: that a learner takes Mathematics; and\n• H: that a learner takes History.\n408\n10.1.\nRevision\n\nWe need to do some calculations before drawing the full Venn diagram, but with the\ninformation above we can already draw the outline.\nS\nM\nH\nStep 2: Write down sizes of the event sets, their union and intersection\nWe are told that 12 learners take neither of the two subjects. Graphically we can\nrepresent this as:\nS\nM\nH\n12\nSince there are 50 elements in the sample space, we can see from this figure that there\nare 50 −12 = 38 elements in (M or H). So far we know\n• n(M) = 35\n• n(H) = 30\n• n(M or H) = 38\nFrom the addition rule,\nn(M or H) = n(M) + n(H) −n(M and H)\n∴n(M and H) = 35 + 30 −38\n= 27\nStep 3: Draw the final Venn diagram\nS\nM\nH\n12\n27\n8\n3\n409\nChapter 10.\nProbability\n\nWorked example 5: Venn diagrams with probabilities\nQUESTION\nDraw a Venn diagram to represent the same information as in the previous example,\nexcept showing the probabilities of the different events, rather than the counts.\nIf a learner is chosen at random from this group, what is the probability that she takes\nboth Mathematics and History?\nSOLUTION\nStep 1: Use counts to compute probabilities\nSince there are 50 elements (learners) in the sample space, we can compute the prob-\nability of any event by dividing the size of the event set by 50. This gives the following\nprobabilities:\n• P(M) = 35\n50 = 7\n10\n• P(H) = 30\n50 = 3\n5\n• P(M or H) = 38\n50 = 19\n25\n• P(M and H) = 27\n50\nStep 2: Draw the Venn diagram\nNext we replace each count from the Venn diagram in the previous example with a\nprobability.\nS\nM\nH\n6\n25\n27\n50\n4\n25\n3\n50\nStep 3: Find the answer\nThe probability that a random learner will take both Mathematics and History is\nP(M and H) = 27\n50.\nSee video: 23B2 at www.everythingmaths.co.za\n410\n10.1.\nRevision\n\nExercise 10 – 2: Venn diagram revision\n1. Given the following information:\n• P(A) = 0,3\n• P(B and A) = 0,2\n• P(B) = 0,7\nFirst draw a Venn diagram to represent this information. Then compute the value\nof P(B and (not A)).\n2. You are given the following information:\n• P(A) = 0,5\n• P(A and B) = 0,2\n• P(not B) = 0,6\nDraw a Venn diagram to represent this information and determine P(A or B).\n3. A study was undertaken to see how many people in Port Elizabeth owned either\na Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60%\nowned a Volkswagen. What percentage of people owned neither car?\n4. Let S denote the set of whole numbers from 1 to 15, X denote the set of even\nnumbers from 1 to 15 and Y denote the set of prime numbers from 1 to 15.\nDraw a Venn diagram depicting S, X and Y .\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B3\n2. 23B4\n3. 23B5\n4. 23B6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.2\nDependent and independent events\nEMBJT\nSometimes the presence or absence of one event tells us something about other events.\nWe call events dependent if knowing whether one of them happened tells us some-\nthing about whether the others happened. Independent events give us no information\nabout one another; the probability of one event occurring does not affect the probabil-\nity of the other events occurring.\nDEFINITION: Independent events\nTwo events, A and B are independent if and only if\nP(A and B) = P(A) × P(B)\nAt first it might not be clear why we should call events that satisfy the equation above\nindependent. We will explore this further using a number of examples.\n411\nChapter 10.\nProbability\n\nInvestigation: Independence\nRoll a single 6-sided die and consider the following two events:\n• E: you get an even number\n• T: you get a number that is divisible by three\nNow answer the following questions:\n• What is the probability of E?\n• What is the probability of getting an even number if you are told that the number\nwas also divisible by three?\n• Does knowing that the number was divisible by three change the probability that\nthe number was even?\nAre the events E and T dependent or independent according to the definition (hint:\ncompute the probabilities in the definition of independence)?\nSee video: 23B7 at www.everythingmaths.co.za\nSo, why do we call it independence when P(A and B) = P(A) × P(B)? For two\nevents, A and B, independence means that knowing the outcome of B does not affect\nthe probability of A.\nConsider the following Venn diagram.\nS\nA\nB\nA and B\nThe probability of A is the ratio between the number of outcomes in A and the number\nof outcomes in the sample space, S.\nP(A) = n(A)\nn(S)\n412\n10.2.\nDependent and independent events\n\nNow, let’s say that we know that event B happened. How does this affect the proba-\nbility of A? Here is how the Venn diagram changes:\nS\nA\nB\nA and B\nA lot of the possible outcomes (all of the outcomes outside B) are now out of the pic-\nture, because we know that they did not happen. Now the probability of A happening,\ngiven that we know that B happened, is the ratio between the size of the region where\nA is present (A and B) and the size of all possible events (B).\nP(A if we know B) = n(A and B)\nn(B)\nIf P(A) = P(A if we know B) we call them independent, because knowing B does\nnot change the probability of A.\nWith some algebra, we can prove that this statement of independence is the same\nas the definition of independence that we saw at the beginning of this section. For\nindependent events\nP(A and B) = P(A) × P(B)\nThis is equivalent to\nP(A) = P(A and B) ÷ P(B)\n= n(A and B)\nn(S)\n÷ n(B)\nn(S)\n= n(A and B)\nn(B)\n= P(A if we know B)\nThat is why we call events independent!\n(For enrichment only):\nThe ratio\nP(A and B)\nP(B)\nis called a conditional probability and written using the notation P(A | B). This\nnotation is read as “the probability of A given B.”\nIf (and only if) A and B are independent: P(A | B) = P(A) and P(B | A) = P(B).\nTry to prove this using the definition of independence.\n413\nChapter 10.\nProbability\n\nWorked example 6: Independent and dependent events\nQUESTION\nA bag contains 5 red and 5 blue balls. We remove a random ball from the bag, record\nits colour and put it back into the bag. We then remove another random ball from the\nbag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Probability of a red ball first\nSince there are a total of 10 balls, of which 5 are red, the probability of getting a red\nball is\nP(first ball red) = 5\n10 = 1\n2\nStep 2: Probability of a blue ball second\nThe problem states that the first ball is placed back into the bag before we take the\nsecond ball. This means that when we draw the second ball, there are again a total of\n10 balls in the bag, of which 5 are blue. Therefore the probability of drawing a blue\nball is\nP(second ball blue) = 5\n10 = 1\n2\nStep 3: Probability of red first and blue second\nWhen drawing two balls from the bag, there are 4 possibilities. We can get\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nWe want to know the probability of the second outcome, where we have to get a red\nball first. Since there are 5 red balls and 10 balls in total, there are\n5\n10 ways to get a\nred ball first. Now we put the first ball back, so there are again 5 red balls and 5 blue\nballs in the bag. Therefore there are\n5\n10 ways to get a blue ball second if the first ball\nwas red. This means that there are\n5\n10 × 5\n10 = 25\n100\n414\n10.2.\nDependent and independent events\n\nways to get a red ball first and a blue ball second. So, the probability of getting a red\nball first and a blue ball second is 1\n4.\nStep 4: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 1\n4\nSince 1\n4 = 1\n2 × 1\n2, the events are independent.\nSee video: 23B8 at www.everythingmaths.co.za\nWorked example 7: Independent and dependent events\nQUESTION\nIn the previous example, we picked a random ball and put it back into the bag before\ncontinuing. This is called sampling with replacement. In this example, we will follow\nthe same process, except that we will not put the first ball back into the bag. This is\ncalled sampling without replacement.\nSo, from a bag with 5 red and 5 blue balls, we remove a random ball and record its\ncolour. Then, without putting back the first ball, we remove another random ball from\nthe bag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Count the number of outcomes\nWe will look directly at the number of possible ways in which we can get the 4 possible\noutcomes when removing 2 balls. In the previous example, we saw that the 4 possible\noutcomes are\n415\nChapter 10.\nProbability\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nFor the first outcome, we have to get a red ball first. Since there are 5 red balls and\n10 balls in total, there are\n5\n10 ways to get a red ball first. After we have taken out a red\nball, there are now 4 red balls and 5 blue balls left. Therefore there are 4\n9 ways to get\na red ball second if the first ball was also red. This means that there are\n5\n10 × 4\n9 = 20\n90\nways to get a red ball first and a red ball second. The probability of the first outcome\nis 2\n9.\nFor the second outcome, we have to get a red ball first. As in the first outcome, there\nare\n5\n10 ways to get a red ball first; and there are now 4 red balls and 5 blue balls left.\nTherefore there are 5\n9 ways to get a blue ball second if the first ball was red. This means\nthat there are\n5\n10 × 5\n9 = 25\n90\nways to get a red ball first and a blue ball second. The probability of the second\noutcome is 5\n18.\nWe can compute the probabilities of the third and fourth outcomes in the same way as\nthe first two, but there is an easier way. Notice that there are only 2 types of ball and\nthat there are exactly equal numbers of them at the start. This means that the problem\nis completely symmetric in red and blue. We can use this symmetry to compute the\nprobabilities of the other two outcomes.\nIn the third outcome, the first ball is blue and the second ball is red. Because of\nsymmetry this outcome must have the same probability as the second outcome (when\nthe first ball is red and the second ball is blue). Therefore the probability of the third\noutcome is 5\n18.\nIn the fourth outcome, the first and second balls are both blue. From symmetry, this\noutcome must have the same probability as the first outcome (when both balls are red).\nTherefore the probability of the fourth outcome is 2\n9.\nTo summarise, these are the possible outcomes and their probabilities:\n• first ball red and second ball red: 2\n9;\n• first ball red and second ball blue:\n5\n18;\n• first ball blue and second ball red:\n5\n18;\n• first ball blue and second ball blue: 2\n9.\nStep 2: Probability of a red ball first\nTo determine the probability of getting a red ball on the first draw, we look at all of the\noutcomes that contain a red ball first. These are\n416\n10.2.\nDependent and independent events\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball.\nThe probability of the first outcome is 2\n9 and the probability of the second outcome is\n5\n18. By adding these two probabilities, we see that the probability of getting a red ball\nfirst is\nP(first ball red) = 2\n9 + 5\n18 = 1\n2\nThis is the same as in the previous exercise, which should not be too surprising since\nthe probability of the first ball being red is not affected by whether or not we put it\nback into the bag before drawing the second ball.\nStep 3: Probability of a blue ball second\nTo determine the probability of getting a blue ball on the second draw, we look at all\nof the outcomes that contain a blue ball second. These are\n• a red ball and then a blue ball;\n• a blue ball and then another blue ball.\nThe probability of the first outcome is 5\n18 and the probability of the second outcome is\n2\n9. By adding these two probabilities, we see that the probability of getting a blue ball\nsecond is\nP(second ball blue) = 5\n18 + 2\n9 = 1\n2\nThis is also the same as in the previous exercise! You might find it surprising that the\nprobability of the second ball is not affected by whether or not we replace the first ball.\nThe reason why this probability is still 1\n2 is that we are computing the probability that\nthe second ball is blue without knowing the colour of the first ball. Because there are\nonly two equal possibilities for the second ball (red and blue) and because we don’t\nknow whether the first ball is red or blue, there is an equal chance that the second ball\nwill be one colour or the other.\nStep 4: Probability of red first and blue second\nWe have already calculated the probability that the first ball is red and the second ball\nis blue. It is 5\n18.\nStep 5: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 5\n18\nSince 5\n18 ̸= 1\n2 × 1\n2, the events are dependent.\n417\nChapter 10.\nProbability\n\nWARNING!\nJust because two events are mutually exclusive does not necessarily mean that they\nare independent. To test whether events are mutually exclusive, always check that\nP(A and B) = 0. To test whether events are independent, always check that P(A and B) =\nP(A) × P(B). See the exercises below for examples of events that are mutually ex-\nclusive and independent in different combinations.\nExercise 10 – 3: Dependent and independent events\n1. Use the following Venn diagram to determine whether events X and Y are\na) mutually exclusive or not mutually exclusive;\nb) dependent or independent.\nS\nX\nY\n11\n7\n3\n14\n2. Of the 30 learners in a class 17 have black hair, 11 have brown hair and 2 have\nred hair. A learner is selected from the class at random.\na) What is the probability that the learner has black hair?\nb) What is the probability that the learner has brown hair?\nc) Are these two events mutually exclusive?\nd) Are these two events independent?\n3. P(M) = 0,45; P(N) = 0,3 and P(M or N) = 0,615. Are the events M and N\nmutually exclusive, independent or neither mutually exclusive nor independent?\n4. (For enrichment)\nProve that if event A and event B are mutually exclusive with P(A) ̸= 0 and\nP(B) ̸= 0, then A and B are always dependent.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B9\n2. 23BB\n3. 23BC\n4. 23BD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n418\n10.2.\nDependent and independent events\n\n10.3\nMore Venn diagrams\nEMBJV\nIn the rest of this chapter we will look at tools and techniques for working with proba-\nbility problems.\nWhen working with more complex problems, we can have three or more events that\nintersect in various ways.\nTo solve these problems, we usually want to count the\nnumber (or percentage) of outcomes in an event, or a combination of events. Venn\ndiagrams are a useful tool for recording and visualising the counts.\nInvestigation: Venn diagram for 3 events\nThe diagram below shows a general Venn diagram for 3 events.\nS\nA\nB\nC\nWrite down the sets corresponding to each of the three coloured regions and also\nto the shaded region. Remember that the intersections between circles represent the\nintersections between the different events.\nWhat is the event for\n• the red region;\n• the green region;\n• the blue region; and\n• the shaded region?\n419\nChapter 10.\nProbability\n\nWorked example 8: Venn diagram for 3 events\nQUESTION\nDraw a Venn diagram that shows the following sample space and events:\n• S: all the integers from 1 to 30\n• P: prime numbers\n• M: multiples of 3\n• F: factors of 30\nSOLUTION\nStep 1: Write down the sample space and event sets\nThe sample space contains all the positive integers up to 30.\nS = {1; 2; 3; . . . ; 30}\nThe prime numbers between 1 and 30 are\nP = {2; 3; 5; 7; 11; 13; 17; 19; 23; 29}\nThe multiples of 3 between 1 and 30 are\nM = {3; 6; 9; 12; 15; 18; 21; 24; 27; 30}\nThe factors of 30 are\nF = {1; 2; 3; 5; 6; 10; 15; 30}\nStep 2: Draw the outline of the Venn diagram\nThere are 3 events, namely P, M and F, and the sample space, S. Put this information\non a Venn diagram:\nS\nP\nM\nF\n420\n10.3.\nMore Venn diagrams\n\nStep 3: Place the outcomes in the appropriate event sets\nS\nP\nM\nF\n3\n2\n5\n6\n15\n30\n1\n10\n7\n11\n13\n17\n19\n23\n29\n9\n12\n21\n24\n18\n27\n4\n8\n14\n16\n20\n22\n25\n26\n28\nWorked example 9: Venn diagram for 3 events\nQUESTION\nAt Dawnview High there are 400 Grade 11 learners. 270 do Computer Science, 300\ndo English and 50 do Business studies. All those doing Computer Science do English,\n20 take Computer Science and Business studies and 35 take English and Business\nstudies. Using a Venn diagram, calculate the probability that a pupil drawn at random\nwill take:\n1. English, but not Business studies or Computer Science\n2. English but not Business studies\n3. English or Business studies but not Computer Science\n4. English or Business studies\nSOLUTION\nStep 1: Draw the outline of the Venn diagram\nWe need to be careful with this problem. In the question statement we are told that all\nthe learners who do Computer Science also do English. This means that the circle for\nComputer Science on the Venn diagram needs to be inside the circle for English.\n421\nChapter 10.\nProbability\n\nS\nE\nC\nB\nStep 2: Fill in the counts on the Venn diagram\nS\nE\nC\nB\n20\n250\n15\n15\n15\n85\nStep 3: Compute probabilities\nTo find the number of learners taking English, but not Business studies or Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 15 and there are a total of 400 learners in the grade. There-\nfore the probability that a learner will take English but not Business studies or Computer\nScience is\n15\n400 = 3\n80.\nTo find the number of learners taking English but not Business studies, we need to look\nat this region of the Venn diagram:\n422\n10.3.\nMore Venn diagrams\n\nThe count in this region is 265. Therefore the probability that a learner will take\nEnglish but not Business studies is 265\n400 = 53\n80.\nTo find the number of learners taking English or Business studies but not Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 45. Therefore the probability that a learner will take English\nor Business studies but not Computer Science is\n45\n400 = 9\n80.\nTo find the number of learners taking English or Business studies, we need to look at\nthis region of the Venn diagram:\nThe count in this region is 315. Therefore the probability that a learner will take\nEnglish or Business studies is 315\n400 = 63\n80.\n423\nChapter 10.\nProbability\n\nThere are some words that tell you which part of the Venn diagram should be filled in.\nThe following table summarises the most important ones:\nWords\nSymbols\nVenn diagram\n“all”\nA and B and C / A ∩B ∩C\n“none”\n“at least one”\nA or B or C / A ∪B ∪C\n“both A and B”\nA and B / A ∩B\n“A or B”\nA or B / A ∪B\nExercise 10 – 4: Venn diagrams\n1. Use the Venn diagram below to answer the following questions. Also given:\nn(S) = 120.\nS\nF\n8\n10\nG\n24\n15\nH\n14\n7\n2\na) Compute P(F).\nb) Compute P(G or H).\nc) Compute P(F and G).\nd) Are F and G dependent or independent?\n424\n10.3.\nMore Venn diagrams\n\n2. The Venn diagram below shows the probabilities of 3 events. Complete the Venn\ndiagram using the additional information provided.\nS\nZ\n1\n25\nY\n17\n100\nX\n17\n100\n3\n20\n• P(Z and (not Y )) =\n31\n100\n• P(Y and X) =\n23\n100\n• P(Y ) =\n39\n100\nAfter completing the Venn diagram, compute the following:\nP (Z and not (X or Y ))\n3. There are 79 Grade 10 learners at school. All of these take some combination of\nMaths, Geography and History. The number who take Geography is 41; those\nwho take History is 36; and 30 take Maths. The number who take Maths and\nHistory is 16; the number who take Geography and History is 6, and there are 8\nwho take Maths only and 16 who take History only.\na) Draw a Venn diagram to illustrate all this information.\nb) How many learners take Maths and Geography but not History?\nc) How many learners take Geography only?\nd) How many learners take all three subjects?\n4. Draw a Venn diagram with 3 mutually exclusive events. Use the diagram to\nshow that for 3 mutually exclusive events, A, B and C, the following is true:\nP(A or B or C) = P(A) + P(B) + P(C)\nThis is the addition rule for 3 mutually exclusive events.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BF\n2. 23BG\n3. 23BH\n4. 23BJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n425\nChapter 10.\nProbability\n\n10.4\nTree diagrams\nEMBJW\nTree diagrams are useful for organising and visualising the different possible outcomes\nof a sequence of events. For each possible outcome of the first event, we draw a line\nwhere we write down the probability of that outcome and the state of the world if that\noutcome happened. Then, for each possible outcome of the second event we do the\nsame thing.\nBelow is an example of a simple tree diagram, showing the possible outcomes of\nrolling a 6-sided die.\n1\n1\n6\n2\n1\n6\n3\n1\n6\n4\n1\n6\n5\n1\n6\n6\n1\n6\noutcomes\nprobabilities\nNote that each outcome (the numbers 1 to 6) is shown at the end of a line; and that\nthe probability of each outcome (all 1\n6 in this case) is shown shown on a line. The\nprobabilities have to add up to 1 in order to cover all of the possible outcomes. In the\nexamples below, we will see how to draw tree diagrams with multiple events and how\nto compute probabilities using the diagrams.\nEarlier in this chapter you learned about dependent and independent events. Tree\ndiagrams are very helpful for analysing dependent events. A tree diagram allows you\nto show how each possible outcome of one event affects the probabilities of the other\nevents.\nTree diagrams are not so useful for independent events since we can just multiply the\nprobabilities of separate events to get the probability of the combined event. Remem-\nber that for independent events:\nP(A and B) = P(A) × P(B)\nSo if you already know that events are independent, it is usually easier to solve a\nproblem without using tree diagrams. But if you are uncertain about whether events\nare independent or if you know that they are not, you should use a tree diagram.\nWorked example 10: Drawing a tree diagram\nQUESTION\nIf it rains on a given day, the probability that it rains the next day is 1\n3. If it does not rain\non a given day, the probability that it rains the next day is 1\n6. The probability that it will\nrain tomorrow is 1\n5. What is the probability that it will rain the day after tomorrow?\nDraw a tree diagram of all the possibilities to determine the answer.\nSOLUTION\nStep 1: Draw the first level of the tree diagram\nBefore we can determine what happens on the day after tomorrow, we first have to\ndetermine what might happen tomorrow. We are told that there is a 1\n5 probability that\n426\n10.4.\nTree diagrams\n\nit will rain tomorrow. Here is how to represent this information using a tree diagram:\n1\n5\nrain\n4\n5\nno rain\ntoday:\ntomorrow:\nStep 2: Draw the second level of the tree diagram\nWe are also told that if it does rain on one day, there is a 1\n3 probability that it will also\nrain on the following day. On the other hand, if it does not rain on one day, there is\nonly a 1\n6 probability that it will also rain on the following day. Using this information\nwe complete the tree diagram:\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nStep 3: Compute the probability\nWe are asked what the probability is that it will rain the day after tomorrow. On the\ntree diagram above we can see that there are 2 situations where it rains on the day\nafter tomorrow. They are marked in red below.\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nTo get the probability for the first situation (that it rains tomorrow and the day after\ntomorrow) we have to multiply the probabilies along the first red line.\nP(rain tomorrow and rain day after tomorrow)\n=1\n5 × 1\n3\n= 1\n15\n427\nChapter 10.\nProbability\n\nTo get the probability for the second situation (that it does not rain tomorrow, but it\ndoes rain the day after tomorrow) we have to multiply the probabilies along the second\nred line.\nP(not rain tomorrow and rain day after tomorrow)\n=4\n5 × 1\n6\n= 2\n15\nTherefore the total probability that it will rain the day after tomorrow is the sum of the\nprobabilities along the two red paths, namely\n1\n15 + 2\n15 = 1\n5\nWorked example 11: Drawing a tree diagram\nQUESTION\nYou play the following game. You flip a coin. If it comes up tails, you get 2 points\nand your turn ends. If it comes up heads, you get only 1 point, but you can flip the\ncoin again. If you flip the coin multiple times in one turn, you add up the points. You\ncan flip the coin at most 3 times in one turn. What is the probability that you will get\nexactly 3 points in one turn? Draw a tree diagram to visualise the different possibilities.\nSOLUTION\nStep 1: Write down the events and their symbols\nEach coin toss has on of two possible outcomes, namely heads (H) and tails (T). Each\noutcome has a probability of 1\n2. We are asked to count the number of points, so we\nwill also indicate how many points we have for each outcome.\nStep 2: Draw the first level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\nThis tree diagram shows the possible outcomes after 1 flip of the coin. Remember that\nwe can have up to 3 flips, so the diagram is not complete yet. If the coin comes up\nheads, we flip the coin again. If the coin comes up tails, we stop.\n428\n10.4.\nTree diagrams\n\nStep 3: Draw the second and third level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\n1\n2\nH\n2 pts\n1\n2\nT\n3 pts\n1\n2\nH\n3 pts\n1\n2\nT\n4 pts\nIn this tree diagram you can see that we add up the points we get with each coin flip.\nAfter three coin flips, the game is over.\nStep 4: Find the relevant outcomes and compute the probability\nWe are interested in getting exactly 3 points during the game. To find these outcomes\nwe look only at the tips of the tree. We end with exactly 3 points when the coin flips\nare\n• (H; T) with probability 1\n2 × 1\n2 = 1\n4;\n• (H; H; H) with probability 1\n2 × 1\n2 × 1\n2 = 1\n8.\nNotice that we compute the probability of an outcome by multiplying all the probabil-\nities along the path from the start of the tree to the tip where the outcome is. We add\nthe above two probabilites to obtain the final probability of getting exactly 3 points as\n1\n4 + 1\n8 = 3\n8.\nWorked example 12: Drawing a tree diagram\nQUESTION\nA person takes part in a medical trial that tests the effect of a medicine on a disease.\nHalf the people are given medicine and the other half are given a sugar pill, which has\nno effect on the disease. The medicine has a 60% chance of curing someone. But,\npeople who do not get the medicine still have a 10% chance of getting well. There are\n50 people in the trial and they all have the disease. Talwar takes part in the trial, but\nwe do not know whether he got the medicine or the sugar pill. Draw a tree diagram\nof all the possible cases. What is the probability that Talwar gets cured?\nSOLUTION\nStep 1: Summarise the information in the problem\nThere are two uncertain events in this problem. Each person either receives medicine\n(probability 1\n2) or a sugar pill (probability 1\n2). Each person also gets cured (probability\n429\nChapter 10.\nProbability\n\n3\n5 with medicine and\n1\n10 without) or stays ill (probability 2\n5 with medicine and\n9\n10\nwithout).\nStep 2: Draw the tree diagram\n1\n2\nmedicine\n1\n2\nsugar pill\n3\n5\ncured\n2\n5\nnot cured\n1\n10\ncured\n9\n10\nnot cured\nIn the first level of the tree diagram we show that Talwar either gets the medicine or\nthe sugar pill. The second level of the tree diagram shows whether Talwar is cured or\nnot, depending on which one of the pills he got.\nStep 3: Compute the required probability\nWe multiply the probabilites along each path in the tree diagram that leads to Talwer\nbeing cured:\n1\n2 × 3\n5 = 3\n10\n1\n2 × 1\n10 = 1\n20\nWe then add these probabilites to get the final answer. The probability that Talwar is\ncured is 7\n20.\nExercise 10 – 5: Tree diagrams\n1. You roll a die twice and add up the dots to get a score. Draw a tree diagram to\nrepresent this experiment. What is the probability that your score is a multiple\nof 5?\n2. What is the probability of throwing at least one five in four rolls of a regular\n6-sided die? Hint: do not show all possible outcomes of each roll of the die. We\nare interested in whether the outcome is 5 or not 5 only.\n3. You flip one coin 4 times.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\n430\n10.4.\nTree diagrams\n\n4. You flip 4 different coins at the same time.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BK\n2. 23BM\n3. 23BN\n4. 23BP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.5\nContingency tables\nEMBJX\nA contingency table is another tool for keeping a record of the counts or percentages\nin a probability problem. Contingency tables are especially helpful for figuring out\nwhether events are dependent or independent.\nWe will be studying two-way contingency tables, where we count the number of out-\ncomes for 2 events and their complements, making 4 events in total. A two-way contin-\ngency table always shows the counts for the 4 possible combinations of events, as well\nas the totals for each event and its complement. We can use a contingency table to\ncompute the probabilities of various events by computing the ratios between counts,\nand to determine whether the events are dependent or independent. The example\nbelow shows a two-way contingency table, representing the outcome of a medical\nstudy.\nWorked example 13: Contingency tables\nQUESTION\nA medical trial into the effectiveness of a new medication was carried out. 120 females\nand 90 males took part in the trial. Out of those people, 50 females and 30 males\nresponded positively to the medication. Given below is a contingency table with the\ngiven information filled in.\nFemale\nMale\nTotals\nPositive\n50\n30\nNegative\nTotals\n120\n90\n1. What is the probability that the medicine gives a positive result for females?\n2. What is the probability that the medicine gives a negative result for males?\n3. Was the medication’s success independent of gender? Explain.\n431\nChapter 10.\nProbability\n\nSOLUTION\nStep 1: Complete the contingency table\nThe best place to start is always to complete the contingency table. Because the each\ncolumn has to sum up to its total, we can work out the number of females and males\nwho responded negatively to the medication. Then we can add each row to get the\ntotals on the right hand side of the table.\nFemale\nMale\nTotals\nPositive\n50\n30\n80\nNegative\n70\n60\n130\nTotals\n120\n90\n210\nStep 2: Compute the required probabilities\nThe way the first question is phrased, we need to determine the probability that a\nperson responds positively if she is female. This means that we do not include males\nin this calculation. So, the probability that the medicine gives a positive result for\nfemales is the ratio between the number of females who got a positive response and\nthe total number of females.\nP(positive if female) = n(positive and female)\nn(female)\n= 50\n120\n= 5\n12\nSimilarly, the probability that the medicine gives a negative result for males is:\nP(negative if male) = n(negative and male)\nn(male)\n= 60\n90\n= 2\n3\nStep 3: Independence\nWe need to determine whether the effect of the medicine and the gender of a par-\nticipant are dependent or independent. According to the definition, two events are\nindependent if and only if\nP(A and B) = P(A) × P(B)\nWe will look at the events that a participant is female and that the participant re-\nsponded positively to the trial.\nP(female) =\nn(female)\nn(total trials)\n= 120\n210\n= 4\n7\n432\n10.5.\nContingency tables\n\nP(positive) =\nn(positive)\nn(total trials)\n= 80\n210\n= 8\n21\nP(female and positive) = n(female and positive)\nn(total trials)\n= 50\n210\n= 5\n21\nFrom these probabilities we can see that\nP(female and positive) ̸= P(female) × P(positive)\nand therefore the gender of a participant and the outcome of a trial are dependent\nevents.\nWorked example 14: Contingency tables\nQUESTION\nUse the contingency table below to answer the following questions.\nGrade 11\nGrade 12\nTotals\nHas cellphone\n59\n50\n109\nNo cellphone\n6\n3\n9\nTotals\n65\n53\n118\n1. What is the probability that a learner from Grade 11 has a cellphone?\n2. What is the probability that a learner who does not have a cellphone is from\nGrade 11.\n3. Are the grade of a learner and whether he has a cellphone or not independent\nevents? Explain your answer.\nSOLUTION\n1. There are 65 learners in Grade 11 and 59 of them have a cellphone. Therefore\nthe probability that a learner from Grade 11 has a cellphone is 59\n65.\n2. There are 9 learners who do not have a cellphone and 6 of them are in Grade\n11. Therefore the probability that a learner who does not have a cellphone is\nfrom from Grade 11 is 6\n9 = 2\n3.\n433\nChapter 10.\nProbability\n\n3. To test for independence, we will consider whether a learner is in Grade 11 and\nwhether a learner has a cellphone. The probability that a learner is in Grade 11\nis\n65\n118. The probability that a learner has a cellphone is 109\n118. The probability that\na learner is in Grade 11 and has a cellphone is\n59\n118 = 1\n2. Since 1\n2 ̸=\n65\n118 × 109\n118\nthe grade of a learner and whether he has a cellphone are dependent.\nExercise 10 – 6: Contingency tables\n1. Use the contingency table below to answer the following questions.\nBrown eyes\nNot brown eyes\nTotals\nBlack hair\n50\n30\n80\nRed hair\n70\n80\n150\nTotals\n120\n110\n230\na) What is the probability that someone with black hair has brown eyes?\nb) What is the probability that someone has black hair?\nc) What is the probability that someone has brown eyes?\nd) Are having black hair and having brown eyes dependent or independent\nevents?\n2. Given the following contingency table, identify the events and determine\nwhether they are dependent or independent.\nLocation A\nLocation B\nTotals\nBuses left late\n15\n40\n55\nBuses left on time\n25\n20\n45\nTotals\n40\n60\n100\n3. You are given the following information.\n• Events A and B are independent.\n• P(not A) = 0,3.\n• P(B) = 0,4.\nComplete the contingency table below.\nA\nnot A\nTotals\nB\nnot B\nTotals\n50\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BQ\n2. 23BR\n3. 23BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n434\n10.5.\nContingency tables\n\n10.6\nSummary\nEMBJY\nSee presentation: 23BT at www.everythingmaths.co.za\n• Terminology:\n– Outcome: a single observation of an experiment.\n– Sample space of an experiment: the set of all possible outcomes of the\nexperiment.\n– Event: a set of outcomes of an experiment.\n– Probability of an event: a real number between 0 and 1 that describes how\nlikely it is that the event will occur.\n– Relative frequency of an event: the number of times that the event occurs\nduring experimental trials, divided by the total number of trials conducted.\n– Union of events: the set of all outcomes that occur in at least one of the\nevents, written as “A or B”.\n– Intersection of events: the set of all outcomes that occur in all of the events,\nwritten as “A and B”.\n– Mutually exclusive events: events with no outcomes in common, that is\n(A and B) = ∅.\n– Complementary events: two mutually exclusive events that together con-\ntain all the outcomes in the sample space. We write the complement as\n“not A”.\n– Independent events: two events where knowing the outcome of one event\ndoes not affect the probability of the other event. Events are independent if\nand only if P(A and B) = P(A) × P(B).\n• Identities:\n– The addition rule: P(A or B) = P(A) + P(B) −P(A and B)\n– The addition rule for 2 mutually exclusive events: P(A or B) = P(A) +\nP(B)\n– The complementary rule: P(not A) = 1 −P(A)\n• A Venn diagram is a visual tool used to show how events overlap. Each region\nin a Venn diagram represents an event and could contain either the outcomes in\nthe event, the number of outcomes in the event or the probability of the event.\n• A tree diagram is a visual tool that helps with computing probabilities for depen-\ndent events. The outcomes of each event are shown along with the probability\nof each outcome. For each event that depends on a previous event, we go one\nlevel deeper into the tree. To compute the probability of some combination of\noutcomes, we\n– find all the paths that contain the outcome of interest;\n– multiply the probabilities along each path;\n– add the probabilities between different paths.\n• A 2-way contingency table is a tool for organising data, especially when we want\nto determine whether two events, each with only two outcomes, are dependent\nor independent. The counts for each possible combination of outcomes are\nentered into the table, along with the totals of each row and column.\n435\nChapter 10.\nProbability\n\nExercise 10 – 7: End of chapter exercises\n1. Jane invested in the stock market. The probability that she will not lose all her\nmoney is 0,32. What is the probability that she will lose all her money? Explain.\n2. If D and F are mutually exclusive events, with P(not D)\n=\n0,3 and\nP(D or F) = 0,94, find P(F).\n3. A car sales person has pink, lime-green and purple models of car A and purple,\norange and multicolour models of car B. One dark night a thief steals a car.\na) What is the experiment and sample space?\nb) What is the probability of stealing either a model of A or a model of B?\nc) What is the probability of stealing both a model of A and a model of B?\n4. The probability of event X is 0,43 and the probability of event Y is 0,24. The\nprobability of both occurring together is 0,10. What is the probability that X or\nY will occur?\n5. P(H) = 0,62; P(J) = 0,39 and P(H and J) = 0,31. Calculate:\na) P(H′)\nb) P(H or J)\nc) P(H′ or J′)\nd) P(H′ or J)\ne) P(H′ and J′)\n6. The last ten letters of the alphabet are placed in a hat and people are asked to\npick one of them. Event D is picking a vowel, event E is picking a consonant\nand event F is picking one of the last four letters. Draw a Venn diagram showing\nthe outcomes in the sample space and the different events. Then calculate the\nfollowing probabilities:\na) P(not F)\nb) P(F or D)\nc) P(neither E nor F)\nd) P(D and E)\ne) P(E and F)\nf) P(E and D′)\n7. Thobeka compares three neighbourhoods (we’ll call them A, B and C) to see\nwhere the best place is to live. She interviews 80 people and asks them whether\nthey like each of the neighbourhoods, or not.\n• 40 people like neighbourhood A.\n• 35 people like neighbourhood B.\n• 40 people like neighbourhood C.\n• 21 people like both neighbourhoods A and C.\n• 18 people like both neighbourhoods B and C.\n• 68 people like at least one neighbourhood.\n• 7 people like all three neighbourhoods.\n436\n10.6.\nSummary\n\na) Use this information to draw a Venn diagram.\nb) How many people like none of the neighbourhoods?\nc) How many people like neighbourhoods A and B, but not C?\nd) What is the probability that a randomly chosen person from the survey likes\nat least one of the neighbourhoods?\n8. Let G and H be two events in a sample space.\nSuppose that P(G) = 0,4;\nP(H) = h; and P(G or H) = 0,7.\na) For what value of h are G and H mutually exclusive?\nb) For what value of h are G and H independent?\n9. The following tree diagram represents points scored by two teams in a soccer\ngame. At each level in the tree, the points are shown as (points for Team 1;\npoints for Team 2).\n0,75\n(3; 0)\n0,25\n(2; 1)\n0,5\n(2; 1)\n0,5\n(1; 2)\n0,65\n(2; 0)\n0,35\n(1; 1)\n0,4\n(1; 1)\n0,6\n(0; 2)\n0,52\n(1; 0)\n0,48\n(0; 1)\n(0; 0)\nUse this diagram to determine the probability that:\na) Team 1 will win\nb) The game will be a draw\nc) The game will end with an even number of total points\n10. A bag contains 10 orange balls and 7 black balls. You draw 3 balls from the bag\nwithout replacement. What is the probability that you will end up with exactly\n2 orange balls? Represent this experiment using a tree diagram.\n11. Complete the following contingency table and determine whether the events are\ndependent or independent.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\nDid not like living there\n140\n340\nTotals\n230\n500\n12. Summarise the following information about a medical trial with 2 types of multi-\nvitamin in a contingency table and determine whether the events are dependent\nor independent.\n• 960 people took part in the medical trial.\n• 540 people used multivitamin A for a month and 400 of those people\nshowed an improvement in their health.\n437\nChapter 10.\nProbability\n\n• 300 people showed an improvement in health when using multivitamin B\nfor a month.\nIf the events are independent, it means that the two multivitamins have the same\neffect on people. If the events are dependent, it means that one multivitamin is\nbetter than the other. Which multivitamin is better than the other, or are the both\nequally effective?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BV\n2. 23BW\n3. 23BX\n4. 23BY\n5. 23BZ\n6. 23C2\n7. 23C3\n8. 23C4\n9. 23C5\n10. 23C6\n11. 23C7\n12. 23C8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n438\n10.6.\nSummary\n\nCHAPTER\n11\nStatistics\n11.1\nRevision\n440\n11.2\nHistograms\n444\n11.3\nOgives\n451\n11.4\nVariance and standard deviation\n455\n11.5\nSymmetric and skewed data\n461\n11.6\nIdentification of outliers\n464\n11.7\nSummary\n467\n\n11\nStatistics\n11.1\nRevision\nEMBJZ\nMeasures of central tendency\nEMBK2\nThe mean and median of a data set both give an indication where the centre of the\ndata distribution is located. The mean, or average, is calculated as\nx =\nPn\ni=1 xi\nn\nwhere the xi are the data and n is the number of data. We read x as “x bar”.\nThe median is the middle value of an ordered data set. To find the median, we first\nsort the data and then pick out the value in the middle of the sorted list. If the middle\nis in between two values, the median is the average of those two values.\nSee video: 23C9 at www.everythingmaths.co.za\nWorked example 1: Computing measures of central tendency\nQUESTION\nCompute the mean and median of the following data set:\n72,5 ; 92,6 ; 15,6 ; 53,0 ; 86,4 ; 89,9 ; 90,9 ; 21,7 ; 46,0 ; 4,1 ; 51,7 ; 2,2\nSOLUTION\nStep 1: Compute the mean\nUsing the formula for the mean, we first compute the sum of the values and then divide\nby the number of values.\nx = 626,6\n12\n≈52,22\nStep 2: Compute the median\nTo find the median, we first have to sort the data:\n2,2 ; 4,1 ; 15,6 ; 21,7 ; 46,0 ; 51,7 ; 53,0 ; 72,5 ; 86,4 ; 89,9 ; 90,9 ; 92,6\nSince there are an even number of values, the median will lie between two values.\nIn this case, the two values in the middle are 51,7 and 53,0. Therefore the median is\n52,35.\n440\n11.1.\nRevision\n\nMeasures of dispersion\nEMBK3\nMeasures of dispersion tell us how spread out a data set is. If a measure of dispersion\nis small, the data are clustered in a small region. If a measure of dispersion is large,\nthe data are spread out over a large region.\nThe range is the difference between the maximum and minimum values in the data\nset.\nThe inter-quartile range is the difference between the first and third quartiles of the\ndata set. The quartiles are computed in a similar way to the median. The median is\nhalfway into the ordered data set and is sometimes also called the second quartile.\nThe first quartile is one quarter of the way into the ordered data set; whereas the third\nquartile is three quarters of the way into the ordered data set.\nSee video: 23CB at www.everythingmaths.co.za\nWorked example 2: Range and inter-quartile range\nQUESTION\nDetermine the range and the inter-quartile range of the following data set.\n14 ; 17 ; 45 ; 20 ; 19 ; 36 ; 7 ; 30 ; 8\nSOLUTION\nStep 1: Sort the values in the data set\nTo determine the range we need to find the minimum and maximum values in the\ndata set. To determine the inter-quartile range we need to compute the first and third\nquartiles of the data set. For both of these requirements, it is easier to order the data\nset first.\nThe sorted data set is\n7 ; 8 ; 14 ; 17 ; 19 ; 20 ; 30 ; 36 ; 45\nStep 2: Find the minimum, maximum and range\nThe minimum value is the first value in the ordered data set, namely 7. The maximum\nis the last value in the ordered data set, namely 45. The range is the difference between\nthe minimum and maximum: 45 −7 = 38.\nStep 3: Find the quartiles and inter-quartile range\nThe diagram below shows how we find the quartiles one quarter, one half and three\nquarters of the way into the ordered list of values.\n441\nChapter 11.\nStatistics\n\n7\n8\n14\n17\n19\n20\n30\n36\n45\n0\n1\n4\n1\n2\n3\n4\n1\nFrom this diagram we can see that the first quartile is at a value of 14, the second\nquartile (median) is at a value of 19 and the third quartile is at a value of 30.\nThe inter-quartile range is the difference between the first and third quartiles. The\nfirst quartile is 14 and the third quartile is 30. Therefore the inter-quartile range is\n30 −14 = 16.\nFive number summary\nEMBK4\nThe five number summary combines a measure of central tendency, namely the me-\ndian, with measures of dispersion, namely the range and the inter-quartile range. This\ngives a good overview of the overall data distribution. More precisely, the five number\nsummary is written in the following order:\n• minimum;\n• first quartile;\n• median;\n• third quartile;\n• maximum.\nThe five number summary is often presented visually using a box and whisker diagram.\nA box and whisker diagram is shown below, with the positions of the five relevant\nnumbers labelled. Note that this diagram is drawn vertically, but that it may also be\ndrawn horizontally.\nmaximum\nupper quartile\nmedian\nlower quartile\nminimum\ninter-quartile range\ndata range\nSee video: 23CC at www.everythingmaths.co.za\n442\n11.1.\nRevision\n\nWorked example 3: Five number summary\nQUESTION\nDraw a box and whisker diagram for the following data set:\n1,25 ; 1,5 ; 2,5 ; 2,5 ; 3,1 ; 3,2 ; 4,1 ; 4,25 ; 4,75 ; 4,8 ; 4,95 ; 5,1\nSOLUTION\nStep 1: Determine the minimum and maximum\nSince the data set is already ordered, we can read off the minimum as the first value\n(1,25) and the maximum as the last value (5,1).\nStep 2: Determine the quartiles\nThere are 12 values in the data set.\n1,25 1,5\n2,5\n2,5\n3,1\n3,2\n4,1 4,25 4,75 4,8 4,95 5,1\n0\n1\n4\n1\n2\n3\n4\n1\nUsing the figure above we can see that the median is between the sixth and seventh\nvalues, making it.\n3,2 + 4,1\n2\n= 3,65\nThe first quartile lies between the third and fourth values, making it\nQ1 = 2,5 + 2,5\n2\n= 2,5\nThe third quartile lies between the ninth and tenth values, making it\nQ3 = 4,75 + 4,8\n2\n= 4,775\nStep 3: Draw the box and whisker diagram\nWe now have the five number summary as (1,25; 2,5; 3,65; 4,775; 5,1). The box and\nwhisker diagram representing the five number summary is given below.\n1,25\n2,5\n3,65\n4,775 5,1\n443\nChapter 11.\nStatistics\n\nExercise 11 – 1: Revision\n1. For each of the following data sets, compute the mean and all the quartiles.\nRound your answers to one decimal place.\na) −3,4 ; −3,1 ; −6,1 ; −1,5 ; −7,8 ; −3,4 ; −2,7 ; −6,2\nb) −6 ; −99 ; 90 ; 81 ; 13 ; −85 ; −60 ; 65 ; −49\nc) 7 ; 45 ; 11 ; 3 ; 9 ; 35 ; 31 ; 7 ; 16 ; 40 ; 12 ; 6\n2. Use the following box and whisker diagram to determine the range and inter-\nquartile range of the data.\n−5,52\n−2,41−1,53\n0,10\n4,08\n3. Draw the box and whisker diagram for the following data.\n0,2 ; −0,2 ; −2,7 ; 2,9 ; −0,2 ; −4,2 ; −1,8 ; 0,4 ; −1,7 ; −2,5 ; 2,7 ; 0,8 ; −0,5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23CD\n1b. 23CF\n1c. 23CG\n2. 23CH\n3. 23CJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.2\nHistograms\nEMBK5\nA histogram is a graphical representation of how many times different, mutually exclu-\nsive events are observed in an experiment. To interpret a histogram, we find the events\non the x-axis and the counts on the y-axis. Each event has a rectangle that shows what\nits count (or frequency) is.\nSee video: 23CK at www.everythingmaths.co.za\nWorked example 4: Reading histograms\nQUESTION\nUse the following histogram to determine the events that were recorded and the rela-\ntive frequency of each event. Summarise your answer in a table.\n444\n11.2.\nHistograms\n\n0\n2\n4\n6\n8\n10\nnot yet\nin school\nin primary\nschool\nin high\nschool\nSOLUTION\nStep 1: Determine the events\nThe events are shown on the x-axis. In this example we have “not yet in school”, “in\nprimary school” and “in high school”.\nStep 2: Read off the count for each event\nThe counts are shown on the y-axis and the height of each rectangle shows the fre-\nquency for each event.\n• not yet in school: 2\n• in primary school: 5\n• in high school: 9\nStep 3: Calculate relative frequency\nThe relative frequency of an event in an experiment is the number of times that the\nevent occurred divided by the total number of times that the experiment was com-\npleted. In this example we add up the frequencies for all the events to get a total\nfrequency of 16. Therefore the relative frequencies are:\n• not yet in school:\n2\n16 = 1\n8\n• in primary school:\n5\n16\n• in high school:\n9\n16\nStep 4: Summarise\nEvent\nCount\nRelative frequency\nnot yet in school\n2\n1\n8\nin primary school\n5\n5\n16\nin high school\n9\n9\n16\n445\nChapter 11.\nStatistics\n\nTo draw a histogram of a data set containing numbers, the numbers first have to be\ngrouped.\nEach group is defined by an interval.\nWe then count how many times\nnumbers from each group appear in the data set and draw a histogram using the counts.\nWorked example 5: Draw a histogram\nQUESTION\nThe following data represent the heights of 16 adults in centimetres.\n162 ; 168 ; 177 ; 147 ; 189 ; 171 ; 173 ; 168\n178 ; 184 ; 165 ; 173 ; 179 ; 166 ; 168 ; 165\nDivide the data into 5 equal length intervals between 140 cm and 190 cm and draw a\nhistogram.\nSOLUTION\nStep 1: Determine intervals\nTo have 5 intervals of the same length between 140 and 190, we need and interval\nlength of 10. Therefore the intervals are (140; 150]; (150; 160]; (160; 170]; (170; 180];\nand (180; 190].\nStep 2: Count data\nThe following table summarises the number of data values in each of the intervals.\nInterval\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\n(180; 190]\nCount\n1\n0\n7\n6\n2\nStep 3: Draw the histogram\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\n446\n11.2.\nHistograms\n\nFrequency polygons\nEMBK6\nA frequency polygon is sometimes used to represent the same information as in a his-\ntogram. A frequency polygon is drawn by using line segments to connect the middle of\nthe top of each bar in the histogram. This means that the frequency polygon connects\nthe coordinates at the centre of each interval and the count in each interval.\nWorked example 6: Drawing a frequency polygon\nQUESTION\nUse the histogram from the previous example to draw a frequency polygon of the same\ndata.\nSOLUTION\nStep 1: Draw the histogram\nWe already know that the histogram looks like this:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\nStep 2: Connect the tops of the rectangles\nWhen we draw line segments between the tops of the rectangles in the histogram, we\nget the following picture:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\n447\nChapter 11.\nStatistics\n\nStep 3: Draw final frequency polygon\nFinally, we remove the histogram to show only the frequency polygon.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\nFrequency polygons are particularly useful for comparing two data sets. Comparing\ntwo histograms would be more difficult since we would have to draw the rectangles of\nthe two data sets on top of each other. Because frequency polygons are just lines, they\ndo not pose the same problem.\nWorked example 7: Drawing frequency polygons\nQUESTION\nHere is another data set of heights, this time of Grade 11 learners.\n132 ; 132 ; 156 ; 147 ; 162 ; 168 ; 152 ; 174\n141 ; 136 ; 161 ; 148 ; 140 ; 174 ; 174 ; 162\nDraw the frequency polygon for this data set using the same interval length as in the\nprevious example. Then compare the two frequency polygons on one graph to see the\ndifferences between the distributions.\nSOLUTION\nStep 1: Frequency table\nWe first create the table of counts for the new data set.\nInterval\n(130; 140]\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\nCount\n4\n3\n2\n4\n3\n448\n11.2.\nHistograms\n\nStep 2: Draw histogram and frequency polygon\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\nStep 3: Compare frequency polygons\nWe draw the two frequency polygons on the same axes. The red line indicates the\ndistribution over heights for adults and the blue line, for Grade 11 learners.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\n190\nFrom this plot we can easily see that the heights for Grade 11 learners are distributed\nmore towards the left (shorter) than adults. The learner heights also seem to be more\nevenly distributed between 130 and 180 cm, whereas the adult heights are mostly\nbetween 160 and 180 cm.\n449\nChapter 11.\nStatistics\n\nExercise 11 – 2: Histograms\n1. Use the histogram below to answer the following questions.\nThe histogram\nshows the number of people born around the world each year. The ticks on\nthe x-axis are located at the start of each year.\npeople (millions)\nyear\n79\n80\n81\n82\n83\n84\n85\n86\n87\n1994 1995 1996 1997 1998 1999 2000 2001\na) How many people were born between the beginning of 1994 and the be-\nginning of 1996?\nb) Is the number people in the world population increasing or decreasing?\n(Ignore the rate at which people are dying for this question.)\nc) How many more people were born in 1994 than in 1997?\n2. In a traffic survey, a random sample of 50 motorists were asked the distance (d)\nthey drove to work daily. The results of the survey are shown in the table below.\nDraw a histogram to represent the data.\nd\n0 < d ≤10\n10 < d ≤20\n20 < d ≤30\n30 < d ≤40\n40 < d ≤50\nf\n9\n19\n15\n5\n4\n3. Below is data for the prevalence of HIV in South Africa. HIV prevalence refers to\nthe percentage of people between the ages of 15 and 49 who are infected with\nHIV.\nyear\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nprevalence (%)\n17,7\n18,0\n18,1\n18,1\n18,1\n18,0\n17,9\n17,9\nDraw a frequency polygon of this data set.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CM\n2. 23CN\n3. 23CP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n450\n11.2.\nHistograms\n\n11.3\nOgives\nEMBK7\nCumulative histograms, also known as ogives, are graphs that can be used to deter-\nmine how many data values lie above or below a particular value in a data set. The\ncumulative frequency is calculated from a frequency table, by adding each frequency\nto the total of the frequencies of all data values before it in the data set. The last value\nfor the cumulative frequency will always be equal to the total number of data values,\nsince all frequencies will already have been added to the previous total.\nAn ogive is drawn by\n• plotting the beginning of the first interval at a y-value of zero;\n• plotting the end of every interval at the y-value equal to the cumulative count for\nthat interval; and\n• connecting the points on the plot with straight lines.\nIn this way, the end of the final interval will always be at the total number of data since\nwe will have added up across all intervals.\nWorked example 8: Cumulative frequencies and ogives\nQUESTION\nDetermine the cumulative frequencies of the following grouped data and complete the\ntable below. Use the table to draw an ogive of the data.\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n20 < n ≤30\n7\n30 < n ≤40\n12\n40 < n ≤50\n10\n50 < n ≤60\n6\nSOLUTION\nStep 1: Compute cumulative frequencies\nTo determine the cumulative frequency, we add up the frequencies going down the\ntable. The first cumulative frequency is just the same as the frequency, because we are\nadding it to zero. The final cumulative frequency is always equal to the sum of all the\nfrequencies. This gives the following table:\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n5\n20 < n ≤30\n7\n12\n30 < n ≤40\n12\n24\n40 < n ≤50\n10\n34\n50 < n ≤60\n6\n40\n451\nChapter 11.\nStatistics\n\nStep 2: Plot the ogive\nThe first coordinate in the plot always starts at a y-value of 0 because we always start\nfrom a count of zero. So, the first coordinate is at (10; 0) — at the beginning of the\nfirst interval. The second coordinate is at the end of the first interval (which is also the\nbeginning of the second interval) and at the first cumulative count, so (20; 5). The third\ncoordinate is at the end of the second interval and at the second cumulative count,\nnamely (30; 12), and so on.\nComputing all the coordinates and connecting them with straight lines gives the fol-\nlowing ogive.\nn\n0\n10\n20\n30\n40\n10\n20\n30\n40\n50\n60\n•\n•\n•\n•\n•\n•\nOgives do look similar to frequency polygons, which we saw earlier. The most impor-\ntant difference between them is that an ogive is a plot of cumulative values, whereas\na frequency polygon is a plot of the values themselves. So, to get from a frequency\npolygon to an ogive, we would add up the counts as we move from left to right in the\ngraph.\nOgives are useful for determining the median, percentiles and five number summary\nof data. Remember that the median is simply the value in the middle when we order\nthe data. A quartile is simply a quarter of the way from the beginning or the end of an\nordered data set. With an ogive we already know how many data values are above or\nbelow a certain point, so it is easy to find the middle or a quarter of the data set.\nWorked example 9: Ogives and the five number summary\nQUESTION\nUse the following ogive to compute the five number summary of the data. Remember\nthat the five number summary consists of the minimum, all the quartiles (including the\nmedian) and the maximum.\n452\n11.3.\nOgives\n\ncount\nvalue\n0\n10\n20\n30\n40\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\nSOLUTION\nStep 1: Find the minimum and maximum\nThe minimum value in the data set is 1 since this is where the ogive starts on the\nhorizontal axis. The maximum value in the data set is 10 since this is where the ogive\nstops on the horizontal axis.\nStep 2: Find the quartiles\nThe quartiles are the values that are 1\n4, 1\n2 and 3\n4 of the way into the ordered data set.\nHere the counts go up to 40, so we can find the quartiles by looking at the values\ncorresponding to counts of 10, 20 and 30. On the ogive a count of\n• 10 corresponds to a value of 3 (first quartile);\n• 20 corresponds to a value of 7 (second quartile); and\n• 30 corresponds to a value of 8 (third quartile).\nStep 3: Write down the five number summary\nThe five number summary is (1; 3; 7; 8; 10). The box-and-whisker plot of this data set\nis given below.\n1\n3\n7\n8\n10\n453\nChapter 11.\nStatistics\n\nExercise 11 – 3: Ogives\n1. Use the ogive to answer the questions below. Note that marks are given as a\npercentage.\nnumber of students\nmark\n0\n10\n20\n30\n40\n50\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n•\n•\n•\n•\n•\n•\n•\n•\n•\na) How many students got between 50% and 70%?\nb) How many students got at least 70%?\nc) Compute the average mark for this class, rounded to the nearest integer.\n2. Draw the histogram corresponding to this ogive.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n−25\n−15\n−5\n5\n15\n25\n•\n•\n•\n•\n•\n•\n3. The following data set lists the ages of 24 people.\n2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53; 38\n4; 28; 5; 73; 79; 17; 15; 5; 34; 37; 45; 56\nUse the data to answer the following questions.\na) Using an interval width of 8 construct a cumulative frequency plot.\nb) How many are below 30?\nc) How many are below 60?\nd) Giving an explanation state below what value the bottom 50% of the ages\nfall.\ne) Below what value do the bottom 40% fall?\nf) Construct a frequency polygon.\n454\n11.3.\nOgives\n\n4. The weights of bags of sand in grams is given below (rounded to the nearest\ntenth):\n50,1; 40,4; 48,5; 29,4; 50,2; 55,3; 58,1; 35,3; 54,2; 43,5\n60,1; 43,9; 45,3; 49,2; 36,6; 31,5; 63,1; 49,3; 43,4; 54,1\na) Decide on an interval width and state what you observe about your choice.\nb) Give your lowest interval.\nc) Give your highest interval.\nd) Construct a cumulative frequency graph and a frequency polygon.\ne) Below what value do 53% of the cases fall?\nf) Below what value of 60% of the cases fall?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CQ\n2. 23CR\n3. 23CS\n4. 23CT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.4\nVariance and standard deviation\nEMBK8\nMeasures of central tendency (mean, median and mode) provide information on the\ndata values at the centre of the data set. Measures of dispersion (quartiles, percentiles,\nranges) provide information on the spread of the data around the centre. In this section\nwe will look at two more measures of dispersion called the variance and the standard\ndeviation.\nSee video: 23CV at www.everythingmaths.co.za\nVariance\nEMBK9\nDEFINITION: Variance\nLet a population consist of n elements, {x1; x2; . . . ; xn}. Write the mean of the data as\nx.\nThe variance of the data is the average squared distance between the mean and each\ndata value.\nσ2 =\nPn\ni=1 (xi −x)2\nn\nNOTE:\nThe variance is written as σ2. It might seem strange that it is written in squared form,\nbut you will see why soon when we discuss the standard deviation.\n455\nChapter 11.\nStatistics\n\nThe variance has the following properties.\n• It is never negative since every term in the variance sum is squared and therefore\neither positive or zero.\n• It has squared units. For example, the variance of a set of heights measured in\ncentimetres will be given in centimeters squared. Since the population variance\nis squared, it is not directly comparable with the mean or the data themselves. In\nthe next section we will describe a different measure of dispersion, the standard\ndeviation, which has the same units as the data.\nWorked example 10: Variance\nQUESTION\nYou flip a coin 100 times and it lands on heads 44 times. You then use the same\ncoin and do another 100 flips. This time in lands on heads 49 times. You repeat this\nexperiment a total of 10 times and get the following results for the number of heads.\n{44; 49; 52; 62; 53; 48; 54; 49; 46; 51}\nCompute the mean and variance of this data set.\nSOLUTION\nStep 1: Compute the mean\nThe formula for the mean is\nx =\nPn\ni=1 xi\nn\nIn this case, we sum the data and divide by 10 to get x = 50,8.\nStep 2: Compute the variance\nThe formula for the variance is\nσ2 =\nPn\ni=1 (xi −x)2\nn\nWe first subtract the mean from each datum and then square the result.\nxi\n44\n49\n52\n62\n53\n48\n54\n49\n46\n51\nxi −x\n−6,8\n−1,8\n1,2\n11,2\n2,2\n−2,8\n3,2\n−1,8\n−4,8\n0,2\n(xi −x)2\n46,24\n3,24\n1,44\n125,44 4,84\n7,84\n10,24\n3,24\n23,04\n0,04\nThe variance is the sum of the last row in this table divided by 10, so σ2 = 22,56.\n456\n11.4.\nVariance and standard deviation\n\nStandard deviation\nEMBKB\nSince the variance is a squared quantity, it cannot be directly compared to the data val-\nues or the mean value of a data set. It is therefore more useful to have a quantity which\nis the square root of the variance. This quantity is known as the standard deviation.\nDEFINITION: Standard deviation\nLet a population consist of n elements, {x1; x2; . . . ; xn}, with a mean of x. The stan-\ndard deviation of the data is\nσ =\nsPn\ni=1 (xi −x)2\nn\nIn statistics, the standard deviation is a very common measure of dispersion. Standard\ndeviation measures how spread out the values in a data set are around the mean. More\nprecisely, it is a measure of the average distance between the values of the data in the\nset and the mean. If the data values are all similar, then the standard deviation will be\nlow (closer to zero). If the data values are highly variable, then the standard variation\nis high (further from zero).\nThe standard deviation is always a positive number and is always measured in the\nsame units as the original data. For example, if the data are distance measurements in\nkilogrammes, the standard deviation will also be measured in kilogrammes.\nThe mean and the standard deviation of a set of data are usually reported together. In\na certain sense, the standard deviation is a natural measure of dispersion if the centre\nof the data is taken as the mean.\nInvestigation: Tabulating results\nIt is often useful to set your data out in a table so that you can apply the for-\nmulae easily.\nComplete the table below to calculate the standard deviation of\n{57; 53; 58; 65; 48; 50; 66; 51}.\n• Firstly, remember to calculate the mean, x.\n• Complete the following table.\nindex: i\ndatum: xi\ndeviation: xi −x\ndeviation\nsquared: (xi −x)2\n1\n57\n2\n53\n3\n58\n4\n65\n5\n48\n6\n50\n7\n66\n8\n51\nP xi = . . .\nP(xi −x) = . . .\nP(xi −x)2 = . . .\n• The sum of the deviations is always zero. Why is this? Find out.\n• Calculate the variance using the completed table.\n• Then calculate the standard deviation.\n457\nChapter 11.\nStatistics\n\nWorked example 11: Variance and standard deviation\nQUESTION\nWhat is the variance and standard deviation of the possibilities associated with rolling\na fair die?\nSOLUTION\nStep 1: Determine all the possible outcomes\nWhen rolling a fair die, the sample space consists of 6 outcomes. The data set is\ntherefore x = {1; 2; 3; 4; 5; 6} and n = 6.\nStep 2: Calculate the mean\nThe mean is:\nx = 1\n6 (1 + 2 + 3 + 4 + 5 + 6)\n= 3,5\nStep 3: Calculate the variance\nThe variance is:\nσ2 =\nP (x −x)2\nn\n= 1\n6 (6,25 + 2,25 + 0,25 + 0,25 + 2,25 + 6,25)\n= 2,917\nStep 4: Calculate the standard deviation\nThe standard deviation is:\nσ =\np\n2,917\n= 1,708\nSee video: 23CW at www.everythingmaths.co.za\n458\n11.4.\nVariance and standard deviation\n\nInterpretation and application\nEMBKC\nA large standard deviation indicates that the data values are far from the mean and a\nsmall standard deviation indicates that they are clustered closely around the mean.\nFor example, consider the following three data sets:\n{65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\n{85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\n{43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nEach of these data sets has the same mean, namely 67. However, they have different\nstandard deviations, namely 8,97, 17,75 and 21,23. The following figures show plots\nof the data sets with the mean and standard deviation indicated on each. You can see\nhow the standard deviation is larger when the data are more spread out.\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 8,97\ndata:\n{xi} = {65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\nmean:\nx = 67\nstandard deviation:\nσ ≈8,97\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 17,75\ndata:\n{xi} = {85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\nmean:\nx = 67\nstandard deviation:\nσ ≈17,75\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 21,23\ndata:\n{xi} = {43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nmean:\nx = 67\nstandard deviation:\nσ ≈21,23\nThe standard deviation may also be thought of as a measure of uncertainty. In the phys-\nical sciences, for example, the reported standard deviation of a group of repeated mea-\nsurements represents the precision of those measurements. When deciding whether\n459\nChapter 11.\nStatistics\n\nmeasurements agree with a theoretical prediction, the standard deviation of those mea-\nsurements is very important: if the mean of the measurements is too far away from the\nprediction (with the distance measured in standard deviations), then we consider the\nmeasurements as contradicting the prediction. This makes sense since they fall outside\nthe range of values that could reasonably be expected to occur if the prediction were\ncorrect.\nExercise 11 – 4: Variance and standard deviation\n1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban.\nThe data, in rands per litre, are given below.\nCape Town\n3,96\n3,76\n4,00\n3,91\n3,69\n3,72\nDurban\n3,97\n3,81\n3,52\n4,08\n3,88\n3,68\na) Find the mean price in each city and then state which city has the lowest\nmean.\nb) Find the standard deviation of each city’s prices.\nc) Which city has the more consistently priced petrol? Give reasons for your\nanswer.\n2. Compute the mean and variance of the following set of values.\n150 ; 300 ; 250 ; 270 ; 130 ; 80 ; 700 ; 500 ; 200 ; 220 ; 110 ; 320 ; 420 ; 140\n3. Compute the mean and variance of the following set of values.\n−6,9 ; −17,3 ; 18,1 ; 1,5 ; 8,1 ; 9,6 ; −13,1 ; −14,0 ; 10,5 ; −14,8 ; −6,5 ; 1,4\n4. The times for 8 athletes who ran a 100 m sprint on the same track are shown\nbelow. All times are in seconds.\n10,2 ; 10,8 ; 10,9 ; 10,3 ; 10,2 ; 10,4 ; 10,1 ; 10,4\na) Calculate the mean time.\nb) Calculate the standard deviation for the data.\nc) How many of the athletes’ times are more than one standard deviation away\nfrom the mean?\n5. The following data set has a mean of 14,7 and a variance of 10,01.\n18 ; 11 ; 12 ; a ; 16 ; 11 ; 19 ; 14 ; b ; 13\nCompute the values of a and b.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CX\n2. 23CY\n3. 23CZ\n4. 23D2\n5. 23D3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n460\n11.4.\nVariance and standard deviation\n\n11.5\nSymmetric and skewed data\nEMBKD\nWe are now going to classify data sets into 3 categories that describe the shape of the\ndata distribution: symmetric, left skewed, right skewed. We can use this classification\nfor any data set, but here we will look only at distributions with one peak. Most of\nthe data distributions that you have seen so far have only one peak, so the plots in this\nsection should look familiar.\nDistributions with one peak are called unimodal distributions.\nUnimodal literally\nmeans having one mode. (Remember that a mode is a maximum in the distribution.)\nSymmetric distributions\nEMBKF\nA symmetric distribution is one where the left and right hand sides of the distribution\nare roughly equally balanced around the mean. The histogram below shows a typical\nsymmetric distribution.\nmean ≈median\nbalanced left and right tails\nFor symmetric distributions, the mean is approximately equal to the median. The tails\nof the distribution are the parts to the left and to the right, away from the mean. The\ntail is the part where the counts in the histogram become smaller. For a symmetric\ndistribution, the left and right tails are equally balanced, meaning that they have about\nthe same length.\nThe figure below shows the box and whisker diagram for a typical symmetric data set.\nmedian halfway\nbetween\nfirst and third quartiles\nAnother property of a symmetric distribution is that its median (second quartile) lies\nin the middle of its first and third quartiles. Note that the whiskers of the plot (the\nminimum and maximum) do not have to be equally far away from the median. In the\nnext section on outliers, you will see that the minimum and maximum values do not\nnecessarily match the rest of the data distribution well.\n461\nChapter 11.\nStatistics\n\nSkewed\nEMBKG\nA distribution that is skewed right (also known as positively skewed) is shown below.\nmean\nmedian\nmean > median\nlong right tail\nshort left tail\nNow the picture is not symmetric around the mean anymore.\nFor a right skewed\ndistribution, the mean is typically greater than the median. Also notice that the tail of\nthe distribution on the right hand (positive) side is longer than on the left hand side.\nmedian closer to first quartile\nFrom the box and whisker diagram we can also see that the median is closer to the first\nquartile than the third quartile. The fact that the right hand side tail of the distribution\nis longer than the left can also be seen.\nA distribution that is skewed left has exactly the opposite characteristics of one that is\nskewed right:\n• the mean is typically less than the median;\n• the tail of the distribution is longer on the left hand side than on the right hand\nside; and\n• the median is closer to the third quartile than to the first quartile.\nThe table below summarises the different categories visually.\nSymmetric\nSkewed right (positive)\nSkewed left (negative)\n462\n11.5.\nSymmetric and skewed data\n\nExercise 11 – 5: Symmetric and skewed data\n1. Is the following data set symmetric, skewed right or skewed left? Motivate your\nanswer.\n27 ; 28 ; 30 ; 32 ; 34 ; 38 ; 41 ; 42 ; 43 ; 44 ; 46 ; 53 ; 56 ; 62\n2. State whether each of the following data sets are symmetric, skewed right or\nskewed left.\na) A data set with this histogram:\nb) A data set with this box and whisker plot:\nc) A data set with this frequency polygon:\n• • • • • • • • •\n•\n•\n•\n• • • •\nd) The following data set:\n11,2 ; 5 ; 9,4 ; 14,9 ; 4,4 ; 18,8 ; −0,4 ; 10,5 ; 8,3 ; 17,8\n3. Two data sets have the same range and interquartile range, but one is skewed\nright and the other is skewed left. Sketch the box and whisker plot for each of\nthese data sets. Then, invent data (6 points in each data set) that matches the\ndescriptions of the two data sets.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23D4\n2a. 23D5\n2b. 23D6\n2c. 23D7\n2d. 23D8\n3. 23D9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n463\nChapter 11.\nStatistics\n\n11.6\nIdentification of outliers\nEMBKH\nAn outlier in a data set is a value that is far away from the rest of the values in the\ndata set. In a box and whisker diagram, outliers are usually close to the whiskers of\nthe diagram. This is because the centre of the diagram represents the data between\nthe first and third quartiles, which is where 50% of the data lie, while the whiskers\nrepresent the extremes — the minimum and maximum — of the data.\nWorked example 12: Identifying outliers\nQUESTION\nFind the outliers in the following data set by drawing a box and whisker diagram and\nlocating the data values on the diagram.\n0,5 ; 1 ; 1,1 ; 1,4 ; 2,4 ; 2,8 ; 3,5 ; 5,1 ; 5,2 ; 6 ; 6,5 ; 9,5\nSOLUTION\nStep 1: Determine the five number summary\nThe minimum of the data set is 0,5. The maximum of the data set is 9,5. Since there\nare 12 values in the data set, the median lies between the sixth and seventh values,\nmaking it equal to 2,8+3,5\n2\n= 3,15. The first quartile lies between the third and fourth\nvalues, making it equal to 1,1+1,4\n2\n= 1,25. The third quartile lies between the ninth\nand tenth values, making it equal to 5,2+6\n2\n= 5,6.\nStep 2: Draw the box and whisker diagram\n0,5 1,25\n3,15\n5,6\n9,5\n• •• •\n• •\n•\n••\n• •\n•\nIn the figure above, each value in the data set is shown with a black dot.\nStep 3: Find the outliers\nFrom the diagram we can see that most of the values are between 1 and 6. The only\nvalue that is very far away from this range is the maximum at 9,5. Therefore 9,5 is the\nonly outlier in the data set.\nYou should also be able to identify outliers in plots of two variables. A scatter plot\nis a graph that shows the relationship between two random variables. We call these\ndata bivariate (literally meaning two variables) and we plot the data for two different\nvariables on one set of axes. The following example shows what a typical scatter plot\nlooks like. For Grade 11 you do not need to learn how to draw these 2-dimensional\n464\n11.6.\nIdentification of outliers\n\nscatter plots, but you should be able to identify outliers on them. As before, an outlier\nis a value that is far removed from the main distribution of data.\nWorked example 13: Scatter plot\nQUESTION\nWe have a data set that relates the heights and weights of a number of people. The\nheight is the first variable and its value is plotted along the horizontal axis. The weight\nis the second variable and its value is plotted along the vertical axis. The data values\nare shown on the plot below. Identify any outliers on the scatter plot.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\nSOLUTION\nWe inspect the plot visually and notice that there are two points that lie far away from\nthe main data distribution. These two points are circled in the plot below.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\n465\nChapter 11.\nStatistics\n\nExercise 11 – 6: Outliers\n1. For each of the following data sets, draw a box and whisker diagram and deter-\nmine whether there are any outliers in the data.\na) 30 ; 21,4 ; 39,4 ; 33,4 ; 21,1 ; 29,3 ; 32,8 ; 31,6 ; 36 ;\n27,9 ; 27,3 ; 29,4 ; 29,1 ; 38,6 ; 33,8 ; 29,1 ; 37,1\nb) 198 ; 166 ; 175 ; 147 ; 125 ; 194 ; 119 ; 170 ; 142 ; 148\nc) 7,1 ; 9,6 ; 6,3 ; −5,9 ; 0,7 ; −0,1 ; 4,4 ; −11,7 ; 10 ; 2,3 ; −3,7 ; 5,8 ; −1,4\n; 1,7 ; −0,7\n2. A class’s results for a test were recorded along with the amount of time spent\nstudying for it. The results are given below. Identify any outliers in the data.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23DB\n1b. 23DC\n1c. 23DD\n2. 23DF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n466\n11.6.\nIdentification of outliers\n\n11.7\nSummary\nEMBKJ\nSee presentation: 23DG at www.everythingmaths.co.za\n• Histograms visualise how many times different events occurred. Each rectangle\nin a histogram represents one event and the height of the rectangle is relative to\nthe number of times that the event occurred.\n• Frequency polygons represent the same information as histograms, but using\nlines and points rather than rectangles. A frequency polygon connects the mid-\ndle of the top edge of each rectangle in a histogram.\n• Ogives (also known as cumulative histograms) show the total number of times\nthat a value or anything less than that value appears in the data set. To draw an\nogive you need to add up all the counts in a histogram from left to right.\n– The first count in an ogive is always zero.\n– The last count in an ogive is always the sum of all the counts in the data\nset.\n• The variance and standard deviation are measures of dispersion.\n– The standard deviation is the square root of the variance.\n– Variance: σ2 = 1\nn\nPn\ni=1(xi −x)2\n– Standard deviation: σ =\nq\n1\nn\nPn\ni=1(xi −x)2\n– The standard deviation is measured in the same units as the mean and the\ndata, but the variance is not. The variance is measured in the square of the\ndata units.\n• In a symmetric distribution\n– the mean is approximately equal to the median; and\n– the tails of the distribution are balanced.\n• In a right (positively) skewed distribution\n– the mean is greater than the median;\n– the tail on the right hand side is longer than the tail on the left hand side;\nand\n– the median is closer to the first quartile than the third quartile.\n• In a left (negatively) skewed distribution\n– the mean is less than the median;\n– the tail on the left hand side is longer than the tail on the right hand side;\nand\n– the median is closer to the third quartile than the first quartile.\n• An outlier is a value that is far away from the rest of the data.\n467\nChapter 11.\nStatistics\n\nExercise 11 – 7: End of chapter exercises\n1. Draw a histogram, frequency polygon and ogive of the following data set. To\ncount the data, use intervals with a width of 1, starting from 0.\n0,4 ; 3,1 ; 1,1 ; 2,8 ; 1,5 ; 1,3 ; 2,8 ; 3,1 ; 1,8 ; 1,3 ;\n2,6 ; 3,7 ; 3,3 ; 5,7 ; 3,7 ; 7,4 ; 4,6 ; 2,4 ; 3,5 ; 5,3\n2. Draw a box and whisker diagram of the following data set and explain whether\nit is symmetric, skewed right or skewed left.\n−4,1 ; −1,1 ; −1 ; −1,2 ; −1,5 ; −3,2 ; −4 ; −1,9 ; −4 ;\n−0,8 ; −3,3 ; −4,5 ; −2,5 ; −4,4 ; −4,6 ; −4,4 ; −3,3\n3. Eight children’s sweet consumption and sleeping habits were recorded. The data\nare given in the following table and scatter plot.\nNumber of sweets\nper week\n15\n12\n5\n3\n18\n23\n11\n4\nAverage sleeping\ntime (hours per day)\n4\n4,5\n8\n8,5\n3\n2\n5\n8\n5\n10\n15\n20\n25\nnumber of sweets\n1\n2\n3\n4\n5\n6\n7\n8\n9\nsleeping time (hours per day)\na) What is the mean and standard deviation of the number of sweets eaten per\nday?\nb) What is the mean and standard deviation of the number of hours slept per\nday?\nc) Make a list of all the outliers in the data set.\n4. The monthly incomes of eight teachers are as follows:\nR 10 050;\nR 14 300;\nR 9800;\nR 15 000;\nR 12 140;\nR 13 800;\nR 11 990;\nR 12 900.\na) What is the mean and standard deviation of their incomes?\nb) How many of the salaries are less than one standard deviation away from\nthe mean?\nc) If each teacher gets a bonus of R 500 added to their pay what is the new\nmean and standard deviation?\nd) If each teacher gets a bonus of 10% on their salary what is the new mean\nand standard deviation?\ne) Determine for both of the above, how many salaries are less than one stan-\ndard deviation away from the mean.\n468\n11.7.\nSummary\n\nf) Using the above information work out which bonus is more beneficial fi-\nnancially for the teachers.\n5. The weights of a random sample of boys in Grade 11 were recorded. The cumu-\nlative frequency graph (ogive) below represents the recorded weights.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100 110 120\n0\n10\n20\n30\n40\n50\n60\nWeight (in kilogrammes)\nCumulative frequency\nCumulative frequency curve showing weight of boys\na) How many of the boys weighed between 90 and 100 kilogrammes?\nb) Estimate the median weight of the boys.\nc) If there were 250 boys in Grade 11, estimate how many of them would\nweigh less than 80 kilogrammes?\n6. Three sets of 12 learners each had their test scores recorded. The test was out of\n50. Use the given data to answer the following questions.\nSet A\nSet B\nSet C\n25\n32\n43\n47\n34\n47\n15\n35\n16\n17\n32\n43\n16\n25\n38\n26\n16\n44\n24\n38\n42\n27\n47\n50\n22\n43\n50\n24\n29\n44\n12\n18\n43\n31\n25\n42\na) For each of the sets calculate the mean and the five number summary.\nb) Make box and whisker plots of the three data sets on the same set of axes.\nc) State, with reasons, whether each of the three data sets are symmetric or\nskewed (either right or left).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DH\n2. 23DJ\n3. 23DK\n4. 23DM\n5. 23DN\n6. 23DP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n469\nChapter 11.\nStatistics\n\n\nCHAPTER\n12\nLinear programming\n12.1\nIntroduction\n472\n\n12\nLinear programming\n12.1\nIntroduction\nEMBKK\nIn everyday life people are interested in knowing the most efficient way of carrying out\na task or achieving a goal. For example, a farmer wants to know how many hectares to\nplant during a season in order to maximise the yield (produce), a stock broker wants to\nknow how much to invest in stocks in order to maximise profit, an entrepreneur wants\nto know how many people to employ to minimise expenditure. These are optimisation\nproblems; we want to to determine either the maximum or the minimum in a specific\nsituation.\nTo describe this mathematically, we assign variables to represent the different factors\nthat influence the situation. Optimisation means finding the combination of variables\nthat gives the best result.\nSee video: 23DQ at www.everythingmaths.co.za\nWorked example 1: Mountees and Roadees\nQUESTION\nInvestigate the following situation and use your knowledge of mathematics to solve the\nproblem:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make the maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nStep 2: Organise the information given\nWrite down a summary of the information given in the problem so that we consider\n472\n12.1.\nIntroduction\n\nall the different components in the situation.\nmaximum number for M\n= 5\nmaximum number for R\n= 3\nnumber of technicians needed for M = 1\nnumber of technicians needed for R = 2\ntotal number of technicians\n= 8\nprofit per M\n= 800\nprofit per R\n= 2400\nStep 3: Draw up a table\nUse the summary to draw up a table of all the possible combinations of the number of\nMountees and Roadees that can be manufactured per day:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n(4; 3)\n5\n(5; 0)\n(5; 1)\n(5; 2)\n(5; 3)\nNote that there are 24 possible combinations.\nStep 4: Consider the limitation of the number of technicians\nIt takes 1 technician to assemble a Mountee and 2 technicians to assemble a Roadee.\nThere are a total of 8 technicians in the assembly department, therefore we can write\nthat 1(M) + 2(R) ≤8.\nWith this limitation, we are able to eliminate some of the combinations in the table\nwhere M + 2R > 8:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n\b\b\b\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n\b\b\b\n(4; 3)\n5\n(5; 0)\n(5; 1)\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nThese combinations have been excluded as possible answers. For example, (5; 3) gives\n5 + 2(3) = 11 technicians.\n473\nChapter 12.\nLinear programming\n\nStep 5: Consider the profit on the bicycles\nWe can express the profit (P) per day as: P = 800(M)+2400(R). Notice that a higher\nprofit is made on a Roadee.\nBy substituting the different combinations for M and R, we can find the values that\ngive the maximum profit:\nFor (5; 0)\nP = 800(5) + 2400(0)\n= R 4000\nFor (3; 1)\nP = 800(3) + 2400(1)\n= R 4800\nM\nR\n0\n1\n2\n3\n0\n(0; 0) ⇒R 0\n(0; 1) ⇒R 2400\n(0; 2) ⇒R 4800\n(0; 3) ⇒R 7200\n1\n(1; 0) ⇒R 800\n(1; 1) ⇒R 3200\n(1; 2) ⇒R 5600\n(1; 3) ⇒R 8000\n2\n(2; 0) ⇒R 1600\n(2; 1) ⇒R 4000\n(2; 2) ⇒R 6400\n(2; 3) ⇒R 8800\n3\n(3; 0) ⇒R 2400\n(3; 1) ⇒R 4800\n(3; 2) ⇒R 7200\n\b\b\b\n(3; 3)\n4\n(4; 0) ⇒R 3200\n(4; 1) ⇒R 5600\n(4; 2) ⇒R 8000\n\b\b\b\n(4; 3)\n5\n(5; 0) ⇒R 4000\n(5; 1) ⇒R 6400\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nStep 6: Write the final answer\nTherefore the maximum profit of R 8800 is obtained if 2 Mountees and 3 Roadees are\nmanufactured per day.\nExercise 12 – 1: Optimisation\n1. Furniture store opening special:\nAs part of their opening special, a furniture store has promised to give away at\nleast 40 prizes with a total value of at least R 4000. They intend to give away\nkettles and toasters. They decide there will be at least 10 units of each prize. A\nkettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the\ncompany. Calculate how much this combination of kettles and toasters will cost.\nUse a suitable strategy to organise the information and solve the problem.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n474\n12.1.\nIntroduction\n\nOptimisation using graphs\nA more efficient way to solve optimisation problems is using graphs.\nWe write the limitations in the situation, called constraints, as inequalities. Some con-\nstraints can be modelled by an equation, which needs to be maximised or minimized.\nWe sketch the inequalities and indicate the region above or below the line that is to be\nconsidered in determining the solution. This method of solving optimisation problems\nis called linear programming.\nSee video: 23DS at www.everythingmaths.co.za\nWorked example 2: Optimisation using graphs\nQUESTION\nConsider again the example of Mr. Hunter who manufactures Mountees and Roadees:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make a maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nNotice that the values of M and R are limited to positive integers; Mr. Hunter cannot\nsell negative numbers of bikes nor can he sell a fraction of a bike.\nStep 2: Organise the information\nWe can write these constraints as inequalities:\nnumber of Mountees: 0 ≤M ≤5\nnumber of Roadees: 0 ≤R ≤3\ntotal number of technicians: M + 2R ≤8\nWe also know that P = 800M + 2400R. This is called the objective function, some-\ntimes also referred to as the search line, because the objective (goal) is to determine\nthe maximum value of P.\n475\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nWe represent the number of Mountees manufactured daily on the horizontal axis and\nthe number of Roadees manufactured daily on the vertical axis. Since M and R are\npositive integers, we only use the first quadrant of the Cartesian plane. Note that the\ngraph only includes the integer values of M between 0 and 5 and R between 0 and 3.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nFor the number of technicians in the assembly department M + 2R ≤8. If we make\nR (represented on the y-axis) the subject of the inequality we get R ≤−1\n2M + 4.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR = −1\n2M + 4\nThe arrows indicate the region in which the solution will lie, where R ≤−1\n2M + 4.\nThis area is called the feasible region.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR ≤−1\n2M + 4\nA\n476\n12.1.\nIntroduction\n\nWe substitute the possible combinations into the profit equation P = 800M + 2400R,\nand find the combination that gives the maximum profit.\nAt A(2; 3) :\nP = 800(2) + 2400(3)\n= R 8800\nStep 4: Write the final answer\nTherefore the maximum profit is obtained if 2 Mountees and 3 Roadees are manufac-\ntured per day.\nSee video: 23DT at www.everythingmaths.co.za\nWorked example 3: Optimisation using graphs\nQUESTION\nSolve the “furniture store opening special” problem using graphs:\nAs part of their opening special, a furniture store has promised to give away at least 40\nprizes. They intend to give away kettles and toasters. They decide there will be at least\n10 units of each prize. A kettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the company.\nCalculate how much this combination of kettles and toasters will cost.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nkettles be k and the number of toasters be t, with k, t ∈Z.\nStep 2: Organise the information\nWe can write the given information as inequalities:\nnumber of kettles: k ≥10\nnumber of toasters: t ≥10\ntotal number of prizes: k + t ≥40\nWe make t the subject of the inequality:\nt ≥−k + 40\n477\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nRepresent the constraints on a set of axes:\nKettles (k)\nToasters (t)\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nt ≥−k + 40\nt ≥10\nk ≥10\nWe shade the feasible region as shown in the diagram. Remember that in this situation\nonly the points with integer coordinates inside or on the border of the feasible region\nare possible solutions. The combination giving the minimum cost will lie towards or\non the lower border of the feasible region, which gives us many points to consider. To\nfind the optimum value of C, we use the graph of the objective function\nC = 120k + 100t\nTo draw the line, we make t the subject of the formula\nt = −6\n5k + C\n100\nWe see that the gradient of the objective function is −6\n5, but we do not know the exact\nvalue of the t-intercept ( C\n100). To find the minimum value of C, we need to determine\nthe position of the objective function where it first touches the feasible region and also\ngives the lowest t-intercept.\nKettles (k)\nToasters (t)\nA\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nWe indicate the gradient of the objective function on the graph (the green search\nline). Keeping the gradient the same, we “slide” the objective function towards the\nlower border of the feasible region and find that it touches the feasible region at point\n478\n12.1.\nIntroduction\n\nA(10; 30). This optimum position of the objective function is indicated on the graph\nby the dotted line passing through point A.\nWe substitute the coordinates of A into the cost equation C = 120k + 100t:\nAt A(10; 30) :\nC = 120(10) + 100(30)\n= R 4200\nThe minimum cost can also be determined graphically by reading off the coordinates\nof the t-intercept of the objective function in the optimum position:\ntint = 42\n∴C\n100 = 42\n∴C = R 4200\nStep 4: Write the final answer\nTherefore the minimum cost to the company is R 4200 with 10 kettles and 30 toast-\ners.\nExercise 12 – 2: Optimisation\n1. You are given a test consisting of two sections. The first section is on algebra and\nthe second section is on geometry. You are not allowed to answer more than 10\nquestions from any section, but you have to answer at least 4 algebra questions.\nThe time allowed is not more than 30 minutes. An algebra problem will take 2\nminutes and a geometry problem will take 3 minutes to solve.\nLet x be the number of algebra questions and y be the number of geometry\nquestions.\na) Formulate the equations and inequalities that satisfy the above constraints.\nb) The algebra questions carry 5 marks each and the geometry questions carry\n10 marks each. If T is the total marks, write down an expression for T.\n2. A local clinic wants to produce a guide to healthy living. The clinic intends to\nproduce the guide in two formats: a short video and a printed book. The clinic\nneeds to decide how many of each format to produce for sale. Estimates show\nthat no more than 10 000 copies of both items together will be sold. At least\n4000 copies of the video and at least 2000 copies of the book could be sold,\nalthough sales of the book are not expected to exceed 4000 copies. Let x be the\nnumber of videos sold, and y the number of printed books sold.\na) Write down the constraint inequalities that can be deduced from the given\ninformation.\nb) Represent these inequalities graphically and indicate the feasible region\nclearly.\n479\nChapter 12.\nLinear programming\n\nc) The clinic is seeking to maximise the income, I, earned from the sales of\nthe two products. Each video will sell for R 50 and each book for R 30.\nWrite down the objective function for the income.\nd) What maximum income will be generated by the two guides?\n3. A certain motorcycle manufacturer produces two basic models, the Super X and\nthe Super Y. These motorcycles are sold to dealers at a profit of R 20 000 per\nSuper X and R 10 000 per Super Y. A Super X requires 150 hours for assembly,\n50 hours for painting and finishing and 10 hours for checking and testing. The\nSuper Y requires 60 hours for assembly, 40 hours for painting and finishing and\n20 hours for checking and testing. The total number of hours available per month\nis: 30 000 in the assembly department, 13 000 in the painting and finishing\ndepartment and 5000 in the checking and testing department.\nThe above information is summarised by the following table:\nDepartment\nHours for\nSuper X\nHours for\nSuper Y\nHours available\nper month\nAssembly\n150\n60\n30 000\nPainting and\nfinishing\n50\n40\n13 000\nChecking and testing\n10\n20\n5000\nLet x be the number of Super X and y be the number of Super Y models manu-\nfactured per month.\na) Write down the set of constraint inequalities.\nb) Use graph paper to represent the set of constraint inequalities.\nc) Shade the feasible region on the graph paper.\nd) Write down the profit generated in terms of x and y.\ne) How many motorcycles of each model must be produced in order to max-\nimise the monthly profit?\nf) What is the maximum monthly profit?\n4. A group of students plan to sell x hamburgers and y chicken burgers at a rugby\nmatch. They have meat for at most 300 hamburgers and at most 400 chicken\nburgers. Each burger of both types is sold in a packet. There are 500 packets\navailable. The demand is likely to be such that the number of chicken burgers\nsold is at least half the number of hamburgers sold.\na) Write the constraint inequalities and draw a graph of the feasible region.\nb) A profit of R 3 is made on each hamburger sold and R 2 on each chicken\nburger sold. Write the equation which represents the total profit P in terms\nof x and y.\nc) The objective is to maximise profit. How many of each type of burger\nshould be sold?\n5. Fashion-Cards is a small company that makes two types of cards, type X and type\nY. With the available labour and material, the company can make at most 150\ncards of type X and at most 120 cards of type Y per week. Altogether they cannot\nmake more than 200 cards per week.\n480\n12.1.\nIntroduction\n\nThere is an order for at least 40 type X cards and 10 type Y cards per week.\nFashion-Cards makes a profit of R 5 for each type X card sold and R 10 for each\ntype Y card.\nLet the number of type X cards manufactured per week be x and the number of\ntype Y cards manufactured per week be y.\na) One of the constraint inequalities which represents the restrictions above is\n0 ≤x ≤150. Write the other constraint inequalities.\nb) Represent the constraints graphically and shade the feasible region.\nc) Write the equation that represents the profit P (the objective function), in\nterms of x and y.\nd) Calculate the maximum weekly profit.\n6. To meet the requirements of a specialised diet a meal is prepared by mixing\ntwo types of cereal, Vuka and Molo. The mixture must contain x packets of\nVuka cereal and y packets of Molo cereal. The meal requires at least 15 g of\nprotein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains\n4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains\n3 g of protein and 24 g of carbohydrates. There are at most 5 packets of cereal\navailable. The feasible region is shaded on the attached graph paper.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\nNumber of packets of Vuka\nNumber of packets of Molo\na) Write down the constraint inequalities.\nb) If Vuka cereal costs R 6 per packet and Molo cereal also costs R 6 per\npacket, use the graph to determine how many packets of each cereal must\nbe used so that the total cost for the mixture is a minimum.\nc) Use the graph to determine how many packets of each cereal must be used\nso that the total cost for the mixture is a maximum (give all possibilities).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DV\n2. 23DW\n3. 23DX\n4. 23DY\n5. 23DZ\n6. 23F2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n481\nChapter 12.\nLinear programming\n\n\nSolutions to exercises\n1\nExponents and surds\nExercise 1 – 1: The number system\n1. R; Q′\n2. R; Q\n3. R; Q\n4. R; Q\n5. R; Q; Z; N0\n6. R′Q′\n7. R; Q\n8. R; Q′\n9. R′\n10. R; Q′\n11. R; Q\n12. R; Q; Z\n13. R; Q\n14. R; Q′\n15. R; Q\n16. R; Q; Z\nExercise 1 – 2: Laws of exponents\n1. 43a+3\n2. 72\n3. 9p10\n4. k2x−2\n5. 52z−2 + 5z\n6. 1\n7. x10\n8.\nb2\na2\n9.\n1\nm+n\n10. 2pts\n11.\n1\na\n12. k\n13. 2a+1\n14. h4\n15.\na4b6\nc6d2\n16. 4\n17.\nm2n2\n2\n18. 400\n19.\n1\ny7\n20. 8\n21. 26a+2\n22. 2pt\n23. 81q2sy8a+2\nExercise 1 – 3: Rational exponents and surds\n1.\na) 7\nb)\n1\n6\nc)\n1\n3√\n36\nd) −4\n3\ne) 8x3\n2.\na) s\n1\n6\nb) 16m4\nc)\n3\n2 m2\nd) 8\n3. x\n31\n16\n483\nChapter 12.\nLinear programming\n\nExercise 1 – 4: Simplification of surds\n1.\na) 4\nb) ab4c2\nc) 2\nd) xy4\n2.\na)\nab\nb−a\nb) −\n\u0010\na\n1\n2 + b\n1\n2\n\u0011\nExercise 1 – 5: Rationalising the denominator\n1. 2\n√\n5\n2.\n√\n6\n2\n3.\n√\n6\n4.\n3\n√\n5 + 3\n4\n5.\nx√y\ny\n6.\n√\n6 +\n√\n14\n2\n7.\n3p −4√p\np\n8.\n√\nt −2\n9.\n1−√m\n1−m\n10.\n√\nab\nExercise 1 – 6: Solving surd equations\n1. x = 4\n2. p = 3\n3. y = 1\n4. t = 3\n5. z = 9 or z = 1\n4\n6. x = 8 or x = −27\n7. n = −1\n4\n8. d = 3 or d = −5\n9. y = 1 or y = 81\n10. f = 5\nExercise 1 – 7: Applications of exponentials\n1. 9,7%\n2. 4 254 691\n3. 7\n4. 26 893\n484\n12.1.\nIntroduction\n\nExercise 1 – 8: End of chapter exercises\n1.\na)\n1\n4\nb) 4 1\n4\n2.\na) x4\nb) s\nc) m\n25\n3\nd) m\n8\n3\ne) −m\n8\n3\nf) 81y\n16\n3\n3.\na)\n3b\n45\n2\n(a12c\n5\n2\nb) 3a3b2\nc) a24b12\nd) x\n7\n2\ne) x\n4\n3 b\n5\n3\n4.\n1\nx\n1\n16\n5. x −2\n6.\n10√x + 10\nx −1\n7.\n3√x + 2x√x\n2x\n8.\na) 6\n√\n2\nb) 7\n√\n5\nc) 2\nd)\n1\n4\n√\n2\ne) 2\nf)\n16\n√\n15\n5\n9.\na) 6 + 4\n√\n2\nb) 6 + 5\n√\n2\nc) 4+2\n√\n2+2\n√\n3+2\n√\n6\n10.\na) 55\nb) 1\n11. 15\n√\n2x3\n12.\na) 1 + 2\n√\n5\n5\nb)\n2y + y√y −4√y −8\ny −4\nc) 2√x + 2\n√\n10\n13.\n3\n2\n15. 3\n16. −\n√\n288\n17.\na) 4\nb) −1\n3\nc) 3\nd) No solution\ne) x = 1\n8 or x = −8\n18.\nb) x = 1\n2\nEquations and inequalities\nExercise 2 – 1: Solution by factorisation\n1. t = 0 or t = −2\n2. y = −1\n3. s = ±5\n4. y = 3 or y = 2\n5. y = 4 or y = −9\n6. p = −2\n7. y = −3 or y = −8\n8. y = 6 or y = 7\n9. x = −7 or x = −2\n10. y = 4k or y = k\n11. y = 9 or y = −9\n12. y = ±\n√\n5\n13. h = ±6\n14. y = ±\n√\n14\n15. p = −2\n16. y = ±6\n√\n2\n17. f = 5\n2 or f = −3\n18. x = 1\n4\n19. y = 1\n7\n20. x ∈R, x ̸= ±3\n21. y = −13 or y = −1\n22. t = 3\n2 or t = −2\n23. m = −6\n24. t = 0 or t = 3\nExercise 2 – 2: Solution by completing the square\n1.\na) x = −5 −3\n√\n3 or x = −5 + 3\n√\n3\nb) x = −1 or x = −3\nc) p = −4 ±\n√\n21\nd) x = −3 ±\n√\n7\ne) No real solution\nf) t = −8 ± 3\n√\n6\ng) x = −1 ±\nq\n5\n3\nh) z = −4 ±\n√\n22\ni) z = 11\n2 or z = 0\nj) z = 5 or z = −1\n2. k = −3 ± √9 −a\n3. y = −q±√\nq2−4pr\n2p\n485\nChapter 12.\nLinear programming\n\nExercise 2 – 3: Solution by the quadratic formula\n1. t = 1 or t = −4\n3\n2. x = 5+\n√\n37\n2\nor t = 5−\n√\n37\n2\n3. No real solution\n4. p = 1\n2 or p = −1\n5. No real solution\n6. t = −3+\n√\n69\n10\nor t = −3−\n√\n69\n10\n7. t = 2 ±\n√\n2\n8. k = 7+\n√\n373\n18\nor k = 7−\n√\n373\n18\n9. f = 1\n2 or f = −2\n10. No real solution\nExercise 2 – 4:\n1. x = −1, x = −4, x = −2 and x = −3\n2. x = 1, x = 4 and x = −2\n3. x = −7, x = 4, x = −1 and x = −2\n4. x = −4, x = 3, x = −3 and x = 2\n5. x = 8±\n√\n40\n4\n6. x = −5, x = 3, x = −1 +\n√\n10 and\nx = −1 −\n√\n10\nExercise 2 – 5: Finding the equation\n1. x2 −x −6 = 0\n2. x2 −16 = 0\n3. 2x2 −5x −3 = 0\n4. k = 3 and x = 3\n4\n5. p = 5 and x = −1\nExercise 2 – 6: Mixed exercises\n1. y = 1\n8 or y = −8\n3\n2. x = 3\n2 or x = −7\n2\n3. t = 2\n3 or t = 2\n4. y = 1 or y = −1\n5. m = 1 or m = 4\n6. y = ± 5\n7\n7. w = 3\n2 or w = 4\n8. y = 6\n5 or y = 1\n4\n9. n = 8\n3 or n = −9\n8\n10. y = −8\n3 or y = 3\n2\n11. x = −1\n2 or x = 3\n12. y = −5\n2 or y = −5\n9\n13. y = 4\n5 or y = 1\n5\n14. g = −1\n4 or g = 1\n15. y = 2 or y = −5\n9\n16. p = 3\n7 or p = −1\n5\n17. y = −2\n9 or y = −1\n18. y = 9\n2 or y = 9\n7\n486\n12.1.\nIntroduction\n\nExercise 2 – 7: From past papers\n1.\na) Real, unequal and rational\nb) Real and equal\nc) Real, unequal and irrational\nd) Real, unequal and rational\ne) Real, unequal and irrational\nf) Non-real\ng) Real, unequal and rational\nh) Real, unequal and irrational\ni) Non-real\nj) Real and equal\n2.\nb) real and unequal\nc) k = −6 ± 2\n√\n6\n4.\na) k = 6\nb) k = 1\n3\n5.\na) k = 4 or k = 1\nb) k = 0 or k = 5\n6.\na) all real values of a, b and p\nb) a = b and p = 0\nExercise 2 – 8: Solving quadratic inequalities\n1.\na) −3 < x < 4\nb) x < −4\n3 or when x > 1\nc) no real solutions\nd) −1 < t < 3\ne) All real values of s.\nf) All real values of x.\ng) x ≤−1\n4 or x ≥0\ni) x < 3 or x > 6 with x ̸= 3\nj) −2 ≤x ≤2 and x > 7 with x ̸= 7\nk) x > 0 with x ̸= 0\n2.\na) x < −3 or x > 3\nb) −\n√\n5 ≤x ≤\n√\n5\nc) no solution\nd) All real values of x\nExercise 2 – 9: Solving simultaneous equations\n1.\na) (0; 5) and (2; 3)\nb) x = 3 ±\n√\n2 and y = 2 ±\n√\n2\nc) (−1; 0) and ( 1\n4 ; 5\n8 )\nd) b = 2 ±\n√\n88\n6\nand a = 11 ±\n√\n88\n6\ne) (−3; −20) and (2; 0)\nf) x = 6 ±\n√\n264\n2\nand y = 70 ±\n√\n264\n2\n2.\na) (−3; 8) and (2; 3)\nb) (−4; 14) and (3; 7)\nc) (3; 4) and (4; 3)\nExercise 2 – 10:\n1. b = 2 m, l = 4 m\n2. 187\n3. t = 10,5 s\n4. t = 5d; 105 minutes; 1,4 km\n5. 24 A; 70 W; 12 A\n487\nChapter 12.\nLinear programming\n\nExercise 2 – 11: End of chapter exercises\n1. x = 1,62 or x = −0,62\n2. x = ±4 or x = −1\n3. y = 0 or y = ±1\n4. x = ±2\n5.\na) x = 7 or x = 2\nb) x = 2,3 or x = −1,3\nc) x = 1,65 or x = −3,65\nd) x = 0 or x = −3\n6. x =\n√\n16+p2−2\n2\n7. a = 3; b = 10 and c = −8\n8. p = ±16\n9. x2 + 2x −15\n10. Undefined:b = −2 Zero:b = 2 or b = 3\n13. a ≥4\n14. x = −3\n2 or x = 1\n15.\na) x < 3 or x ≥7:\nb) x < 1 or x > 5:\nc) 3 < x < 7:\nd) x < −1 or x > 3\ne) 0,5 < x < 2,5\nf) x ≤−3 or 0 < x ≤5\n2\ng) x < 2\n3\nh) −1 ≤x < 0 or x ≥3\ni) −4 ≤x ≤1\nj) 2 1\n2 ≤x < 3\n16.\na) x = ±\n√\n3 and y = ±2\n√\n3\nb) a = −3 and b = −1 or a = 12 and b = 4\nc) x = −5 and y = 0 or x = 2 and y = 14\nd) p = 5\n3 and q = 2\n9 or p = −1 and q = −2\n3\ne) b = 3±\n√\n5\n2\nand a = 7±3\n√\n5\n2\nf) b = −10±\n√\n140\n4\nand a = −12±\n√\n140\n2\ng) x = 3,4 and y = 5,4 or x = 3 and y = 5\nh) b = −1,4 and a = 23,6 or\nb = 3 and a = 6\n17.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\nx\n0\ny\nb\nb\nb)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n1\n2\n3\n4\n5\n6\n−1\n−2\nx\n0\ny\nb\nb\n18. 35 m\n20.\na) y = −5\n4 or y = −9\nb) x = −9\n4 or x = 1\nc) p = −8\n7 or p = −4\n3\nd) y = −1\n4 or y = 1\n2\ne) y = −2\n9 or y = −1\nf) y = 7\n3 or y = −1\n2\ng) y = 9\n4 or y = −9\n4\nh) y = 8\n3 or y = −6\ni) y = 9\n5 or y = −7\nj) x = ±4\nk) y = ±7\n21. k = 76 and 4\n9\n22. x = 3 or x = −2 and y = 1±√−7\n2\n23. x = 4 or x = −1\n24. y = 3\n2 , y = 1\n2 and p = 9\n2 , p = 7\n2\n25.\n69\n4\n26. 7\n27.\n2±\n√\n12\n2\n28. t = 1\n2 , t = 1 or t = 3±\n√\n33\n4\n488\n12.1.\nIntroduction\n\n3\nNumber patterns\nExercise 3 – 1: Linear sequences\n1. −19; −35; −51\n2.\na) −19\nb) T2 = 15; T4 = 33\n3.\na) Tn = 10 + 3n; T10 = 40; T15 = 55;\nT30 = 100\nb) Tn = 12 + 6n; T10 = 72; T15 = 102;\nT30 = 192\nc) Tn = −5 −5n; T10 = −55; T15 = −80;\nT30 = −155\n4. T9 = 36\n5.\na) 44; 66; 121\nExercise 3 – 2: Quadratic sequences\n1.\na) 10\nb) 2\nc) 2\nd) −2\ne) 2\nf) −4\ng) 4\nh) −2\ni) 6a\nj) 6\nk) 2t\n2.\na) T4 = 53\nb) T2 = 30\nc) T1 = 17\nd) T2 = −3\ne) T4 = 63\nf) T1 = 2\n3.\na) 3; 9; 17; 27\nb) −6; −9; −14; −21\nc) 1; 8; 21; 40\nd) 0; −5; −14; −27\nExercise 3 – 3: Quadratic sequences\n1.\na) 1\nb) 2\nc) 4\nd) 8\ne) −2\n2. 12; 30; 58; 96; 144\n3. T9 = 379\n4. n = 4\n5.\na) T5 = 84; T6 = 111\nb) Tn = 2n2 + 5n + 9\n489\nChapter 12.\nLinear programming\n\nExercise 3 – 4: End of chapter exercises\n1. −4; 9; 16; 25; 36\n2.\na) Quadratic sequence\nb) Quadratic sequence\nc) Quadratic sequence\nd) Quadratic sequence\ne) Quadratic sequence\nf) Quadratic sequence\ng) Linear sequence\nh) Linear sequence\ni) Quadratic sequence\nj) Quadratic sequence\nk) Quadratic sequence\nl) Linear sequence\nm) Quadratic sequence\n3. x = 31\n4. n = 11\n5. T11 = 363\n6. n = 9\n7. T5 = 114\n8. n = 8\n9.\na) T5 = 19;\nTn = 4n −1;\nT10 = 39\nb) T5 = −3;\nTn = 22 −5n;\nT10 = −28\nc) T5 = 2 1\n2 ; Tn = 1\n2 n;\nT10 = 5\nd) T5 = a + 4b;\nTn = a −b + bn;\nT10 = a + 9b\ne) T5 = −7;\nTn = 3 −2n;\nT10 = −17\n10.\na) Tn = n2 + 3;\nT100 = 10 003\nb) Tn = 6n −4;\nT100 = 596\nc) Tn = 2n2 + 5;\nT100 = 20 005\nd) Tn = 3n2 + 2;\nT100 = 30 002\n11.\na) 2; 5; 8; 11; 14\nb) Constant difference,\nd = 3\nc) Yes\n12.\na) Tn = 4n −19\nb) n = 48\n13.\na) Incorrect\nb) Correct\n14.\nc) Linear\n15.\nb) Linear\nd) Quadratic\ne) Tn = 1\n2 n2 + 3\n2 n + 1\nf) T21 = 253\ng) 31 cm\n16.\na) −1\nb) 7\n17.\nb) 2\nc) Tn = n2 −n\nd) 210\ne) 25\n18. 4; 14; 34; 64; 104; 154\n4\nAnalytical geometry\nExercise 4 – 1: Revision\n1.\na) 2\n√\n26units\nb) 7 units\nc) x + 1units\n2. p = 6 or p = 2\n3.\na) −1\n2\nb) 3\n5. 2\n6.\na) (1; 2)\nb)\n\u0000 −1\n2 ; −1\n2\n\u0001\n7. B(4; 2)\n8.\na) y = −4x + 3 and\ny = −4x + 19\nc) AD =\n√\n17units and\nBC =\n√\n17units\nd) y = 4\n3 x −7\n3\ne) Parallelogram (one\nopposite side equal\nand parallel)\n9. N(0; 3)\n10.\na) PQ =\n√\n20 and\nSR =\n√\n20\nb) M( 3\n2 ; 1)\nd) PS: y = −2\n5 x −1\n5\nand SR: y = 1\n2 x −2\ne) No\nf) Parallelogram\nExercise 4 – 2: The two-point form of the straight line equation\n1. y = 2\n3 x + 5\n2. y = −3x + 1\n4\n3. y = x + 3\n4. y = 2x −1\n5. y = −5\n6. y = 3\n4 x + 3\n7. y = −x + (s + t)\n8. y = 5x + 2\n9. y = q\npx −q\n490\n12.1.\nIntroduction\n\nExercise 4 – 3: Gradient–point form of a straight line equation\n1. y = 2\n3 x + 4\n2. y = −x −2\n3. y = −1\n3 x\n4. y = 11\n5. y = −2x + 7\n6. x = −3\n2\n7. y = −4\n5 x + 1\n8. x = 4\n9. y = 3ax + b\nExercise 4 – 4: The gradient–intercept form of a straight line equation\n1. y = 2x + 3\n2. y = 4x −4\n3. y = −x −1\n4. y = −3\n7 x\n5. y = 1\n2 x −1\n5\n6. y = 2x −2\n7. y = −3\n2\n8. y = 3x + 4\n9. y = −5x\nExercise 4 – 5: Angle of inclination\n1.\na) 1,7\nb) −1\nc) 0\nd) 1,4\ne) Undefined\nf) 1\ng) −0,8\nh) 0\ni) 3,7\n2.\na) 36,8◦\nb) 26,6◦\nc) 45◦\nd) Horizontal line\ne) 18,4◦\nf) Vertical line\ng) 71,6◦\nh) 30◦\nExercise 4 – 6: Inclination of a straight line\n1.\na) 38,7◦\nb) 135◦\nc) 80◦\nd) 80◦\ne) 102,5◦\nf) 45◦\ng) 56,3◦\nh) 63,4◦\ni) 161,6◦\nj) Gradient undefined\n2. 85,2◦\n3. 90◦\n4. 81,8◦\nExercise 4 – 7: Parallel lines\n1.\na) Parallel\nb) Parallel\nc) Parallel\nd) Not parallel\ne) Parallel\nf) Parallel\n2. y = −2x −3\n3. y = 3x\n4. y = 3\n2 x + 1\n5. y = −7\n10 x −1\n491\nChapter 12.\nLinear programming\n\nExercise 4 – 8: Perpendicular lines\n1.\na) Perpendicular\nb) Not perpendicular\nc) Perpendicular\nd) Perpendicular\ne) Perpendicular\nf) Not perpendicular\ng) Not perpendicular\n2. y = 1\n2 x −3\n3. y = −5x + 3\n4. y = −x + 2\n5. x = −2\nExercise 4 – 9: End of chapter exercises\n1.\na) y = 1\n2 x + 7\n2\nb) y = −x + 4\nc) y = 1\n2 x + 4\nd) y = 2x + 4\ne) y = 3x\n2.\na) θ = 63,4◦\nb) θ = 18,4◦\nc) θ = 36,9◦\nd) θ = 146,3◦\ne) θ = 161,6◦\n3.\na) y = −2x + 7\nb)\n\u0000 7\n2 ; 0\n\u0001\nc) θ = 116,6◦\nd) m = 1\n2\ne) Q ˆPR = 90◦\nf) y = −2x\ng)\n\u0000 1\n2 ; −3\n2\n\u0001\nh) y = −2x −1\n2\n4.\na)\n1\n2\n3\n4\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\nb\nb\nb\ny\nx\nA(−3; 5)\nB(−7; −4)\nC(2; 0)\nD(x; y)\nb) D (6; 9)\n5.\na) (−1; −2)\nb) (8; 3)\nc) x = −1\nd) MN = 5 units\ne) M ˆ\nNP = 21,8◦\nf) y = 5\n2 x + 11\n2\n6.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nA(−2; 3)\nB(2; 4)\nC(3; 0)\ny\nx\n0\nb\nb\nb\nc) y = 1\n4 x + 7\n2\nd) D(−1; −1)\ne) E\n\u0000 5\n2 ; 2\n\u0001\n7.\na) y = 3\n2 x + 2\nb) T ˆSV = 49,6◦\n8.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nb\nb\nb\nF(−1; 3)\nH(4; 4)\nG(2; 1)\ny\nx\n0\nc) y = −5x + 11\nd) Yes\ne) y = 3\n2 x + 9\n2\n9.\na)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nb\nb\nb\nA(−1; 5)\nB(5; −3)\nC(0; −6)\nx\ny\nM\nN\n492\n12.1.\nIntroduction\n\n5\nFunctions\nExercise 5 – 1: Revision\n1.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nc)\n1\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nd)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nExercise 5 – 2: Domain and range\n1. {x : x ∈R} ; {y : y ≥−1, y ∈R}\n2. {x : x ∈R} ; {y : y ≤4, y ∈R}\n3. {x : x ∈R} ; {y : y ≥0, y ∈R}\n4. {x : x ∈R} ; {y : y ≤0, y ∈R}\n5. {x : x ∈R} ; {y : y ≤2, y ∈R}\nExercise 5 – 3: Intercepts\n1. (0; 15) and (−5; 0); (−3; 0)\n2. (0; 16) and (4; 0)\n3. (0; −3) and (1; 0); (3; 0)\n4. (0; 35) and (−7\n2 ; 0); (−5\n2 ; 0)\n5. (0; 37) and no x-intercepts\n6. (0; −4) and\n(−0,85; 0); (−2,35; 0)\nExercise 5 – 4: Turning points\n1. (3; −1)\n2. (2; 1)\n3. (−2; −1)\n4. (−1\n2 ; 1\n2 )\n5. (1; 21)\n6. (−1; −6)\nExercise 5 – 5: Axis of symmetry\n1.\na) Axis of symmetry:\nx = 5\n4\nb) Axis of symmetry:\nx = 2\nc) Axis of symmetry:\nx = 2\n2. y = ax2 + q\n493\nChapter 12.\nLinear programming\n\nExercise 5 – 6: Sketching parabolas\n1.\na)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n1\n2\n3\n4\n5\n6\n−1\ny\nx\n0\nIntercepts: (−1; 0), (5; 0), (0; 5)\nTurning point: (2; 9)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≤9, y ∈R}\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−1; 0), (0; 2)\nTurning point: (−1; 0)\nAxes of symmetry: x = −1\nDomain: {x : x ∈R}\nRange: {y : y ≥0, y ∈R}\nc)\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−0,87; 0), (1,54; 0), (0; −4)\nTurning point: (0,33; −4,33)\nAxes of symmetry: x = −0,33\nDomain: {x : x ∈R}\nRange: {y : y ≥4,33, y ∈R}\nd)\n1\n2\n3\n4\n5\n6\n−1\n1\n2\n3\n4\n−1\ny\nx\n0\nIntercepts: (0; 13) Turning point: (2; 1)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≥1, y ∈R}\n3.\na)\ny\nx\n0\nb)\ny\nx\n0\nc)\ny\nx\n0\nd)\ny\nx\n0\ne)\ny\nx\n0\nf)\ny\nx\n0\n4.\na) yshifted = 2x2 + 16x + 32\nb) yshifted = −x2 −2x\nc) yshifted = 3x2 −16x + 22\n494\n12.1.\nIntroduction\n\nExercise 5 – 7: Finding the equation\n1. y = −3(x + 1)2 + 6 or y = −3x2 −6x + 3\n2. y = 1\n2 x2 −5\n2 x\n3. y = 2\n3 (x + 2)2\n4. y = −x2 + 3x + 4\nExercise 5 – 8:\n1.\na) 11\n2.\na)\n1\n2\n3\n4\n1\n2\n−1\n−2\nf(x)\nx\n0\nA(1; 3)\nb\nb) 6\nc) y = 6x −3\n3.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n−1\n−2\ng(x)\nx\n0\nb) 1\nc) 4\nd) 0\nExercise 5 – 10: Domain and range\n1. {x : x ∈R, x ̸= 0} ; {y : y ∈R, y ̸= 1}\n2. {x : x ∈R, x ̸= 8} ; {y : y ∈R, y ̸= 4}\n3. {x : x ∈R, x ̸= −1} ; {y : y ∈R, y ̸= −3}\n4. {x : x ∈R, x ̸= 5} ; {y : y ∈R, y ̸= 3}\n5. {x : x ∈R, x ̸= −2} ; {y : y ∈R, y ̸= 2}\nExercise 5 – 11: Intercepts\n1. (0; −1 3\n4 ) and\n\u0000−3 1\n2 ; 0\n\u0001\n2.\n\u0000 5\n2 ; 0\n\u0001\n3. (0; 1) and\n\u0000 1\n3 ; 0\n\u0001\n4.\n\u00000; 3\n2\n\u0001\nand\n\u0000 1\n3 ; 0\n\u0001\n5. (0; 2) and (8; 0)\nExercise 5 – 12: Asymptotes\n1. y = −2 and x = −4\n2. y = 0 and x = 0\n3. y = 1 and x = 2\n4. y = −8 and x = 0\n5. y = 0 and x = 2\n495\nChapter 12.\nLinear programming\n\nExercise 5 – 13: Axes of symmetry\n1.\na) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (0; 1); y1 = x + 1 and\ny2 = −x + 1\nb) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (−1; 0); y1 = x + 1 and\ny2 = −x −1\nc) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (1; −1); y1 = x −2 and\ny2 = −x\n2. k(x) =\n5\nx+1 + 2\nExercise 5 – 14: Sketching graphs\n1.\na) Asymptotes: x = 0; y = 2\nIntercepts:\n\u0000−1\n2 ; 0\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= 0}\nRange: {y : y ∈R, y ̸= 2}\nb) Asymptotes: x = −4; y = −2\nIntercepts:\n\u0000−3 1\n2 ; 0\n\u0001\nand\n\u00000; −1 3\n4\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x −6\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\nc) Asymptotes: x = −1; y = 3\nIntercepts:\n\u0000−2\n3 ; 0\n\u0001\nand (0; 2)\nAxes of symmetry: y = x + 4 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 3}\nd) Asymptotes: x = −2 1\n2 ; y = −2\nIntercepts: (0; 0)\nAxes of symmetry: y = x −4 1\n2 and\ny = −x + 1\n2\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\ne) Asymptotes: x = 8; y = 4\nIntercepts: (6; 0) and (0; 3)\nAxes of symmetry: y = x −4 and\ny = −x + 12\nDomain: {x : x ∈R, x ̸= 8}\nRange: {y : y ∈R, y ̸= 4}\n2. y =\n1\nx+2 −1\n3. y = −4\nx + 2\n4.\na)\nb) Average gradient = 1\nc) Average gradient = 12\nExercise 5 – 16: Domain and range\n1. {x : x ∈R} ; {y : y > 0, y ∈R}\n2. {x : x ∈R} ; {y : y < 1, y ∈R}\n3. {x : x ∈R} ; {y : y > −3, y ∈R}\n4. {x : x ∈R} ; {y : y > n, y ∈R}\n5. {x : x ∈R} ; {y : y > 2, y ∈R}\nExercise 5 – 17: Intercepts\n1. (0; −6) and (2; 0)\n2. (0; −17 1\n3 ) and (3; 0)\n3. (0; −20) and (−1; 0)\n4. (0; 15\n16 ) and (−2; 0)\n496\n12.1.\nIntroduction\n\nExercise 5 – 18: Asymptote\n1. y = 0\n2. y = 1\n3. y = −2\n3\n4. y = −2\n5. y = −2\nExercise 5 – 19: Mixed exercises\n1.\nb)\ni. y = 3\nx + 3\nii. y =\n3\nx−3\niii. y = −3\nx\niv. y = 3\nx −1\n4\nv. y = 3\nx + 4\nvi. y =\n3\nx+2 −1\n2.\na) M(−2; 2)\nb) g(x) = −4\nx\nc) f(x) = 2(x + 1)2\nd) −2 < x < 0\ne) Range: {y : y ∈R, y ≥0}\n3.\na) For k(x) :\nIntercepts:\n(−2; 0), (1; 0) and (0; −4)\nTurning point:\n\u0000−1\n2 ; −4 1\n2\n\u0001\nAsymptote:\nnone\nFor h(x) :\nIntercepts:\n(1,41; 0)\nTurning point:\nnone\nAsymptote:\ny = 0\n6.\na) f(x) = −3\n4 (x −2)2 + 3 ;\nAxes of symmetry: x = 2 ;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≤3}\nb) g(x) = 1\n4 x2 −2;\nAxes of symmetry: x = 0;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≥−2}; h(x) = 2\nx ;\nAxes of symmetry: y = x\nDomain: {x : x ∈R, x < 0};\nRange: {y : y ∈R, y < 0};\nc) k(x) =\n\u0000 1\n2\n\u0001x + 1\n2 ;\nDomain: {x : x ∈R};\nRange:\n\b\ny : y ∈R, y > 1\n2\n\t\n7.\nb) p = 9\nc) Average gradient = −2 8\n9\nd) y =\n\u0000 1\n3\n\u0001x+2 −2\n8.\na) f(x) = 2x −3\n2 and g(x) = −1\n4 x −1\n2\nb) h(x) = −\n3\nx+2 + 1\n9.\na) AO = 2 units OB = 5 units\nOC = 10 units DE = 12,25 units\nb) DE = 12 1\n4\nc) h(x) = −2x + 10\nd) {x : x ∈R, x < −2 and x > 5}\ne) {x : x ∈R, 0 ≤x ≤5}\nf) 5,25 units\n497\nChapter 12.\nLinear programming\n\nExercise 5 – 20: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\nPeriod: = 360◦\nAmplitude: = 1\nDomain: = [0◦; 360◦]\nRange: = [−1; 1]\nx-intercepts: = (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: = (0◦; 0)\nMax. turning point: = (90◦; 1)\nMin. turning point: = (270◦; −1)\n2.\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny2 = −2 sin θ\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 2\nDomain: [0◦; 360◦]\nRange: [−2; 2]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nMax. turning point: (270◦; 2)\nMin. turning point: (90◦; −2)\n3.\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny3 = sin θ + 1\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [0; 2]\nx-intercepts: (270◦; 0)\ny-intercepts: (0◦; 1)\nMax. turning point: (90◦; 2)\nMin. turning point: (270◦; 0)\n4.\n1\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny4 = 1\n2 sin θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (90◦; 1\n2 )\nMin. turning point: (270◦; −3\n2 )\nExercise 5 – 21: Sine functions of the form y = sin kθ\n2.\na) k = 2\nb) k = −3\n4\n498\n12.1.\nIntroduction\n\nExercise 5 – 23: The sine function\n1.\na)\n1\n2\n−1\n−2\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 2 sin( θ\n2 )\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 1\n2 sin(θ −45◦)\nc)\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(θ + 90◦) + 1\nd)\n1\n−1\n60◦\n120◦\n180◦\n−60◦\n−120◦\n−180◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(−3θ\n2 )\ne)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(30◦−θ)\n2. a = 2; p = 90◦∴y = 2 sin(θ + 90◦) and\ny = 2 cos θ\nExercise 5 – 24: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\n2.\n1\n2\n3\n−1\n−2\n−3\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny2 = −3 cos θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−3; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; −3)\n3.\n1\n2\n3\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny3 = cos θ + 2\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [1; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; 1)\n4.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny4 = 1\n2 cos θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (0◦; −1\n2 ); (360◦; −1\n2 )\nMin. turning point: (180◦; −3\n2 )\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n2.\na) k = 3\n2\nb) k = 2\n3\n499\nChapter 12.\nLinear programming\n\nExercise 5 – 27: The cosine function\n1.\na)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nθ\n0◦\ny\ny = cos θ\ny = cos(θ + 15◦)\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny = cos θ\nf(θ) = 1\n3 cos(θ −60◦)\nc)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ny = −2 cos θ\nd)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(30◦−θ)\ne)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ng(θ) = 1 + cos(θ −90◦)\nf)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(2θ + 60◦)\n2.\na) a = −1\nb) p = −180◦\nc) cos(θ −180◦) = −cos θ\nExercise 5 – 28: Revision\n1.\n1\n2\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1\n2 )\nAsymptotes: 90◦; 270◦\n2.\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: 90◦; 270◦\n3.\n1\n2\n3\n4\n5\n−1\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (116,6◦; 0); (296,6◦; 0)\ny-intercepts: (0◦; 2)\nAsymptotes: 90◦; 270◦\n4.\n1\n2\n−1\n−2\n−3\n−4\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1)\nAsymptotes: 90◦; 270◦\n500\n12.1.\nIntroduction\n\nExercise 5 – 29: Tangent functions of the form y = tan kθ\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦] Range: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 240◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −120◦; 120◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n4.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 270◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; 135◦\n501\nChapter 12.\nLinear programming\n\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−225◦; 0); (−45◦; 0); (135◦; 0);\n(315◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−330◦; 0); (−150◦; 0); (30◦; 0);\n(210◦; 0)\ny-intercepts: (0◦; −0,58)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−240◦; 0); (−60◦; 0); (120◦; 0);\n(300◦; 0)\ny-intercepts: (0◦; 1.73)\nAsymptotes: −330◦; −150◦; 30◦; 210◦\nExercise 5 – 31: The tangent function\n1.\na)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n−45◦\n−90◦\nθ\ny\nb)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\nθ\ny\nc)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nd)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\ny\n2. a = −1; k = 1\n2\n502\n12.1.\nIntroduction\n\nExercise 5 – 32: Mixed exercises\n1.\na) f(θ) = 3\n2 sin 2θ and g(θ) = −3\n2 tan θ\nb) f(θ) = −2 sin θ and\ng(θ) = 2 cos(θ + 360◦\nc) y = 3 tan θ\n2\nd) y = y = 2 cos θ + 2\n2.\na)\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\nθ\n0\ng\nf\ny\nb\nb\nb\nb\n(90◦; 2)\n(270◦; −2)\nb) 360◦\nc) 1\nd) At θ = 180◦\n3.\na) a = 2, b = −1 and c = 240◦\nb) 180◦\nc) θ = 60◦; 300◦\nd) y = −tan(θ −45◦)\n4.\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny1\ny2\ny\nb\nb\nExercise 5 – 33: End of chapter exercises\n2. a = −2; k = −1\n4.\na) y = x + 22 + 2\nb) y = x −12 + 5\n5. (−1; 0)\n6.\ny\nx\n0\n4\n−4\n4\n−4\ny =\n2\nx−3 −1\n7. y =\n1\n(x−1) + 2\n9.\na) a = −1\nb) f(−15) = 0,99997\nc) x = −1\nd) h(x) = −2(x−2) + 1\n10.\na) a = 256\nb) f(x) = 256\n\u0000 3\n4\n\u0001x\nc) f(13) = 6,08\n11.\na)\n1\n−1\n90◦\n180◦\n−90◦\n−180◦\nθ\n0\ny\nb)\n1\n−1\n90◦\n180◦\nθ\n0\ny\nd)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\n0\ny\ne)\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny\nf)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\n503\nChapter 12.\nLinear programming\n\n6\nTrigonometry\nExercise 6 – 1: Revision\n1.\na) True\nb) True\nc) False\nd) True\n2.\na) 50,2◦\nb) 40,5◦\nc) 26,6◦\nd) 109,8◦\ne) No solution\nf) 17,7◦\ng) 69,4◦\n3.\na) 17,3 cm\nb) 10 cm\nc) 64,8◦\n4.\na) 10 cm\nb) 5,2 cm and 19,3 cm\nc) 50 cm2\n5.\na) 2\nb) 0\nc) −1 1\n2\nd) 1\ne) 1\n6.\na) 60◦\nb)\n1\n2\nc) 1\n7. No\nExercise 6 – 2: Trigonometric identities\n1.\na) cos α\nb) tan2 θ\nc) cos2 θ\nd) 0\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1.\na)\n√\n3\n3\nb)\n1\n8\nc) 1\n2.\na)\n1−cos2 θ\ncos θ\nb) −1\n3.\na) 2t\nb) −1\nt\nExercise 6 – 4: Using reduction formula\n1.\na) −tan θ\nb) 1\nc) 1\n2. −cos β\n3.\na)\n1\n√\n3\nb) 2\nc) 2\nd) −3\n2\ne) −4\n√\n3\n5\n5.\na) −t\nb) 1 −t2\nc) ±\nt\n√\n1−t2\nExercise 6 – 5: Co-functions\n1.\na) cos θ\nb)\n3\n2\n2.\na) p\nb)\np\n1 −p2\nc) −\np\n√\n1−p2\nd) p\n504\n12.1.\nIntroduction\n\nExercise 6 – 6: Reduction formulae\n1.\na) sin2 θ\nb) cos2 θ\nc)\ni. 1\nii. tan2 θ\n2.\na) sin 17◦\nb) cos 33◦\nc) tan 68◦\nd) −cos 33◦\n3.\na)\n√\n3\nb)\n√\n3\n2\nc)\n1\n4\nd) 1\nExercise 6 – 7: Solving trigonometric equations\n1.\na) α = 60◦; 300◦\nb) α = 220,5◦; 319,5◦\nc) α = 79,2◦; 259,2◦\nd) α = 200,1◦; 339,9◦\ne) α = 36,9◦; 143,1◦\nf) α = 109,7◦; 289,7◦\n2.\na) θ = −323,1◦; −216,9◦; 36,9◦; 143,1◦\nb) θ = −221,4◦; −138,6◦; 138,6◦; 221,4◦\nc) θ = −278,5◦; −98,5◦; 81,5◦; 261,5◦\nd) θ = −90◦; 270◦\ne) θ = −293,6◦; −66,4◦; 66,4◦; 293,6◦\nExercise 6 – 8: General solution\n1.\na) θ = −128,36◦; −101,64◦; 51,64◦\nb) θ = −80,45◦; −9,54◦; 99,55◦; 170,46◦\nc) θ = −53,27◦; 126,73◦\nd) α = 0◦\ne) θ = −180◦; 0◦; 180◦\nf) θ = −180◦; 180◦\ng) θ = 84◦\nh) θ = −120◦; 120◦\ni) θ = −60◦; −30◦; 120◦; 150◦\n2.\na) θ = −20◦+ n . 360◦\nb) α = 30◦+ n . 120◦\nc) β = 10,25◦+ n . 45◦or\nβ = 55,25◦+ n . 45◦\nd) α = 70◦+ n . 360◦or\nα = 340◦+ n . 360◦\ne) θ = 140◦+ n . 240◦or\nθ = 220◦+ n . 240◦\nf) β = 15◦+ n . 180◦\nExercise 6 – 9: Solving trigonometric equations\n1.\na) θ = 45◦+ k . 180◦or\nθ = 135◦+ k . 180◦\nb) α = 50◦+ k . 360◦or\nα = 110◦+ k . 360◦\nc) θ = 60◦+ k . 720◦or\nθ = 660◦+ k . 720◦\nd) β = 146,6◦+ k . 180◦\ne) θ = 110,27◦+ k . 360◦or\nθ = 249,73◦+ k . 360◦\nf) α = 210◦+ k . 360◦or\nα = 330◦+ k . 360◦\ng) β = 23,3◦+ k . 120◦\nh) θ = 122◦+ k . 180◦\ni) α = 21◦+ k . 180◦or\nα = 39,5◦+ k . 90◦\nj) β = 22,5◦+ k . 90◦\n2. θ = 0◦, 180◦, 210◦, 330◦or 360◦\n3.\na) θ = 120◦+ k . 360◦or\nθ = 240◦+ k . 360◦\nb) θ = 0◦+ k . 180◦or\nθ = 146,3◦+ k . 180◦\nc) α = 36,9◦+ k . 360◦or\nα = 143,1◦+ k . 360◦or\nα = 216,9◦+ k . 360◦or\nα = 323,1◦+ k . 360◦\nd) β = 15◦+ k . 120◦or\nβ = 75◦+ k . 120◦\ne) α = 48,4◦+ k . 180◦\nf) θ = 63,4◦+ k . 180◦or\nθ = 116,6◦+ k . 180◦\ng) θ = 54,8◦+ k . 180◦or\nθ = 95,25◦+ k . 180◦\n4. β = −70,5◦or β = 109,5◦\n505\nChapter 12.\nLinear programming\n\nExercise 6 – 10: The area rule\n1.\na)\nP\nQ\nR\n30◦\n10\n7\nArea △PQR = 17,5 square units\nb)\nP\nQ\nR\n110◦\n9\n8\nArea △PQR = 33,8 square units\n2. Area △XY Z = 645,6 square units\n3. Area = 106,5 square units\n4.\nˆC = 72,2◦or ˆC = 107,8◦\nExercise 6 – 11: Sine rule\n1.\na)\nˆP = 92◦, q = 6,6, p = 7,4\nb)\nˆL = 87◦, l = 1,3, k = 0,89\nc)\nˆB = 76,8◦, b = 94,3, c = 91,3\nd)\nˆY = 84◦, y = 60, z = 38,8\n2.\nˆB = 32◦, AB = 23, BC = 39\n3. ST = 78,1 km\n4. m = 26,2\n5. BC = 3,2\nExercise 6 – 12: The cosine rule\n1.\na) a = 8,5, ˆC = 83,9◦, ˆB = 26,1◦\nb)\nˆR = 120◦, ˆS = 32,2◦, ˆT = 27,8◦\nc)\nˆ\nM = 27,7◦, ˆL = 40,5◦, ˆ\nK = 111,8◦\nd) h = 19,1, ˆJ = 18,2◦, ˆ\nK = 31,8◦\ne)\nˆD = 34◦, ˆE = 44,4◦, ˆF = 101,6◦\n2.\na) x = 4,4 km\nb) y = 63,5 cm\n3.\na)\nˆ\nK = 117,3◦\nb)\nˆQ = 78,5◦\nExercise 6 – 13: Area, sine and cosine rule\n1.\na) 7,78 km\nb) 6 km\n2. XZ = 1,73 km, XY = 0,87 km\n3.\na) 1053 km\nb) 4,42◦\n4. DC = x sin a sin(b+c)\nsin(a+c) sin b\n5.\nb) 438,5 km\n6. 9,38 m2\n7. DC = x sin α\nsin β\n506\n12.1.\nIntroduction\n\nExercise 6 – 14: End of chapter exercises\n1. sin2 A\n2. 1 1\n4\n3. cos α\n4. 3\n7.\na) −1\nb) θ = 135◦or θ = 315◦\n8.\na)\nb\nx\ny\n0\nθ\n(−12; −5)\nb) −5\n13 and 12\n13\nc) θ = 202,62◦\n9.\na) a = 1 and b = −\n√\n3\nb) −\n√\n3\n2\n10.\na) x = 50,9◦or x = 309,1◦\nb) x = 127,3◦or x = 307,3◦\nc) x = 26,6◦; 153,4◦206,6◦or 333,4◦\n11.\na) x = 55◦+ k . 360◦or\nx = 175◦+ k . 360◦\nb) x = 180◦+ k . 360◦\n12.\na) x = 28,6◦+ k . 180◦or\nx = 61,4◦+ k . 180◦\nb)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nb\nb\nb\nb\nθ\n0◦\ny\ny = sin 2α\nc) 28,6◦; 61,4◦; 208,6◦; 241,4◦\n13.\na) A ˆGN = α −β\nb)\nˆ\nA = 90◦−α\nd) H = 5 m\n14.\na) AC = 9,43 m\nb) AD = 6,2 m\nc) Area = 49,25 m2\nd) Area = 49,23 m2\n7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "3.1" }, { "title": "Quadratic sequences", "content": "", "chapter_id": "3.2" }, { "title": "Summary", "content": "4.6\nSummary\n142\n\n4\nAnalytical geometry\nAnalytical geometry, also referred to as coordinate or Cartesian geometry, is the study\nof geometric properties and relationships between points, lines and angles in the Carte-\nsian plane. Geometrical shapes are defined using a coordinate system and algebraic\nprinciples. In this chapter we deal with the equation of a straight line, parallel and\nperpendicular lines and inclination of a line.\n4.1\nRevision\nEMBG7\nPoints A(x1; y1), B(x2; y2) and C(x2; y1) are shown in the diagram below:\nb\nb\nA(x1; y1)\nC(x2; y1)\nB(x2; y2)\nx\ny\n0\nTheorem of Pythagoras\nAB2 = AC2 + BC2\nDistance formula\nDistance between two points:\nAB =\np\n(x2 −x1)2 + (y2 −y1)2\nNotice that (x1 −x2)2 = (x2 −x1)2.\nSee video: 22JD at www.everythingmaths.co.za\nGradient\nGradient (m) describes the slope or steepness of the line joining two points. The\ngradient of a line is determined by the ratio of vertical change to horizontal change.\nmAB = y2 −y1\nx2 −x1\nor\nmAB = y1 −y2\nx1 −x2\nRemember to be consistent: m ̸= y1 −y2\nx2 −x1\n.\n104\n4.1.\nRevision\n\nHorizontal lines\nx\ny\n0\nm = 0\nVertical lines\nx\ny\n0\nm is undefined\nParallel lines\nθ\nθ\nx\ny\n0\nm1 = m2\nPerpendicular lines\nθ2\nθ1\nx\ny\n0\nm1 × m2 = −1\nMid-point of a line segment\nA(x1; y1)\nM(x; y)\nB(x2; y2)\nx\ny\n0\nThe coordinates of the mid-point M(x; y) of a line between any two points A(x1; y1)\nand B(x2; y2):\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nSee video: 22JF at www.everythingmaths.co.za\n105", "chapter_id": "3.3" }, { "title": "Analytical geometry", "content": "Chapter 4.\nAnalytical geometry\n\nPoints on a straight line\nThe diagram shows points P(x1; y1), Q(x2; y2) and R(x; y) on a straight line.\nb\nb\nx\ny\nR(x; y)\n0\nb\nQ(x2; y2)\nP(x1; y1)\nWe know that mPR = mQR = mPQ.\nUsing mPR = mPQ, we obtain the following for any point (x; y) on a straight line\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 1: Revision\nQUESTION\nGiven the points P(−5; −4) and Q(0; 6):\n1. Determine the length of the line segment PQ.\n2. Determine the mid-point T(x; y) of the line segment PQ.\n3. Show that the line passing through R(1; −3\n4) and T(x; y) is perpendicular to the\nline PQ.\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n−2\n−4\n−6\n2\n−2\n−4\n−6\nb\nb\nb\nP(−5; −4)\nT(x; y)\nQ(0; 6)\nx\ny\n0\n106\n4.1.\nRevision\n\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q (x2; y2)\nx1 = −5;\ny1 = −4;\nx2 = 0;\ny2 = 6\nWrite down the distance formula\nPQ =\np\n(x2 −x1)2 + (y2 −y1)2\n=\np\n(0 −(−5))2 + (6 −(−4))2\n=\n√\n25 + 100\n=\n√\n125\n= 5\n√\n5\nThe length of the line segment PQ is 5\n√\n5 units.\nStep 3: Write down the mid-point formula and substitute the values\nT(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nx = x1 + x2\n2\n= −5 + 0\n2\n= −5\n2\ny = y1 + y2\n2\n= −4 + 6\n2\n= 2\n2\n= 1\nThe mid-point of PQ is T(−5\n2; 1).\nStep 4: Determine the gradients of PQ and RT\nm = y2 −y1\nx2 −x1\nmPQ = 6 −(−4)\n0 −(−5)\n= 10\n5\n= 2\n107\nChapter 4.\nAnalytical geometry\n\nmRT = −3\n4 −1\n1 −(−5\n2)\n= −7\n4\n7\n2\n= −7\n4 × 2\n7\n= −1\n2\nCalculate the product of the two gradients:\nmRT × mPQ = −1\n2 × 2\n= −1\nTherefore PQ is perpendicular to RT.\nQuadrilaterals\n• A quadrilateral is a closed shape consisting of four straight line segments.\n• A parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n– Both pairs of opposite sides are equal in length.\n– Both pairs of opposite angles are equal.\n– The diagonals bisect each other.\n• A rectangle is a parallelogram that has all four angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other.\n– The diagonals are equal in length.\n108\n4.1.\nRevision\n\n• A rhombus is a parallelogram that has all four sides equal in length.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\n×\n×\n××\n•\n•\n•\n•\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals of a rhombus bisect both pairs of opposite angles.\n• A square is a rhombus that has all four interior angles equal to 90◦.\nb\nA\nb B\nb C\nb\nD\nb\n/\n/\n/\n/\n•\n•\n••\n••\n••\n– Both pairs of opposite sides are equal and parallel.\n– The diagonals bisect each other at 90◦.\n– The diagonals are equal in length.\n– The diagonals bisect both pairs of interior opposite angles (that is, all angles\nare 45◦).\n• A trapezium is a quadrilateral with one pair of opposite sides parallel.\n• A kite is a quadrilateral with two pairs of adjacent sides equal.\nA\nB\nC\nD\nb\nb\n××\n– One pair of opposite angles are equal (the angles are between unequal\nsides).\n– The diagonal between equal sides bisects the other diagonal.\n– The diagonal between equal sides bisects the interior angles.\n– The diagonals intersect at 90◦.\n109\nChapter 4.\nAnalytical geometry\n\nWorked example 2: Quadrilaterals\nQUESTION\nPoints A (−1; 0), B (0; 3), C (8; 11) and D (x; y) are points on the Cartesian plane.\nDetermine D (x; y) if ABCD is a parallelogram.\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n−1\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\nb\nb\nx\ny\nC(8; 11)\n0\nD(x; y)\nA(−1; 0)\nb\nB(0; 3)\nM\nThe mid-point of AC will be the same as the mid-point of BD. We first find the\nmid-point of AC and then use it to determine the coordinates of point D.\nStep 2: Assign values to (x1; y1) and (x2; y2)\nLet the mid-point of AC be M(x; y)\nx1 = −1;\ny1 = 0;\nx2 = 8;\ny2 = 11\nStep 3: Write down the mid-point formula\nM(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\nStep 4: Substitute the values and calculate the coordinates of M\nM(x; y) =\n\u0012−1 + 8\n2\n; 0 + 11\n2\n\u0013\n=\n\u00127\n2; 11\n2\n\u0013\n110\n4.1.\nRevision\n\nStep 5: Use the coordinates of M to determine D\nM is also the mid-point of BD so we use M\n\u0000 7\n2; 11\n2\n\u0001\nand B (0; 3) to find D (x; y)\nStep 6: Substitute values and determine x and y\nM =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n∴\n\u00127\n2; 11\n2\n\u0013\n=\n\u00120 + x\n2\n; 3 + y\n2\n\u0013\n7\n2 = 0 + x\n2\n7 = 0 + x\n∴x = 7\n11\n2 = 3 + y\n2\n11 = 3 + y\n∴y = 8\nStep 7: Alternative method: inspection\nSince we are given that ABCD is a parallelogram, we can use the properties of a\nparallelogram and the given points to determine the coordinates of D.\nFrom the sketch we expect that point D will lie below C.\nConsider the given points A, B and C:\n• Opposite sides of a parallelogram are parallel, therefore BC must be parallel to\nAD and their gradients must be equal.\n• The vertical change from B to C is 8 units up.\n• Therefore the vertical change from A to D is also 8 units up (y = 0 + 8 = 8).\n• The horizontal change from B to C is 8 units to the right.\n• Therefore the horizontal change from A to D is also 8 units to the right (x =\n−1 + 8 = 7).\nor\n• Opposite sides of a parallelogram are parallel, therefore AB must be parallel to\nDC and their gradients must be equal.\n• The vertical change from A to B is 3 units up.\n111\nChapter 4.\nAnalytical geometry\n\n• Therefore the vertical change from C to D is 3 units down (y = 11 −3 = 8).\n• The horizontal change from A to B is 1 unit to the right.\n• Therefore the horizontal change from C to D is 1 unit to the left (x = 8 −1 = 7).\nStep 8: Write the final answer\nThe coordinates of D are (7; 8).\nExercise 4 – 1: Revision\n1. Determine the length of the line segment between the following points:\na) P(−3; 5) and Q(−1; −5)\nb) R(0,75; 3) and S(0,75; −4)\nc) T(2x; y −2) and U(3x + 1; y −2)\n2. Given Q(4; 1), T(p; 3) and length QT =\n√\n8 units, determine the value of p.\n3. Determine the gradient of the line AB if:\na) A(−5; 3) and B(−7; 4)\nb) A(3; −2) and B(1; −8)\n4. Prove that the line PQ, with P(0; 3) and Q(5; 5), is parallel to the line 5y + 5 =\n2x.\n5. Given the points A(−1; −1), B(2; 5), C(−1; −5\n2) and D(x; −4) and AB ⊥CD,\ndetermine the value of x.\n6. Calculate the coordinates of the mid-point P(x; y) of the line segment between\nthe points:\na) M(3; 5) and N(−1; −1)\nb) A(−3; −4) and B(2; 3)\n7. The line joining A(−2; 4) and B(x; y) has the mid-point C(1; 3). Determine the\nvalues of x and y.\n8. Given\nquadrilateral\nABCD\nwith\nvertices\nA(0; 3), B(4; 3), C(5; −1)\nand\nD(1; −1).\na) Determine the equation of the line AD and the line BC.\nb) Show that AD ∥BC.\nc) Calculate the lengths of AD and BC.\nd) Determine the equation of the diagonal BD.\ne) What type of quadrilateral is ABCD?\n112\n4.1.\nRevision\n\n9. MPQN is a parallelogram with points M(−5; 3), P(−1; 5) and Q(4; 5). Draw a\nsketch and determine the coordinates of N(x; y).\n10. PQRS is a quadrilateral with points P(−3; 1), Q(1; 3), R(6; 1) and S(2; −1) in\nthe Cartesian plane.\na) Determine the lengths of PQ and SR.\nb) Determine the mid-point of PR.\nc) Show that PQ ∥SR.\nd) Determine the equations of the line PS and the line SR.\ne) Is PS ⊥SR? Explain your answer.\nf) What type of quadrilateral is PQRS?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22JG\n1b. 22JH\n1c. 22JJ\n2. 22JK\n3a. 22JM\n3b. 22JN\n4. 22JP\n5. 22JQ\n6a. 22JR\n6b. 22JS\n7. 22JT\n8. 22JV\n9. 22JW\n10. 22JX\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.2\nEquation of a line\nEMBG8\nWe can derive different forms of the straight line equation. The different forms are\nused depending on the information provided in the problem:\n• The two-point form of the straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• The gradient–point form of the straight line equation: y −y1 = m(x −x1)\n• The gradient–intercept form of the straight line equation: y = mx + c\nThe two-point form of the straight line equation\nEMBG9\nb\nb\n(x1; y1)\n(x2; y2)\nx\ny\n0\n113\nChapter 4.\nAnalytical geometry\n\nGiven any two points (x1; y1) and (x2; y2), we can determine the equation of the line\npassing through the two points using the equation:\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nWorked example 3: The two-point form of the straight line equation\nQUESTION\nFind the equation of the straight line passing through P (−1; −5) and Q (5; 4).\nSOLUTION\nStep 1: Draw a sketch\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nP(−1; −5)\nQ(5; 4)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nLet the coordinates of P be (x1; y1) and Q(x2; y2)\nx1 = −1;\ny1 = −5;\nx2 = 5;\ny2 = 4\nStep 3: Write down the two-point form of the straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\n114\n4.2.\nEquation of a line\n\nStep 4: Substitute the values and make y the subject of the equation\ny −(−5)\nx −(−1) = 4 −(−5)\n5 −(−1)\ny + 5\nx + 1 = 9\n6\ny + 5 = 3\n2(x + 1)\ny + 5 = 3\n2x + 3\n2\ny = 3\n2x −7\n2\nStep 5: Write the final answer\ny = 3\n2x −31\n2\nExercise 4 – 2: The two-point form of the straight line equation\nDetermine the equation of the straight line passing through the points:\n1. (3; 7) and (−6; 1)\n2. (1; −11\n4 ) and (2\n3; −7\n4)\n3. (−2; 1) and (3; 6)\n4. (2; 3) and (3; 5)\n5. (1; −5) and (−7; −5)\n6. (−4; 0) and (1; 15\n4 )\n7. (s; t) and (t; s)\n8. (−2; −8) and (1; 7)\n9. (2p; q) and (0; −q)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22JY\n2. 22JZ\n3. 22K2\n4. 22K3\n5. 22K4\n6. 22K5\n7. 22K6\n8. 22K7\n9. 22K8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n115\nChapter 4.\nAnalytical geometry\n\nThe gradient–point form of the straight line equation\nEMBGB\nWe derive the gradient–point form of the straight line equation using the definition of\ngradient and the two-point form of a straight line equation\ny −y1\nx −x1\n= y2 −y1\nx2 −x1\nSubstitute m = y2 −y1\nx2 −x1\non the right-hand side of the equation\ny −y1\nx −x1\n= m\nMultiply both sides of the equation by (x −x1)\ny −y1 = m(x −x1)\nTo use this equation, we need to know the gradient of the line and the coordinates of\none point on the line.\nSee video: 22K9 at www.everythingmaths.co.za\nWorked example 4: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −1\n3 and passing through\nthe point (−1; 1).\nSOLUTION\nStep 1: Draw a sketch\nWe notice that m < 0, therefore the graph decreases as x increases.\n1\n2\n3\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\n(−1; 1)\nx\ny\n0\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\n116\n4.2.\nEquation of a line\n\nSubstitute the value of the gradient\ny −y1 = −1\n3(x −x1)\nSubstitute the coordinates of the given point\ny −1 = −1\n3(x −(−1))\ny −1 = −1\n3(x + 1)\ny = −1\n3x −1\n3 + 1\n= −1\n3x + 2\n3\nStep 3: Write the final answer\nThe equation of the straight line is y = −1\n3x + 2\n3.\nIf we are given two points on a straight line, we can also use the gradient–point form\nto determine the equation of a straight line. We first calculate the gradient using the\ntwo given points and then substitute either of the two points into the gradient–point\nform of the equation.\nWorked example 5: The gradient–point form of the straight line equation\nQUESTION\nDetermine the equation of the straight line passing through (−3; 2) and (5; 8).\nSOLUTION\nStep 1: Draw a sketch\n2\n4\n6\n8\n2\n4\n6\n−2\n−4\nb\nb\n(−3; 2)\n(5; 8)\nx\ny\n0\n117\nChapter 4.\nAnalytical geometry\n\nStep 2: Assign variables to the coordinates of the given points\nx1 = −3;\ny1 = 2;\nx2 = 5;\ny2 = 8\nStep 3: Calculate the gradient using the two given points\nm = y2 −y1\nx2 −x1\n=\n8 −2\n5 −(−3)\n= 6\n8\n= 3\n4\nStep 4: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the value of the gradient\ny −y1 = 3\n4(x −x1)\nSubstitute the coordinates of a given point\ny −y1 = 3\n4(x −x1)\ny −2 = 3\n4(x −(−3))\ny −2 = 3\n4(x + 3)\ny = 3\n4x + 9\n4 + 2\n= 3\n4x + 17\n4\nStep 5: Write the final answer\nThe equation of the straight line is y = 3\n4x + 41\n4.\nSee video: 22KB at www.everythingmaths.co.za\n118\n4.2.\nEquation of a line\n\nExercise 4 – 3: Gradient–point form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (−1; 10\n3 ) and with m = 2\n3.\n2. with m = −1 and passing through the point (−2; 0).\n3. passing through the point (3; −1) and with m = −1\n3.\n4. parallel to the x-axis and passing through the point (0; 11).\n5. passing through the point (1; 5) and with m = −2.\n6. perpendicular to the x-axis and passing through the point (−3\n2; 0).\n7. with m = −0,8 and passing through the point (10; −7).\n8. with undefined gradient and passing through the point (4; 0).\n9. with m = 3a and passing through the point (−2; −6a + b).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KC\n2. 22KD\n3. 22KF\n4. 22KG\n5. 22KH\n6. 22KJ\n7. 22KK\n8. 22KM\n9. 22KN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe gradient–intercept form of a straight line equation EMBGC\nUsing the gradient–point form, we can also derive the gradient–intercept form of the\nstraight line equation.\nStarting with the equation\ny −y1 = m(x −x1)\nExpand the brackets and make y the subject of the formula\ny −y1 = mx −mx1\ny = mx −mx1 + y1\ny = mx + (y1 −mx1)\nWe define constant c such that c = y1 −mx1 so that we get the equation\ny = mx + c\nThis is also called the standard form of the straight line equation.\n119\nChapter 4.\nAnalytical geometry\n\nNotice that when x = 0, we have\ny = m(0) + c\n= c\nTherefore c is the y-intercept of the straight line.\nWorked example 6: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line with gradient m = −2 and passing through\nthe point (−1; 7).\nSOLUTION\nStep 1: Slope of the line\nWe notice that m < 0, therefore the graph decreases as x increases.\n2\n4\n6\n8\n−2\n2\n4\n−2\n−4\nb\n(−1; 7)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\nSubstitute the value of the gradient\ny = −2x + c\n120\n4.2.\nEquation of a line\n\nSubstitute the coordinates of the given point and find c\ny = −2x + c\n7 = −2(−1) + c\n7 −2 = c\n∴c = 5\nThis gives the y-intercept (0; 5).\nStep 3: Write the final answer\nThe equation of the straight line is y = −2x + 5.\nIf we are given two points on a straight line, we can also use the gradient–intercept\nform to determine the equation of a straight line. We solve for the two unknowns m\nand c using simultaneous equations — using the methods of substitution or elimina-\ntion.\nWorked example 7: The gradient–intercept form of straight line equation\nQUESTION\nDetermine the equation of the straight line passing through the points (−2; −7) and\n(3; 8).\nSOLUTION\nStep 1: Draw a sketch\n4\n8\n−4\n−8\n2\n4\n−2\n−4\nb\nb\n(−2; −7)\n(3; 8)\nx\ny\n0\nStep 2: Write down the gradient–intercept form of straight line equation\ny = mx + c\n121\nChapter 4.\nAnalytical geometry\n\nStep 3: Substitute the coordinates of the given points\n−7 = m(−2) + c\n−7 = −2m + c\n. . . (1)\n8 = m(3) + c\n8 = 3m + c\n. . . (2)\nWe have two equations with two unknowns; we can therefore solve using simultane-\nous equations.\nStep 4: Make the coefficient of one of the variables the same in both equations\nWe notice that the coefficient of c in both equations is 1, therefore we can subtract\none equation from the other to eliminate c:\n−7 = −2m + c\n−(8 = 3m + c)\n−15 = −5m\n∴3 = m\nSubstitute m = 3 into either of the two equations and determine c:\n−7 = −2m + c\n−7 = −2(3) + c\n∴c = −1\nor\n8 = 3m + c\n8 = 3(3) + c\n∴c = −1\nStep 5: Write the final answer\nThe equation of the straight line is y = 3x −1.\n122\n4.2.\nEquation of a line\n\nExercise 4 – 4: The gradient–intercept form of a straight line equation\nDetermine the equation of the straight line:\n1. passing through the point (1\n2; 4) and\nwith m = 2.\n2. passing through the points (1\n2; −2)\nand (2; 4).\n3. passing through the points (2; −3)\nand (−1; 0).\n4. passing through the point (2; −6\n7)\nand with m = −3\n7.\n5. which cuts the y-axis at y = −1\n5 and\nwith m = 1\n2.\n6.\nb\nb\n(−1; −4)\n(2; 2)\nx\ny\n0\n7.\nb −3\n2\nx\ny\n0\n8.\nb\n(−2; −2)\n4\nx\ny\n0\n9.\nb (−2; 10)\nx\ny\n0\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22KP\n2. 22KQ\n3. 22KR\n4. 22KS\n5. 22KT\n6. 22KV\n7. 22KW\n8. 22KX\n9. 22KY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n123\nChapter 4.\nAnalytical geometry\n\n4.3\nInclination of a line\nEMBGD\n1\n2\n3\n1\n2\n3\nθ\n∆y\n∆x\nx\ny\nThe diagram shows that a straight line makes an angle θ with the positive x-axis. This\nis called the angle of inclination of a straight line.\nWe notice that if the gradient changes, then the value of θ also changes, therefore the\nangle of inclination of a line is related to its gradient. We know that gradient is the\nratio of a change in the y-direction to a change in the x-direction:\nm = ∆y\n∆x\nFrom trigonometry we know that the tangent function is defined as the ratio:\ntan θ = opposite side\nadjacent side\nAnd from the diagram we see that\ntan θ = ∆y\n∆x\n∴m = tan θ\nfor 0◦≤θ < 180◦\nTherefore the gradient of a straight line is equal to the tangent of the angle formed\nbetween the line and the positive direction of the x-axis.\nVertical lines\n• θ = 90◦\n• Gradient is undefined since there is no change in the x-values (∆x = 0).\n• Therefore tan θ is also undefined (the graph of tan θ has an asymptote at θ =\n90◦).\n124\n4.3.\nInclination of a line\n\nHorizontal lines\n• θ = 0◦\n• Gradient is equal to 0 since there is no change in the y-values (∆y = 0).\n• Therefore tan θ is also equal to 0 (the graph of tan θ passes through the origin\n(0◦; 0).\nLines with negative gradients\nIf a straight line has a negative gradient (m < 0, tan θ < 0), then the angle formed\nbetween the line and the positive direction of the x-axis is obtuse.\nθ\nx\ny\n0\nFrom the CAST diagram in trigonometry, we know that the tangent function is negative\nin the second and fourth quadrant. If we are calculating the angle of inclination for a\nline with a negative gradient, we must add 180◦to change the negative angle in the\nfourth quadrant to an obtuse angle in the second quadrant:\nIf we are given a straight line with gradient m = −0,7, then we can determine the\nangle of inclination using a calculator:\ntan θ = m\n= −0,7\n∴θ = tan−1(−0,7)\n= −35,0◦\nThis negative angle lies in the fourth quadrant. We must add 180◦to get an obtuse\nangle in the second quadrant:\nθ = −35,0◦+ 180◦\n= 145◦\n125\nChapter 4.\nAnalytical geometry\n\nAnd we can always use our calculator to check that the obtuse angle θ = 145◦gives a\ngradient of m = −0,7.\n35◦\n180◦−35◦= 145◦\nx\ny\n0\nExercise 4 – 5: Angle of inclination\n1. Determine the gradient (correct to 1 decimal place) of each of the following\nstraight lines, given that the angle of inclination is equal to:\na) 60◦\nb) 135◦\nc) 0◦\nd) 54◦\ne) 90◦\nf) 45◦\ng) 140◦\nh) 180◦\ni) 75◦\n2. Determine the angle of inclination (correct to 1 decimal place) for each of the\nfollowing:\na) a line with m = 3\n4\nb) 2y −x = 6\nc) the line passes through the points (−4; −1) and (2; 5)\nd) y = 4\ne) x = 3y + 1\n2\nf) x = −0,25\ng) the line passes through the points (2; 5) and (2\n3; 1)\nh) a line with gradient equal to 0,577\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22KZ\n1b. 22M2\n1c. 22M3\n1d. 22M4\n1e. 22M5\n1f. 22M6\n1g. 22M7\n1h. 22M8\n1i. 22M9\n2a. 22MB\n2b. 22MC\n2c. 22MD\n2d. 22MF\n2e. 22MG\n2f. 22MH\n2g. 22MJ\n2h. 22MK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n126\n4.3.\nInclination of a line\n\nWorked example 8: Inclination of a straight line\nQUESTION\nDetermine the angle of inclination (correct to 1 decimal place) of the straight line\npassing through the points (2; 1) and (−3; −9).\nSOLUTION\nStep 1: Draw a sketch\n2\n−2\n−4\n−6\n−8\n1\n2\n−1\n−2\n−3\nb\nb\n(2; 1)\n(−3; −9)\nx\ny\n0\nStep 2: Assign variables to the coordinates of the given points\nx1 = 2;\ny1 = 1;\nx2 = −3;\ny2 = −9\nStep 3: Determine the gradient of the line\nm = y2 −y1\nx2 −x1\n= −9 −1\n−3 −2\n= −10\n−5\n∴m = 2\nStep 4: Use the gradient to determine the angle of inclination of the line\ntan θ = m\n= 2\n∴θ = tan−1 2\n= 63,4◦\nImportant: make sure your calculator is in DEG (degrees) mode.\nStep 5: Write the final answer\nThe angle of inclination of the straight line is 63,4◦.\n127\nChapter 4.\nAnalytical geometry\n\nWorked example 9: Inclination of a straight line\nQUESTION\nDetermine the equation of the straight line passing through the point (3; 1) and with\nan angle of inclination of 135◦.\nSOLUTION\nStep 1: Use the angle of inclination to determine the gradient of the line\nm = tan θ\n= tan 135◦\n∴m = −1\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = −1\ny −y1 = −(x −x1)\nSubstitute the given point (3; 1)\ny −1 = −(x −3)\ny = −x + 3 + 1\n= −x + 4\nStep 3: Write the final answer\nThe equation of the straight line is y = −x + 4.\nWorked example 10: Inclination of a straight line\nQUESTION\nDetermine the acute angle (correct to 1 decimal place) between the line passing\nthrough the points M(−1; 13\n4) and N(4; 3) and the straight line y = −3\n2x + 4.\nSOLUTION\nStep 1: Draw a sketch\nDraw the line through points M(−1; 13\n4) and N(4; 3) and the line y = −3\n2x + 4 on a\nsuitable system of axes. Label α and β, the angles of inclination of the two lines. Label\nθ, the acute angle between the two straight lines.\n128\n4.3.\nInclination of a line\n\n2\n4\n6\n−2\n2\n4\n−2\n−4\n−6\n−8\nb\nb\nβ\nˆB1\nα\nθ\nx\ny\n0\nM(−1; 7\n4)\nN(4; 3)\nNotice that α and θ are acute angles and β is an obtuse angle.\nˆB1 = 180◦−β\n(∠on str. line)\nand θ = α + ˆB1\n(ext. ∠of △= sum int. opp)\n∴θ = α + (180◦−β)\n= 180◦+ α −β\nStep 2: Use the gradient to determine the angle of inclination β\nFrom the equation y = −3\n2x + 4 we see that m < 0, therefore β is an obtuse angle\nsuch that 90◦< β < 180◦.\ntan β = m\n= −3\n2\ntan−1\n\u0012\n−3\n2\n\u0013\n= −56,3◦\nThis negative angle lies in the fourth quadrant. We know that the angle of inclination\nβ is an obtuse angle that lies in the second quadrant, therefore\nβ = −56,3◦+ 180◦\n= 123,7◦\nStep 3: Determine the gradient and angle of inclination of the line through M and\nN\n129\nChapter 4.\nAnalytical geometry\n\nDetermine the gradient\nm = y2 −y1\nx2 −x1\n=\n3 −7\n4\n4 −(−1)\n=\n5\n4\n5\n= 1\n4\nDetermine the angle of inclination\ntan α = m\n= 1\n4\n∴α = tan−1\n\u00121\n4\n\u0013\n= 14,0◦\nStep 4: Write the final answer\nθ = 180◦+ α −β\n= 180◦+ 14,0◦−123,7◦\n= 70,3◦\nThe acute angle between the two straight lines is 70,3◦.\nExercise 4 – 6: Inclination of a straight line\n1. Determine the angle of inclination for each of the following:\na) a line with m = 4\n5\nb) x + y + 1 = 0\nc) a line with m = 5,69\nd) the line that passes through (1; 1) and (−2; 7)\ne) 3 −2y = 9x\nf) the line that passes through (−1; −6) and (−1\n2; −11\n2 )\ng) 5 = 10y −15x\n130\n4.3.\nInclination of a line\n\nh)\nb\nx\ny\n(2; 3)\n−1\n0\ni)\nb\nx\ny\n(6; 0)\n2\n0\nj)\nb\nx\ny\n(−3; 3)\n−3\n0\n2. Determine the acute angle between the line passing through the points A(−2; 1\n5)\nand B(0; 1) and the line passing through the points C(1; 0) and D(−2; 6).\n3. Determine the angle between the line y + x = 3 and the line x = y + 1\n2.\n4. Find the angle between the line y = 2x and the line passing through the points\n(−1; 7\n3) and (0; 2).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22MM\n1b. 22MN\n1c. 22MP\n1d. 22MQ\n1e. 22MR\n1f. 22MS\n1g. 22MT\n1h. 22MV\n1i. 22MW\n1j. 22MX\n2. 22MY\n3. 22MZ\n4. 22N2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n131\nChapter 4.\nAnalytical geometry\n\n4.4\nParallel lines\nEMBGF\nInvestigation: Parallel lines\n1. Draw a sketch of the line passing through the points P(−1; 0) and Q(1; 4) and\nthe line passing through the points R(1; 2) and S(2; 4).\n2. Label and measure α and β, the angles of inclination of straight lines PQ and\nRS respectively.\n3. Describe the relationship between α and β.\n4. “α and β are alternate angles, therefore PQ ∥RS.” Is this a true statement? If\nnot, provide a correct statement.\n5. Use your calculator to determine tan α and tan β.\n6. Complete the sentence: . . . . . . lines have . . . . . . angles of inclination.\n7. Determine the equations of the straight lines PQ and RS.\n8. What do you notice about mPQ and mRS?\n9. Complete the sentence: . . . . . . lines have . . . . . . gradients.\nAnother method of determining the equation of a straight line is to be given a point on\nthe unknown line, (x1; y1), and the equation of a line which is parallel to the unknown\nline.\nLet the equation of the unknown line be y = m1x + c1 and the equation of the given\nline be y = m2x + c2.\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nθ\nθ\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are parallel then\nm1 = m2\n132\n4.4.\nParallel lines\n\nImportant: when determining the gradient of a line using the coefficient of x, make\nsure the given equation is written in the gradient–intercept (standard) form. y = mx+c\nSubstitute the value of m2 and the given point (x1; y1), into the gradient–intercept form\nof a straight line equation\ny −y1 = m(x −x1)\nand determine the equation of the unknown line.\nWorked example 11: Parallel lines\nQUESTION\nDetermine the equation of the line that passes through the point (−1; 1) and is parallel\nto the line y −2x + 1 = 0.\nSOLUTION\nStep 1: Write the equation in gradient–intercept form\nWe write the given equation in gradient–intercept form and determine the value of m.\ny = 2x −1\nWe know that the two lines are parallel, therefore m1 = m2 = 2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m = 2\ny −y1 = 2(x −x1)\nSubstitute the given point (−1; 1)\ny −1 = 2(x −(−1))\ny −1 = 2x + 2\ny = 2x + 2 + 1\n= 2x + 3\n133\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\n(−1; 1)\ny = 2x −1\ny = 2x + 3\nx\ny\n0\nA sketch was not required, but it is always helpful and can be used to check answers.\nStep 3: Write the final answer\nThe equation of the straight line is y = 2x + 3.\nWorked example 12: Parallel lines\nQUESTION\nLine AB passes through the point A(0; 3) and has an angle of inclination of 153,4◦.\nDetermine the equation of the line CD which passes through the point C(2; −3) and\nis parallel to AB.\nSOLUTION\nStep 1: Use the given angle of inclination to determine the gradient\nmAB = tan θ\n= tan 153,4◦\n= −0,5\nStep 2: Parallel lines have equal gradients\nSince we are given AB ∥CD,\nmCD = mAB = −0,5\n134\n4.4.\nParallel lines\n\nStep 3: Write down the gradient–point form of a straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient mCD = −0,5.\ny −y1 = −1\n2(x −x1)\nSubstitute the given point (2; −3).\ny −(−3) = −1\n2(x −2)\ny + 3 = −1\n2x + 1\ny = −1\n2x −2\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\nb C(2; −3)\ny = −1\n2x −2\ny = −1\n2x + 3\nx\ny\n0\nA sketch was not required, but it is always useful.\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n2x −2.\nSee video: 22N3 at www.everythingmaths.co.za\n135\nChapter 4.\nAnalytical geometry\n\nExercise 4 – 7: Parallel lines\n1. Determine whether or not the following two lines are parallel:\na) y + 2x = 1 and −2x + 3 = y\nb)\ny\n3 + x + 5 = 0 and 2y + 6x = 1\nc) y = 2x −7 and the line passing through (1; −2) and (1\n2; −1)\nd) y + 1 = x and x + y = 3\ne) The line passing through points (−2; −1) and (−4; −3) and the line −y +\nx −4 = 0\nf) y −1 = 1\n3x and the line passing through points (−2; 4) and (1; 5)\n2. Determine the equation of the straight line that passes through the point (1; −5)\nand is parallel to the line y + 2x −1 = 0.\n3. Determine the equation of the straight line that passes through the point (−2; −6)\nand is parallel to the line 2y + 1 = 6x.\n4. Determine the equation of the straight line that passes through the point (−2; −2)\nand is parallel to the line with angle of inclination θ = 56,31◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is parallel to the line with angle of inclination θ = 145◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22N4\n1b. 22N5\n1c. 22N6\n1d. 22N7\n1e. 22N8\n1f. 22N9\n2. 22NB\n3. 22NC\n4. 22ND\n5. 22NF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.5\nPerpendicular lines\nEMBGG\nInvestigation: Perpendicular lines\n1. Draw a sketch of the line passing through the points A(−2; −3) and B(2; 5) and\nthe line passing through the points C(−1; 1\n2) and D(4; −2).\n2. Label and measure α and β, the angles of inclination of straight lines AB and\nCD respectively.\n3. Label and measure θ, the angle between the lines AB and CD.\n4. Describe the relationship between the lines AB and CD.\n5. “θ is a reflex angle, therefore AB ⊥CD.” Is this a true statement? If not, provide\na correct statement.\n136\n4.5.\nPerpendicular lines\n\n6. Determine the equation of the straight line AB and the line CD.\n7. Use your calculator to determine tan α × tan β.\n8. Determine mAB × mCD.\n9. What do you notice about these products?\n10. Complete the sentence: if two lines are . . . . . . to each other, then the product of\ntheir . . . . . . is equal . . . . . .\n11. Complete the sentence: if the gradient of a straight line is equal to the negative\n. . . . . . of the gradient of another straight line, then the two lines are . . . . . .\nDeriving the formula: m1 × m2 = −1\nb\nb\nA(4; 3)\nB(−3; 4)\nθ\n90◦+ θ\nO\ny\nx\nConsider the point A(4; 3) on the Cartesian plane with an angle of inclination A ˆOX =\nθ. Rotate through an angle of 90◦and place point B at (−3; 4) so that we have the\nangle of inclination B ˆOX = 90◦+ θ.\nWe determine the gradient of OA:\nmOA = y2 −y1\nx2 −x1\n= 3 −0\n4 −0\n= 3\n4\nAnd determine the gradient of OB:\nmOB = y2 −y1\nx2 −x1\n= 4 −0\n−3 −0\n= 4\n−3\n137\nChapter 4.\nAnalytical geometry\n\nBy rotating through an angle of 90◦we know that OB ⊥OA:\nmOA × mOB = 3\n4 × 4\n−3\n= −1\nWe can also write that\nmOA = −\n1\nmOB\nb\nb\nA(x; y)\nB(−y; x)\nθ\n90◦+ θ\nO\ny\nx\nIf we have the general point A(x; y) with an angle of inclination A ˆOX = θ and point\nB(−y; x) such that B ˆOX = 90◦+ θ, then we know that\nmOA = y\nx\nmOB = −x\ny\n∴mOA × mOB = y\nx × −x\ny\n= −1\nAnother method of determining the equation of a straight line is to be given a point on\nthe line, (x1; y1), and the equation of a line which is perpendicular to the unknown\nline. Let the equation of the unknown line be y = m1x + c1 and the equation of the\ngiven line be y = m2x + c2.\nθ2\nθ1\nx\ny\n0\ny = m1x + c1\ny = m2x + c2\nIf the two lines are perpendicular then\nm1 × m2 = −1\nNote: this rule does not apply to vertical or horizontal lines.\n138\n4.5.\nPerpendicular lines\n\nWhen determining the gradient of a line using the coefficient of x, make sure the\ngiven equation is written in the gradient–intercept (standard) form y = mx + c. Then\nwe know that\nm1 = −1\nm2\nSubstitute the value of m1 and the given point (x1; y1), into the gradient–intercept form\nof the straight line equation y −y1 = m(x −x1) and determine the equation of the\nunknown line.\nWorked example 13: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point T(2; 2) and per-\npendicular to the line 3y + 2x −6 = 0.\nSOLUTION\nStep 1: Write the equation in standard form\nLet the gradient of the unknown line be m1 and the given gradient be m2. We write\nthe given equation in gradient–intercept form and determine the value of m2.\n3y + 2x −6 = 0\n3y = −2x + 6\ny = −2\n3x + 2\n∴m2 = −2\n3\nWe know that the two lines are perpendicular, therefore m1 × m2 = −1. Therefore\nm1 = 3\n2.\nStep 2: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute m1 = 3\n2.\ny −y1 = 3\n2(x −x1)\nSubstitute the given point T(2; 2).\ny −2 = 3\n2(x −2)\ny −2 = 3\n2x −3\ny = 3\n2x −1\n139\nChapter 4.\nAnalytical geometry\n\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\nb T(2; 2)\ny = −2\n3x + 2\ny = 3\n2x −1\nx\ny\n0\nA sketch was not required, but it is useful for checking the answer.\nStep 3: Write the final answer\nThe equation of the straight line is y = 3\n2x −1.\nWorked example 14: Perpendicular lines\nQUESTION\nDetermine the equation of the straight line passing through the point (2; 1\n3) and per-\npendicular to the line with an angle of inclination of 71,57◦.\nSOLUTION\nStep 1: Use the given angle of inclination to determine gradient\nLet the gradient of the unknown line be m1 and let the given gradient be m2.\nm2 = tan θ\n= tan 71,57◦\n= 3,0\nStep 2: Determine the unknown gradient\nSince we are given that the two lines are perpendicular,\nm1 × m2 = −1\n∴m1 = −1\n3\n140\n4.5.\nPerpendicular lines\n\nStep 3: Write down the gradient–point form of the straight line equation\ny −y1 = m(x −x1)\nSubstitute the gradient m1 = −1\n3.\ny −y1 = −1\n3(x −x1)\nSubstitute the given point (2; 1\n3).\ny −\n\u00121\n3\n\u0013\n= −1\n3(x −2)\ny −1\n3 = −1\n3x + 2\n3\ny = −1\n3x + 1\nStep 4: Write the final answer\nThe equation of the straight line is y = −1\n3x + 1.\nSee video: 22NG at www.everythingmaths.co.za\nExercise 4 – 8: Perpendicular lines\n1. Calculate whether or not the following two lines are perpendicular:\na) y −1 = 4x and 4y + x + 2 = 0\nb) 10x = 5y −1 and 5y −x −10 = 0\nc) x = y −5 and the line passing through (−1; 5\n4) and (3; −11\n4 )\nd) y = 2 and x = 1\ne)\ny\n3 = x and 3y + x = 9\nf) 1 −2x = y and the line passing through (2; −1) and (−1; 5)\ng) y = x + 2 and 2y + 1 = 2x\n2. Determine the equation of the straight line that passes through the point (−2; −4)\nand is perpendicular to the line y + 2x = 1.\n3. Determine the equation of the straight line that passes through the point (2; −7)\nand is perpendicular to the line 5y −x = 0.\n141\nChapter 4.\nAnalytical geometry\n\n4. Determine the equation of the straight line that passes through the point (3; −1)\nand is perpendicular to the line with angle of inclination θ = 135◦.\n5. Determine the equation of the straight line that passes through the point (−2; 2\n5)\nand is perpendicular to the line y = 4\n3.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NH\n1b. 22NJ\n1c. 22NK\n1d. 22NM\n1e. 22NN\n1f. 22NP\n1g. 22NQ\n2. 22NR\n3. 22NS\n4. 22NT\n5. 22NV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n4.6\nSummary\nEMBGH\nSee presentation: 22NW at www.everythingmaths.co.za\n• Distance between two points: d =\np\n(x2 −x1)2 + (y2 −y1)2\n• Gradient of a line between two points: m = y2 −y1\nx2 −x1\n• Mid-point of a line: M(x; y) =\n\u0012x1 + x2\n2\n; y1 + y2\n2\n\u0013\n• Parallel lines: m1 = m2\n• Perpendicular lines: m1 × m2 = −1\n• General form of a straight line equation: ax + by + c = 0\n• Two-point form of a straight line equation: y −y1\nx −x1\n= y2 −y1\nx2 −x1\n• Gradient–point form of a straight line equation: y −y1 = m(x −x1)\n• Gradient–intercept form of a straight line equation (standard form): y = mx + c\n• Angle of inclination of a straight line: θ, the angle formed between the line and\nthe positive x-axis; m = tan θ\n142\n4.6.\nSummary\n\nExercise 4 – 9: End of chapter exercises\n1. Determine the equation of the line:\na) through points (−1; 3) and (1; 4)\nb) through points (7; −3) and (0; 4)\nc) parallel to y = 1\n2x + 3 and passing through (−2; 3)\nd) perpendicular to y = −1\n2x + 3 and passing through (−1; 2)\ne) perpendicular to 3y + x = 6 and passing through the origin\n2. Determine the angle of inclination of the following lines:\na) y = 2x −3\nb) y = 1\n3x −7\nc) 4y = 3x + 8\nd) y = −2\n3x + 3\ne) 3y + x −3 = 0\n3. P(2; 3), Q(−4; 0) and R(5; −3) are the vertices of △PQR in the Cartesian plane.\nPR intersects the x-axis at S. Determine the following:\na) the equation of the line PR\nb) the coordinates of point S\nc) the angle of inclination of PR (correct to two decimal places)\nd) the gradient of line PQ\ne) Q ˆPR\nf) the equation of the line perpendicular to PQ and passing through the origin\ng) the mid-point M of QR\nh) the equation of the line parallel to PR and passing through point M\n4. Points A(−3; 5), B(−7; −4) and C(2; 0) are given.\na) Plot the points on the Cartesian plane.\nb) Determine the coordinates of D if ABCD is a parallelogram.\nc) Prove that ABCD is a rhombus.\n5.\nb\nb\nb\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\nx\ny\n0\nM\nN\nP\n143\nChapter 4.\nAnalytical geometry\n\nConsider the sketch above, with the following lines shown:\ny = −x −3\ny = 3\ny = 5\n2x + 1\n2\na) Determine the coordinates of the point N.\nb) Determine the coordinates of the point P.\nc) Determine the equation of the vertical line MN.\nd) Determine the length of the vertical line MN.\ne) Find M ˆNP.\nf) Determine the equation of the line parallel to NP and passing through the\npoint M.\n6. The following points are given: A(−2; 3), B(2; 4), C(3; 0).\na) Plot the points on the Cartesian plane.\nb) Prove that △ABC is a right-angled isosceles triangle.\nc) Determine the equation of the line AB.\nd) Determine the coordinates of D if ABCD is a square.\ne) Determine the coordinates of E, the mid-point of BC.\n7. Given points S(2; 5), T(−3; −4) and V (4; −2).\na) Determine the equation of the line ST.\nb) Determine the size of T ˆSV .\n8. Consider triangle FGH with vertices F(−1; 3), G(2; 1) and H(4; 4).\na) Sketch △FGH on the Cartesian plane.\nb) Show that △FGH is an isosceles triangle.\nc) Determine the equation of the line PQ, perpendicular bisector of FH.\nd) Does G lie on the line PQ?\ne) Determine the equation of the line parallel to GH and passing through\npoint F.\n9. Given the points A(−1; 5), B(5; −3) and C(0; −6). M is the mid-point of AB\nand N is the mid-point of AC.\na) Draw a sketch on the Cartesian plane.\nb) Show that the coordinates of M and N are (2; 1) and (−1\n2; −1\n2) respectively.\nc) Use analytical geometry methods to prove the mid-point theorem. (Prove\nthat NM ∥CB and NM = 1\n2CB.)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22NX\n1b. 22NY\n1c. 22NZ\n1d. 22P2\n1e. 22P3\n2a. 22P4\n2b. 22P5\n2c. 22P6\n2d. 22P7\n2e. 22P8\n3. 22P9\n4. 22PB\n5. 22PC\n6. 22PD\n7. 22PF\n8. 22PG\n9. 22PH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n144\n4.6.\nSummary\n\nCHAPTER\n5\nFunctions\n5.1\nQuadratic functions\n146\n5.2\nAverage gradient\n164\n5.3\nHyperbolic functions\n170\n5.4\nExponential functions\n184\n5.5\nThe sine function\n197\n5.6\nThe cosine function\n209\n5.7\nThe tangent function\n222\n5.8\nSummary\n235\n\n5\nFunctions\nA function describes a specific relationship between two variables; where an indepen-\ndent (input) variable has exactly one dependent (output) variable. Every element in the\ndomain maps to only one element in the range. Functions can be one-to-one relations\nor many-to-one relations. A many-to-one relation associates two or more values of the\nindependent variable with a single value of the dependent variable. Functions allow\nus to visualise relationships in the form of graphs, which are much easier to read and\ninterpret than lists of numbers.\n5.1\nQuadratic functions\nEMBGJ\nRevision\nEMBGK\nFunctions of the form y = ax2 + q\nFunctions of the general form y = ax2 + q are called parabolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = ax2 + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\nThe turning point of f(x) is\nabove the x-axis.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\nThe turning point of f(x) is be-\nlow the x-axis.\n– q is also the y-intercept of the\nparabola.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\n• The effect of a on shape\n– For a > 0; the graph of f(x) is a “smile” and has a minimum turning point\n(0; q). As the value of a becomes larger, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n– For a < 0; the graph of f(x) is a “frown” and has a maximum turning point\n(0; q). As the value of a becomes smaller, the graph becomes narrower.\nAs a gets closer to 0, f(x) becomes wider.\n146\n5.1.\nQuadratic functions\n\nExercise 5 – 1: Revision\n1. On separate axes, accurately draw each of the following functions.\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = x2\nb) y2 = 1\n2x2\nc) y3 = −x2 −1\nd) y4 = −2x2 + 4\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nyint = 0\nvalue of a\na = 1\neffect of a\nstandard\nparabola\nturning point\n(0; 0)\naxis of symmetry\nx = 0\n(y-axis)\ndomain\n{x : x ∈R}\nrange\n{y : y ≥0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22PJ\n1b. 22PK\n1c. 22PM\n1d. 22PN\n2. 22PP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22PQ at www.everythingmaths.co.za\n147\nChapter 5.\nFunctions\n\nFunctions of the form y = a(x + p)2 + q\nEMBGM\nWe now consider parabolic functions of the form y = a(x + p)2 + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a parabolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = x2\nb) y2 = (x −2)2\nc) y3 = (x −1)2\nd) y4 = (x + 1)2\ne) y5 = (x + 2)2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = x2 + 2\nb) y2 = (x −2)2 −1\nc) y3 = (x −1)2 + 1\nd) y4 = (x + 1)2 + 1\ne) y5 = (x + 2)2 −1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nx-intercept(s)\ny-intercept\nturning point\naxis of symmetry\ndomain\nrange\neffect of q\n3. Consider the three functions given below and answer the questions that follow:\n• y1 = (x −2)2 + 1\n• y2 = 2(x −2)2 + 1\n• y3 = −1\n2(x −2)2 + 1\na) What is the value of a for y2?\nb) Does y1 have a minimum or maximum turning point?\n148\n5.1.\nQuadratic functions\n\nc) What are the coordinates of the turning point of y2?\nd) Compare the graphs of y1 and y2. Discuss the similarities and differences.\ne) What is the value of a for y3?\nf) Will the graph of y3 be narrower or wider than the graph of y1?\ng) Determine the coordinates of the turning point of y3.\nh) Compare the graphs of y1 and y3. Describe any differences.\nSee video: 22PR at www.everythingmaths.co.za\nThe effect of the parameters on y = a(x + p)2 + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects whether the turning point is to the left of the y-axis (p > 0)\nor to the right of the y-axis (p < 0). The axis of symmetry is the line x = −p.\nThe effect of q is a vertical shift. The value of q affects whether the turning point of the\ngraph is above the x-axis (q > 0) or below the x-axis (q < 0).\nThe value of a affects the shape of the graph. If a < 0, the graph is a “frown” and has\na maximum turning point. If a > 0 then the graph is a “smile” and has a minimum\nturning point. When a = 0, the graph is a horizontal line y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\nSee simulation: 22PS at www.everythingmaths.co.za\n149\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form f(x) = y = a(x + p)2 + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative. If a > 0\nwe have:\n(x + p)2 ≥0\n(perfect square is always positive)\n∴a(x + p)2 ≥0\n(a is positive)\n∴a(x + p)2 + q ≥q\n∴f(x) ≥q\nThe range is therefore {y : y ≥q, y ∈R} if a > 0. Similarly, if a < 0, the range is\n{y : y ≤q, y ∈R}.\nWorked example 1: Domain and range\nQUESTION\nState the domain and range for g(x) = −2(x −1)2 + 3.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n(x −1)2 ≥0\n−2(x −1)2 ≤0\n−2(x −1)2 + 3 ≤3\ng(x) ≤3\nTherefore the range is {g(x) : g(x) ≤3} or in interval notation (−∞; 3].\nNotice in the example above that it helps to have the function in the form y = a(x +\np)2 + q.\nWe use the method of completing the square to write a quadratic function of the\ngeneral form y = ax2 + bx + c in the form y = a(x + p)2 + q (see Chapter 2).\n150\n5.1.\nQuadratic functions\n\nExercise 5 – 2: Domain and range\nGive the domain and range for each of the following functions:\n1. f(x) = (x −4)2 −1\n2. g(x) = −(x −5)2 + 4\n3. h(x) = x2 −6x + 9\n4. j(x) = −2(x + 1)2\n5. k(x) = −x2 + 2x −3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PT\n2. 22PV\n3. 22PW\n4. 22PX\n5. 22PY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nEvery point on the y-axis has an x-coordinate of 0, therefore to calculate the y-intercept\nwe let x = 0.\nFor example, the y-intercept of g(x) = (x −1)2 + 5 is determined by setting x = 0:\ng(x) = (x −1)2 + 5\ng(0) = (0 −1)2 + 5\n= 6\nThis gives the point (0; 6).\nThe x-intercept:\nEvery point on the x-axis has a y-coordinate of 0, therefore to calculate the x-intercept\nwe let y = 0.\nFor example, the x-intercept of g(x) = (x −1)2 + 5 is determined by setting y = 0:\ng(x) = (x −1)2 + 5\n0 = (x −1)2 + 5\n−5 = (x −1)2\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n151\nChapter 5.\nFunctions\n\nExercise 5 – 3: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = (x + 4)2 −1\n2. g(x) = 16 −8x + x2\n3. h(x) = −x2 + 4x −3\n4. j(x) = 4(x −3)2 −1\n5. k(x) = 4(x −3)2 + 1\n6. l(x) = 2x2 −3x −4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22PZ\n2. 22Q2\n3. 22Q3\n4. 22Q4\n5. 22Q5\n6. 22Q6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTurning point\nThe turning point of the function f(x) = a(x+p)2 +q is determined by examining the\nrange of the function:\n• If a > 0, f(x) has a minimum turning point and the range is [q; ∞):\nThe minimum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\n• If a < 0, f(x) has a maximum turning point and the range is (−∞; q]:\nThe maximum value of f(x) is q.\nIf f(x) = q, then a(x + p)2 = 0, and therefore x = −p.\nThis gives the turning point (−p; q).\nTherefore the turning point of the quadratic function f(x) = a(x + p)2 + q is (−p; q).\nAlternative form for quadratic equations:\nWe can also write the quadratic equation in the form\ny = a(x −p)2 + q\nThe effect of p is still a horizontal shift, however notice that:\n• For p > 0, the graph is shifted to the right by p units.\n• For p < 0, the graph is shifted to the left by p units.\nThe turning point is (p; q) and the axis of symmetry is the line x = p.\n152\n5.1.\nQuadratic functions\n\nWorked example 2: Turning point\nQUESTION\nDetermine the turning point of g(x) = 3x2 −6x −1.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q\nWe use the method of completing the square:\ng(x) = 3x2 −6x −1\n= 3(x2 −2x) −1\n= 3\n\u0000(x −1)2 −1\n\u0001\n−1\n= 3(x −1)2 −3 −1\n= 3(x −1)2 −4\nStep 2: Determine turning point (−p; q)\nFrom the equation g(x) = 3(x −1)2 −4 we know that the turning point for g(x) is\n(1; −4).\nWorked example 3: Turning point\nQUESTION\n1. Show that the x-value for the turning point of h(x) = ax2 + bx + c is given by\nx = −b\n2a.\n2. Hence, determine the turning point of k(x) = 2 −10x + 5x2.\nSOLUTION\nStep 1: Write the equation in the form y = a(x + p)2 + q and show that p =\nb\n2a\nWe use the method of completing the square:\nh(x) = ax2 + bx + c\n= a\n\u0012\nx2 + b\nax + c\na\n\u0013\nTake half the coefficient of the x term and square it; then add and subtract it from the\n153\nChapter 5.\nFunctions\n\nexpression.\nh(x) = a\n \nx2 + b\nax +\n\u0012 b\n2a\n\u00132\n−\n\u0012 b\n2a\n\u00132\n+ c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2\n4a2 + c\na\n!\n= a\n \u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a2\n!\n= a\n\u0012\nx + b\n2a\n\u00132\n−b2 −4ac\n4a\nFrom the above we have that the turning point is at x = −p = −b\n2a and y = q =\n−b2−4ac\n4a\n.\nStep 2: Determine the turning point of k(x)\nWrite the equation in the general form y = ax2 + bx + c.\nk(x) = 5x2 −10x + 2\nTherefore a = 5; b = −10; c = 2.\nUse the results obtained above to determine x = −b\n2a:\nx = −\n\u0012−10\n2(5)\n\u0013\n= 1\nSubstitute x = 1 to obtain the corresponding y-value :\ny = 5x2 −10x + 2\n= 5(1)2 −10(1) + 2\n= 5 −10 + 2\n= −3\nThe turning point of k(x) is (1; −3).\nExercise 5 – 4: Turning points\nDetermine the turning point of each of the following:\n1. y = x2 −6x + 8\n2. y = −x2 + 4x −3\n3. y = 1\n2(x + 2)2 −1\n4. y = 2x2 + 2x + 1\n154\n5.1.\nQuadratic functions\n\n5. y = 18 + 6x −3x2\n6. y = −2[(x + 1)2 + 3]\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22Q7\n2. 22Q8\n3. 22Q9\n4. 22QB\n5. 22QC\n6. 22QD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxis of symmetry\nThe axis of symmetry for f(x) = a(x + p)2 + q is the vertical line x = −p. The axis of\nsymmetry passes through the turning point (−p; q) and is parallel to the y-axis.\ny\nx\n0\nb\nx = −p\nf(x) = a(x + p)2 + q\nExercise 5 – 5: Axis of symmetry\n1. Determine the axis of symmetry of each of the following:\na) y = 2x2 −5x −18\nb) y = 3(x −2)2 + 1\nc) y = 4x −x2\n2. Write down the equation of a parabola where the y-axis is the axis of symmetry.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QF\n1b. 22QG\n1c. 22QH\n2. 22QJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n155\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) = a(x + p)2 + q\nIn order to sketch graphs of the form f(x) = a(x + p)2 + q, we need to determine five\ncharacteristics:\n• sign of a\n• turning point\n• y-intercept\n• x-intercept(s) (if they exist)\n• domain and range\nSee video: 22QK at www.everythingmaths.co.za\nWorked example 4: Sketching a parabola\nQUESTION\nSketch the graph of y = −1\n2(x + 1)2 −3.\nMark the intercepts, turning point and the axis of symmetry. State the domain and\nrange of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = a(x + p)2 + q\nWe notice that a < 0, therefore the graph is a “frown” and has a maximum turning\npoint.\nStep 2: Determine the turning point (−p; q)\nFrom the equation we know that the turning point is (−1; −3).\nStep 3: Determine the axis of symmetry x = −p\nFrom the equation we know that the axis of symmetry is x = −1.\nStep 4: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = −1\n2 ((0) + 1)2 −3\n= −1\n2 −3\n= −31\n2\nThis gives the point (0; −31\n2).\n156\n5.1.\nQuadratic functions\n\nStep 5: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = −1\n2 (x + 1)2 −3\n3 = −1\n2 (x + 1)2\n−6 = (x + 1)2\nwhich has no real solutions. Therefore, there are no x-intercepts and the graph lies\nbelow the x-axis.\nStep 6: Plot the points and sketch the graph\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n1\n2\n3\n4\n−1\n−2\n−3\n−4\nb\nb\ny\nx\n0\n(0; −3 1\n2)\n(−1; −3)\nStep 7: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≤−3, y ∈R}\nSee video: 22QM at www.everythingmaths.co.za\nWorked example 5: Sketching a parabola\nQUESTION\nSketch the graph of y = 1\n2x2 −4x + 7\n2.\nDetermine the intercepts, turning point and the axis of symmetry. Give the domain\nand range of the function.\nSOLUTION\n157\nChapter 5.\nFunctions\n\nStep 1: Examine the equation of the form y = ax2 + bx + c\nWe notice that a > 0, therefore the graph is a “smile” and has a minimum turning\npoint.\nStep 2: Determine the turning point and the axis of symmetry\nCheck that the equation is in standard form and identify the coefficients.\na = 1\n2;\nb = −4;\nc = 7\n2\nCalculate the x-value of the turning point using\nx = −b\n2a\n= −\n \n−4\n2\n\u0000 1\n2\n\u0001\n!\n= 4\nTherefore the axis of symmetry is x = 4.\nSubstitute x = 4 into the original equation to obtain the corresponding y-value.\ny = 1\n2x2 −4x + 7\n2\n= 1\n2(4)2 −4(4) + 7\n2\n= 8 −16 + 7\n2\n= −41\n2\nThis gives the point\n\u00004; −41\n2\n\u0001\n.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny = 1\n2(0)2 −4(0) + 7\n2\n= 7\n2\nThis gives the point\n\u00000; 7\n2\n\u0001\n.\nStep 4: Determine the x-intercepts\nThe x-intercepts are obtained by letting y = 0:\n0 = 1\n2x2 −4x + 7\n2\n= x2 −8x + 7\n= (x −1)(x −7)\nTherefore x = 1 or x = 7. This gives the points (1; 0) and (7; 0).\nStep 5: Plot the points and sketch the graph\n158\n5.1.\nQuadratic functions\n\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\nb\nb\ny\nx\n0\n(4; −41\n2)\n(0; 3 1\n2)\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y ≥−41\n2, y ∈R}\nSee video: 22QN at www.everythingmaths.co.za\nInvestigation: Shifting the equation of a parabola\nCarl and Eric are doing their Mathematics homework and decide to check each others\nanswers.\nHomework question:\nIf the parabola y = 3x2 + 1 is shifted 2 units to the right, determine the equation of the\nnew parabola.\n• Carl’s answer:\nA shift to the right means moving in the positive x direction, therefore x is re-\nplaced with x + 2 and the new equation is y = 3(x + 2)2 + 1.\n• Eric’s answer:\nWe replace x with x −2, therefore the new equation is y = 3(x −2)2 + 1.\nWork together in pairs. Discuss the two different answers and decide which one is\ncorrect. Use calculations and sketches to help explain your reasoning.\n159\nChapter 5.\nFunctions\n\nWriting an equation of a shifted parabola\nThe parabola is shifted horizontally:\n• If the parabola is shifted m units to the right, x is replaced by (x −m).\n• If the parabola is shifted m units to the left, x is replaced by (x + m).\nThe parabola is shifted vertically:\n• If the parabola is shifted n units down, y is replaced by (y + n).\n• If the parabola is shifted n units up, y is replaced by (y −n).\nWorked example 6: Shifting a parabola\nQUESTION\nGiven y = x2 −2x −3.\n1. If the parabola is shifted 1 unit to the right, determine the new equation of the\nparabola.\n2. If the parabola is shifted 3 units down, determine the new equation of the\nparabola.\nSOLUTION\nStep 1: Determine the new equation of the shifted parabola\n1. The parabola is shifted 1 unit to the right, so x must be replaced by (x −1).\ny = x2 −2x −3\n= (x −1)2 −2(x −1) −3\n= x2 −2x + 1 −2x + 2 −3\n= x2 −4x\nBe careful not to make a common error: replacing x with x + 1 for a shift to the\nright.\n2. The parabola is shifted 3 units down, so y must be replaced by (y + 3).\ny + 3 = x2 −2x −3\ny = x2 −2x −3 −3\n= x2 −2x −6\n160\n5.1.\nQuadratic functions\n\nExercise 5 – 6: Sketching parabolas\n1. Sketch graphs of the following functions and determine:\n• intercepts\n• turning point\n• axes of symmetry\n• domain and range\na) y = −x2 + 4x + 5\nb) y = 2(x + 1)2\nc) y = 3x2 −2(x + 2)\nd) y = 3(x −2)2 + 1\n2. Draw the following graphs on the same system of axes:\nf(x) = −x2 + 7\ng(x) = −(x −2)2 + 7\nh(x) = (x −2)2 −7\n3. Draw a sketch of each of the following graphs:\na) y = ax2 + bx + c if a > 0, b > 0, c < 0.\nb) y = ax2 + bx + c if a < 0, b = 0, c > 0.\nc) y = ax2 + bx + c if a < 0, b < 0, b2 −4ac < 0.\nd) y = (x + p)2 + q if p < 0, q < 0 and the x-intercepts have different signs.\ne) y = a(x + p)2 + q if a < 0, p < 0, q > 0 and one root is zero.\nf) y = a(x + p)2 + q if a > 0, p = 0, b2 −4ac > 0.\n4. Determine the new equation (in the form y = ax2 + bx + c) if:\na) y = 2x2 + 4x + 2 is shifted 3 units to the left.\nb) y = −(x + 1)2 is shifted 1 unit up.\nc) y = 3(x −1)2 + 2\n\u0000x −1\n2\n\u0001\nis shifted 2 units to the right.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22QP\n1b. 22QQ\n1c. 22QR\n1d. 22QS\n2. 22QT\n3a. 22QV\n3b. 22QW\n3c. 22QX\n3d. 22QY\n3e. 22QZ\n3f. 22R2\n4a. 22R3\n4b. 22R4\n4c. 22R5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n161\nChapter 5.\nFunctions\n\nFinding the equation of a parabola from the graph\nIf the intercepts are given, use y = a(x −x1)(x −x2).\nExample:\ny\nx\n0\n(0; 2)\n(4; 0)\n(−1; 0)\nx-intercepts: (−1; 0) and (4; 0)\ny = a(x −x1)(x −x2)\n= a(x + 1)(x −4)\n= ax2 −3ax −4a\ny-intercept: (0; 2)\n−4a = 2\na = −1\n2\nEquation of the parabola:\ny = ax2 −3ax −4a\n= −1\n2x2 −3\n\u0012\n−1\n2\n\u0013\nx −4\n\u0012\n−1\n2\n\u0013\n= −1\n2x2 + 3\n2x + 2\nIf the x-intercepts and another point are given, use y = a(x −x1)(x −x2).\nExample:\nb\ny\nx\n0\n(−1; 12)\n(1; 0)\n(5; 0)\nx-intercepts: (1; 0) and (5; 0)\ny = a(x −x1)(x −x2)\n= a(x −1)(x −5)\n= ax2 −6ax + 5a\nSubstitute the point: (−1; 12)\n12 = a(−1)2 −6a(−1) + 5a\n12 = a + 6a + 5a\n12 = 12a\n1 = a\nEquation of the parabola:\ny = ax2 −6ax + 5a\n= x2 −6x + 5\nIf the turning point and another point are given, use y = a(x + p)2 + q.\nExample:\nb\nb\ny\nx\n0\n(1; 5)\n(−3; 1)\nTurning point: (−3; 1)\ny = a(x + p)2 + q\n= a(x + 3)2 + 1\n= ax2 + 6ax + 9a + 1\nSubstitute the point: (1; 5)\n5 = a(1)2 + 6a(1) + 9a + 1\n4 = 16a\n1\n4 = a\nEquation of the parabola:\ny = 1\n4(x + 3)2 + 1\n162\n5.1.\nQuadratic functions\n\nExercise 5 – 7: Finding the equation\nDetermine the equations of the following graphs. Write your answers in the form\ny = a(x + p)2 + q.\n1.\nb\nb\ny\nx\n0\n3\n(−1; 6)\n2.\nb\ny\nx\n0\n(−1; 3)\n5\n3.\nb\nb\ny\nx\n0\n(1; 6)\n−2\n4.\nb\nb\nb\ny\nx\n0\n(1; 6)\n(3; 4)\n4\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22R6\n2. 22R7\n3. 22R8\n4. 22R9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n163\nChapter 5.\nFunctions\n\n5.2\nAverage gradient\nEMBGN\nWe notice that the gradient of a curve changes at every point on the curve, therefore\nwe need to work with the average gradient. The average gradient between any two\npoints on a curve is the gradient of the straight line passing through the two points.\ny\nx\n0\nb\nb\nA(−3; 7)\nC(−1; −1)\nFor the diagram above, the gradient of the line AC is\nGradient = yA −yC\nxA −xC\n= 7 −(−1)\n−3 −(−1)\n= 8\n−2\n= −4\nThis is the average gradient of the curve between the points A and C.\nWhat happens to the gradient if we fix the position of one point and move the second\npoint closer to the fixed point?\nSee video: 22RB at www.everythingmaths.co.za\nInvestigation: Gradient at a single point on a curve\nThe curve shown here is defined by y = −2x2 −5.\nPoint B is fixed at (0; −5) and the position of point\nA varies.\nComplete the table below by calculating the y-\ncoordinates of point A for the given x-coordinates\nand then calculating the average gradient between\npoints A and B.\ny\nx\n0\nb\nb\nA\nB(0; −5)\n164\n5.2.\nAverage gradient\n\nxA\nyA\nAverage gradient\n−2\n−1,5\n−1\n−0,5\n0\n0,5\n1\n1,5\n2\n1. What happens to the average gradient as A moves towards B?\n2. What happens to the average gradient as A moves away from B?\n3. What is the average gradient when A overlaps with B?\nIn the example above, the gradient of the straight line that passes through points A and\nC changes as A moves closer to C. At the point where A and C overlap, the straight\nline only passes through one point on the curve. This line is known as a tangent to the\ncurve.\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb\nb\nA\nC\ny\nx\n0\nb A\nC\ny\nx\n0\nb\nb\nA\nC\nWe therefore introduce the idea of the gradient at a single point on a curve. The\ngradient at a point on a curve is the gradient of the tangent to the curve at the given\npoint.\n165\nChapter 5.\nFunctions\n\nWorked example 7: Average gradient\nQUESTION\ny\nx\n0\nb\nb\nP(a; g(a))\nQ(a + h; g(a + h))\ng(x) = x2\n1. Find the average gradient between two points P (a; g(a)) and Q (a + h; g(a + h))\non a curve g(x) = x2.\n2. Determine the average gradient between P (2; g(2)) and Q (5; g(5)).\n3. Explain what happens to the average gradient if Q moves closer to P.\nSOLUTION\nStep 1: Assign labels to the x-values for the given points\nx1 = a\nx2 = a + h\nStep 2: Determine the corresponding y-coordinates\nUsing the function g(x) = x2, we can determine:\ny1 = g(a)\n= a2\ny2 = g(a + h)\n= (a + h)2\n= a2 + 2ah + h2\n166\n5.2.\nAverage gradient\n\nStep 3: Calculate the average gradient\ny2 −y1\nx2 −x1\n=\n\u0000a2 + 2ah + h2\u0001\n−\n\u0000a2\u0001\n(a + h) −(a)\n= a2 + 2ah + h2 −a2\na + h −a\n= 2ah + h2\nh\n= h(2a + h)\nh\n= 2a + h\nThe average gradient between P (a; g(a)) and Q (a + h; g(a + h)) on the curve g(x) =\nx2 is 2a + h.\nStep 4: Calculate the average gradient between P (2; g(2)) and Q (5; g(5))\nThe x-coordinate of P is a and the x-coordinate of Q is a+h therefore if we know that\na = 2 and a + h = 5, then h = 3.\nThe average gradient is therefore 2a + h = 2 (2) + (3) = 7\nStep 5: When Q moves closer to P\nWhen point Q moves closer to point P, h gets smaller.\nWhen the point Q overlaps with the point P, h = 0 and the gradient is given by 2a.\nWe can write the equation for average gradient in another form. Given a curve f(x)\nwith two points P and Q with P (a; f(a)) and Q (a + h; f(a + h)). The average gradi-\nent between P and Q is:\nAverage gradient = yQ −yP\nxQ −xP\n= f(a + h) −f(a)\n(a + h) −(a)\n= f(a + h) −f(a)\nh\nThis result is important for calculating the gradient at a point on a curve and will be\nexplored in greater detail in Grade 12.\n167\nChapter 5.\nFunctions\n\nWorked example 8: Average gradient\nQUESTION\nGiven f(x) = −2x2.\n1. Draw a sketch of the function and determine the average gradient between the\npoints A, where x = 1, and B, where x = 3.\n2. Determine the gradient of the curve at point A.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that a < 0, therefore the graph is a “frown” and has a\nmaximum turning point. We also see that when x = 0, y = 0, therefore the graph\npasses through the origin.\nStep 2: Draw a rough sketch\ny = −2x2\nx\ny\n0\nb\nb\nA\nB\nStep 3: Calculate the average gradient between A and B\nAverage gradient = f(3) −f(1)\n3 −1\n= −2(3)2 −(−2(1)2)\n2\n= −18 + 2\n2\n= −16\n2\n= −8\n168\n5.2.\nAverage gradient\n\nStep 4: Calculate the average gradient for f(x)\nAverage gradient = f(a + h) −f(a)\n(a + h) −a\n= −2(a + h)2 −(−2a2)\nh\n= −2a2 −4ah −2h2 + 2a2\nh\n= −4ah −2h2\nh\n= h(−4a −2h)\nh\n= −4a −2h\nAt point A, h = 0 and a = 1. Therefore\nAverage gradient = −4a −2h\n= −4(1) −2(0)\n= −4\nExercise 5 – 8:\n1.\na) Determine the average gradient of the curve f(x) = x (x + 3) between\nx = 5 and x = 3.\nb) Hence, state what you can deduce about the function f between x = 5 and\nx = 3.\n2. A (1; 3) is a point on f(x) = 3x2.\na) Draw a sketch of f(x) and label point A.\nb) Determine the gradient of the curve at point A.\nc) Determine the equation of the tangent line at A.\n3. Given: g(x) = −x2 + 1.\na) Draw a sketch of g(x).\nb) Determine the average gradient of the curve between x = −2 and x = 1.\nc) Determine the gradient of g at x = 2.\nd) Determine the gradient of g at x = 0.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RC\n2. 22RD\n3. 22RF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n169\nChapter 5.\nFunctions\n\n5.3\nHyperbolic functions\nEMBGP\nRevision\nEMBGQ\nFunctions of the form y = a\nx + q\nFunctions of the general form y = a\nx + q are called hyperbolic functions, where a and\nq are constants.\nThe effects of a and q on f(x) = a\nx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted vertically upwards by q units.\n– For q < 0, f(x) is shifted vertically downwards by q units.\n– The horizontal asymptote is the line y = q.\n– The vertical asymptote is the y-axis, the line x = 0.\n• The effect of a on shape and quad-\nrants\n– For a > 0, f(x) lies in the first\nand third quadrants.\n– For a > 1, f(x) will be further\naway from both axes than y =\n1\nx.\n– For 0 < a < 1, as a tends to\n0, f(x) moves closer to the axes\nthan y = 1\nx.\n– For a < 0, f(x) lies in the sec-\nond and fourth quadrants.\n– For a < −1, f(x) will be further\naway from both axes than y =\n−1\nx.\n– For −1 < a < 0, as a tends to\n0, f(x) moves closer to the axes\nthan y = −1\nx.\na < 0\na > 0\nq > 0\nq = 0\nq < 0\nExercise 5 – 9: Revision\n1. Consider the following hyperbolic functions:\n• y1 = 1\nx\n• y2 = −4\nx\n• y3 = 4\nx −2\n• y4 = −4\nx + 1\n170\n5.3.\nHyperbolic functions\n\nComplete the table to summarise the properties of the hyperbolic function:\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nlies in I and III quad\nasymptotes\ny-axis, x = 0\nx-axis, y = 0\naxes of symmetry\ny = x\ny = −x\ndomain\n{x : x ∈R, x ̸= 0}\nrange\n{y : y ∈R, y ̸= 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSee video: 22RH at www.everythingmaths.co.za\nFunctions of the form y =\na\nx+p + q\nEMBGR\nWe now consider hyperbolic functions of the form y =\na\nx+p + q and the effects of\nparameter p.\nInvestigation: The effects of a, p and q on a hyperbolic graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 1\nx\nb) y2 =\n1\nx−2\nc) y3 =\n1\nx−1\nd) y4 =\n1\nx+1\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nintercept(s)\nasymptotes\naxes of symmetry\ndomain\nrange\neffect of p\n171\nChapter 5.\nFunctions\n\n2. Complete the following sentences for functions of the form y =\na\nx+p + q:\na) A change in p causes a . . . . . . shift.\nb) If the value of p increases, the graph and the vertical asymptote . . . . . .\nc) If the value of q changes, then the . . . . . . asymptote of the hyperbola will\nshift.\nd) If the value of p decreases, the graph and the vertical asymptote . . . . . .\nThe effect of the parameters on y =\na\nx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe value of p also affects the vertical asymptote, the line x = −p.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position on the Cartesian plane.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n172\n5.3.\nHyperbolic functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y =\na\nx+p + q:\nDomain and range\nThe domain is {x : x ∈R, x ̸= −p}. If x = −p, the dominator is equal to zero and the\nfunction is undefined.\nWe see that\ny =\na\nx + p + q\ncan be re-written as:\ny −q =\na\nx + p\nIf x ̸= −p then:\n(y −q) (x + p) = a\nx + p =\na\ny −q\nThe range is therefore {y : y ∈R, y ̸= q}.\nThese restrictions on the domain and range determine the vertical asymptote x = −p\nand the horizontal asymptote y = q.\nWorked example 9: Domain and range\nQUESTION\nDetermine the domain and range for g(x) =\n2\nx+1 + 2.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R, x ̸= −1} since g(x) is undefined for x = −1.\nStep 2: Determine the range\nLet g(x) = y:\ny =\n2\nx + 1 + 2\ny −2 =\n2\nx + 1\n(y −2)(x + 1) = 2\nx + 1 =\n2\ny −2\nTherefore the range is {g(x) : g(x) ∈R, g(x) ̸= 2}.\n173\nChapter 5.\nFunctions\n\nExercise 5 – 10: Domain and range\nDetermine the domain and range for each of the following functions:\n1. y = 1\nx + 1\n2. g(x) =\n8\nx−8 + 4\n3. y = −\n4\nx+1 −3\n4. x =\n2\n3−y + 5\n5. (y −2)(x + 2) = 3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RJ\n2. 22RK\n3. 22RM\n4. 22RN\n5. 22RP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n2\nx+1+2\nis determined by setting x = 0:\ng(x) =\n2\nx + 1 + 2\ng(0) =\n2\n0 + 1 + 2\n= 2 + 2\n= 4\nThis gives the point (0; 4).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n2\nx+1+2\nis determined by setting y = 0:\ng(x) =\n2\nx + 1 + 2\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\n174\n5.3.\nHyperbolic functions\n\nExercise 5 – 11: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) =\n1\nx+4 −2\n2. g(x) = −5\nx + 2\n3. j(x) =\n2\nx−1 + 3\n4. h(x) =\n3\n6−x + 1\n5. k(x) =\n5\nx+2 −1\n2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RQ\n2. 22RR\n3. 22RS\n4. 22RT\n5. 22RV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptotes\nThere are two asymptotes for functions of the form y =\na\nx+p + q. The asymptotes\nindicate the values of x for which the function does not exist. In other words, the\nvalues that are excluded from the domain and the range. The horizontal asymptote is\nthe line y = q and the vertical asymptote is the line x = −p.\nExercise 5 – 12: Asymptotes\nDetermine the asymptotes for each of the following functions:\n1. y =\n1\nx+4 −2\n2. y = −5\nx\n3. y =\n3\n2−x + 1\n4. y = 1\nx −8\n5. y = −\n2\nx−2\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22RW\n2. 22RX\n3. 22RY\n4. 22RZ\n5. 22S2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAxes of symmetry\nThere are two lines about which a hyperbola is symmetrical.\nFor the standard hyperbola y = 1\nx, we see that if we replace x ⇒y and y ⇒x, we get\ny = 1\nx. Similarly, if we replace x ⇒−y and y ⇒−x, the function remains the same.\nTherefore the function is symmetrical about the lines y = x and y = −x.\nFor the shifted hyperbola y =\na\nx+p + q, the axes of symmetry intersect at the point\n(−p; q).\n175\nChapter 5.\nFunctions\n\nTo determine the axes of symmetry we define the two straight lines y1 = m1x + c1 and\ny2 = m2x + c2. For the standard and shifted hyperbolic function, the gradient of one\nof the lines of symmetry is 1 and the gradient of the other line of symmetry is −1. The\naxes of symmetry are perpendicular to each other and the product of their gradients\nequals −1. Therefore we let y1 = x+c1 and y2 = −x+c2. We then substitute (−p; q),\nthe point of intersection of the axes of symmetry, into both equations to determine the\nvalues of c1 and c2.\nWorked example 10: Axes of symmetry\nQUESTION\nDetermine the axes of symmetry for y =\n2\nx+1 −2.\nSOLUTION\nStep 1: Determine the point of intersection (−p; q)\nFrom the equation we see that p = 1 and q = −2. So the axes of symmetry will\nintersect at (−1; −2).\nStep 2: Define two straight line equations\ny1 = x + c1\ny2 = −x + c2\nStep 3: Solve for c1 and c2\nUse (−1; −2) to solve for c1:\ny1 = x + c1\n−2 = −1 + c1\n−1 = c1\nUse (−1; −2) to solve for c2:\ny2 = −x + c2\n−2 = −(−1) + c2\n−3 = c2\nStep 4: Write the final answer\nThe axes of symmetry for y =\n2\nx+1 −2 are the lines\ny1 = x −1\ny2 = −x −3\n176\n5.3.\nHyperbolic functions\n\n1\n2\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny1 = x −1\ny2 = −x −3\nExercise 5 – 13: Axes of symmetry\n1. Complete the following for f(x) and g(x):\n• Sketch the graph.\n• Determine (−p; q).\n• Find the axes of symmetry.\nCompare f(x) and g(x) and also their axes of symmetry. What do you notice?\na) f(x) = 2\nx\ng(x) = 2\nx + 1\nb) f(x) = −3\nx\ng(x) = −\n3\nx+1\nc) f(x) = 5\nx\ng(x) =\n5\nx−1 −1\n2. A hyperbola of the form k(x) =\na\nx+p + q passes through the point (4; 3). If the\naxes of symmetry intersect at (−1; 2), determine the equation of k(x).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S3\n1b. 22S4\n1c. 22S5\n2. 22S6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n177\nChapter 5.\nFunctions\n\nSketching graphs of the form f(x) =\na\nx+p + q\nIn order to sketch graphs of functions of the form, f(x) =\na\nx+p +q, we need to calculate\nfive characteristics:\n• quadrants\n• asymptotes\n• y-intercept\n• x-intercept\n• domain and range\nWorked example 11: Sketching a hyperbola\nQUESTION\nSketch the graph of y =\n2\nx+1 + 2. Determine the intercepts, asymptotes and axes of\nsymmetry. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Determine the asymptotes\nFrom the equation we know that p = 1 and q = 2.\nTherefore the horizontal asymptote is the line y = 2 and the vertical asymptote is the\nline x = −1.\nStep 3: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n2\n0 + 1 + 2\n= 4\nThis gives the point (0; 4).\nStep 4: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n2\nx + 1 + 2\n−2 =\n2\nx + 1\n178\n5.3.\nHyperbolic functions\n\n−2(x + 1) = 2\n−2x −2 = 2\n−2x = 4\nx = −2\nThis gives the point (−2; 0).\nStep 5: Determine the axes of symmetry\nUsing (−1; 2) to solve for c1:\ny1 = x + c1\n2 = −1 + c1\n3 = c1\ny2 = −x + c2\n2 = −(−1) + c2\n1 = c2\nTherefore the axes of symmetry are y = x + 3 and y = −x + 1.\nStep 6: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\ny = x + 3\ny = −x + 1\nStep 7: State the domain and range\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 2}\n179\nChapter 5.\nFunctions\n\nWorked example 12: Sketching a hyperbola\nQUESTION\nUse horizontal and vertical shifts to sketch the graph of f(x) =\n1\nx−2 + 3.\nSOLUTION\nStep 1: Examine the equation of the form y =\na\nx+p + q\nWe notice that a > 0, therefore the graph will lie in the first and third quadrants.\nStep 2: Sketch the standard hyperbola y = 1\nx\nStart with a sketch of the standard hyperbola g(x) = 1\nx. The vertical asymptote is x = 0\nand the horizontal asymptote is y = 0.\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 3: Determine the vertical shift\nFrom the equation we see that q = 3, which means g(x) must shifted 3 units up. The\nhorizontal asymptote is also shifted 3 units up to y = 3 .\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 4: Determine the horizontal shift\nFrom the equation we see that p = −2, which means g(x) must shifted 2 units to the\nright. The vertical asymptote is also shifted 2 units to the right.\n180\n5.3.\nHyperbolic functions\n\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n4\n5\n6\n−1\n−2\ny\nx\n0\nStep 5: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\ny =\n1\n0 −2 + 3\n= 21\n2\nThis gives the point (0; 21\n2).\nStep 6: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 =\n1\nx −2 + 3\n−3 =\n1\nx −2\n−3(x −2) = 1\n−3x + 6 = 1\n−3x = −5\nx = 5\n3\nThis gives the point (5\n3; 0).\nStep 7: Determine the domain and range\nDomain: {x : x ∈R, x ̸= 2}\nRange: {y : y ∈R, y ̸= 3}\n181\nChapter 5.\nFunctions\n\nWorked example 13: Finding the equation of a hyperbola from the graph\nQUESTION\nUse the graph below to determine the values of a, p and q for y =\na\nx+p + q.\n1\n2\n3\n4\n5\n6\n7\n−1\n1\n2\n3\n−1\n−2\n−3\n−4\n−5\ny\nx\n0\nSOLUTION\nStep 1: Examine the graph and deduce the sign of a\nWe notice that the graph lies in the second and fourth quadrants, therefore a < 0.\nStep 2: Determine the asymptotes\nFrom the graph we see that the vertical asymptote is x = −1, therefore p = 1. The\nhorizontal asymptote is y = 3, and therefore q = 3.\ny =\na\nx + 1 + 3\nStep 3: Determine the value of a\nTo determine the value of a we substitute a point on the graph, namely (0; 0):\ny =\na\nx + 1 + 3\n0 =\na\n0 + 1 + 3\n∴−3 = a\nStep 4: Write the final answer\ny = −\n3\nx + 1 + 3\n182\n5.3.\nHyperbolic functions\n\nExercise 5 – 14: Sketching graphs\n1. Draw the graphs of the following functions and indicate:\n• asymptotes\n• intercepts, where applicable\n• axes of symmetry\n• domain and range\na) y = 1\nx + 2\nb) y =\n1\nx+4 −2\nc) y = −\n1\nx+1 + 3\nd) y = −\n5\nx−2 1\n2 −2\ne) y =\n8\nx−8 + 4\n2. Given the graph of the hyperbola of the form y =\n1\nx+p + q, determine the values\nof p and q.\ny\nx\n−2\n−1\n3. Given a sketch of the function of the form y =\na\nx+p + q, determine the values of\na, p and q.\ny\nx\n2\n2\n4.\na) Draw the graph of f(x) = −3\nx, x > 0.\nb) Determine the average gradient of the graph between x = 1 and x = 3.\nc) Is the gradient at (1\n2; −6) less than or greater than the average gradient be-\ntween x = 1 and x = 3? Illustrate this on your graph.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22S7\n1b. 22S8\n1c. 22S9\n1d. 22SB\n1e. 22SC\n2. 22SD\n3. 22SF\n4. 22SG\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n183\nChapter 5.\nFunctions\n\n5.4\nExponential functions\nEMBGS\nRevision\nEMBGT\nFunctions of the form y = abx + q\nFunctions of the general form y = abx + q, for b > 0, are called exponential functions,\nwhere a, b and q are constants.\nThe effects of a, b and q on f(x) = abx + q:\n• The effect of q on vertical shift\n– For q > 0, f(x) is shifted verti-\ncally upwards by q units.\n– For q < 0, f(x) is shifted verti-\ncally downwards by q units.\n– The horizontal asymptote is the\nline y = q.\n• The effect of a on shape\n– For a > 0, f(x) is increasing.\n– For a < 0, f(x) is decreasing.\nThe graph is reflected about the\nhorizontal asymptote.\n• The effect of b on direction\nAssuming a > 0:\n– If b > 1, f(x) is an increasing\nfunction.\n– If 0 < b < 1, f(x) is a decreas-\ning function.\n– If b ≤0, f(x) is not defined.\nb > 1\na < 0\na > 0\nq > 0\nq < 0\n0 < b < 1\na < 0\na > 0\nq > 0\nq < 0\nExercise 5 – 15: Revision\n1. On separate axes, accurately draw each of the following functions:\n• Use tables of values if necessary.\n• Use graph paper if available.\na) y1 = 3x\nb) y2 = −2 × 3x\nc) y3 = 2 × 3x + 1\nd) y4 = 3x −2\n2. Use your sketches of the functions given above to complete the following table\n(the first column has been completed as an example):\n184\n5.4.\nExponential functions\n\ny1\ny2\ny3\ny4\nvalue of q\nq = 0\neffect of q\nno vertical shift\nvalue of a\na = 1\neffect of a\nincreasing\nasymptote\nx-axis, y = 0\ndomain\n{x : x ∈R}\nrange\n{y : y ∈R, y > 0}\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22SH\n1b. 22SJ\n1c. 22SK\n1d. 22SM\n2. 22SN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = ab(x+p) + q\nEMBGV\nWe now consider exponential functions of the form y = ab(x+p) + q and the effects of\nparameter p.\nSee video: 22SP at www.everythingmaths.co.za\nInvestigation: The effects of a, p and q on an exponential graph\n1. On the same system of axes, plot the following graphs:\na) y1 = 2x\nb) y2 = 2(x−2)\nc) y3 = 2(x−1)\nd) y4 = 2(x+1)\ne) y5 = 2(x+2)\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptote\ndomain\nrange\neffect of p\n2. On the same system of axes, plot the following graphs:\na) y1 = 2(x−1) + 2\n185\nChapter 5.\nFunctions\n\nb) y2 = 3 × 2(x−1) + 2\nc) y3 = 1\n2 × 2(x−1) + 2\nd) y4 = 0 × 2(x−1) + 2\ne) y5 = −3 × 2(x−1) + 2\nUse your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nintercept(s)\nasymptotes\ndomain\nrange\neffect of a\nThe effect of the parameters on y = abx+p + q\nThe effect of p is a horizontal shift because all points are moved the same distance in\nthe same direction (the entire graph slides to the left or to the right).\n• For p > 0, the graph is shifted to the left by p units.\n• For p < 0, the graph is shifted to the right by p units.\nThe effect of q is a vertical shift. The value of q also affects the horizontal asymptotes,\nthe line y = q.\nThe value of a affects the shape of the graph and its position relative to the horizontal\nasymptote.\n• For a > 0, the graph lies above the horizontal asymptote, y = q.\n• For a < 0, the graph lies below the horizontal asymptote, y = q.\np > 0\np < 0\na < 0\na > 0\na < 0\na > 0\nq > 0\nq < 0\n186\n5.4.\nExponential functions\n\nDiscovering the characteristics\nFor functions of the general form: f(x) = y = ab(x+p) + q:\nDomain and range\nThe domain is {x : x ∈R} because there is no value of x for which f(x) is undefined.\nThe range of f(x) depends on whether the value for a is positive or negative.\nIf a > 0 we have:\nb(x+p) > 0\nab(x+p) > 0\nab(x+p) + q > q\nf(x) > q\nThe range is therefore {y : y > q, y ∈R}.\nSimilarly, if a < 0, the range is {y : y < q, y ∈R}.\nWorked example 14: Domain and range\nQUESTION\nState the domain and range for g(x) = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the domain\nThe domain is {x : x ∈R} because there is no value of x for which g(x) is undefined.\nStep 2: Determine the range\nThe range of g(x) can be calculated from:\n3(x+1) > 0\n5 × 3(x+1) > 0\n5 × 3(x+1) −1 > −1\n∴g(x) > −1\nTherefore the range is {g(x) : g(x) > −1} or in interval notation (−1; ∞).\n187\nChapter 5.\nFunctions\n\nExercise 5 – 16: Domain and range\nGive the domain and range for each of the following functions:\n1. y =\n\u0000 3\n2\n\u0001(x+3)\n2. f(x) = −5(x−2) + 1\n3. y + 3 = 2(x+1)\n4. y = n + 3(x−m)\n5.\ny\n2 = 3(x−1) −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SQ\n2. 22SR\n3. 22SS\n4. 22ST\n5. 22SV\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIntercepts\nThe y-intercept:\nTo calculate the y-intercept we let x = 0. For example, the y-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting x = 0:\ng(0) = 3 × 2(0+1) + 2\n= 3 × 2 + 2\n= 8\nThis gives the point (0; 8).\nThe x-intercept:\nTo calculate the x-intercept we let y = 0. For example, the x-intercept of g(x) =\n3 × 2(x+1) + 2 is determined by setting y = 0:\n0 = 3 × 2(x+1) + 2\n−2 = 3 × 2(x+1)\n−2\n3 = 2(x+1)\nwhich has no real solutions. Therefore, the graph of g(x) lies above the x-axis and\ndoes not have any x-intercepts.\n188\n5.4.\nExponential functions\n\nExercise 5 – 17: Intercepts\nDetermine the x- and y-intercepts for each of the following functions:\n1. f(x) = 2(x+1) −8\n2. y = 2 × 3(x−1) −18\n3. y + 5(x+2) = 5\n4. y = 1\n2\n\u0000 3\n2\n\u0001(x+3) −0,75\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22SW\n2. 22SX\n3. 22SY\n4. 22SZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nAsymptote\nExponential functions of the form y = ab(x+p) + q have a horizontal asymptote, the\nline y = q.\nWorked example 15: Asymptote\nQUESTION\nDetermine the asymptote for y = 5 × 3(x+1) −1.\nSOLUTION\nStep 1: Determine the asymptote\nThe asymptote of g(x) can be calculated as:\n3(x+1) ̸= 0\n5 × 3(x+1) ̸= 0\n5 × 3(x+1) −1 ̸= −1\n∴y ̸= −1\nTherefore the asymptote is the line y = −1.\n189\nChapter 5.\nFunctions\n\nExercise 5 – 18: Asymptote\nGive the asymptote for each of the following functions:\n1. y = −5(x+1)\n2. y = 3(x−2) + 1\n3.\n\u0010\n3y\n2\n\u0011\n= 5(x+3) −1\n4. y = 7(x+1) −2\n5.\ny\n2 + 1 = 3(x+2)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T2\n2. 22T3\n3. 22T4\n4. 22T5\n5. 22T6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching graphs of the form f(x) = ab(x+p) + q\nIn order to sketch graphs of functions of the form, f(x) = ab(x+p) + q, we need to\ndetermine five characteristics:\n• shape\n• y-intercept\n• x-intercept\n• asymptote\n• domain and range\nWorked example 16: Sketching an exponential graph\nQUESTION\nSketch the graph of 2y = 10 × 2(x+1) −5.\nMark the intercept(s) and asymptote. State the domain and range of the function.\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nWe notice that a > 0 and b > 1, therefore the function is increasing.\n190\n5.4.\nExponential functions\n\nStep 2: Determine the y-intercept\nThe y-intercept is obtained by letting x = 0:\n2y = 10 × 2(0+1) −5\n= 10 × 2 −5\n= 15\n∴y = 71\n2\nThis gives the point (0; 71\n2).\nStep 3: Determine the x-intercept\nThe x-intercept is obtained by letting y = 0:\n0 = 10 × 2(x+1) −5\n5 = 10 × 2(x+1)\n1\n2 = 2(x+1)\n2−1 = 2(x+1)\n∴−1 = x + 1\n(same base)\n−2 = x\nThis gives the point (−2; 0).\nStep 4: Determine the asymptote\nThe horizontal asymptote is the line y = −5\n2.\nStep 5: Plot the points and sketch the graph\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n−2\n−3\n1\n2\n−1\n−2\n−3\n−4\ny\nx\n0\nStep 6: State the domain and range\nDomain: {x : x ∈R}\nRange: {y : y > −5\n2, y ∈R}\n191\nChapter 5.\nFunctions\n\nWorked example 17: Finding the equation of an exponential function from a\ngraph\nQUESTION\nUse the given graph of y = −2 × 3(x+p) + q to determine the values of p and q.\n1\n2\n3\n4\n5\n6\n7\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nSOLUTION\nStep 1: Examine the equation of the form y = ab(x+p) + q\nFrom the graph we see that the function is decreasing. We also note that a = −2 and\nb = 3. We need to solve for p and q.\nStep 2: Use the asymptote to determine q\nThe horizontal asymptote y = 6 is given, therefore we know that q = 6.\ny = −2 × 3(x+p) + 6\nStep 3: Use the x-intercept to determine p\nSubstitute (2; 0) into the equation and solve for p:\ny = −2 × 3(x+p) + 6\n0 = −2 × 3(2+p) + 6\n−6 = −2 × 3(2+p)\n3 = 3(2+p)\n∴1 = 2 + p\n(same base)\n∴p = −1\nStep 4: Write the final answer\ny = −2 × 3(x−1) + 6\n192\n5.4.\nExponential functions\n\nExercise 5 – 19: Mixed exercises\n1. Given the graph of the hyperbola of the form h(x) = k\nx, x < 0, which passes\nthough the point A(−1\n2; −6).\nb\ny\nx\n0\nA(−1\n2; −6)\na) Show that k = 3.\nb) Write down the equation for the new function formed if h(x):\ni. is shifted 3 units vertically upwards\nii. is shifted to the right by 3 units\niii. is reflected about the y-axis\niv. is shifted so that the asymptotes are x = 0 and y = −1\n4\nv. is shifted upwards to pass through the point (−1; 1)\nvi. is shifted to the left by 2 units and 1 unit vertically downwards (for\nx < 0)\n2. Given the graphs of f(x) = a(x + p)2 and g(x) = a\nx.\nThe axis of symmetry for f(x) is x = −1 and f(x) and g(x) intersect at point M.\nThe line y = 2 also passes through M.\nb\ny\nx\n0\nM\n−1\n2\nf\ng\n193\nChapter 5.\nFunctions\n\nDetermine:\na) the coordinates of M\nb) the equation of g(x)\nc) the equation of f(x)\nd) the values for which f(x) < g(x)\ne) the range of f(x)\n3. On the same system of axes, sketch:\na) the graphs of k(x) = 2(x + 1\n2)2 −41\n2 and h(x) = 2(x+ 1\n2 ). Determine all\nintercepts, turning point(s) and asymptotes.\nb) the reflection of h(x) about the x-axis. Label this function as j(x).\n4. Sketch the graphs of y = ax2 + bx + c for:\na) a < 0, b > 0, b2 < 4ac\nb) a > 0, b > 0, one root = 0\n5. On separate systems of axes, sketch the graphs:\ny =\n2\nx−2\ny = 2\nx −2\ny = −2(x−2)\n6. For the diagrams shown below, determine:\n• the equations of the functions; f(x) = a(x + p)2 + q, g(x) = ax2 + q,\nh(x) = a\nx, x < 0 and k(x) = bx + q\n• the axes of symmetry of each function\n• the domain and range of each function\na)\nb\ny\nx\n0\n(2; 3)\nf\n194\n5.4.\nExponential functions\n\nb)\nb\ny\nx\n0\n(−2; −1)\ng\nh\n−2\nc)\ny\nx\n0\nk\ny = 2x + 1\n2\n7. Given the graph of the function Q(x) = ax.\nb\nb\nb\ny\nx\n0\nQ = ax\n(−2; p)\n1\n(1; 1\n3)\na) Show that a = 1\n3.\nb) Find the value of p if the point (−2; p) is on Q.\nc) Calculate the average gradient of the curve between x = −2 and x = 1.\nd) Determine the equation of the new function formed if Q is shifted 2 units\nvertically downwards and 2 units to the left.\n8. Find the equation for each of the functions shown below:\na) f(x) = 2x + q\ng(x) = mx + c\n195\nChapter 5.\nFunctions\n\nb\ny\nx\n0\nf\ng\n−1\n2\n−2\nb) h(x) =\nk\nx+p + q\nb\nb\ny\nx\n0\n1\n−2\n−1\n2\nh\n9. Given: the graph of k(x) = −x2 + 3x + 10 with turning point at D. The graph of\nthe straight line h(x) = mx + c passing through points B and C is also shown.\nb\nb\nb\ny\nx\n0\nB\nA\nE\nF\nD\nC\nk\nh\nDetermine:\na) the lengths AO, OB, OC and DE\nb) the equation of DE\nc) the equation of h(x)\nd) the x-values for which k(x) < 0\n196\n5.4.\nExponential functions\n\ne) the x-values for which k(x) ≥h(x)\nf) the length of DF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22T7\n2. 22T8\n3. 22T9\n4. 22TB\n5. 22TC\n6a. 22TD\n6b. 22TF\n6c. 22TG\n7. 22TH\n8a. 22TJ\n8b. 22TK\n9. 22TM\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nIMPORTANT: Trigonometric functions are examined in PAPER 2.\n5.5\nThe sine function\nEMBGW\nRevision\nEMBGX\nFunctions of the form y = sin θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• Period of one complete wave is 360◦.\n• Amplitude is the maximum height of the wave above and below the x-axis and\nis always positive. Amplitude = 1.\n• Domain: [0◦; 360◦]\nFor y = sin θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Maximum turning point: (90◦; 1)\n• Minimum turning point: (270◦; −1)\n197\nChapter 5.\nFunctions\n\nFunctions of the form y = a sin θ + q\nThe effects of a and q on f(θ) = a sin θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 20: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦.\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function also determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n198\n5.5.\nThe sine function\n\n1. y1 = sin θ\n2. y2 = −2 sin θ\n3. y3 = sin θ + 1\n4. y4 = 1\n2 sin θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TN\n2. 22TP\n3. 22TQ\n4. 22TR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin kθ\nEMBGY\nWe now consider cosine functions of the form y = sin kθ and the effects of k.\nInvestigation: The effects of k on a sine graph\n1. Complete the following table for y1 = sin θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−270◦\n−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n2. Use the table of values to plot the graph of y1 = sin θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = sin(−θ)\nb) y3 = sin 2θ\nc) y4 = sin θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n199\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = sin θ and y2 = sin(−θ)?\n6. Is sin(−θ) = −sin θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = sin kθ?\nThe effect of the parameter on y = sin kθ\nThe value of k affects the period of the sine function. If k is negative, then the graph is\nreflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the sine function decreases.\nFor 0 < k < 1, the period of the sine function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\nsin(−θ) = −sin θ\nCalculating the period:\nTo determine the period of y = sin kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k (this means that k is always considered to be\npositive).\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n200\n5.5.\nThe sine function\n\nWorked example 18: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = sin θ\nb) y2 = sin 3θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin kθ\nNotice that k > 1 for y2 = sin 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\nsin θ\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\nsin 3θ\n2\n1\n0,38\n−0,71\n−0,92\n0\n0,92\n0,71\n−0,38\n−1\nStep 3: Sketch the sine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin 3\n2θ\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin 3θ\n2\nperiod\n360◦\n240◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(90◦; 1)\n(−180◦; 1) and (60◦; 1)\nminimum turning points\n(−90◦; −1)\n(−60◦; −1) and (180◦; 1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\n201\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = sin kθ\n= sin 0◦\n= 0\nThis gives the point (0◦; 0).\nExercise 5 – 21: Sine functions of the form y = sin kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦and for each graph deter-\nmine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = sin 3θ\nb) g(θ) = sin θ\n3\nc) h(θ) = sin(−2θ)\nd) k(θ) = sin 3θ\n4\n2. For each graph of the form f(θ) = sin kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n202\n5.5.\nThe sine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22TS\n1b. 22TT\n1c. 22TV\n1d. 22TW\n2a. 22TX\n2b. 22TY\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = sin(θ + p)\nEMBGZ\nInvestigation: The effects of p on a sine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = sin θ\nb) y2 = sin(θ −90◦)\nc) y3 = sin(θ −60◦)\nd) y4 = sin(θ + 90◦)\ne) y5 = sin(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of p\n203\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = sin(θ + p)\nThe effect of p on the sine function is a horizontal shift, also called a phase shift; the\nentire graph slides to the left or to the right.\n• For p > 0, the graph of the sine function shifts to the left by p.\n• For p < 0, the graph of the sine function shifts to the right by p.\np > 0\np < 0\nWorked example 19: Sine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = sin θ\nb) y2 = sin(θ −30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = sin(θ + p)\nNotice that for y1 = sin θ we have p = 0 (no phase shift) and for y2 = sin(θ −30◦),\np < 0 therefore the graph shifts to the right by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\nsin θ\n0\n1\n0\n−1\n0\n1\n0\n−1\n0\nsin(θ −30◦)\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\n−0,5\n204\n5.5.\nThe sine function\n\nStep 3: Sketch the sine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = sin θ\ny2 = sin(θ −30◦)\nStep 4: Complete the table\ny1 = sin θ\ny2 = sin(θ −30◦)\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−270◦; 1) and (90◦; 1)\n(−240◦; 1) and\n(120◦; 1)\nminimum turning points\n(−90◦; −1) and\n(270◦; −1)\n(−60◦; −1) and\n(300◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; −1\n2)\nx-intercept(s)\n(−360◦; 0), (−180◦; 0),\n(0◦; 0), (180◦; 0) and\n(360◦; 0)\n(−330◦; 0), (−150◦; 0),\n(30◦; 0) and (210◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = sin(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\n205\nChapter 5.\nFunctions\n\nExercise 5 – 22: Sine functions of the form y = sin(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = sin(θ + 30◦)\n2. g(θ) = sin(θ −45◦)\n3. h(θ) = sin(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22TZ\n2. 22V2\n3. 22V3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching sine graphs\nEMBH2\nWorked example 20: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(45◦−θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin(θ + p).\nf(θ) = sin(45◦−θ)\n= sin(−θ + 45◦)\n= sin (−(θ −45◦))\n= −sin(θ −45◦)\nTo draw a graph of the above function, we know that the standard sine graph, y = sin θ,\n206\n5.5.\nThe sine function\n\nmust:\n• be reflected about the x-axis\n• be shifted to the right by 45◦\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n0,71\n0\n−0,71\n−1\n−0,71\n0\n0,71\n1\n0,71\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = −sin(θ −45◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 360◦\nAmplitude: 1\nDomain: [−360◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (315◦; 1)\nMinimum turning point: (135◦; −1)\ny-intercepts: (0◦; 0,71)\nx-intercept: (45◦; 0) and (225◦; 0)\nWorked example 21: Sketching a sine graph\nQUESTION\nSketch the graph of f(θ) = sin(3θ + 60◦) for 0◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = sin k(θ + p).\nf(θ) = sin(3θ + 60◦)\n= sin 3(θ + 20◦)\n207\nChapter 5.\nFunctions\n\nTo draw a graph of the above equation, the standard sine graph, y = sin θ, must be\nchanged in the following ways:\n• decrease the period by a factor of 3;\n• shift to the left by 20◦.\nStep 2: Complete a table of values\nθ\n0◦\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nf(θ)\n0,87\n0,5\n−0,87\n−0,5\n0,87\n0,5\n−0,87\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n30◦\n60◦\n90◦\n120◦\n150◦\n180◦\nθ\nf(θ)\nf(θ) = sin 3(θ + 20◦)\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 180◦]\nRange: [−1; 1]\nMaximum turning point: (10◦; 1) and (130◦; 1)\nMinimum turning point: (70◦; −1)\ny-intercept: (0◦; 0,87)\nx-intercepts: (40◦; 0), (100◦; 0) and (160◦; 0)\nExercise 5 – 23: The sine function\n1. Sketch the following graphs on separate axes:\na) y = 2 sin θ\n2 for −360◦≤θ ≤360◦\nb) f(θ) = 1\n2 sin(θ −45◦) for −90◦≤θ ≤90◦\nc) y = sin(θ + 90◦) + 1 for 0◦≤θ ≤360◦\nd) y = sin(−3θ\n2 ) for −180◦≤θ ≤180◦\ne) y = sin(30◦−θ) for −360◦≤θ ≤360◦\n2. Given the graph of the function y = a sin(θ + p), determine the values of a and\np.\n208\n5.5.\nThe sine function\n\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nθ\nf(θ)\nCan you describe this graph in terms of cos θ?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22V4\n1b. 22V5\n1c. 22V6\n1d. 22V7\n1e. 22V8\n2. 22V9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n5.6\nThe cosine function\nEMBH3\nRevision\nEMBH4\nFunctions of the form y = cos θ for 0◦≤θ ≤360◦\n0\n1\n−1\n30◦60◦90◦120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\ny\n• The period is 360◦and the amplitude is 1.\n• Domain: [0◦; 360◦]\nFor y = cos θ, the domain is {θ : θ ∈R}, however in this case, the domain has\nbeen restricted to the interval 0◦≤θ ≤360◦.\n• Range: [−1; 1]\n• x-intercepts: (90◦; 0), (270◦; 0)\n• y-intercept: (0◦; 1)\n• Maximum turning points: (0◦; 1), (360◦; 1)\n• Minimum turning point: (180◦; −1)\n209\nChapter 5.\nFunctions\n\nFunctions of the form y = a cos θ + q\nCosine functions of the general form y = a cos θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a cos θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted\nvertically\nupwards\nby\nq\nunits.\n– For q < 0, f(θ) is shifted\nvertically downwards by q\nunits.\n• The effect of a on shape\n– For a > 1, the amplitude of\nf(θ) increases.\n– For 0 < a < 1, the ampli-\ntude of f(θ) decreases.\n– For a < 0, there is a reflec-\ntion about the x-axis.\n– For −1 < a < 0, there\nis a reflection about the x-\naxis and the amplitude de-\ncreases.\n– For a\n<\n−1,\nthere is\na reflection about the x-\naxis and the amplitude in-\ncreases.\nq = 0\nq > 0\nq < 0\nθ\ny\na > 1\na = 1\n0 < a < 1\n−1 < a < 0\na < −1\nθ\ny\nExercise 5 – 24: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function in the previous problem determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n210\n5.6.\nThe cosine function\n\n1. y1 = cos θ\n2. y2 = −3 cos θ\n3. y3 = cos θ + 2\n4. y4 = 1\n2 cos θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VB\n2. 22VC\n3. 22VD\n4. 22VF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos(kθ)\nEMBH5\nWe now consider cosine functions of the form y = cos kθ and the effects of k.\nInvestigation: The effects of k on a cosine graph\n1. Complete the following table for y1 = cos θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ncos θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ncos θ\n2. Use the table of values to plot the graph of y1 = cos θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = cos(−θ)\nb) y3 = cos 3θ\nc) y4 = cos 3θ\n4\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\namplitude\ndomain\nrange\nmaximum turning\npoints\nminimum turning\npoints\ny-intercept(s)\nx-intercept(s)\neffect of k\n211\nChapter 5.\nFunctions\n\n5. What do you notice about y1 = cos θ and y2 = cos(−θ)?\n6. Is cos(−θ) = −cos θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = cos kθ?\nThe effect of the parameter k on y = cos kθ\nThe value of k affects the period of the cosine function.\n• For k > 0:\nFor k > 1, the period of the cosine function decreases.\nFor 0 < k < 1, the period of the cosine function increases.\n• For k < 0:\nFor −1 < k < 0, the period increases.\nFor k < −1, the period decreases.\nNegative angles:\ncos(−θ) = cos θ\nNotice that for negative values of θ, the graph is not reflected about the x-axis.\nCalculating the period:\nTo determine the period of y = cos kθ we use,\nPeriod = 360◦\n|k|\nwhere |k| is the absolute value of k.\n0 < k < 1\n−1 < k < 0\nk > 1\nk < −1\n212\n5.6.\nThe cosine function\n\nWorked example 22: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = cos θ\nb) y2 = cos θ\n2\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos kθ\nNotice that for y2 = cos θ\n2, k < 1 therefore the period of the graph increases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ncos θ\n−1\n−0,71\n0\n0,71\n1\n0,71\n0\n−0,71\n−1\ncos θ\n2\n0\n0,38\n0,71\n0,92\n1\n0,92\n0,71\n0,38\n0\nStep 3: Sketch the cosine graphs\n1\n−1\n30◦60◦90◦120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\n0◦\ny\ny1 = cos θ\n2\ny2 = cos θ\nStep 4: Complete the table\ny1 = cos θ\ny2 = cos θ\n2\nperiod\n360◦\n720◦\namplitude\n1\n1\ndomain\n[−180◦; 180◦]\n[−180◦; 180◦]\nrange\n[−1; 1]\n[0; 1]\nmaximum turning points\n(0◦; 1)\n(0◦; 1)\nminimum turning points\n(−180◦; −1) and (180◦; −1)\nnone\ny-intercept(s)\n(0◦; 1)\n(0◦; 1)\nx-intercept(s)\n(−90◦; 0) and (90◦; 0)\n(−180◦; 0) and (180◦; 0)\n213\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos kθ:\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R} or [−1; 1].\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos kθ\n= cos 0◦\n= 1\nThis gives the point (0◦; 1).\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n1. Sketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\na) f(θ) = cos 2θ\nb) g(θ) = cos θ\n3\nc) h(θ) = cos(−2θ)\nd) k(θ) = cos 3θ\n4\n2. For each graph of the form f(θ) = cos kθ, determine the value of k:\na)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\ny\n0◦\n214\n5.6.\nThe cosine function\n\nb)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nbA(135◦; 0)\nθ\ny\n0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VG\n1b. 22VH\n1c. 22VJ\n1d. 22VK\n2a. 22VM\n2b. 22VN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = cos (θ + p)\nEMBH6\nWe now consider cosine functions of the form y = cos(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a cosine graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = cos θ\nb) y2 = cos(θ −90◦)\nc) y3 = cos(θ −60◦)\nd) y4 = cos(θ + 90◦)\ne) y5 = cos(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\namplitude\ndomain\nrange\nmaximum turning points\nminimum turning points\ny-intercept(s)\nx-intercept(s)\neffect of p\n215\nChapter 5.\nFunctions\n\nThe effect of the parameter on y = cos(θ + p)\nThe effect of p on the cosine function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the cosine function shifts to the left by p degrees.\n• For p < 0, the graph of the cosine function shifts to the right by p degrees.\np > 0\np < 0\nWorked example 23: Cosine function\nQUESTION\n1. Sketch the following functions on the same set of axes for −360◦≤θ ≤360◦.\na) y1 = cos θ\nb) y2 = cos(θ + 30◦)\n2. For each function determine the following:\na) Period\nb) Amplitude\nc) Domain and range\nd) x- and y-intercepts\ne) Maximum and minimum turning points\nSOLUTION\nStep 1: Examine the equations of the form y = cos(θ + p)\nNotice that for y1 = cos θ we have p = 0 (no phase shift) and for y2 = cos(θ + 30◦),\np < 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−360◦−270◦−180◦\n−90◦\n0◦\n90◦\n180◦\n270◦\n360◦\ncos θ\n1\n0\n−1\n0\n1\n0\n−1\n0\n1\ncos(θ +30◦)\n0,87\n−0,5\n−0,87\n0,5\n0,87\n−0,5\n−0,87\n0,5\n0,87\n216\n5.6.\nThe cosine function\n\nStep 3: Sketch the cosine graphs\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nθ\ny\ny1 = cos θ\ny2 = cos(θ + 30◦)\nStep 4: Complete the table\ny1\ny2\nperiod\n360◦\n360◦\namplitude\n1\n1\ndomain\n[−360◦; 360◦]\n[−360◦; 360◦]\nrange\n[−1; 1]\n[−1; 1]\nmaximum turning points\n(−360◦; 1), (0◦; 1) and\n(360◦; 1)\n(−30◦; 1) and (330◦; 1)\nminimum turning points\n(−180◦; −1) and\n(180◦; −1)\n(−210◦; −1) and\n(150◦; −1)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,87)\nx-intercept(s)\n(−270◦; 0), (−90◦; 0),\n(90◦; 0) and (270◦; 0)\n(−300◦; 0), (−120◦; 0),\n(60◦; 0) and (240◦; 0)\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = cos(θ + p):\nDomain and range\nThe domain is {θ : θ ∈R} because there is no value for θ for which f(θ) is undefined.\nThe range is {f(θ) : −1 ≤f(θ) ≤1, f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = cos(θ + p)\n= cos(0◦+ p)\n= cos p\nThis gives the point (0◦; cos p).\n217\nChapter 5.\nFunctions\n\nExercise 5 – 26: Cosine functions of the form y = cos(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Amplitude\n• Domain and range\n• x- and y-intercepts\n• Maximum and minimum turning points\n1. f(θ) = cos(θ + 45◦)\n2. g(θ) = cos(θ −30◦)\n3. h(θ) = cos(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22VP\n2. 22VQ\n3. 22VR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSketching cosine graphs\nEMBH7\nWorked example 24: Sketching a cosine graph\nQUESTION\nSketch the graph of f(θ) = cos(180◦−3θ) for 0◦≤θ ≤360◦.\nSOLUTION\nStep 1: Examine the form of the equation\nWrite the equation in the form y = cos k(θ + p).\nf(θ) = cos(180◦−3θ)\n= cos(−3θ + 180◦)\n= cos (−3(θ −60◦))\n= cos 3(θ −60◦)\nTo draw a graph of the above function, the standard cosine graph, y = cos θ, must be\nchanged in the following ways:\n218\n5.6.\nThe cosine function\n\n• decrease the period by a factor of 3\n• shift to the right by 60◦.\nStep 2: Complete a table of values\nθ\n0◦\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nf(θ)\n−1\n0,71\n0\n−0,71\n1\n−0,71\n0\n0,71\n−1\nStep 3: Plot the points and join with a smooth curve\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nf(θ) = cos 3(θ −60◦)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nPeriod: 120◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−1; 1]\nMaximum turning point: (60◦; 1), (180◦; 1) and (300◦; 1)\nMinimum turning point: (0◦; −1), (120◦; −1), (240◦; −1) and (360◦; −1)\ny-intercepts: (0◦; −1)\nx-intercept: (30◦; 0), (90◦; 0), (150◦; 0), (210◦; 0), (270◦; 0) and (330◦; 0)\nWorked example 25: Finding the equation of a cosine graph\nQUESTION\nGiven the graph of y = a cos(kθ+p), determine the values of a, k, p and the minimum\nturning point.\nθ\ny\ny = a cos(θ + p)\nb\n(45◦; 2)\n−45◦\n315◦\n219\nChapter 5.\nFunctions\n\nSOLUTION\nStep 1: Determine the value of k\nFrom the sketch we see that the period of the graph is 360◦, therefore k = 1.\ny = a cos(θ + p)\nStep 2: Determine the value of a\nFrom the sketch we see that the maximum turning point is (45◦; 2), so we know that\nthe amplitude of the graph is 2 and therefore a = 2.\ny = 2 cos(θ + p)\nStep 3: Determine the value of p\nCompare the given graph with the standard cosine function y = cos θ and notice the\ndifference in the maximum turning points. We see that the given function has been\nshifted to the right by 45◦, therefore p = 45◦.\ny = 2 cos(θ −45◦)\nStep 4: Determine the minimum turning point\nAt the minimum turning point, y = −2:\ny = 2 cos(θ −45◦)\n−2 = 2 cos(θ −45◦)\n−1 = cos(θ −45◦)\ncos−1(−1) = θ −45◦\n180◦= θ −45◦\n225◦= θ\nThis gives the point (225◦; −2).\n220\n5.6.\nThe cosine function\n\nExercise 5 – 27: The cosine function\n1. Sketch the following graphs on separate axes:\na) y = cos(θ + 15◦) for −180◦≤θ ≤180◦\nb) f(θ) = 1\n3 cos(θ −60◦) for −90◦≤θ ≤90◦\nc) y = −2 cos θ for 0◦≤θ ≤360◦\nd) y = cos(30◦−θ) for −360◦≤θ ≤360◦\ne) g(θ) = 1 + cos(θ −90◦) for 0◦≤θ ≤360◦\nf) y = cos(2θ + 60◦) for −360◦≤θ ≤360◦\n2. Two girls are given the following graph:\n0\n1\n−1\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\ny\nb\na) Audrey decides that the equation for the graph is a cosine function of the\nform y = a cos θ. Determine the value of a.\nb) Megan thinks that the equation for the graph is a cosine function of the\nform y = cos(θ + p). Determine the value of p.\nc) What can they conclude?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22VS\n1b. 22VT\n1c. 22VV\n1d. 22VW\n1e. 22VX\n1f. 22VY\n2. 22VZ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n221\nChapter 5.\nFunctions\n\n5.7\nThe tangent function\nEMBH8\nRevision\nEMBH9\nFunctions of the form y = tan θ for 0◦≤θ ≤360◦\n0\n1\n2\n−1\n−2\n90◦\n180◦\n270◦\n360◦\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nf(θ)\nθ\nThe dashed vertical lines are called the asymptotes. The asymptotes are at the values\nof θ where tan θ is not defined.\n• Period: 180◦\n• Domain: {θ : 0◦≤θ ≤360◦, θ ̸= 90◦; 270◦}\n• Range: {f(θ) : f(θ) ∈R}\n• x-intercepts: (0◦; 0), (180◦; 0), (360◦; 0)\n• y-intercept: (0◦; 0)\n• Asymptotes: the lines θ = 90◦and θ = 270◦\nFunctions of the form y = a tan θ + q\nTangent functions of the general form y = a tan θ + q, where a and q are constants.\nThe effects of a and q on f(θ) = a tan θ + q:\n• The effect of q on vertical shift\n– For q > 0, f(θ) is shifted vertically upwards by q units.\n– For q < 0, f(θ) is shifted vertically downwards by q units.\n• The effect of a on shape\n– For a > 1, branches of f(θ) are steeper.\n– For 0 < a < 1, branches of f(θ) are less steep and curve more.\n222\n5.7.\nThe tangent function\n\n– For a < 0, there is a reflection about the x-axis.\n– For −1 < a < 0, there is a reflection about the x-axis and the branches of\nthe graph are less steep.\n– For a < −1, there is a reflection about the x-axis and the branches of the\ngraph are steeper.\na < 0\na > 0\nq > 0\nb\n0\nb\n0\nq = 0\nb\n0\nb\n0\nq < 0\nb\n0\nb\n0\nExercise 5 – 28: Revision\nOn separate axes, accurately draw each of the following functions for 0◦≤θ ≤360◦:\n• Use tables of values if necessary.\n• Use graph paper if available.\nFor each function determine the following:\n• Period\n• Domain and range\n223\nChapter 5.\nFunctions\n\n• x- and y-intercepts\n• Asymptotes\n1. y1 = tan θ −1\n2\n2. y2 = −3 tan θ\n3. y3 = tan θ + 2\n4. y4 = 2 tan θ −1\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W2\n2. 22W3\n3. 22W4\n4. 22W5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFunctions of the form y = tan(kθ)\nEMBHB\nInvestigation: The effects of k on a tangent graph\n1. Complete the following table for y1 = tan θ for −360◦≤θ ≤360◦:\nθ\n−360◦\n−300◦\n−240◦\n−180◦\n−120◦\n−60◦\n0◦\ntan θ\nθ\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\ntan θ\n2. Use the table of values to plot the graph of y1 = tan θ for −360◦≤θ ≤360◦.\n3. On the same system of axes, plot the following graphs:\na) y2 = tan(−θ)\nb) y3 = tan 3θ\nc) y4 = tan θ\n2\n4. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of k\n224\n5.7.\nThe tangent function\n\n5. What do you notice about y1 = tan θ and y2 = tan(−θ)?\n6. Is tan(−θ) = −tan θ a true statement? Explain your answer.\n7. Can you deduce a formula for determining the period of y = tan kθ?\nThe effect of the parameter on y = tan kθ\nThe value of k affects the period of the tangent function. If k is negative, then the\ngraph is reflected about the y-axis.\n• For k > 0:\nFor k > 1, the period of the tangent function decreases.\nFor 0 < k < 1, the period of the tangent function increases.\n• For k < 0:\nFor −1 < k < 0, the graph is reflected about the y-axis and the period increases.\nFor k < −1, the graph is reflected about the y-axis and the period decreases.\nNegative angles:\ntan(−θ) = −tan θ\nCalculating the period:\nTo determine the period of y = tan kθ we use,\nPeriod = 180◦\n|k|\nwhere |k| is the absolute value of k.\nk > 0\nk < 0\n225\nChapter 5.\nFunctions\n\nWorked example 26: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan 3θ\n2\n2. For each function determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan kθ\nNotice that k > 1 for y2 = tan 3θ\n2 , therefore the period of the graph decreases.\nStep 2: Complete a table of values\nθ\n−180◦\n−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan 3θ\n2\nUNDEF\n−0,41\n1\n−2,41\n0\n2,41\n−1\n0,41\nUNDEF\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan 3θ\n2\nperiod\n180◦\n120◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦< θ < 180◦, θ ̸=\n−60◦; 60◦}\nrange\n{f(θ) : f(θ) ∈R}\n{f(θ) : f(θ) ∈R}\ny-intercept(s)\n(0◦; 0)\n(0◦; 0)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and\n(180◦; 0)\n(−120◦; 0), (0◦; 0) and\n(120◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −180◦; −60◦and 180◦\n226\n5.7.\nThe tangent function\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan kθ:\nDomain and range\nThe domain of one branch is {θ : −90◦\nk\n< θ < 90◦\nk , θ ∈R} because f(θ) is undefined\nfor θ = −90◦\nk and θ = 90◦\nk .\nThe range is {f(θ) : f(θ) ∈R} or (−∞; ∞).\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0 and solving for f(θ).\ny = tan kθ\n= tan 0◦\n= 0\nThis gives the point (0◦; 0).\nAsymptotes\nThese are the values of kθ for which tan kθ is undefined.\nExercise 5 – 29: Tangent functions of the form y = tan kθ\nSketch the following functions for −180◦≤θ ≤180◦. For each graph determine:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan 2θ\n2. g(θ) = tan 3θ\n4\n3. h(θ) = tan(−2θ)\n4. k(θ) = tan 2θ\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22W6\n2. 22W7\n3. 22W8\n4. 22W9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n227\nChapter 5.\nFunctions\n\nFunctions of the form y = tan (θ + p)\nEMBHC\nWe now consider tangent functions of the form y = tan(θ + p) and the effects of\nparameter p.\nInvestigation: The effects of p on a tangent graph\n1. On the same system of axes, plot the following graphs for −360◦≤θ ≤360◦:\na) y1 = tan θ\nb) y2 = tan(θ −60◦)\nc) y3 = tan(θ −90◦)\nd) y4 = tan(θ + 60◦)\ne) y5 = tan(θ + 180◦)\n2. Use your sketches of the functions above to complete the following table:\ny1\ny2\ny3\ny4\ny5\nperiod\ndomain\nrange\ny-intercept(s)\nx-intercept(s)\nasymptotes\neffect of p\nThe effect of the parameter on y = tan(θ + p)\nThe effect of p on the tangent function is a horizontal shift (or phase shift); the entire\ngraph slides to the left or to the right.\n• For p > 0, the graph of the tangent function shifts to the left by p.\n• For p < 0, the graph of the tangent function shifts to the right by p.\np > 0\np < 0\n228\n5.7.\nThe tangent function\n\nWorked example 27: Tangent function\nQUESTION\n1. Sketch the following functions on the same set of axes for −180◦≤θ ≤180◦.\na) y1 = tan θ\nb) y2 = tan(θ + 30◦)\nFor each function determine the following:\n2.\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\nSOLUTION\nStep 1: Examine the equations of the form y = tan(θ + p)\nNotice that for y1 = tan θ we have p = 0◦(no phase shift) and for y2 = tan(θ + 30◦),\np > 0 therefore the graph shifts to the left by 30◦.\nStep 2: Complete a table of values\nθ\n−180◦−135◦\n−90◦\n−45◦\n0◦\n45◦\n90◦\n135◦\n180◦\ntan θ\n0\n1\nUNDEF\n−1\n0\n1\nUNDEF\n−1\n0\ntan(θ+30◦)\n0,58\n3,73\n−1,73\n−0,27\n0,58\n3,73\n−1,73\n−0,27\n0,58\nStep 3: Sketch the tangent graphs\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nStep 4: Complete the table\ny1 = tan θ\ny2 = tan(θ + 30◦)\nperiod\n180◦\n180◦\ndomain\n{θ : −180◦≤θ ≤180◦, θ ̸=\n−90◦; 90◦}\n{θ : −180◦≤θ ≤\n180◦, θ ̸= −120◦; 60◦}\nrange\n(−∞; ∞)\n(−∞; ∞)\ny-intercept(s)\n(0◦; 0)\n(0◦; 0,58)\nx-intercept(s)\n(−180◦; 0), (0◦; 0) and (180◦; 0)\n(−30◦; 0) and (150◦; 0)\nasymptotes\nθ = −90◦and θ = 90◦\nθ = −120◦and θ = 60◦\n229\nChapter 5.\nFunctions\n\nDiscovering the characteristics\nFor functions of the general form: f(θ) = y = tan(θ + p):\nDomain and range\nThe domain of one branch is {θ : θ ∈(−90◦−p; 90◦−p)} because the function is\nundefined for θ = −90◦−p and θ = 90◦−p.\nThe range is {f(θ) : f(θ) ∈R}.\nIntercepts\nThe x-intercepts are determined by letting f(θ) = 0 and solving for θ.\nThe y-intercept is calculated by letting θ = 0◦and solving for f(θ).\ny = tan(θ + p)\n= tan(0◦+ p)\n= tan p\nThis gives the point (0◦; tan p).\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\nSketch the following functions for −360◦≤θ ≤360◦.\nFor each function, determine the following:\n• Period\n• Domain and range\n• x- and y-intercepts\n• Asymptotes\n1. f(θ) = tan(θ + 45◦)\n2. g(θ) = tan(θ −30◦)\n3. h(θ) = tan(θ + 60◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WB\n2. 22WC\n3. 22WD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n230\n5.7.\nThe tangent function\n\nSketching tangent graphs\nEMBHD\nWorked example 28: Sketching a tangent graph\nQUESTION\nSketch the graph of f(θ) = tan 1\n2(θ −30◦) for −180◦≤θ ≤180◦.\nSOLUTION\nStep 1: Examine the form of the equation\nFrom the equation we see that 0 < k < 1, therefore the branches of the graph will be\nless steep than the standard tangent graph y = tan θ. We also notice that p < 0 so the\ngraph will be shifted to the right on the x-axis.\nStep 2: Determine the period\nThe period for f(θ) = tan 1\n2(θ −30◦) is:\nPeriod = 180◦\n|k|\n= 180◦\n1\n2\n= 360◦\nStep 3: Determine the asymptotes\nThe standard tangent graph, y = tan θ, for −180◦≤θ ≤180◦is undefined at θ = −90◦\nand θ = 90◦. Therefore we can determine the asymptotes of f(θ) = tan 1\n2(θ −30◦):\n•\n−90◦\n0,5 + 30◦= −150◦\n•\n90◦\n0,5 + 30◦= 210◦\nThe asymptote at θ = 210◦lies outside the required interval.\n231\nChapter 5.\nFunctions\n\nStep 4: Plot the points and join with a smooth curve\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 360◦\nDomain: {θ : −180◦≤θ ≤180◦, θ ̸= −150◦}\nRange: (−∞; ∞)\ny-intercepts: (0◦; −0,27)\nx-intercept: (30◦; 0)\nAsymptotes: θ = −150◦\nExercise 5 – 31: The tangent function\n1. Sketch the following graphs on separate axes:\na) y = tan θ −1 for −90◦≤θ ≤90◦\nb) f(θ) = −tan 2θ for 0◦≤θ ≤90◦\nc) y = 1\n2 tan(θ + 45◦) for 0◦≤θ ≤360◦\nd) y = tan(30◦−θ) for −180◦≤θ ≤180◦\n2. Given the graph of y = a tan kθ, determine the values of a and k.\nθ\nf(θ)\nb\nb\n(90◦; −1)\n360◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WF\n1b. 22WG\n1c. 22WH\n1d. 22WJ\n2. 22WK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n232\n5.7.\nThe tangent function\n\nExercise 5 – 32: Mixed exercises\n1. Determine the equation for each of the following:\na) f(θ) = a sin kθ and g(θ) = a tan θ\nθ\ny\nb\nb\nf\n(45◦; −3\n2 )\ng\n(180◦; 0)\n(135◦; −1 1\n2 )\nb) f(θ) = a sin kθ and g(θ) = a cos(θ + p)\nθ\n0\ny\nb\n(−90◦; 2)\n−180◦\n180◦\nf and g\nc) y = a tan kθ\nθ\n0\ny\nb\n(90◦; 3)\n360◦\n180◦\nd) y = a cos θ + q\nθ\n0\ny\n4\n360◦\n180◦\n233\nChapter 5.\nFunctions\n\n2. Given the functions f(θ) = 2 sin θ and g(θ) = cos θ + 1:\na) Sketch the graphs of both functions on the same system of axes, for 0◦≤\nθ ≤360◦. Indicate the turning points and intercepts on the diagram.\nb) What is the period of f?\nc) What is the amplitude of g?\nd) Use your sketch to determine how many solutions there are for the equation\n2 sin θ −cos θ = 1. Give one of the solutions.\ne) Indicate on your sketch where on the graph the solution to 2 sin θ = −1 is\nfound.\n3. The sketch shows the two functions f(θ) = a cos θ and g(θ) = tan θ for 0◦≤θ ≤\n360◦. Points P(135◦; b) and Q(c; −1) lie on g(θ) and f(θ) respectively.\nθ\n0\ny\nb\nb\n360◦\n180◦\nP\nQ\ng\nf\n−1\n2\n−2\na) Determine the values of a, b and c.\nb) What is the period of g?\nc) Solve the equation cos θ = 1\n2 graphically and show your answer(s) on the\ndiagram.\nd) Determine the equation of the new graph if g is reflected about the x-axis\nand shifted to the right by 45◦.\n4. Sketch the graphs of y1 = −1\n2 sin(θ + 30◦) and y2 = cos(θ −60◦), on the same\nsystem of axes for 0◦≤θ ≤360◦.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22WM\n1b. 22WN\n1c. 22WP\n1d. 22WQ\n2. 22WR\n3. 22WS\n4. 22WT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n234\n5.7.\nThe tangent function\n\n5.8\nSummary\nEMBHF\nSee presentation: 22WV at www.everythingmaths.co.za\n1. Parabolic functions:\nStandard form: y = ax2 + bx + c\n• y-intercept: (0; c)\n• x-intercept: x = −b±\n√\nb2−4ac\n2a\n• Turning point:\n\u0010\n−b\n2a; −b2\n4a + c\n\u0011\n• Axis of symmetry: x = −b\n2a\nCompleted square form: y = a(x + p)2 + q\n• Turning point: (−p; q)\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n2. Average gradient:\n• Average gradient = y2−y1\nx2−x1\n3. Hyperbolic functions:\nStandard form: y = k\nx\n• k > 0: first and third quadrant\n• k < 0: second and fourth quadrant\nShifted form: y =\nk\nx+p + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: x = −p and y = q\n4. Exponential functions:\nStandard form: y = abx\n• a > 0: above x-axis\n• a < 0: below x-axis\n• b > 1: increasing function if a > 0; decreasing function if a < 0\n• 0 < b < 1: decreasing function if a > 0; increasing function if a < 0\nShifted form: y = ab(x+p) + q\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n235\nChapter 5.\nFunctions\n\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• Asymptotes: y = q\n5. Sine functions:\nShifted form: y = a sin(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• sin(−θ) = −sin θ\n6. Cosine functions:\nShifted form: y = a cos(kθ + p) + q\n• Period = 360◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• cos(−θ) = cos θ\n7. Tangent functions:\nShifted form: y = a tan(kθ + p) + q\n• Period = 180◦\n|k|\n• k > 1 or k < −1: period decreases\n• 0 < k < 1 or −1 < k < 0: period increases\n• p > 0: horizontal shift left\n• p < 0: horizontal shift right\n• q > 0: vertical shift up\n• q < 0: vertical shift down\n• tan(−θ) = −tan θ\n• Asymptotes: 90◦−p\nk\n± 180◦n\nk\n, n ∈Z\n236\n5.8.\nSummary\n\nExercise 5 – 33: End of chapter exercises\n1. Show that if a\n<\n0,\nthen the range of f(x)\n=\na(x + p)2 + q is\n{f(x) : f(x) ∈(−∞, q]}.\n2. If (2; 7) is the turning point of f(x) = −2x2 −4ax + k, find the values of the\nconstants a and k.\n3. The following graph is represented by the equation f(x) = ax2 + bx. The coor-\ndinates of the turning point are (3; 9). Show that a = −1 and b = 6.\nb (3; 9)\nx\n0\ny\n4. Given: f(x) = x2 −2x + 3. Give the equation of the new graph originating if:\na) the graph of f is moved three units to the left.\nb) the x-axis is moved down three units.\n5. A parabola with turning point (−1; −4) is shifted vertically by 4 units upwards.\nWhat are the coordinates of the turning point of the shifted parabola?\n6. Plot the graph of the hyperbola defined by y = 2\nx for −4 ≤x ≤4. Suppose\nthe hyperbola is shifted 3 units to the right and 1 unit down. What is the new\nequation then?\n7. Based on the graph of y =\nk\n(x+p) + q, determine the equation of the graph with\nasymptotes y = 2 and x = 1 and passing through the point (2; 3).\ny\nx\n0\n2\n1\nb (2; 3)\n237\nChapter 5.\nFunctions\n\n8. The columns in the table below give the y-values for the following functions:\ny = ax, y = ax+1 and y = ax + 1. Match each function to the correct column.\nx\nA\nB\nC\n−2\n7,25\n6,25\n2,5\n−1\n3,5\n2,5\n1\n0\n2\n1\n0,4\n1\n1,4\n0,4\n0,16\n2\n1,16\n0,16\n0,064\n9. The graph of f(x) = 1 + a . 2x (a is a constant) passes through the origin.\na) Determine the value of a.\nb) Determine the value of f(−15) correct to five decimal places.\nc) Determine the value of x, if P (x; 0,5) lies on the graph of f.\nd) If the graph of f is shifted 2 units to the right to give the function h, write\ndown the equation of h.\n10. The graph of f(x) = a . bx (a ̸= 0) has the point P (2; 144) on f.\na) If b = 0,75, calculate the value of a.\nb) Hence write down the equation of f.\nc) Determine, correct to two decimal places, the value of f(13).\nd) Describe the transformation of the curve of f to h if h(x) = f(−x).\n11. Using your knowledge of the effects of p and k draw a rough sketch of the fol-\nlowing graphs without a table of values.\na) y = sin 3θ for −180◦≤θ ≤180◦\nb) y = −cos 2θ for 0◦≤θ ≤180◦\nc) y = tan 1\n2θ for 0◦≤θ ≤360◦\nd) y = sin(θ −45◦) for −360◦≤θ ≤360◦\ne) y = cos(θ + 45◦) for 0◦≤θ ≤360◦\nf) y = tan(θ −45◦) for 0◦≤θ ≤360◦\ng) y = 2 sin 2θ for −180◦≤θ ≤180◦\nh) y = sin(θ + 30◦) + 1 for −360◦≤θ ≤0◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22WW\n2. 22WX\n3. 22WY\n4. 22WZ\n5. 22X2\n6. 22X3\n7. 22X4\n8. 22X5\n9. 22X6\n10. 22X7\n11a. 22X8\n11b. 22X9\n11c. 22XB\n11d. 22XC\n11e. 22XD\n11f. 22XF\n11g. 22XG\n11h. 22XH\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n238\n5.8.\nSummary\n\nCHAPTER\n6\nTrigonometry", "chapter_id": "22" }, { "title": "Revision", "content": "6.1\nRevision\n240\n6.2\nTrigonometric identities\n247\n6.3\nReduction formula\n253\n6.4\nTrigonometric equations\n266\n6.5\nArea, sine, and cosine rules\n280\n6.6\nSummary\n301\n\n6\nTrigonometry\n6.1\nRevision\nEMBHG\nTrigonometric ratios\nb\nb\nb\ny\nx\nP(x; y)\nQ(−x; y)\nO\nα\nβ\nr\nr\nWe plot the points P(x; y) and Q(−x; y) in the Cartesian plane and measure the angles\nfrom the positive x-axis to the terminal arms (OP and OQ).\nP(x; y) lies in the first quadrant with P ˆOX = α and Q(−x; y) lies in the second\nquadrant with Q ˆOX = β.\nUsing the theorem of Pythagoras we have that\nOP 2 = x2 + y2\nAnd OQ2 = (−x)2 + y2\n= x2 + y2\n∴OP = OQ\nLet OP = OQ = r.\nTrigonometric ratios\nsin α = y\nr\ncos α = x\nr\ntan α = y\nx\nIn the second quadrant we notice that −x < 0\nsin β = y\nr\ncos β = −x\nr\ntan β = −y\nx\n240\n6.1.\nRevision\n\nSimilarily, in the third and fourth quadrants the sign of the trigonometric ratios depends\non the signs of x and y:\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nSpecial angles\n30◦\n60◦\n1\n√\n3\n2\n45◦\n45◦\n1\n1\n√\n2\nθ\n0◦\n30◦\n45◦\n60◦\n90◦\ncos θ\n1\n√\n3\n2\n1\n√\n2\n1\n2\n0\nsin θ\n0\n1\n2\n1\n√\n2\n√\n3\n2\n1\ntan θ\n0\n1\n√\n3\n1\n√\n3\nundef\nSee video: 22XJ at www.everythingmaths.co.za\n241\nChapter 6.\nTrigonometry\n\nSolving equations\nWorked example 1: Solving equations\nQUESTION\nDetermine the values of a and b in the right-angled triangle TUW (correct to one\ndecimal place):\nT\nW\nU\n47◦\nb\n30\na\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of a\nsin θ = opposite side\nhypotenuse\nsin 47◦= 30\na\na =\n30\nsin 47◦\n∴a = 41,0\nStep 3: Determine the value of b\nAlways try to use the information that is given for calculations and not answers that you\nhave worked out in case you have made an error. For example, avoid using a = 41,0\nto determine the value of b.\ntan θ = opposite side\nadjacent side\ntan 47◦= 30\nb\nb =\n30\ntan 47◦\n∴b = 28,0\nStep 4: Write the final answer\na = 41,0 units and b = 28,0 units.\n242\n6.1.\nRevision\n\nFinding an angle\nWorked example 2: Finding an angle\nQUESTION\nCalculate the value of θ in the right-angled triangle MNP (correct to one decimal\nplace):\nP\nN\nM\n41\nθ\n24\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of θ\ntan θ = opposite side\nadjacent side\ntan θ = 41\n24\n∴θ = tan−1\n\u001241\n24\n\u0013\nθ = 59,7◦\n243\nChapter 6.\nTrigonometry\n\nWorked example 3: Finding an angle\nQUESTION\nGiven 2 sin θ\n2 = cos 43◦, for θ ∈[0◦; 90◦], determine the value of θ (correct to one\ndecimal place).\nSOLUTION\nStep 1: Simplify the equation\nAvoiding rounding off in calculations until you have determined the final answer. In\nthe calculation below, the dots indicate that the number has not been rounded so that\nthe answer is as accurate as possible.\n2 sin θ\n2 = cos 43◦\nsin θ\n2 = cos 43◦\n2\nθ\n2 = sin−1(0,365 . . .)\nθ = 2(21,449 . . .)\n∴θ = 42,9◦\nTwo-dimensional problems\nWorked example 4: Flying a kite\nQUESTION\nThelma flies a kite on a 22 m piece of string and the height of the kite above the\nground is 20,4 m. Determine the angle of inclination of the string (correct to one\ndecimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the opposite and adjacent sides and the hy-\npotenuse\nLet the angle of inclination of the string be θ.\n244\n6.1.\nRevision\n\nKite\nThelma\n20,4\nθ\n22\nStep 2: Use an appropriate trigonometric ratio to find θ\nsin θ = opposite side\nhypotenuse\n= 20,4\n22\nθ = sin−1(0,927 . . .)\n∴θ = 68,0◦\nExercise 6 – 1: Revision\n1. If p = 49◦and q = 32◦, use a calculator to determine whether the following\nstatements are true of false:\na) sin p + 3 sin p = 4 sin p\nb) sin q\ncos q = tan q\nc) cos(p −q) = cos p −cos q\nd) sin(2p) = 2 sin p cos p\n2. Determine the following angles (correct to one decimal place):\na) cos α = 0,64\nb) sin θ + 2 = 2,65\nc) 1\n2 cos 2β = 0,3\nd) tan θ\n3 = sin 48◦\ne) cos 3p = 1,03\nf) 2 sin 3β + 1 = 2,6\ng) sin θ\ncos θ = 42\n3\n3. In △ABC, A ˆCB = 30◦, AC = 20 cm and BC = 22 cm. The perpendicular\nline from A intersects BC at T.\n245\nChapter 6.\nTrigonometry\n\nDetermine:\nA\nC\nB\nT\n20 cm\n22 cm\n30◦\na) the length TC\nb) the length AT\nc) the angle B ˆAT\n4. A rhombus has a perimeter of 40 cm and one of the internal angles is 30◦.\na) Determine the length of the sides.\nb) Determine the lengths of the diagonals.\nc) Calculate the area of the rhombus.\n5. Simplify the following without using a calculator:\na) 2 sin 45◦× 2 cos 45◦\nb) cos2 30◦−sin2 60◦\nc) sin 60◦cos 30◦−cos 60◦sin 30◦−tan 45◦\nd) 4 sin 60◦cos 30◦−2 tan 45◦+ tan 60◦−2 sin 60◦\ne) sin 60◦×\n√\n2 tan 45◦+ 1 −sin 30◦\n6. Given the diagram below.\nb\nx\ny\n0\nB(2; 2\n√\n3)\nβ\nDetermine the following without using a calculator:\na) β\nb) cos β\nc) cos2 β + sin2 β\n7. The 10 m ladder of a fire truck leans against the wall of a burning building at an\nangle of 60◦. The height of an open window is 9 m from the ground. Will the\nladder reach the window?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22XK\n2a. 22XM\n2b. 22XN\n2c. 22XP\n2d. 22XQ\n2e. 22XR\n2f. 22XS\n2g. 22XT\n3. 22XV\n4. 22XW\n5a. 22XX\n5b. 22XY\n5c. 22XZ\n5d. 22Y2\n5e. 22Y3\n6. 22Y4\n7. 22Y5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n246\n6.1.\nRevision\n\n6.2\nTrigonometric identities\nEMBHH\nAn identity is a mathematical statement that equates one quantity with another. Trigono-\nmetric identities allow us to simplify a given expression so that it contains sine and co-\nsine ratios only. This enables us to solve equations and also to prove other identities.\nQuotient identity\nInvestigation: Quotient identity\n1. Complete the table without using a calculator, leaving your answer in surd form\nwhere applicable:\nθ = 45◦\nθ\n3\n5\nx\ny\n(3; 2)\nθ\nb\nsin θ\ncos θ\nsin θ\ncos θ\ntan θ\n2. Examine the last two rows of the table and make a conjecture.\n3. Are there any values of θ for which your conjecture would not be true? Explain\nyour answer.\nWe know that tan θ is defined as:\ntan θ = opposite side\nadjacent side\nUsing the diagram below and the theorem of Pythagoras, we can write the tangent\nfunction in terms of x, y and r:\nx\ny\n(x; y)\nθ\nb\nO\n247\nChapter 6.\nTrigonometry\n\ntan θ = y\nx\n= y\nx × r\nr\n= y\nr × r\nx\n= y\nr ÷ x\nr\n= sin θ ÷ cos θ\n= sin θ\ncos θ\nThis is the quotient identity:\ntan θ = sin θ\ncos θ\nNotice that tan θ is undefined if cos θ = 0, therefore θ ̸= k × 90◦, where k is an odd\ninteger.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\ntan θ\nSquare identity\nInvestigation: Square identity\n1. Use a calculator to complete the following table:\nsin2 80◦+ cos2 80◦=\ncos2 23◦+ sin2 23◦=\nsin 50◦+ cos 50◦=\nsin2 67◦−cos2 67◦=\nsin2 67◦+ cos2 67◦=\n2. What do you notice? Make a conjecture.\n3. Draw a sketch and prove your conjecture in general terms, using x, y and r.\n248\n6.2.\nTrigonometric identities\n\nx\ny\n(x; y)\nα\nb\nO\nr\nUsing the theorem of Pythagoras, we can write the sine and cosine functions in terms\nof x, y and r:\nsin2 θ + cos2 θ =\n\u0010y\nr\n\u00112\n+\n\u0010x\nr\n\u00112\n= y2\nr2 + x2\nr2\n= y2 + x2\nr2\n= r2\nr2\n= 1\nThis is the square identity:\nsin2 θ + cos2 θ = 1\nOther forms of the square identity\nComplete the following:\n1. sin2 θ = 1 −. . . . . .\n2. cos θ = ±√. . . . . .\n3. sin2 θ = (1 + . . . . . .)(1 −. . . . . .)\n4. cos2 θ −1 = . . . . . .\nHere are some useful tips for proving identities:\n• Change all trigonometric ratios to sine and cosine.\n• Choose one side of the equation to simplify and show that it is equal to the other\nside.\n• Usually it is better to choose the more complicated side to simplify.\n• Sometimes we need to simplify both sides of the equation to show that they are\nequal.\n• A square root sign often indicates that we need to use the square identity.\n• We can also add to the expression to make simplifying easier:\n– replace 1 with sin2 θ + cos2 θ.\n– multiply by 1 in the form of a suitable fraction, for example 1 + sin θ\n1 + sin θ.\n249\nChapter 6.\nTrigonometry\n\nSee video: 22Y6 at www.everythingmaths.co.za\nWorked example 5: Trigonometric identities\nQUESTION\nSimplify the following:\n1. tan2 θ × cos2 θ\n2.\n1\ncos2 θ −tan2 θ\nSOLUTION\nStep 1: Write the expression in terms of sine and cosine only\nWe use the square and quotient identities to write the given expression in terms of sine\nand cosine and then simplify as far as possible.\n1.\ntan2 θ × cos2 θ =\n\u0012 sin θ\ncos θ\n\u00132\n× cos2 θ\n= sin2 θ\ncos2 θ × cos2 θ\n= sin2 θ\n2.\n1\ncos2 θ −tan2 θ =\n1\ncos2 θ −\n\u0012 sin θ\ncos θ\n\u00132\n=\n1\ncos2 θ −sin2 θ\ncos2 θ\n= 1 −sin2 θ\ncos2 θ\n= cos2 θ\ncos2 θ\n= 1\n250\n6.2.\nTrigonometric identities\n\nWorked example 6: Trigonometric identities\nQUESTION\nProve: 1 −sin α\ncos α\n=\ncos α\n1 + sin α\nSOLUTION\nStep 1: Note restrictions\nWhen working with fractions, we must be careful that the denominator does not equal\n0. Therefore cos θ ̸= 0 for the fraction on the left-hand side and sin θ + 1 ̸= 0 for the\nfraction on the right-hand side.\nStep 2: Simplify the left-hand side\nThis is not an equation that needs to be solved. We are required to show that one side\nof the equation is equal to the other. We can choose either of the two sides to simplify.\nLHS = 1 −sin α\ncos α\n= 1 −sin α\ncos α\n× 1 + sin α\n1 + sin α\nNotice that we have not changed the equation — this is the same as multiplying by 1\nsince the numerator and the denominator are the same.\nStep 3: Determine the lowest common denominator and simplify\nLHS =\n1 −sin2 α\ncos α(1 + sin α)\n=\ncos2 α\ncos α(1 + sin α)\n=\ncos α\n1 + sin α\n= RHS\n251\nChapter 6.\nTrigonometry\n\nExercise 6 – 2: Trigonometric identities\n1. Reduce the following to one trigonometric ratio:\na) sin α\ntan α\nb) cos2 θ tan2 θ + tan2 θ sin2 θ\nc) 1 −sin θ cos θ tan θ\nd)\n\u00121 −cos2 β\ncos2 β\n\u0013\n−tan2 β\n2. Prove the following identities and state restrictions where appropriate:\na) 1 + sin θ\ncos θ\n=\ncos θ\n1 −sin θ\nb) sin2α + (cos α −tan α) (cos α + tan α) = 1 −tan2α\nc)\n1\ncos θ −cos θtan2θ\n1\n= cos θ\nd)\n2 sin θ cos θ\nsin θ + cos θ = sin θ + cos θ −\n1\nsin θ + cos θ\ne)\n\u0012cos β\nsin β + tan β\n\u0013\ncos β =\n1\nsin β\nf)\n1\n1 + sin θ +\n1\n1 −sin θ = d\n2 tan θ\nsin θ cos θ\ng) (1 + tan2 α) cos α\n(1 −tan α)\n=\n1\ncos α −sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Y7\n1b. 22Y8\n1c. 22Y9\n1d. 22YB\n2a. 22YC\n2b. 22YD\n2c. 22YF\n2d. 22YG\n2e. 22YH\n2f. 22YJ\n2g. 22YK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n252\n6.2.\nTrigonometric identities\n\n6.3\nReduction formula\nEMBHJ\nAny trigonometric function whose argument is 90◦± θ; 180◦± θ and 360◦± θ can be\nwritten simply in terms of θ.\nDeriving reduction formulae\nEMBHK\nInvestigation: Reduction formulae for function values of 180◦± θ\n1. Function values of 180◦−θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the y-axis, determine the coordi-\nnates of P ′.\nb) Write down values for sin θ, cos θ and tan θ.\nc) Use the coordinates for P ′ to determine sin(180◦−θ), cos(180◦−θ),\ntan(180◦−θ).\nd) From your results determine a relationship between the trigonometric func-\ntion values of (180◦−θ) and θ.\n253\nChapter 6.\nTrigonometry\n\n2. Function values of 180◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦+ θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the origin (the two points are sym-\nmetrical about both the x-axis and the y-axis), determine the coordinates of\nP ′.\nb) Use the coordinates for P ′ to determine sin(180◦+ θ), cos(180◦+ θ) and\ntan(180◦+ θ).\nc) From your results determine a relationship between the trigonometric func-\ntion values of (180◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(180◦−θ) = . . . . . .\nb) cos(180◦−θ) = . . . . . .\nc) tan(180◦−θ) = . . . . . .\nd) sin(180◦+ θ) = . . . . . .\ne) cos(180◦+ θ) = . . . . . .\nf) tan(180◦+ θ) = . . . . . .\n254\n6.3.\nReduction formula\n\nWorked example 7: Reduction formulae for function values of 180◦± θ\nQUESTION\nWrite the following as a single trigonometric ratio:\nsin 163◦\ncos 197◦+ tan 17◦+ cos(180◦−θ) × tan(180◦+ θ)\nSOLUTION\nStep 1: Use reduction formulae to write the trigonometric function values in terms\nof acute angles and θ\n= sin(180◦−17◦)\ncos(180◦+ 17◦) + tan 17◦+ (−cos θ) × tan θ\nStep 2: Simplify\n=\nsin 17◦\n−cos 17◦+ tan 17◦−cos θ × sin θ\ncos θ\n= −tan 17◦+ tan 17◦−sin θ\n= −sin θ\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1. Determine the value of the following expressions without using a calculator:\na) tan 150◦sin 30◦−cos 210◦\nb) (1 + cos 120◦)(1 −sin2 240◦)\nc) cos2 140◦+ sin2 220◦\n2. Write the following in terms of a single trigonometric ratio:\na) tan(180◦−θ) × sin(180◦+ θ)\nb) tan(180◦+ θ) cos(180◦−θ)\nsin(180◦−θ)\n255\nChapter 6.\nTrigonometry\n\n3. If t = tan 40◦, express the following in terms of t:\na) tan 140◦+ 3 tan 220◦\nb) cos 220◦\nsin 140◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YM\n1b. 22YN\n1c. 22YP\n2a. 22YQ\n2b. 22YR\n3. 22YS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of (360◦± θ) and (−θ)\n1. Function values of (360◦−θ) and (−θ)\nIn the Cartesian plane we measure angles from the positive x-axis to the terminal\narm, which means that an anti-clockwise rotation gives a positive angle. We can\ntherefore measure negative angles by rotating in a clockwise direction.\nFor an acute angle θ, we know that −θ will lie in the fourth quadrant.\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n360◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the x-axis (y = 0), determine the\ncoordinates of P ′.\nb) Use the coordinates of P ′ to determine sin(360◦−θ), cos(360◦−θ) and\ntan(360◦−θ).\nc) Use the coordinates of P ′ to determine sin(−θ), cos(−θ) and tan(−θ).\nd) From your results determine a relationship between the function values of\n(360◦−θ) and −θ.\n256\n6.3.\nReduction formula\n\ne) Complete the following reduction formulae:\ni. sin(360◦−θ) = . . . . . .\nii. cos(360◦−θ) = . . . . . .\niii. tan(360◦−θ) = . . . . . .\niv. sin(−θ) = . . . . . .\nv. cos(−θ) = . . . . . .\nvi. tan(−θ) = . . . . . .\n2. Function values of 360◦+ θ\nWe can also have an angle that is larger than 360◦. The angle completes a\nrevolution of 360◦and then continues to give an angle of θ.\nComplete the following reduction formulae:\na) sin(360◦+ θ) = . . . . . .\nb) cos(360◦+ θ) = . . . . . .\nc) tan(360◦+ θ) = . . . . . .\nFrom working with functions, we know that the graph of y = sin θ has a period of\n360◦. Therefore, one complete wave of a sine graph is the same as one complete\nrevolution for sin θ in the Cartesian plane.\n0\n1\n−1\n90◦\n180◦\n270◦\n360◦\n1st\n2nd\n3rd\n4th\npositive\npositive\nnegative\nnegative\n0◦/360◦\n90◦\n180◦\n270◦\n2nd\npos.\nneg.\n3rd\nneg.\n4th\n1st\npos.\nWe can also have multiple revolutions. The periodicity of the trigonometric graphs\nshows this clearly. A complete sine or cosine curve is completed in 360◦.\ny = cos θ\ny = sin θ\nθ\ny\n257\nChapter 6.\nTrigonometry\n\nIf k is any integer, then\nsin(k . 360◦+ θ) = sin θ\ncos(k . 360◦+ θ) = cos θ\ntan(k . 360◦+ θ) = tan θ\nWorked example 8: Reduction formulae for function values of 360◦± θ\nQUESTION\nIf f = tan 67◦, express the following in terms of f\nsin 293◦\ncos 427◦+ tan(−67◦) + tan 1147◦\nSOLUTION\nStep 1: Using reduction formula\n= sin(360◦−67◦)\ncos(360◦+ 67◦) −tan(67◦) + tan (3(360◦) + 67◦)\n= −sin 67◦\ncos 67◦−tan 67◦+ tan 67◦\n= −tan 67◦\n= −f\nWorked example 9: Using reduction formula\nQUESTION\nEvaluate without using a calculator:\ntan2 210◦−(1 + cos 120◦) sin2 405◦\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and special angles\n258\n6.3.\nReduction formula\n\n= tan2(180◦+ 30◦) −(1 + cos(180◦−60◦)) sin2(360◦+ 45◦)\n= tan2 30◦−(1 + (−cos 60◦)) sin2 45◦\n=\n\u0012 1\n√\n3\n\u00132\n−\n\u0012\n1 −1\n2\n\u0013 \u0012 1\n√\n2\n\u00132\n= 1\n3 −\n\u00121\n2\n\u0013 \u00121\n2\n\u0013\n= 1\n3 −1\n4\n= 1\n12\nExercise 6 – 4: Using reduction formula\n1. Simplify the following:\na) tan(180◦−θ) sin(360◦+ θ)\ncos(180◦+ θ) tan(360◦−θ)\nb) cos2(360◦+ θ) + cos(180◦+ θ) tan(360◦−θ) sin(360◦+ θ)\nc)\nsin(360◦+ α) tan(180◦+ α)\ncos(360◦−α) tan2(360◦+ α)\n2. Write the following in terms of cos β:\ncos(360◦−β) cos(−β) −1\nsin(360◦+ β) tan(360◦−β)\n3. Simplify the following without using a calculator:\na)\ncos 300◦tan 150◦\nsin 225◦cos(−45◦)\nb) 3 tan 405◦+ 2 tan 330◦cos 750◦\nc) cos 315◦cos 405◦+ sin 45◦sin 135◦\nsin 750◦\nd) tan 150◦cos 390◦−2 sin 510◦\ne) 2 sin 120◦+ 3 cos 765◦−2 sin 240◦−3 cos 45◦\n5 sin 300◦+ 3 tan 225◦−6 cos 60◦\n4. Given 90◦< α < 180◦, use a sketch to help explain why:\na) sin(−α) = −sin α\nb) cos(−α) = −cos α\n259\nChapter 6.\nTrigonometry\n\n5. If t = sin 43◦, express the following in terms of t:\na) sin 317◦\nb) cos2 403◦\nc) tan(−43◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YT\n1b. 22YV\n1c. 22YW\n2. 22YX\n3a. 22YY\n3b. 22YZ\n3c. 22Z2\n3d. 22Z3\n3e. 22Z4\n4. 22Z5\n5. 22Z6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of 90◦± θ\nIn any right-angled triangle, the two acute angles are complements of each other, ˆA +\nˆC = 90◦\nA\nB\nC\nb\na\nc\nComplete the following:\nIn △ABC\nsin ˆC = c\nb = cos . . .\ncos ˆC = a\nb = sin . . .\nComplementary angles are positive acute angles that add up to 90◦. For example 20◦\nand 70◦are complementary angles.\n260\n6.3.\nReduction formula\n\nIn the figure P(\n√\n3; 1) and P ′ lie on a circle with radius 2. OP makes an angle of\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\nP ′\n2\n2\n90◦−θ\n1. Function values of 90◦−θ\na) If points P and P ′ are symmetrical about the line y = x, determine the\ncoordinates of P ′.\nb) Use the coordinates for P ′ to determine sin(90◦−θ) and cos(90◦−θ).\nc) From your results determine a relationship between the function values of\n(90◦−θ) and θ.\n2. Function values of 90◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nθ\nb\nP\nO\nx\ny\nθ\nb\nP ′\n90◦+ θ\n2\n2\n261\nChapter 6.\nTrigonometry\n\na) If point P is rotated through 90◦to get point P ′, determine the coordinates\nof P ′.\nb) Use the coordinates for P ′ to determine sin(90◦+ θ) and cos(90◦+ θ).\nc) From your results determine a relationship between the function values of\n(90◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(90◦−θ) = . . . . . .\nb) cos(90◦−θ) = . . . . . .\nc) sin(90◦+ θ) = . . . . . .\nd) cos(90◦+ θ) = . . . . . .\nSine and cosine are known as co-functions. Two functions are called co-functions if\nf (A) = g (B) whenever A + B = 90◦(that is, A and B are complementary angles).\nThe function value of an angle is equal to the co-function of its complement.\nThus for sine and cosine we have\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nThe sine and cosine graphs illustrate this clearly: the two graphs are identical except\nthat they have a 90◦phase difference.\nθ\ny\ny = cos θ\ny = sin θ\n262\n6.3.\nReduction formula\n\nWorked example 10: Using the co-function rule\nQUESTION\nWrite each of the following in terms of sin 40◦:\n1. cos 50◦\n2. sin 320◦\n3. cos 230◦\n4. cos 130◦\nSOLUTION\n1. cos 50◦= sin(90◦−50◦) = sin 40◦\n2. sin 320◦= sin(360◦−40◦) = −sin 40◦\n3. cos 230◦= cos(180◦+ 50◦) = −cos 50◦= −cos(90◦−40◦) = −sin 40◦\n4. cos 130◦= cos(90◦+ 40◦) = −sin 40◦\nFunction values of θ −90◦\nWe can write sin(θ −90◦) as\nsin(θ −90◦) = sin [−(90◦−θ)]\n= −sin(90◦−θ)\n= −cos θ\nsimilarly, we can show that cos (θ −90◦) = sin θ\nTherefore, sin (θ −90◦) = −cos θ and cos (θ −90◦) = sin θ.\nWorked example 11: Co-functions\nQUESTION\nExpress the following in terms of t if t = sin θ:\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and co-functions\n263\nChapter 6.\nTrigonometry\n\nUse the CAST diagram to check in which quadrants the trigonometric ratios are positive\nand negative.\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\n=cos[−(90◦−θ)] cos[2(360◦) + θ] tan[−(360◦−θ)]\nsin2(360◦+ θ) cos(90◦+ θ)\n=sin θ cos θ tan θ\nsin2 θ(−sin θ)\n= −cos θ\n\u0000 sin θ\ncos θ\n\u0001\nsin2 θ\n= −\n1\nsin θ\n= −1\nt\nExercise 6 – 5: Co-functions\n1. Simplify the following:\na) cos(90◦+ θ) sin(θ + 90◦)\nsin(−θ)\nb) 2 sin(90◦−x) + sin(90◦+ x)\nsin(90◦−x) + cos(180◦+ x)\n2. Given cos 36◦= p, express the following in terms on p:\na) sin 54◦\nb) sin 36◦\nc) tan 126◦\nd) cos 324◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Z7\n1b. 22Z8\n2. 22Z9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n264\n6.3.\nReduction formula\n\nReduction formulae and co-functions:\n1. The reduction formulae hold for any angle θ. For convenience, we assume θ is\nan acute angle (0◦< θ < 90◦).\n2. When determining function values of (180◦±θ), (360◦±θ) and (−θ) the function\ndoes not change.\n3. When determining function values of (90◦±θ) and (θ±90◦) the function changes\nto its co-function.\nsecond quadrant (180◦−θ) or (90◦+ θ)\nfirst quadrant (θ) or (90◦−θ)\nsin(180◦−θ) = + sin θ\nall trig functions are positive\ncos(180◦−θ) = −cos θ\nsin(360◦+ θ) = sin θ\ntan(180◦−θ) = −tan θ\ncos(360◦+ θ) = cos θ\nsin(90◦+ θ) = + cos θ\ntan(360◦+ θ) = tan θ\ncos(90◦+ θ) = −sin θ\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nthird quadrant (180◦+ θ)\nfourth quadrant (360◦−θ)\nsin(180◦+ θ) = −sin θ\nsin(360◦−θ) = −sin θ\ncos(180◦+ θ) = −cos θ\ncos(360◦−θ) = + cos θ\ntan(180◦+ θ) = + tan θ\ntan(360◦−θ) = −tan θ\nExercise 6 – 6: Reduction formulae\n1. Write A and B as a single trigonometric ratio:\na) A = sin(360◦−θ) cos(180◦−θ) tan(360◦+ θ)\nb) B = cos(360◦+ θ) cos(−θ) sin(−θ)\ncos(90◦+ θ)\nc) Hence, determine:\ni. A + B = . . .\nii.\nA\nB = . . .\n2. Write the following as a function of an acute angle:\na) sin 163◦\nb) cos 327◦\nc) tan 248◦\nd) cos(−213◦)\n3. Determine the value of the following, without using a calculator:\na) sin(−30◦)\ntan(150◦) + cos 330◦\nb) tan 300◦cos 120◦\nc) (1 −cos 30◦)(1 −cos 210◦)\nd) cos 780◦−(sin 315◦)(cos 405◦)\n4. Prove that the following identity is true and state any restrictions:\nsin(180◦+ α) tan(360◦+ α) cos α\ncos(90◦−α)\n= sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22ZB\n2a. 22ZC\n2b. 22ZD\n2c. 22ZF\n2d. 22ZG\n3a. 22ZH\n3b. 22ZJ\n3c. 22ZK\n3d. 22ZM\n4. 22ZN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n265\nChapter 6.\nTrigonometry\n\n6.4\nTrigonometric equations\nEMBHM\nSolving trigonometric equations requires that we find the value of the angles that satisfy\nthe equation. If a specific interval for the solution is given, then we need only find the\nvalue of the angles within the given interval that satisfy the equation. If no interval is\ngiven, then we need to find the general solution. The periodic nature of trigonometric\nfunctions means that there are many values that satisfy a given equation, as shown in\nthe diagram below.\n1\n−1\n90◦180◦270◦360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nθ\n0\ny\ny = 0,5\ny = sin θ\nWorked example 12: Solving trigonometric equations\nQUESTION\nSolve for θ (correct to one decimal place), given tan θ = 5 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to solve for θ\ntan θ = 5\n∴θ = tan−1 5\n= 78,7◦\nThis value of θ is an acute angle which lies in the first quadrant and is called the\nreference angle.\nStep 2: Use the CAST diagram to determine in which quadrants tan θ is positive\nThe CAST diagram indicates that tan θ is positive in the first and third quadrants, there-\nfore we must determine the value of θ such that 180◦< θ < 270◦.\nUsing reduction formulae, we know that tan(180◦+ θ) = tan θ\nθ = 180◦+ 78,7◦\n∴θ = 258,7◦\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 78,7◦or θ = 258,7◦.\n266\n6.4.\nTrigonometric equations\n\nWorked example 13: Solving trigonometric equations\nQUESTION\nSolve for α (correct to one decimal place), given cos α = −0,7 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we do not include the negative sign. The reference\nangle must be an acute angle in the first quadrant, where all the trigonometric functions\nare positive.\nref ∠= cos−1 0,7\n= 45,6◦\nStep 2: Use the CAST diagram to determine in which quadrants cos α is negative\nThe CAST diagram indicates that cos α is negative in the second and third quadrants,\ntherefore we must determine the value of α such that 90◦< α < 270◦.\nUsing reduction formulae, we know that cos(180◦−α) = −cos α and cos(180◦+α) =\n−cos α\nIn the second quadrant:\nα = 180◦−45,6◦\n= 134,4◦\nIn the third quadrant:\nα = 180◦+ 45,6◦\n= 225,6◦\nNote: the reference angle (45,6◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nα = 134,4◦or α = 225,6◦.\n267\nChapter 6.\nTrigonometry\n\nWorked example 14: Solving trigonometric equations\nQUESTION\nSolve for β (correct to one decimal place), given sin β = −0,5 and β ∈[−360◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we use a positive value.\nref ∠= sin−1 0,5\n= 30◦\nStep 2: Use the CAST diagram to determine in which quadrants sin β is negative\nThe CAST diagram indicates that sin β is negative in the third and fourth quadrants.\nWe also need to find the values of β such that −360◦≤β ≤360◦.\nUsing reduction formulae, we know that sin(180◦+β) = −sin β and sin(360◦−β) =\n−sin β\nIn the third quadrant:\nβ = 180◦+ 30◦\n= 210◦\nor β = −180◦+ 30◦\n= −150◦\nIn the fourth quadrant:\nβ = 360◦−30◦\n= 330◦\nor β = 0◦−30◦\n= −30◦\nNotice: the reference angle (30◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nβ = −150◦, −30◦, 210◦or 330◦.\n268\n6.4.\nTrigonometric equations\n\nExercise 6 – 7: Solving trigonometric equations\n1. Determine the values of α for α ∈[0◦; 360◦] if:\na) 4 cos α = 2\nb) sin α + 3,65 = 3\nc) tan α = 51\n4\nd) cos α + 0,939 = 0\ne) 5 sin α = 3\nf)\n1\n2 tan α = −1,4\n2. Determine the values of θ for θ ∈[−360◦; 360◦] if:\na) sin θ = 0,6\nb) cos θ + 3\n4 = 0\nc) 3 tan θ = 20\nd) sin θ = cos 180◦\ne) 2 cos θ = 4\n5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22ZP\n1b. 22ZQ\n1c. 22ZR\n1d. 22ZS\n1e. 22ZT\n1f. 22ZV\n2a. 22ZW\n2b. 22ZX\n2c. 22ZY\n2d. 22ZZ\n2e. 2322\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe general solution\nEMBHN\nIn the previous worked example, the solution was restricted to a certain interval. How-\never, the periodicity of the trigonometric functions means that there are an infinite\nnumber of positive and negative angles that satisfy an equation. If we do not restrict\nthe solution, then we need to determine the general solution to the equation. We know\nthat the sine and cosine functions have a period of 360◦and the tangent function has\na period of 180◦.\nMethod for finding the general solution:\n1. Determine the reference angle (use a positive value).\n2. Use the CAST diagram to determine where the function is positive or negative\n(depending on the given equation).\n3. Find the angles in the interval [0◦; 360◦] that satisfy the equation and add multi-\nples of the period to each answer.\n4. Check answers using a calculator.\n269\nChapter 6.\nTrigonometry\n\nWorked example 15: Finding the general solution\nQUESTION\nDetermine the general solution for sin θ = 0,3 (correct to one decimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nsin θ = 0,3\n∴ref ∠= sin−1 0,3\n= 17,5◦\nStep 2: Use CAST diagram to determine in which quadrants sin θ is positive\nThe CAST diagram indicates that sin θ is positive in the first and second quadrants.\nUsing reduction formulae, we know that sin(180◦−θ) = sin θ.\nIn the first quadrant:\nθ = 17,5◦\n∴θ = 17,5◦+ k . 360◦\nIn the second quadrant:\nθ = 180◦−17,5◦\n∴θ = 162,5◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 4:\nθ = 17,5◦+ 4(360)◦\n∴θ = 1457,5◦\nAnd sin 1457,5◦= 0,3007 . . .\nThis solution is correct.\nSimilarly, if we let k = −2:\nθ = 162,5◦−2(360)◦\n∴θ = −557,5◦\nAnd sin(−557,5◦) = 0,3007 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 17,5◦+ k . 360◦or θ = 162,5◦+ k . 360◦.\n270\n6.4.\nTrigonometric equations\n\nWorked example 16: Finding the general solution\nQUESTION\nDetermine the general solution for cos 2θ = −0,6427 (give answers correct to one\ndecimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nref ∠= sin−1 0,6427\n= 50,0◦\nStep 2: Use CAST diagram to determine in which quadrants cos θ is negative\nThe CAST diagram shows that cos θ is negative in the second and third quadrants.\nTherefore we use the reduction formulae cos(180◦−θ) = −cos θ and cos(180◦+θ) =\n−cos θ.\nIn the second quadrant:\n2θ = 180◦−50◦+ k . 360◦\n= 130◦+ k . 360◦\n∴θ = 65◦+ k . 180◦\nIn the third quadrant:\n2θ = 180◦+ 50◦+ k . 360◦\n= 230◦+ k . 360◦\n∴θ = 115◦+ k . 180◦\nwhere k ∈Z.\nRemember: also divide the period (360◦) by the coefficient of θ.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 2:\nθ = 65◦+ 2(180◦)\n∴θ = 425◦\nAnd cos 2(425)◦= −0,6427 . . .\nThis solution is correct.\n271\nChapter 6.\nTrigonometry\n\nSimilarly, if we let k = −5:\nθ = 115◦−5(180◦)\n∴θ = −785◦\nAnd cos 2(−785◦) = −0,6427 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 65◦+ k . 180◦or θ = 115◦+ k . 180◦.\nWorked example 17: Finding the general solution\nQUESTION\nDetermine the general solution for tan(2α −10◦) = 2,5 such that −180◦≤α ≤180◦\n(give answers correct to one decimal place).\nSOLUTION\nStep 1: Make a substitution\nTo solve this equation, it can be useful to make a substitution: let x = 2α −10◦.\ntan(x) = 2,5\nStep 2: Use a calculator to find the reference angle\ntan x = 2,5\n∴ref ∠= tan−1 2,5\n= 68,2◦\nStep 3: Use CAST diagram to determine in which quadrants the tangent function is\npositive\nWe see that tan x is positive in the first and third quadrants, so we use the reduction\nformula tan(180◦+ x) = tan x. It is also important to remember that the period of the\ntangent function is 180◦.\n272\n6.4.\nTrigonometric equations\n\nIn the first quadrant:\nx = 68,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 68,2◦+ k . 180◦\n2α = 78,2◦+ k . 180◦\n∴α = 39,1◦+ k . 90◦\nIn the third quadrant:\nx = 180◦+ 68,2◦+ k . 180◦\n= 248,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 248,2◦+ k . 180◦\n2α = 258,2◦+ k . 180◦\n∴α = 129,1◦+ k . 90◦\nwhere k ∈Z.\nRemember: to divide the period (180◦) by the coefficient of α.\nStep 4: Find the answers within the given interval\nSubstitute suitable values of k to determine the values of α that lie within the interval\n(−180◦≤α ≤180◦).\nI: α = 39,1◦+ k . 90◦\nIII: α = 129,1◦+ k . 90◦\nk = 0\n39,1◦\n129,1◦\nk = 1\n129,1◦\n219,1◦\n(outside)\nk = 2\n219,1◦\n(outside)\nk = −1\n−50,9◦\n39,1◦\nk = −2\n−140,9◦\n−50,9◦\nk = −3\n−230,9◦\n(outside)\n−140,9◦\nk = −4\n−230,9◦\n(outside)\nNotice how some of the values repeat. This is because of the periodic nature of the\ntangent function. Therefore we need only determine the solution:\nα = 39,1◦+ k . 90◦\nfor k ∈Z.\nStep 5: Write the final answer\nα = −140,9◦; −50,9◦; 39,1◦or 129,1◦.\n273\nChapter 6.\nTrigonometry\n\nWorked example 18: Finding the general solution using co-functions\nQUESTION\nDetermine the general solution for sin(θ −20◦) = cos 2θ.\nSOLUTION\nStep 1: Use co-functions to simplify the equation\nsin(θ −20◦) = cos 2θ\n= sin(90◦−2θ)\n∴θ −20◦= 90◦−2θ + k . 360◦,\nk ∈Z\n3θ = 110◦+ k . 360◦\n∴θ = 36,7◦+ k . 120◦\nStep 2: Use the CAST diagram to determine the correct quadrants\nSince the original equation equates a sine and cosine function, we need to work in the\nquadrant where both functions are positive or in the quadrant where both functions\nare negative so that the equation holds true. We therefore determine the solution using\nthe first and third quadrants.\nIn the first quadrant: θ = 36,7◦+ k . 120◦.\nIn the third quadrant:\n3θ = 180◦+ 110◦+ k . 360◦\n= 290◦+ k . 360◦\n∴θ = 96,6◦+ k . 120◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 36,7◦+ k . 120◦or θ = 96,6◦+ k . 120◦\n274\n6.4.\nTrigonometric equations\n\nExercise 6 – 8: General solution\n1.\n• Find the general solution for each equation.\n• Hence, find all the solutions in the interval [−180◦; 180◦].\na) cos(θ + 25◦) = 0,231\nb) sin 2α = −0,327\nc) 2 tan β = −2,68\nd) cos α = 1\ne) 4 sin θ = 0\nf) cos θ = −1\ng) tan θ\n2 = 0,9\nh) 4 cos θ + 3 = 1\ni) sin 2θ = −\n√\n3\n2\n2. Find the general solution for each equation.\na) cos(θ + 20◦) = 0\nb) sin 3α = −1\nc) tan 4β = 0,866\nd) cos(α −25◦) = 0,707\ne) 2 sin 3θ\n2 = −1\nf) 5 tan(β + 15◦) =\n5\n√\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2323\n1b. 2324\n1c. 2325\n1d. 2326\n1e. 2327\n1f. 2328\n1g. 2329\n1h. 232B\n1i. 232C\n2a. 232D\n2b. 232F\n2c. 232G\n2d. 232H\n2e. 232J\n2f. 232K\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSolving quadratic trigonometric equations\nWe can use our knowledge of algebraic equations to solve quadratic trigonometric\nequations.\nWorked example 19: Quadratic trigonometric equations\nQUESTION\nFind the general solution of 4 sin2 θ = 3.\nSOLUTION\nStep 1: Simplify the equation and determine the reference angle\n4 sin2 θ = 3\nsin2 θ = 3\n4\n∴sin θ = ±\nr\n3\n4\n= ±\n√\n3\n2\n∴ref ∠= 60◦\n275\nChapter 6.\nTrigonometry\n\nStep 2: Determine in which quadrants the sine function is positive and negative\nThe CAST diagram shows that sin θ is positive in the first and second quadrants and\nnegative in the third and fourth quadrants.\nPositive in the first and second quadrants:\nθ = 60◦+ k . 360◦\nor θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nNegative in the third and fourth quadrants:\nθ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor θ = 360◦−60◦+ k . 360◦\n= 300◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 60◦+ k . 360◦or 120◦+ k . 360◦or 240◦+ k . 360◦or 300◦+ k . 360◦\nWorked example 20: Quadratic trigonometric equations\nQUESTION\nFind θ if 2 cos2 θ −cos θ −1 = 0 for θ ∈[−180◦; 180◦].\nSOLUTION\nStep 1: Factorise the equation\n2 cos2 θ −cos θ −1 = 0\n(2 cos θ + 1)(cos θ −1) = 0\n∴2 cos θ + 1 = 0 or cos θ −1 = 0\n276\n6.4.\nTrigonometric equations\n\nStep 2: Simplify the equations and solve for θ\n2 cos θ + 1 = 0\n2 cos θ = −1\ncos θ = −1\n2\n∴ref ∠= 60◦\nII quadrant: θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nIII quadrant: θ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor\ncos θ −1 = 0\ncos θ = 1\n∴ref ∠= 0◦\nII and IV quadrants: θ = k . 360◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of θ that lie within the the given interval θ ∈[−180◦; 180◦] by\nsubstituting suitable values of k.\nIf k = −1,\nθ = 240◦+ k . 360◦\n= 240◦−(360◦)\n= −120◦\nIf k = 0,\nθ = 120◦+ k . 360◦\n= 120◦+ 0(360◦)\n= 120◦\nIf k = 1,\nθ = k . 360◦\n= 0(360◦)\n= 0◦\n277\nChapter 6.\nTrigonometry\n\nStep 4: Alternative method: substitution\nWe can simplify the given equation by letting y = cos θ and then factorising as:\n2y2 −y −1 = 0\n(2y + 1)(y −1) = 0\n∴y = −1\n2 or y = 1\nWe substitute y = cos θ back into these two equations and solve for θ.\nStep 5: Write the final answer\nθ = −120◦; 0◦; 120◦\nWorked example 21: Quadratic trigonometric equations\nQUESTION\nFind α if 2 sin2 α −sin α cos α = 0 for α ∈[0◦; 360◦].\nSOLUTION\nStep 1: Factorise the equation by taking out a common factor\n2 sin2 α −sin α cos α = 0\nsin α(2 sin α −cos α) = 0\n∴sin α = 0 or 2 sin α −cos α = 0\nStep 2: Simplify the equations and solve for α\nsin α = 0\n∴ref ∠= 0◦\n∴α = 0◦+ k . 360◦\nor α = 180◦+ k . 360◦\nand since 360◦= 2 × 180◦\nwe therefore have α = k . 180◦\n278\n6.4.\nTrigonometric equations\n\nor\n2 sin α −cos α = 0\n2 sin α = cos α\nTo simplify further, we divide both sides of the equation by cos α.\n2 sin α\ncos α = cos α\ncos α\n(cos α ̸= 0)\n2 tan α = 1\ntan α = 1\n2\n∴ref ∠= 26,6◦\n∴α = 26,6◦+ k . 180◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of α that lie within the the given interval α ∈[0◦; 360◦] by\nsubstituting suitable values of k.\nIf k = 0:\nα = 0◦\nor α = 26,6◦\nIf k = 1:\nα = 180◦\nor α = 26,6◦+ 180◦\n= 206,6◦\nIf k = 2:\nα = 360◦\nStep 4: Write the final answer\nα = 0◦; 26,6◦; 180◦; 206,6◦; 360◦\n279\nChapter 6.\nTrigonometry\n\nExercise 6 – 9: Solving trigonometric equations\n1. Find the general solution for each of the following equations:\na) cos 2θ = 0\nb) sin(α + 10◦) =\n√\n3\n2\nc) 2 cos θ\n2 −\n√\n3 = 0\nd)\n1\n2 tan(β −30◦) = −1\ne) 5 cos θ = tan 300◦\nf) 3 sin α = −1,5\ng) sin 2β = cos(β + 20◦)\nh) 0,5 tan θ + 2,5 = 1,7\ni) sin(3α −10◦) = sin(α + 32◦)\nj) sin 2β = cos 2β\n2. Find θ if sin2 θ + 1\n2 sin θ = 0 for θ ∈[0◦; 360◦].\n3. Determine the general solution for each of the following:\na) 2 cos2 θ −3 cos θ = 2\nb) 3 tan2 θ + 2 tan θ = 0\nc) cos2 α = 0,64\nd) sin(4β + 35◦) = cos(10◦−β)\ne) sin(α + 15◦) = 2 cos(α + 15◦)\nf) sin2 θ −4 cos2 θ = 0\ng) cos(2θ + 30◦)\n2\n+ 0,38 = 0\n4. Find β if 1\n3 tan β = cos 200◦for β ∈[−180◦; 180◦].\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 232M\n1b. 232N\n1c. 232P\n1d. 232Q\n1e. 232R\n1f. 232S\n1g. 232T\n1h. 232V\n1i. 232W\n1j. 232X\n2. 232Y\n3a. 232Z\n3b. 2332\n3c. 2333\n3d. 2334\n3e. 2335\n3f. 2336\n3g. 2337\n4. 2338\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n6.5\nArea, sine, and cosine rules\nEMBHP\nThere are three identities relating to the trigonometric functions that make working\nwith triangles easier:\n1. the area rule\n2. the sine rule\n3. the cosine rule\n280\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nEMBHQ\nInvestigation: The area rule\n1. Consider △ABC:\nB\nA\nC\n10\n54◦\n7\nComplete the following:\na) Area △ABC = 1\n2 × . . . × AC\nb) sin ˆB = . . . and AC = . . . × . . .\nc) Therefore area △ABC = . . . × . . . × . . . × . . .\n2. Consider △A′B′C′:\nB′\nA′\nC′\n10\n54◦\n7\nComplete the following:\na) How is △A′B′C′ different from △ABC?\nb) Calculate area △A′B′C′.\n3. Use your results to write a general formula for determining the area of △PQR:\nQ\nP\nR\nr\np\nq\n281\nChapter 6.\nTrigonometry\n\nFor any △ABC with AB = c, BC = a and AC = b, we can construct a perpendicular\nheight (h) from vertex A to the line BC:\nB\nA\nC\nc\na\nb\nh\nIn △ABC:\nsin ˆB = h\nc\n∴h = c sin ˆB\nAnd we know that\nArea △ABC = 1\n2 × a × h\n= 1\n2 × a × c sin ˆB\n∴Area △ABC = 1\n2ac sin ˆB\nAlternatively, we could write that\nsin ˆC = h\nb\n∴h = b sin ˆC\nAnd then we would have that\nArea △ABC = 1\n2 × a × h\n= 1\n2ab sin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nThe area rule\nIn any △ABC:\nArea △ABC = 1\n2bc sin ˆA\n= 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n282\n6.5.\nArea, sine, and cosine rules\n\nWorked example 22: The area rule\nQUESTION\nFind the area of △ABC (correct to two decimal places):\nA\n7\nB\nC\n50◦\nSOLUTION\nStep 1: Use the given information to determine unknown angles and sides\nAB = AC = 7\n(given)\n∴ˆB = ˆC = 50◦\n(∠s opp. equal sides)\nAnd ˆA = 180◦−50◦−50◦\n(∠s sum of △ABC)\n∴ˆA = 80◦\nStep 2: Use the area rule to calculate the area of △ABC\nNotice that we do not know the length of side a and must therefore choose the form\nof the area rule that does not include this side of the triangle.\nIn △ABC:\nArea = 1\n2bc sin ˆA\n= 1\n2(7)(7) sin 80◦\n= 24,13\nStep 3: Write the final answer\nArea of △ABC = 24,13 square units.\n283\nChapter 6.\nTrigonometry\n\nWorked example 23: The area rule\nQUESTION\nShow that the area of △DEF = 1\n2df sin ˆE.\nD\nF\nE\nH\ne\nd\nf\nh\n1\n2\nSOLUTION\nStep 1: Construct a perpendicular height h\nDraw DH such that DH ⊥EF and let DH = h, D ˆEF = ˆE1 and D ˆEH = ˆE2.\nIn △DHE:\nsin ˆE2 = h\nf\nh = f sin(180◦−ˆE1)\n(∠s on str. line)\n= f sin ˆE1\nStep 2: Use the area rule to calculate the area of △DEF\nIn △DEF:\nArea = 1\n2d × h\n= 1\n2df sin ˆE1\n284\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nIn any △PQR:\nP\nP\nQ\nQ\nR\nR\nq\nq\nr\nr\np\np\nArea △PQR = 1\n2qr sin ˆP\n= 1\n2pr sin ˆQ\n= 1\n2pq sin ˆR\nThe area rule states that the area of any triangle is equal to half the product of the\nlengths of the two sides of the triangle multiplied by the sine of the angle included by\nthe two sides.\nExercise 6 – 10: The area rule\n1. Draw a sketch and calculate the area of △PQR given:\na) ˆQ = 30◦; r = 10 and p = 7\nb) ˆR = 110◦; p = 8 and q = 9\n2. Find the area of △XY Z given XZ = 52 cm, XY = 29 cm and ˆX = 58,9◦.\n3. Determine the area of a parallelogram in which two adjacent sides are 10 cm\nand 13 cm and the angle between them is 55◦.\n4. If the area of △ABC is 5000 m2 with a = 150 m and b = 70 m, what are the two\npossible sizes of ˆC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2339\n1b. 233B\n2. 233C\n3. 233D\n4. 233F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n285\nChapter 6.\nTrigonometry\n\nThe sine rule\nEMBHR\nSo far we have only applied the trigonometric ratios to right-angled triangles. We now\nexpand the application of the trigonometric ratios to triangles that do not have a right\nangle:\nInvestigation: The sine rule\nIn △ABC, AC = 15, BC = 11 and ˆA = 48◦. Find ˆB.\nA\nC\nB\nb = 15\nF\na = 11\n48◦\n1. Method 1: using the sine ratio\na) Draw a sketch of △ABC.\nb) Construct CF ⊥AB.\nc) In △CBF:\nCF\n. . . = sin ˆB\n∴CF = . . . × sin ˆB\nd) In △CAF:\nCF\n15 = . . .\n∴CF = 15 × . . .\ne) Therefore we have that:\nCF = 15 × . . .\nand CF = . . . × sin ˆB\n∴15 × . . . = . . . × sin ˆB\n∴sin ˆB = . . . . . . . . .\n∴ˆB = . . .\n2. Method 2: using the area rule\n286\n6.5.\nArea, sine, and cosine rules\n\na) In △ABC:\nArea △ABC = 1\n2AB × AC × . . .\n= 1\n2AB × . . . × . . .\nb) And we also know that\nArea △ABC = 1\n2AB × . . . × sin ˆB\nc) We can equate these two equations and solve for ˆB:\n1\n2AB × . . . × sin ˆB = 1\n2AB × . . . × . . .\n∴. . . × sin ˆB = . . . × . . .\n∴sin ˆB = . . . × . . .\n∴ˆB = . . .\n3. Use your results to write a general formula for the sine rule given △PQR:\nP\nQ\nR\nq\nr\np\nFor any triangle ABC with AB = c, BC = a and AC = b, we can construct a perpen-\ndicular height (h) at F:\nA\nC\nB\nb\nF\na\nh\nc\nMethod 1: using the sine ratio\nIn △ABF:\nsin ˆB = h\nc\n∴h = c sin ˆB\n287\nChapter 6.\nTrigonometry\n\nIn △ACF:\nsin ˆC = h\nb\n∴h = b sin ˆC\nWe can equate the two equations\nc sin ˆB = b sin ˆC\n∴sin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that:\nsin ˆA\na\n= sin ˆC\nc\nor\na\nsin ˆA\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nMethod 2: using the area rule\nIn △ABC:\nArea △ABC = 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n∴1\n2ac sin ˆB = 1\n2ab sin ˆC\nc sin ˆB = b sin ˆC\nsin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\n288\n6.5.\nArea, sine, and cosine rules\n\nThe sine rule\nIn any △ABC:\nA\nC\nB\nb\na\nc\nsin ˆA\na\n= sin ˆB\nb\n= sin ˆC\nc\na\nsin ˆA\n=\nb\nsin ˆB\n=\nc\nsin ˆC\nSee video: 233G at www.everythingmaths.co.za\nWorked example 24: The sine rule\nQUESTION\nGiven △TRS with S ˆTR = 55◦, TR = 30 and R ˆST = 40◦, determine RS, ST and\nT ˆRS.\nSOLUTION\nStep 1: Draw a sketch\nLet RS = t, ST = r and TR = s.\nR\nS\nT\n55◦\n30\n40◦\nStep 2: Find T ˆRS using angles in a triangle\nT ˆRS + R ˆST + S ˆTR = 180◦\n(∠s sum of △TRS)\n∴T ˆRS = 180◦−40◦−55◦\n= 85◦\n289\nChapter 6.\nTrigonometry\n\nStep 3: Determine t and r using the sine rule\nt\nsin ˆT\n=\ns\nsin ˆS\nt\nsin 55◦=\n30\nsin 40◦\n∴t =\n30\nsin 40◦× sin 55◦\n= 38,2\nr\nsin ˆR\n=\ns\nsin ˆS\nr\nsin 85◦=\n30\nsin 40◦\n∴r =\n30\nsin 40◦× sin 85◦\n= 46,5\nWorked example 25: The sine rule\nQUESTION\nProve the sine rule for △MNP with MS ⊥NP.\nM\nP\nN\nS\nn\nm\np\nh\n1\n2\nSOLUTION\nStep 1: Use the sine ratio to express the angles in the triangle in terms of the length\nof the sides\nIn △MSN:\nsin ˆN2 = h\np\n∴h = p sin ˆN2\nand ˆN2 = 180◦−ˆN1\n∠s on str. line\n∴h = p sin(180◦−ˆN1)\n= p sin ˆN1\n290\n6.5.\nArea, sine, and cosine rules\n\nIn △MSP:\nsin ˆP = h\nn\n∴h = n sin ˆP\nStep 2: Equate the two equations to derive the sine rule\np sin ˆN1 = n sin ˆP\n∴sin ˆN1\nn\n= sin ˆP\np\nor\nn\nsin ˆN1\n=\np\nsin ˆP\nThe ambiguous case\nIf two sides and an interior angle of a triangle are given, and the side opposite the given\nangle is the shorter of the two sides, then we can draw two different triangles (△NMP\nand △NMP ′), both having the given dimensions. We call this the ambiguous case\nbecause there are two ways of interpreting the given information and it is not certain\nwhich is the required solution.\nM\nP ′\np\nN\nP\nn\nn\nWorked example 26: The ambiguous case\nQUESTION\nIn △ABC, AB = 82, BC = 65 and ˆA = 50◦. Draw △ABC and find ˆC (correct to\none decimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the ambiguous case\nWe notice that for the given dimensions of △ABC, the side BC opposite ˆA is shorter\nthan AB. This means that we can draw two different triangles with the given dimen-\nsions.\nA\nB\nC\n50◦\n65\n82\nA′\nC′\nB′\n65\n82\n50◦\n291\nChapter 6.\nTrigonometry\n\nStep 2: Solve for unknown angle using the sine rule\nIn △ABC:\nsin ˆA\nBC = sin ˆC\nAB\nsin 50◦\n65\n= sin ˆC\n82\n∴sin 50◦\n65\n× 82 = sin ˆC\n∴ˆC = 75,1◦\nIn △A′B′C′:\nWe know that sin(180 −ˆC) = sin ˆC, which means we can also have the solution\nˆC′ = 180◦−75,1◦\n= 104,9◦\nBoth solutions are correct.\nWorked example 27: Lighthouses\nQUESTION\nThere is a coastline with two lighthouses, one on either side of a beach. The two\nlighthouses are 0,67 km apart and one is exactly due east of the other. The lighthouses\ntell how close a boat is by taking bearings to the boat (a bearing is an angle measured\nclockwise from north). These bearings are shown on the diagram below.\nCalculate how far the boat is from each lighthouse.\nˆA = 127◦\nˆB = 255◦\nC\nSOLUTION\nWe see that the two lighthouses and the boat form a triangle. Since we know the\ndistance between the lighthouses and we have two angles we can use trigonometry\n292\n6.5.\nArea, sine, and cosine rules\n\nto find the remaining two sides of the triangle, the distance of the boat from the two\nlighthouses.\nb\nA\nb B\nb\nC\n15◦\n37◦\n128◦\n0,67 km\nWe need to determine the lengths of the two sides AC and BC. We can use the sine\nrule to find the missing lengths.\nBC\nsin ˆA\n= AB\nsin ˆC\nBC = AB . sin ˆA\nsin ˆC\n= (0,67 km) sin 37◦\nsin 128◦\n= 0,51 km\nAC\nsin ˆB\n= AB\nsin ˆC\nAC = AB . sin ˆB\nsin ˆC\n= (0,67 km) sin 15◦\nsin 128◦\n= 0,22 km\nExercise 6 – 11: Sine rule\n1. Find all the unknown sides and angles of the following triangles:\na) △PQR in which ˆQ = 64◦; ˆR = 24◦and r = 3\nb) △KLM in which ˆK = 43◦; ˆ\nM = 50◦and m = 1\nc) △ABC in which ˆA = 32,7◦; ˆC = 70,5◦and a = 52,3\nd) △XY Z in which ˆX = 56◦; ˆZ = 40◦and x = 50\n2. In △ABC, ˆA = 116◦; ˆC = 32◦and AC = 23 m. Find the lengths of the sides\nAB and BC.\n3. In △RST, ˆR = 19◦; ˆS = 30◦and RT = 120 km. Find the length of the side\nST.\n4. In △KMS, ˆK = 20◦; ˆ\nM = 100◦and s = 23 cm. Find the length of the side m.\n293\nChapter 6.\nTrigonometry\n\n5. In △ABD, ˆB = 90◦, AB = 10 cm and A ˆDB = 40◦. In △BCD, ˆC = 106◦and\nC ˆDB = 15◦. Determine BC.\nA\nB\nD\nC\n10\n106◦\n15◦\n40◦\n6. In △ABC, ˆA = 33◦, AC = 21 mm and AB = 17 mm. Can you determine BC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233H\n1b. 233J\n1c. 233K\n1d. 233M\n2. 233N\n3. 233P\n4. 233Q\n5. 233R\n6. 233S\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe cosine rule\nEMBHS\nInvestigation: The cosine rule\nIf a triangle is given with two sides and the included angle known, then we can not\nsolve for the remaining unknown sides and angles using the sine rule. We therefore\ninvestigate the cosine rule:\nIn △ABC, AB = 21, AC = 17 and ˆA = 33◦. Find ˆB.\nA\nC\nB\nH\n21\nc\n17\n33◦\n1. Determine CB:\na) Construct CH ⊥AB.\nb) Let AH = c and therefore HB = . . .\n294\n6.5.\nArea, sine, and cosine rules\n\nc) Applying the theorem of Pythagoras in the right-angled triangles:\nIn△CHB:\nCB2 = BH2 + CH2\n= (. . .)2 + CH2\n= 212 −(2)(21)c + c2 + CH2 . . . . . . (1)\nIn △CHA:\nCA2 = c2 + CH2\n172 = c2 + CH2 . . . . . . (2)\nSubstitute equation (2) into equation (1):\nCB2 = 212 −(2)(21)c + 172\nNow c is the only remaining unknown. In △CHA:\nc\n17 = cos 33◦\n∴c = 17 cos 33◦\nTherefore we have that\nCB2 = 212 −(2)(21)c + 172\n= 212 −(2)(21)(17 cos 33◦) + 172\n= 212 + 172 −(2)(21)(17) cos 33◦\n= 131,189 . . .\n∴CB = 11,5\n2. Use your results to write a general formula for the cosine rule given △PQR:\nP\nQ\nR\nq\nr\np\nThe cosine rule relates the length of a side of a triangle to the angle opposite it and the\nlengths of the other two sides.\n295\nChapter 6.\nTrigonometry\n\nConsider △ABC with CD ⊥AB:\nb\nD\nb\nA\nb\nB\nbC\nh\na\nb\nc\nc −d\nd\nIn △DCB: a2 = (c −d)2 + h2 from the theorem of Pythagoras.\nIn △ACD: b2 = d2 + h2 from the theorem of Pythagoras.\nSince h2 is common to both equations we can write:\na2 = (c −d)2 + h2\n∴h2 = a2 −(c −d)2\nAnd b2 = d2 + h2\n∴h2 = b2 −d2\n∴b2 −d2 = a2 −(c −d)2\na2 = b2 + (c2 −2cd + d2) −d2\n= b2 + c2 −2cd\nIn order to eliminate d we look at △ACD, where we have: cos ˆA = d\nb. So, d = b cos ˆA.\nSubstituting back we get: a2 = b2 + c2 −2bc cos ˆA.\nThe cosine rule\nIn any △ABC:\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\na2 = b2 + c2 −2bc cos ˆA\nb2 = a2 + c2 −2ac cos ˆB\nc2 = a2 + b2 −2ab cos ˆC\nSee video: 233T at www.everythingmaths.co.za\n296\n6.5.\nArea, sine, and cosine rules\n\nWorked example 28: The cosine rule\nQUESTION\nDetermine the length of QR.\nP\nR\nQ\n13 cm\n4 cm\n70◦\nSOLUTION\nStep 1: Use the cosine rule to solve for the unknown side\nQR2 = PR2 + QP 2 −2(PR)(QP) cos ˆP\n= 42 + 132 −2(4)(13) cos 70◦\n= 149,42 . . .\n∴QR = 12,2\nStep 2: Write the final answer\nQR = 12,2 cm\nWorked example 29: The cosine rule\nQUESTION\nDetermine ˆA.\n5\n7\n8\nA\nB\nC\nSOLUTION\nApplying the cosine rule:\na2 = b2 + c2 −2bc cos ˆA\n∴cos ˆA = b2 + c2 −a2\n2bc\n= 82 + 52 −72\n2 . 8 . 5\n= 0,5\n∴ˆA = 60◦\n297\nChapter 6.\nTrigonometry\n\nIt is very important:\n• not to round off before the final answer as this will affect accuracy;\n• to take the square root;\n• to remember to give units where applicable.\nHow to determine which rule to use:\n1. Area rule:\n• if no perpendicular height is given\n2. Sine rule:\n• if no right angle is given\n• if two sides and an angle are given (not the included angle)\n• if two angles and a side are given\n3. Cosine rule:\n• if no right angle is given\n• if two sides and the included angle are given\n• if three sides are given\nExercise 6 – 12: The cosine rule\n1. Solve the following triangles (that is, find all unknown sides and angles):\na) △ABC in which ˆA = 70◦; b = 4 and c = 9\nb) △RST in which RS = 14; ST = 26 and RT = 16\nc) △KLM in which KL = 5; LM = 10 and KM = 7\nd) △JHK in which ˆH = 130◦; JH = 13 and HK = 8\ne) △DEF in which d = 4; e = 5 and f = 7\n2. Find the length of the third side of the △XY Z where:\na) ˆX = 71,4◦; y = 3,42 km and z = 4,03 km\nb) x = 103,2 cm; ˆY = 20,8◦and z = 44,59 cm\n3. Determine the largest angle in:\na) △JHK in which JH = 6; HK = 4 and JK = 3\nb) △PQR where p = 50; q = 70 and r = 60\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233V\n1b. 233W\n1c. 233X\n1d. 233Y\n1e. 233Z\n2a. 2342\n2b. 2343\n3a. 2344\n3b. 2345\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n298\n6.5.\nArea, sine, and cosine rules\n\nSee video: 2346 at www.everythingmaths.co.za\nExercise 6 – 13: Area, sine and cosine rule\n1. Q is a ship at a point 10 km due south of another ship P. R is a lighthouse on\nthe coast such that ˆP = ˆQ = 50◦.\n10 km\nP\nQ\nR\n50◦\n50◦\nDetermine:\na) the distance QR\nb) the shortest distance from the lighthouse to the line joining the two ships\n(PQ).\n2. WXY Z is a trapezium, WX ∥Y Z with WX = 3 m; Y Z = 1,5 m; ˆZ = 120◦\nand ˆW = 30◦.\nDetermine the distances XZ and XY .\n1,5 m\n3 m\n30◦\n120◦\nW\nX\nY\nZ\n3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has\nbeen flying 3◦off course. At this point the plane is 500 km from Johannesburg.\nThe direct distance between Cape Town and Johannesburg airports is 1552 km.\nDetermine, to the nearest km:\na) The distance the plane has to travel to get to Cape Town and hence the\nextra distance that the plane has had to travel due to the pilot’s error.\nb) The correction, to one hundredth of a degree, to the plane’s heading (or\ndirection).\n4. ABCD is a trapezium (meaning that AB ∥CD). AB = x; B ˆAD = a; B ˆCD = b\nand B ˆDC = c.\nFind an expression for the length of CD in terms of x, a, b and c.\nA\nB\nC\nD\na\nb\nc\nx\n299\nChapter 6.\nTrigonometry\n\n5. A surveyor is trying to determine the distance between points X and Z. However\nthe distance cannot be determined directly as a ridge lies between the two points.\nFrom a point Y which is equidistant from X and Z, he measures the angle X ˆY Z.\nY\nX\nZ\nx\nθ\na) If XY = x and X ˆY Z = θ, show that XZ = x\np\n2(1 −cos θ).\nb) Calculate XZ (to the nearest kilometre) if x = 240 km and θ = 132◦.\n6. Find the area of WXY Z (to two decimal places):\nW\nX\nY\nZ\n120◦\n3\n4\n3,5\n7. Find the area of the shaded triangle in terms of x, α, β, θ and φ:\nA\nB\nC\nD\nE\nx\nα\nβ\nθ\nφ\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2347\n2. 2348\n3. 2349\n4. 234B\n5. 234C\n6. 234D\n7. 234F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n300\n6.5.\nArea, sine, and cosine rules\n\n6.6\nSummary\nEMBHT\nSee presentation: 234G at www.everythingmaths.co.za\nsquare identity\nquotient identity\ncos2 θ + sin2 θ = 1\ntan θ = sin θ\ncos θ\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\nnegative angles\nperiodicity identities\nco-function identities\nsin(−θ) = −sin θ\nsin(θ ± 360◦) = sin θ\nsin(90◦−θ) = cos θ\ncos(−θ) = cos θ\ncos(θ ± 360◦) = cos θ\ncos(90◦−θ) = sin θ\nsine rule\narea rule\ncosine rule\nsin A\na\n= sin B\nb\n= sin C\nc\narea △ABC = 1\n2bc sin A\na2 = b2 + c2 −2bc cos A\na\nsin A =\nb\nsin B =\nc\nsin C\narea △ABC = 1\n2ac sin B\nb2 = a2 + c2 −2ac cos B\narea △ABC = 1\n2ab sin C\nc2 = a2 + b2 −2ab cos C\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nGeneral solution:\n301\nChapter 6.\nTrigonometry\n\n1.\nIf sin θ = x\nθ = sin−1 x + k . 360◦\nor θ =\n\u0000180◦−sin−1 x\n\u0001\n+ k . 360◦\n2.\nIf cos θ = x\nθ = cos−1 x + k . 360◦\nor θ =\n\u0000360◦−cos−1 x\n\u0001\n+ k . 360◦\n3.\nIf tan θ = x\nθ = tan−1 x + k . 180◦\nfor k ∈Z.\nHow to determine which rule to use:\n1. Area rule:\n• no perpendicular height is given\n2. Sine rule:\n• no right angle is given\n• two sides and an angle are given (not the included angle)\n• two angles and a side are given\n3. Cosine rule:\n• no right angle is given\n• two sides and the included angle angle are given\n• three sides are given\nExercise 6 – 14: End of chapter exercises\n1. Write the following as a single trigonometric ratio:\ncos(90◦−A) sin 20◦\nsin(180◦−A) cos 70◦+ cos(180◦+ A) sin(90◦+ A)\n2. Determine the value of the following expression without using a calculator:\nsin 240◦cos 210◦−tan2 225◦cos 300◦cos 180◦\n302\n6.6.\nSummary\n\n3. Simplify:\nsin(180◦+ θ) sin(θ + 360◦)\nsin(−θ) tan(θ −360◦)\n4. Without the use of a calculator, evaluate:\n3 sin 55◦sin2 325◦\ncos(−145◦)\n−3 cos 395◦sin 125◦\n5. Prove the following identities:\na)\n1\n(cos x −1)(cos x + 1) =\n−1\ntan2 x cos2 x\nb) (1 −tan α) cos α = sin(90 + α) + cos(90 + α)\n6.\na) Prove: tan y +\n1\ntan y =\n1\ncos2 y tan y\nb) For which values of y ∈[0◦; 360◦] is the identity above undefined?\n7.\na) Simplify: sin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\nb) Hence, solve the equation\nsin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\n= tan θ\nfor θ ∈[0◦; 360◦].\n8. Given 12 tan θ = 5 and θ > 90◦.\na) Draw a sketch.\nb) Determine without using a calculator sin θ and cos(180◦+ θ).\nc) Use a calculator to find θ (correct to two decimal places).\n9.\nθ\nP(a; b)\n2\nx\ny\nO\nb\nIn the figure, P is a point on the Cartesian plane such that OP = 2 units and\nθ = 300◦. Without the use of a calculator, determine:\na) the values of a and b\nb) the value of sin(180◦−θ)\n10. Solve for x with x ∈[−180◦; 180◦] (correct to one decimal place):\na) 2 sin x\n2 = 0,86\n303\nChapter 6.\nTrigonometry\n\nb) tan(x + 10◦) = cos 202,6◦\nc) cos2 x −4 sin2 x = 0\n11. Find the general solution for the following equations:\na)\n1\n2 sin(x −25◦) = 0,25\nb) sin2 x + 2 cos x = −2\n12. Given the equation: sin 2α = 0,84\na) Find the general solution of the equation.\nb) Illustrate how this equation could be solved graphically for α ∈[0◦; 360◦].\nc) Write down the solutions for sin 2α = 0,84 for α ∈[0◦; 360◦].\n13.\nA\nT\nG\nN\nH\nn\nα\nβ\nA is the highest point of a vertical tower AT. At point N on the tower, n metres\nfrom the top of the tower, a bird has made its nest. The angle of inclination from\nG to point A is α and the angle of inclination from G to point N is β.\na) Express A ˆGN in terms of α and β.\nb) Express ˆA in terms of α and/or β.\nc) Show that the height of the nest from the ground (H) can be determined by\nthe formula\nH = n cos α sin β\nsin(α −β)\nd) Calculate the height of the nest H if n = 10 m, α = 68◦and β = 40◦(give\nyour answer correct to the nearest metre).\n304\n6.6.\nSummary\n\n14.\nA\nD\nB\nC\n11\n8\n5\nMr. Collins wants to pave his trapezium-shaped backyard, ABCD. AB ∥DC\nand ˆB = 90◦. DC = 11 m, AB = 8 m and BC = 5 m.\na) Calculate the length of the diagonal AC.\nb) Calculate the length of the side AD.\nc) Calculate the area of the patio using geometry.\nd) Calculate the area of the patio using trigonometry.\n15.\nA\nC\nB\n2t\nF\nt\nn\nn\n2n\nα\nIn △ABC, AC = 2A, AF = BF, A ˆFB = α and FC = 2AF. Prove that\ncos α = 1\n4.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 234H\n2. 234J\n3. 234K\n4. 234M\n5a. 234N\n5b. 234P\n6. 234Q\n7. 234R\n8. 234S\n9. 234T\n10a. 234V\n10b. 234W\n10c. 234X\n11a. 234Y\n11b. 234Z\n12. 2352\n13. 2353\n14. 2354\n15. 2355\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n305\nChapter 6.\nTrigonometry\n\n\nCHAPTER\n7\nMeasurement\n7.1\nArea of a polygon\n308\n7.2\nRight prisms and cylinders\n311\n7.3\nRight pyramids, right cones and spheres\n318\n7.4\nMultiplying a dimension by a constant factor\n322\n7.5\nSummary\n326\n\n7\nMeasurement\nThis chapter is a revision of perimeters and areas of two dimensional objects and\nvolumes of three dimensional objects. We also examine different combinations of\ngeometric objects and calculate areas and volumes in a variety of real-life contexts.\nSee video: 2356 at www.everythingmaths.co.za\n7.1\nArea of a polygon\nEMBHV\nSquare\ns\ns\nArea = s2\nRectangle\nh\nb\nArea = b × h\nTriangle\nh\nb\nArea = 1\n2b × h\nSee video: 2357 at www.everythingmaths.co.za\nTrapezium\nh\nb\na\nArea = 1\n2 (a + b) × h\nParallelogram\nh\nb\nArea = b × h\nCircle\nb r\nArea = πr2\n(Circumference = 2πr)\nSee video: 2358 at www.everythingmaths.co.za\n308\n7.1.\nArea of a polygon\n\nWorked example 1: Finding the area of a polygon\nQUESTION\nABCD is a parallelogram with DC = 15 cm, h = 8 cm and BF = 9 cm.\nA\nB\nC\nD\nH\n9 cm\n15 cm\nh\nF\nCalculate:\n1. the area of ABCD\n2. the perimeter of ABCD\nSOLUTION\nStep 1: Determine the area\nThe area of a parallelogram ABCD = base × height:\nArea = 15 × 8\n= 120 cm2\nStep 2: Determine the perimeter\nThe perimeter of a parallelogram ABCD = 2DC + 2BC.\nTo find the length of BC, we use AF ⊥BC and the theorem of Pythagoras.\nIn △ABF:\nAF 2 = AB2 −BF 2\n= 152 −92\n= 144\n∴AF = 12 cm\nAreaABCD = BC × AF\n120 = BC × 12\n∴BC = 10 cm\n∴PerimeterABCD = 2(15) + 2(10)\n= 50 cm\n309\nChapter 7.\nMeasurement\n\nExercise 7 – 1: Area of a polygon\n1. Vuyo and Banele are having a competition to see who can build the best kite\nusing balsa wood (a lightweight wood) and paper. Vuyo decides to make his kite\nwith one diagonal 1 m long and the other diagonal 60 cm long. The intersection\nof the two diagonals cuts the longer diagonal in the ratio 1 : 3.\nBanele also uses diagonals of length 60 cm and 1 m, but he designs his kite to\nbe rhombus-shaped.\na) Draw a sketch of Vuyo’s kite and write down all the known measurements.\nb) Determine how much balsa wood Vuyo will need to build the outside frame\nof the kite (give answer correct to the nearest cm).\nc) Calculate how much paper he will need to cover the frame of the kite.\nd) Draw a sketch of Banele’s kite and write down all the known measure-\nments.\ne) Determine how much wood and paper Banele will need for his kite.\nf) Compare the two designs and comment on the similarities and differences.\nWhich do you think is the better design? Motivate your answer.\n2. O is the centre of the bigger semi-circle with a radius of 10 units. Two smaller\nsemi-circles are inscribed into the bigger one, as shown on the diagram. Calcu-\nlate the following (in terms of π):\nO\nb\nb\na) The area of the shaded figure.\nb) The perimeter enclosing the shaded area.\n3. Karen’s engineering textbook is 30 cm long and 20 cm wide. She notices that\nthe dimensions of her desk are in the same proportion as the dimensions of her\ntextbook.\na) If the desk is 90 cm wide, calculate the area of the top of the desk.\nb) Karen uses some cardboard to cover each corner of her desk with an isosce-\nles triangle, as shown in the diagram:\n150 mm\n150 mm\ndesk\nCalculate the new perimeter and area of the visible part of the top of her\ndesk.\n310\n7.1.\nArea of a polygon\n\nc) Use this new area to calculate the dimensions of a square desk with the\nsame desk top area.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2359\n2. 235B\n3. 235C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.2\nRight prisms and cylinders\nEMBHW\nA right prism is a geometric solid that has a polygon as its base and vertical sides\nperpendicular to the base. The base and top surface are the same shape and size. It is\ncalled a “right” prism because the angles between the base and sides are right angles.\nA triangular prism has a triangle as its base, a rectangular prism has a rectangle as its\nbase, and a cube is a rectangular prism with all its sides of equal length. A cylinder is\nanother type of right prism which has a circle as its base. Examples of right prisms are\ngiven below: a rectangular prism, a cube, a triangular prism and a cylinder.\nSurface area of prisms and cylinders\nEMBHX\nSurface area is the total area of the exposed or outer surfaces of a prism. This is easier\nto understand if we imagine the prism to be a cardboard box that we can unfold. A\nsolid that is unfolded like this is called a net. When a prism is unfolded into a net, we\ncan clearly see each of its faces. In order to calculate the surface area of the prism, we\ncan then simply calculate the area of each face, and add them all together.\nFor example, when a triangular prism is unfolded into a net, we can see that it has\ntwo faces that are triangles and three faces that are rectangles. To calculate the surface\narea of the prism, we find the area of each triangle and each rectangle, and add them\ntogether.\nIn the case of a cylinder the top and bottom faces are circles and the curved surface\nflattens into a rectangle with a length that is equal to the circumference of the circular\nbase. To calculate the surface area we therefore find the area of the two circles and the\nrectangle and add them together.\n311\nChapter 7.\nMeasurement\n\nBelow are examples of right prisms that have been unfolded into nets. A rectangular\nprism unfolded into a net is made up of six rectangles.\nA cube unfolded into a net is made up of six identical squares.\nA triangular prism unfolded into a net is made up of two triangles and three rectangles.\nThe sum of the lengths of the rectangles is equal to the perimeter of the triangles.\nA cylinder unfolded into a net is made up of two identical circles and a rectangle with\nlength equal to the circumference of the circles.\n312\n7.2.\nRight prisms and cylinders\n\nWorked example 2: Calculating surface area\nQUESTION\nA box of chocolates has the following dimensions:\nlength = 25 cm\nwidth = 20 cm\nheight = 4 cm\n25 cm\n20 cm\n4 cm\nAnd a cylindrical tin of biscuits has the following dimensions:\ndiameter = 20 cm\nheight = 20 cm\nb\n20 cm\n20 cm\n1. Calculate the area of the wrapping paper needed to cover the entire box (assume\nno overlapping at the corners).\n2. Determine if this same sheet of wrapping paper would be enough to cover the\ntin of biscuits.\nSOLUTION\nStep 1: Determine the area of the rectangular box\nSurface area = 2 × (25 × 20) + 2 × (20 × 4) + 2 × (25 × 4)\n= 1360 cm2\n313\nChapter 7.\nMeasurement\n\nStep 2: Determine the area of the cylindrical tin\nThe radius of the cylinder = 20\n2 = 10 cm.\nSurface area = 2 × π(10)2 + 2π(10)(20)\n= 1885 cm2\nStep 3: Write the final answer\nNo, the area of the sheet of wrapping paper used to cover the box is not big enough to\ncover the tin.\nExercise 7 – 2: Calculating surface area\n1. A popular chocolate container is an equilateral right triangular prism with sides\nof 34 mm. The box is 170 mm long. Calculate the surface area of the box (to the\nnearest square centimetre).\n34 mm\n34 mm\n34 mm\n170 mm\n2. Gordon buys a cylindrical water tank to catch rain water off his roof. He discov-\ners a full 2 ℓtin of green paint in his garage and decides to paint the tank (not the\nbase). If he uses 250 ml to cover 1 m2, will he have enough green paint to cover\nthe tank with one layer of paint?\nDimensions of the tank:\ndiameter = 1,1 m\nheight = 1,4 m\nb\n1,1 m\n1,4 m\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235D\n2. 235F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n314\n7.2.\nRight prisms and cylinders\n\nVolume of prisms and cylinders\nEMBHY\nVolume, sometimes also called capacity, is the three dimensional space occupied by\nan object, or the contents of an object. It is measured in cubic units.\nThe volume of a right prism is simply calculated by multiplying the area of the base of\na solid by the height of the solid.\nRectangular\nprism\nl\nb\nh\nVolume = area of base × height\n= area of rectangle × height\n= l × b × h\nTriangular\nprism\nH\nb\nh\nVolume = area of base × height\n= area of triangle × height\n=\n\u00121\n2b × h\n\u0013\n× H\nCylinder\nh\nr\nVolume = area of base × height\n= area of circle × height\n= πr2 × h\nSee video: 235G at www.everythingmaths.co.za\n315\nChapter 7.\nMeasurement\n\nWorked example 3: Calculating volume\nQUESTION\nA rectangular glass vase with dimensions 28 cm × 18 cm × 8 cm is used for flower\narrangements. A florist uses a platic cylindrical jug to pour water into the glass vase.\nThe jug has a diameter of 142 mm and a height of 28 cm.\n28 cm\n18 cm\n8 cm\n142 mm\n28 cm\njug\nvase\n1. Will the plastic jug hold 5 ℓof water?\n2. Will a full jug of water be enough to fill the glass vase?\nSOLUTION\nStep 1: Determine the volume of the plastic jug\nThe diameter of the jug is 142 mm, therefore the radius =\n142\n2×10 = 7,1 cm.\nVolume of a cylinder = area of the base × height\nVolume of the jug = πr2 × h\n= π × (7,1)2 × 28\n= 4434 cm3\nAnd 1000 cm3 = 1 ℓ\n∴Volume of the jug = 4434\n1000\n= 4,434 ℓ\nNo, the capacity of the jug is not enough to hold 5 ℓof water.\n316\n7.2.\nRight prisms and cylinders\n\nStep 2: Determine the volume of the glass vase\nVolume of a rectangular prism = area of the base × height\nVolume of the vase = l × b × h\n= 28 × 18 × 8\n= 4032 cm3\n∴Volume of the vase = 4032\n1000\n= 4,032 ℓ\nYes, the volume of the jug is greater than the volume of the vase.\nExercise 7 – 3: Calculating volume\n1. The roof of Phumza’s house is the shape of a right-angled trapezium. A cylindri-\ncal water tank is positioned next to the house so that the rain on the roof runs\ninto the tank. The diameter of the tank is 140 cm and the height is 2,2 m.\n10 m\n8 m\n7,5 m\n2,2 m\n140 cm\na) Determine the area of the roof.\nb) Determine how many litres of water the tank can hold.\n2. The length of a side of a hexagonal sweet tin is 8 cm and its height is equal to\nhalf of the side length.\nA\nB\nC\nD\nE\nF\n8 cm\nh\na) Show that the interior angles are equal to 120◦.\n317\nChapter 7.\nMeasurement\n\nb) Determine the length of the line AE.\nc) Calculate the volume of the tin.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235H\n2. 235J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.3\nRight pyramids, right cones and spheres\nEMBHZ\nA pyramid is a geometric solid that has a polygon as its base and sides that converge\nat a point called the apex. In other words the sides are not perpendicular to the base.\nb\nThe triangular pyramid and square pyramid take their names from the shape of their\nbase. We call a pyramid a “right pyramid” if the line between the apex and the centre\nof the base is perpendicular to the base. Cones are similar to pyramids except that\ntheir bases are circles instead of polygons. Spheres are solids that are perfectly round\nand look the same from any direction.\nSurface area of pyramids, cones and spheres\nEMBJ2\nSquare\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n= b2 + 4\n\u0000 1\n2bhs\n\u0001\n= b (b + 2hs)\n318\n7.3.\nRight pyramids, right cones and spheres\n\nTriangular\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n=\n\u0000 1\n2b × hb\n\u0001\n+ 3\n\u0000 1\n2b × hs\n\u0001\n= 1\n2b (hb + 3hs)\nRight cone\nh\nr\nH\nSurface area = area of base +\narea of walls\n= πr2 + 1\n2 × 2πrh\n= πr (r + h)\nSphere\nb\nr\nSurface area = 4πr2\nVolume of pyramids, cones and spheres\nEMBJ3\nSquare\npyramid\nb\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × b2 × H\nTriangular\npyramid\nb\nh\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × 1\n2bh × H\n319\nChapter 7.\nMeasurement\n\nRight cone\nr\nH\nVolume = 1\n3 × area of base ×\nheight of cone\n= 1\n3 × πr2 × H\nSphere\nb\nr\nVolume = 4\n3πr3\nSee video: 235K at www.everythingmaths.co.za\nWorked example 4: Finding surface area and volume\nQUESTION\nThe Southern African Large Telescope (SALT) is housed in a cylindrical building with\na domed roof in the shape of a hemisphere. The height of the building wall is 17 m\nand the diameter is 26 m.\n17 m\n26 m\n1. Calculate the total surface area of the building.\n2. Calculate the total volume of the building.\n320\n7.3.\nRight pyramids, right cones and spheres\n\nSOLUTION\nStep 1: Calculate the total surface area\nTotal surface area = area of the dome + area of the cylinder\nSurface area =\n\u00141\n2(4πr2)\n\u0015\n+ [2πr × h]\n= 1\n2(4π)(13)2 + 2π(13)(17)\n= 2450 m2\nStep 2: Calculate the total volume\nTotal volume = volume of the dome + volume of the cylinder\nVolume =\n\u00141\n2 ×\n\u00124\n3πr3\n\u0013\u0015\n+\n\u0002\nπr2h\n\u0003\n= 2\n3π(13)3 + π(11)2(13)\n= 9543 m3\nExercise 7 – 4: Finding surface area and volume\n1. An ice-cream cone has a diameter of 52,4 mm and a total height of 146 mm.\n52,4 mm\n146 mm\na) Calculate the surface area of the ice-cream and the cone.\nb) Calculate the total volume of the ice-cream and the cone.\nc) How many ice-cream cones can be made from a 5 ℓtub of ice-cream (as-\nsume the cone is completely filled with ice-cream)?\n321\nChapter 7.\nMeasurement\n\nd) Consider the net of the cone given below. R is the length from the tip of\nthe cone to its perimeter, P.\nP\nR\nb\nM\ni. Determine the value of R.\nii. Calculate the length of arc P.\niii. Determine the length of arc M.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.4\nMultiplying a dimension by a constant factor\nEMBJ4\nWhen one or more of the dimensions of a prism or cylinder is multiplied by a constant,\nthe surface area and volume will change. The new surface area and volume can be\ncalculated by using the formulae from the preceding section.\nIt is important to see a relationship between the change in dimensions and the resulting\nchange in surface area and volume. These relationships make it simpler to calculate\nthe new volume or surface area of an object when its dimensions are scaled up or\ndown.\nConsider a rectangular prism of dimensions l, b and h. Below we multiply one, two\nand three of its dimensions by a constant factor of 5 and calculate the new volume and\nsurface area.\n322\n7.4.\nMultiplying a dimension by a constant factor\n\nDimensions\nVolume\nSurface\nOriginal dimensions\nl\nb\nh\nV = l × b × h\n= lbh\nA\n= 2 [(l × h) + (l × b) + (b × h)]\n= 2 (lh + lb + bh)\nMultiply one\ndimension by 5\nl\nb\n5h\nV1 = l × b × 5h\n= 5 (lbh)\n= 5V\nA1\n= 2 [(l × 5h) + (l × b) + (b × 5h)]\n= 2 (5lh + lb + 5bh)\nMultiply two\ndimensions by 5\n5l\nb\n5h\nV = 5l × b × 5h\n= 5 . 5(lbh)\n= 52V\nA2\n= 2 [(5l × 5h) + (5l × b) + (b × 5h)]\n= 2 × 5(5lh + lb + bh)\nMultiply all three\ndimensions by 5\n5l\n5b\n5h\nV = 5l × 5b × 5h\n= 53(lbh)\n= 53V\nA3\n= 2 [(5l × 5h) + (5l × 5b) + (5b × 5h)]\n= 2 × (52lh + 52lb + 52bh)\n= 52 × 2(lh + lb + bh)\n= 52A\nMultiply all three\ndimensions by k\nkl\nkb\nkh\nV = kl × kb × kh\n= k3(lbh)\n= k3V\nAk\n= 2 [(kl × kh) + (kl × kb) + (kb × kh)]\n= 2 × (k2lh + k2lb + k2bh)\n= k2 × 2(lh + lb + bh)\n= k2A\n323\nChapter 7.\nMeasurement\n\nWorked example 5: The effects of k\nQUESTION\nThe Nash family wants to build a television room onto their house. The dad draws up\nthe plans for the new square room of length k metres. The mum looks at the plans and\ndecides that the area of the room needs to be doubled. To achieve this:\n• the mum suggests doubling the length of the sides of the room\n• the dad recommends adding 2 m to the length of the sides\n• the daughter suggests multiplying the length of the sides by a factor of\n√\n2\n• the son suggests doubling only the width of the room\nWho’s suggestion will double the area of the square room? Show all calculations.\nSOLUTION\nStep 1: Draw a sketch\nk\nk\n2k\n2k\nk + 2\nk + 2\n√\n2k\n√\n2k\n2k\nk\nArea O\nArea M\nArea D\nArea d\nArea s\nStep 2: Calculate and compare\nFirst calculate the area of the square room in the original plan:\nArea O = length × length\n= k2\nTherefore, double the area of the room would be 2k2.\n324\n7.4.\nMultiplying a dimension by a constant factor\n\nConsider the mum’s suggestion of doubling the length of the sides of the room:\nArea M = length × length\n= 2k × 2k\n= 4k2\nThis area would be 4 times the original area.\nThe dad suggests adding 2 m to the length of the sides of the room:\nArea D = length × length\n= (k + 2) × (k + 2)\n= k2 + 4k + 2\n̸= 2k2\nThis is not double the original area.\nThe daughter suggests multiplying the length of the sides by a factor of\n√\n2:\nArea d = length × length\n=\n√\n2k ×\n√\n2k\n= 2k2\nThe daughter’s suggestion would double the area of the room. Practically, the length\nof the room could be multiplied by\n√\n2 ≈1,41 which would given an area of 1,96 m2.\nThe son suggests doubling only the width of the room:\nArea s = length × length\n= 2k × k\n= 2k2\nThe son’s suggestion would double the area of the room, however the room would no\nlonger be a square.\nStep 3: Write the final answer\nThe daughter’s suggestion of multiplying the length of the sides of the room by a factor\nof\n√\n2 would keep the shape of the room a square and would double the area of the\nroom.\nExercise 7 – 5: The effects of k\n1. Complete the following sentences:\na) If one dimension of a cube is multiplied by a factor 1\n2, the volume of the\ncube . . .\nb) If two dimensions of a cube are multiplied by a factor 7, the volume of the\ncube . . .\n325\nChapter 7.\nMeasurement\n\nc) If three dimensions of a cube are multiplied by a factor 3, then:\ni. each side of the cube will . . .\nii. the outer surface area of the cube will . . .\niii. the volume of the cube will . . .\nd) If each side of a cube is halved, then:\ni. the outer surface area of the cube will . . .\nii. the volume of the cube will . . .\n2. The municipality intends building a swimming pool of volume W 3 cubic metres.\nHowever, they realise that it will be very expensive to fill the pool with water, so\nthey decide to make the pool smaller.\na) The length and breadth of the pool are reduced by a factor of\n7\n10. Express\nthe new volume in terms of W.\nb) The dimensions of the pool are reduced so that the volume of the pool\ndecreases by a factor of 0,8. Determine the new dimensions of the pool in\nterms of W (remember that the pool must be a cube).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235N\n2. 235P\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.5\nSummary\nEMBJ5\nSee presentation: 235Q at www.everythingmaths.co.za\n1. Area is the two dimensional space inside the boundary of a flat object.\n2. Area formulae:\n• square: s2\n• rectangle: b × h\n• triangle: 1\n2b × h\n• trapezium: 1\n2 (a + b) × h\n• parallelogram: b × h\n• circle: πr2\n3. Surface area is the total area of the exposed or outer surfaces of a prism.\n4. A net is the unfolded “plan” of a solid.\n5. Volume is the three dimensional space occupied by an object, or the contents\nof an object.\n• Volume of a rectangular prism: l × b × h\n326\n7.5.\nSummary\n\n• Volume of a triangular prism:\n\u0000 1\n2b × h\n\u0001\n× H\n• Volume of a square prism or cube: s3\n• Volume of a cylinder: πr2 × h\n6. A pyramid is a geometric solid that has a polygon as its base and sides that\nconverge at a point called the apex. The sides are not perpendicular to the base.\n7. Surface area formulae:\n• square pyramid: b (b + 2h)\n• triangular pyramid: 1\n2b (hb + 3hs)\n• right cone: πr (r + hs)\n• sphere: 4πr2\n8. Volume formulae:\n• square pyramid: 1\n3 × b2 × H\n• triangular pyramid: 1\n3 × 1\n2bh × H\n• right cone: 1\n3 × πr2 × H\n• sphere: 4\n3πr3\nExercise 7 – 6: End of chapter exercises\n1.\na) Describe this figure in terms of a prism.\nb) Draw a net of this figure.\n2. Which of the following is a net of a cube?\na)\nb)\nc)\nd)\ne)\n327\nChapter 7.\nMeasurement\n\n3. Name and draw the following figures:\na) A prism with the least number of sides.\nb) A pyramid with the least number of vertices.\nc) A right prism with a kite base.\n4.\na)\ni. Determine how much paper is needed to make a box of width 16 cm,\nheight 3 cm and length 20 cm (assume no overlapping at corners).\nii. Give a mathematical name for the shape of the box.\niii. Calculate the volume of the box.\nb) Determine how much paper is needed to make a cube with a capacity of\n1 ℓ.\nc) Compare the box and the cube. Which has the greater volume and which\nrequires the most paper to make?\n5. ABCD is a rhombus with sides of length 3\n2x millimetres. The diagonals intersect\nat O and length DO = x millimetres. Express the area of ABCD in terms of x.\nO\nB\nD\nx\nC\nA\n3\n2x\n6. The diagram shows a rectangular pyramid with a base of length 80 cm and\nbreadth 60 cm. The vertical height of the pyramid is 45 cm.\n60 cm\n80 cm\n45 cm\nb\nh\nH\na) Calculate the volume of the pyramid.\nb) Calculate H and h.\nc) Calculate the surface area of the pyramid.\n7. A group of children are playing soccer in a field. The soccer ball has a capacity\nof 5000 cc (cubic centimetres). A drain pipe in the corner of the field has a\ndiameter of 20 cm. Is it possible for the children to lose their ball down the pipe?\nShow your calculations.\n328\n7.5.\nSummary\n\n8. A litre of washing powder goes into a standard cubic container at the factory.\na) Determine the length of the sides of the container.\nb) Determine the dimensions of the cubic container required to hold double\nthe volume of washing powder.\n9. A cube has sides of length k units.\na) Describe the effect on the volume of the cube if the height is tripled.\nb) If all three dimensions of the cube are tripled, determine the effect on the\nouter surface area.\nc) If all three dimensions of the cube are tripled, determine the effect on the\nvolume.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235R\n2. 235S\n3a. 235T\n3b. 235V\n3c. 235W\n4. 235X\n5. 235Y\n6. 235Z\n7. 2362\n8. 2363\n9. 2364\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n329\nChapter 7.\nMeasurement\n\n\nCHAPTER\n8\nEuclidean geometry\n8.1\nRevision\n332\n8.2\nCircle geometry\n333\n8.3\nSummary\n363\n\n8\nEuclidean geometry\n8.1\nRevision\nEMBJ6\nParallelogram\nEMBJ7\nA parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nSummary of the properties of a parallelogram:\n• Both pairs of opposite sides are parallel.\n• Both pairs of opposite sides are equal in length.\n• Both pairs of opposite angles are equal.\n• Both diagonals bisect each other.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\nThe mid-point theorem\nEMBJ8\nThe line joining the mid-points of two sides of a triangle is parallel to the third side\nand equal to half the length of the third side.\nA\nB\nC\nD\nE\nGiven: AD = DB and AE = EC, we can conclude that DE ∥BC and DE = 1\n2BC.\n332\n8.1.\nRevision\n\n8.2\nCircle geometry\nEMBJ9\nTerminology\nThe following terms are regularly used when referring to circles:\n• Arc — a portion of the circumference of a circle.\n• Chord — a straight line joining the ends of an arc.\n• Circumference — the perimeter or boundary line of a circle.\n• Radius (r) — any straight line from the centre of the circle to a point on the\ncircumference.\n• Diameter — a special chord that passes through the centre of the circle. A di-\nameter is a straight line segment from one point on the circumference to another\npoint on the circumference that passes through the centre of the circle.\n• Segment — part of the circle that is cut off by a chord. A chord divides a circle\ninto two segments.\n• Tangent — a straight line that makes contact with a circle at only one point on\nthe circumference.\nb\nb\nA\nB\nO\nP\na\nr\nc\nchord\ntangent\ndiameter\nradius\nsegment\nSee video: 2365 at www.everythingmaths.co.za\nAxioms\nAn axiom is an established or accepted principle. For this section, the following are\naccepted as axioms.\n333\nChapter 8.\nEuclidean geometry\n\n1. The theorem of Pythagoras states that the square of the hypotenuse of a right-\nangled triangle is equal to the sum of the squares of the other two sides.\n(AC)2 = (AB)2 + (BC)2\nC\nB\nA\n(AC)2\n(AB)2\n(BC)2\n2. A tangent is perpendicular to the radius (OT ⊥ST), drawn at the point of contact\nwith the circle.\nT\nS\nb\nO\nTheorems\nEMBJB\nA theorem is a hypothesis (proposition) that can be shown to be true by accepted\nmathematical operations and arguments. A proof is the process of showing a theorem\nto be correct.\nThe converse of a theorem is the reverse of the hypothesis and the conclusion. For\nexample, given the theorem “if A, then B”, the converse is “if B, then A”.\n334\n8.2.\nCircle geometry\n\nTheorem: Perpendicular line from circle centre bisects chord\nSTATEMENT\nIf a line is drawn from the centre of a circle perpendicular to a chord, then it bisects\nthe chord.\n(Reason: ⊥from centre bisects chord)\nGiven:\nCircle with centre O and line OP perpendicular to chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = PB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA2 = OP 2 + AP 2\n(Pythagoras)\nOB2 = OP 2 + BP 2\n(Pythagoras)\nand\nOA = OB\n(equal radii)\n∴AP 2 = BP 2\n∴AP = BP\nTherefore OP bisects AB.\nAlternative proof:\nIn △OPA and in △OPB,\nO ˆPA = O ˆPB\n(given OP ⊥AB)\nOA = OB\n(equal radii)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(RHS)\n∴AP = PB\nTherefore OP bisects AB.\n335\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Line from circle centre to mid-point of\nchord is perpendicular\nSTATEMENT\nIf a line is drawn from the centre of a circle to the mid-point of a chord, then the line\nis perpendicular to the chord.\n(Reason: line from centre to mid-point ⊥)\nGiven:\nCircle with centre O and line OP to mid-point P on chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nOP ⊥AB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA = OB\n(equal radii)\nAP = PB\n(given)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(SSS)\n∴O ˆPA = O ˆPB\nand O ˆPA + O ˆPB = 180◦\n(∠on str. line)\n∴O ˆPA = O ˆPB = 90◦\nTherefore OP ⊥AB.\nSee video: 2366 at www.everythingmaths.co.za\n336\n8.2.\nCircle geometry\n\nTheorem: Perpendicular bisector of chord passes through circle centre\nSTATEMENT\nIf the perpendicular bisector of a chord is drawn, then the line will pass through the\ncentre of the circle.\n(Reason: ⊥bisector through centre)\nGiven:\nCircle with mid-point P on chord AB.\nLine QP is drawn such that Q ˆPA = Q ˆPB = 90◦.\nLine RP is drawn such that R ˆPA = R ˆPB = 90◦.\nb\nb\nA\nB\nQ\nP\nR\nRequired to prove:\nCircle centre O lies on the line PR\nPROOF\nDraw lines QA and QB.\nDraw lines RA and RB.\nIn △QPA and in △QPB,\nAP = PB\n(given)\nQP = QP\n(common side)\nQ ˆPA = Q ˆPB = 90◦\n(given)\n∴△QPA ≡△QPB\n(SAS)\n∴QA = QB\nSimilarly it can be shown that in △RPA and in △RPB, RA = RB.\nWe conclude that all the points that are equidistant from A and B will lie on the\nline PR extended. Therefore the centre O, which is equidistant to all points on the\ncircumference, must also lie on the line PR.\n337\nChapter 8.\nEuclidean geometry\n\nWorked example 1: Perpendicular line from circle centre bisects chord\nQUESTION\nGiven OQ ⊥PR and PR = 8 units, determine the value of x.\nO\nx\n5\nP\nQ\nR\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nPQ = QR = 4\n(⊥from centre bisects chord)\nStep 2: Solve for x\nIn △OQP:\nPQ = 4\n(⊥from centre bisects chord)\nOP 2 = OQ2 + QP 2\n(Pythagoras)\n52 = x2 + 42\n∴x2 = 25 −16\nx2 = 9\nx = 3\nStep 3: Write the final answer\nx = 3 units.\n338\n8.2.\nCircle geometry\n\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. In the circle with centre O, OQ ⊥PR,\nOQ = 4 units and PR = 10. Determine\nx.\nO\n4\nP\nQ\nR\nx\n2. In the circle with centre O and radius\n= 10 units, OQ ⊥PR and PR = 8. De-\ntermine x.\nO\nx\n10\nP\nQ\nR\n3. In the circle with centre O, OQ ⊥PR,\nPR = 12 units and SQ = 2 units. Deter-\nmine x.\nO\nx\nP\nQ\nR\nS\n4. In the circle with centre O, OT ⊥SQ,\nOT ⊥PR, OP = 10 units, ST = 5 units\nand PU = 8 units. Determine TU.\nO\nV\n8\nP\nR\nU\n10\n5\nT\nQ\nS\n5. In the circle with centre O, OT ⊥QP,\nOS ⊥PR, OT = 5 units, PQ = 24 units\nand PR = 25 units. Determine OS = x.\nO\nx\nP\nS\n5\nT\nQ\nR\n25\n24\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2367\n2. 2368\n3. 2369\n4. 236B\n5. 236C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n339\nChapter 8.\nEuclidean geometry\n\nInvestigation: Angles subtended by an arc at the centre and the circumference of\na circle\n1. Measure angles x and y in each of the following graphs:\nb\nx1\ny1\nb\nx2\ny2\nb\nx3\ny3\n2. Complete the table:\nx\ny\n3. Use your results to make a conjecture about the relationship between angles\nsubtended by an arc at the centre of a circle and angles at the circumference of\na circle.\n4. Now draw three of your own similar diagrams and measure the angles to check\nyour conjecture.\n340\n8.2.\nCircle geometry\n\nTheorem: Angle at the centre of a circle is twice the size of the angle at the cir-\ncumference\nSTATEMENT\nIf an arc subtends an angle at the centre of a circle and at the circumference, then the\nangle at the centre is twice the size of the angle at the circumference.\n(Reason: ∠at centre = 2∠at circum.)\nGiven:\nCircle with centre O, arc AB subtending A ˆOB at the centre of the circle, and A ˆPB at\nthe circumference.\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nRequired to prove:\nA ˆOB = 2A ˆPB\nPROOF\nDraw PO extended to Q and let A ˆOQ = ˆO1 and B ˆOQ = ˆO2.\nˆO1 = A ˆPO + P ˆAO\n(ext. ∠△= sum int. opp. ∠s)\nand A ˆPO = P ˆAO\n(equal radii, isosceles △APO)\n∴ˆO1 = A ˆPO + A ˆPO\nˆO1 = 2A ˆPO\nSimilarly, we can also show that ˆO2 = 2B ˆPO.\nFor the first two diagrams shown above we have that:\nA ˆOB = ˆO1 + ˆO2\n= 2A ˆPO + 2B ˆPO\n= 2(A ˆPO + B ˆPO)\n∴A ˆOB = 2(A ˆPB)\nAnd for the last diagram:\nA ˆOB = ˆO2 −ˆO1\n= 2B ˆPO −2A ˆPO\n= 2(B ˆPO −A ˆPO)\n∴A ˆOB = 2(A ˆPB)\n341\nChapter 8.\nEuclidean geometry\n\nWorked example 2: Angle at the centre of circle is twice angle at circumference\nQUESTION\nGiven HK, the diameter of the circle passing through centre O.\nb\nJ\nH\nK\nO\na\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nStep 2: Solve for a\nIn △HJK:\nH ˆOK = 180◦\n(∠on str. line)\n= 2a\n(∠at centre = 2∠at circum.)\n∴2a = 180◦\na = 180◦\n2\n= 90◦\nStep 3: Conclusion\nThe diameter of a circle subtends a right angle at the circumference (angles in a semi-\ncircle).\n342\n8.2.\nCircle geometry\n\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\nGiven O is the centre of the circle, determine the unknown angle in each of the fol-\nlowing diagrams:\n1.\nb\nJ\nH\nK\nO\nb\n45◦\n2.\nbO\nJ\nK\nH\n45◦\nc\n3.\nb\nO\nK\nJ\n100◦\nH\nd\n4.\nb\nO\nH\nJ\ne\nK\n35◦\n5.\nb\nO\nJ\nK\nH\n120◦\nf\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236D\n2. 236F\n3. 236G\n4. 236H\n5. 236J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n343\nChapter 8.\nEuclidean geometry\n\nInvestigation: Subtended angles in the same segment of a circle\n1. Measure angles a, b, c, d and e in the diagram below:\na\ne\nd\nb\nc\nP\nQ\n2. Choose any two points on the circumference of the circle and label them A and\nB.\n3. Draw AP and BP, and measure A ˆPB.\n4. Draw AQ and BQ, and measure A ˆQB.\n5. What do you observe? Make a conjecture about these types of angles.\nTheorem: Subtended angles in the same segment of a circle are equal\nSTATEMENT\nIf the angles subtended by a chord of the circle are on the same side of the chord, then\nthe angles are equal.\n(Reason: ∠s in same seg.)\nGiven:\nCircle with centre O, and points P and Q on the circumference of the circle. Arc AB\nsubtends A ˆPB and A ˆQB in the same segment of the circle.\n344\n8.2.\nCircle geometry\n\nbO\nA\nB\nP\nQ\nRequired to prove:\nA ˆPB = A ˆQB\nPROOF\nA ˆOB = 2A ˆPB\n(∠at centre = 2∠at circum.)\nA ˆOB = 2A ˆQB\n(∠at centre = 2∠at circum.)\n∴2A ˆPB = 2A ˆQB\nA ˆPB = A ˆQB\nEqual arcs subtend equal angles\nFrom the theorem above we can deduce that if angles at the circumference of a circle\nare subtended by arcs of equal length, then the angles are equal. In the figure below,\nnotice that if we were to move the two chords with equal length closer to each other,\nuntil they overlap, we would have the same situation as with the theorem above. This\nshows that the angles subtended by arcs of equal length are also equal.\nb\nb\n345\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Concyclic points\nSTATEMENT\nIf a line segment subtends equal angles at two other points on the same side of the line\nsegment, then these four points are concyclic (lie on a circle).\nGiven:\nLine segment AB subtending equal angles at points P and Q on the same side of the\nline segment AB.\nA\nB\nR\nQ\nP\nRequired to prove:\nA, B, P and Q lie on a circle.\nPROOF\nProof by contradiction:\nPoints on the circumference of a circle: we know that there are only two possible\noptions regarding a given point — it either lies on circumference or it does not.\nWe will assume that point P does not lie on the circumference.\nWe draw a circle that cuts AP at R and passes through A, B and Q.\nA ˆQB = A ˆRB\n(∠s in same seg.)\nbut A ˆQB = A ˆPB\n(given)\n∴A ˆRB = A ˆPB\nbut A ˆRB = A ˆPB + R ˆBP\n(ext. ∠△= sum int. opp.)\n∴R ˆBP = 0◦\nTherefore the assumption that the circle does not pass through P must be false.\nWe can conclude that A, B, Q and P lie on a circle (A, B, Q and P are concyclic).\n346\n8.2.\nCircle geometry\n\nWorked example 3: Concyclic points\nQUESTION\nGiven FH ∥EI and E ˆIF = 15◦, determine the value of b.\nE\nF\nG\nH\nI\n15◦\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nH ˆFI = 15◦\n(alt. ∠, FH ∥EI)\nand b = H ˆFI\n(∠s in same seg.)\n∴b = 15◦\nExercise 8 – 3: Subtended angles in the same segment\n1. Find the values of the unknown angles.\na)\nA\nB\nC\nD\n21◦\na\nb)\nJ\nK\nL\nM\n24◦\nc\n102◦\nd\nc)\nN\nO\nP\nQ\n17◦\nd\n347\nChapter 8.\nEuclidean geometry\n\n2.\nR\nS\nT\nU\nV\n45◦\n35◦\n15◦\ne\na) Given T ˆV S = S ˆV R, deter-\nmine the value of e.\nb) Is TV a diameter of the cir-\ncle? Explain your answer.\n3.\nb\nW\nX\nY\nZ\nO\n35◦\nf\nT\n1\n2\nGiven circle with centre O, WT =\nTY and X ˆWT = 35◦. Determine\nf.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236K\n1b. 236M\n1c. 236N\n2. 236P\n3. 236Q\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nCyclic quadrilaterals\nCyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference\nof a circle (concyclic).\nInvestigation: Cyclic quadrilaterals\nConsider the diagrams given below:\nCircle 1\nCircle 2\nCircle 3\nA\nB\nC\nD\nA\nB\nC\nD\nA\nB\nC\nD\n348\n8.2.\nCircle geometry\n\n1. Complete the following:\nABCD is a cyclic quadrilateral because . . . . . .\n2. Complete the table:\nCircle 1\nCircle 2\nCircle 3\nˆA =\nˆB =\nˆC =\nˆD =\nˆA + ˆC =\nˆB + ˆD =\n3. Use your results to make a conjecture about the relationship between angles of\ncyclic quadrilaterals.\nTheorem: Opposite angles of a cyclic quadrilateral\nSTATEMENT\nThe opposite angles of a cyclic quadrilateral are supplementary.\n(Reason: opp. ∠s cyclic quad.)\nGiven:\nCircle with centre O with points A, B, P and Q on the circumference such that ABPQ\nis a cyclic quadrilateral.\nbO\nA\nB\nP\nQ\n1\n2\nRequired to prove:\nA ˆBP + A ˆQP = 180◦and Q ˆAB + Q ˆPB = 180◦\n349\nChapter 8.\nEuclidean geometry\n\nPROOF\nDraw AO and OP. Label ˆO1 and ˆO2.\nˆO1 = 2A ˆBP\n(∠at centre = 2∠at circum.)\nˆO2 = 2A ˆQP\n(∠at centre = 2∠at circum.)\nand ˆO1 + ˆO2 = 360◦\n(∠s around a point)\n∴2A ˆBP + 2A ˆQP = 360◦\nA ˆBP + A ˆQP = 180◦\nSimilarly, we can show that Q ˆAB + Q ˆPB = 180◦.\nConverse: interior opposite angles of a quadrilateral\nIf the interior opposite angles of a quadrilateral are supplementary, then the quadrilat-\neral is cyclic.\nExterior angle of a cyclic quadrilateral\nIf a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle.\nb\nb\nWorked example 4: Opposite angles of a cyclic quadrilateral\nQUESTION\nGiven the circle with centre O and cyclic quadrilateral PQRS. SQ is drawn and\nS ˆPQ = 34◦. Determine the values of a, b and c.\nbO\nP\nQ\nR\nS\na\nb\nc\n34◦\n350\n8.2.\nCircle geometry\n\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nS ˆPQ + c = 180◦\n(opp. ∠s cyclic quad supp.)\n∴c = 180◦−34◦\n= 146◦\na = 90◦\n(∠in semi circle)\nIn △PSQ:\na + b + 34◦= 180◦\n(∠sum of △)\n∴b = 180◦−90◦−34◦\n= 56◦\nMethods for proving a quadrilateral is cyclic\nThere are three ways to prove that a quadrilateral is a cyclic quadrilateral:\nMethod of proof\nReason\nR\nQ\nS\nP\nIf ˆP + ˆR = 180◦or ˆS +\nˆQ = 180◦, then PQRS is\na cyclic quad.\nopp.\nint.\nangles\nsuppl.\nR\nQ\nS\nP\nIf ˆP = ˆQ or ˆS = ˆR, then\nPQRS is a cyclic quad.\nangles in the same\nseg.\nR\nQ\nS\nP\nT\nIf T ˆQR = ˆS, then PQRS\nis a cyclic quad.\next.\nangle equal to\nint. opp. angle\n351\nChapter 8.\nEuclidean geometry\n\nWorked example 5: Proving a quadrilateral is a cyclic quadrilateral\nQUESTION\nProve that ABDE is a cyclic quadrilateral.\nbO\nE\nC\nD\nA\nB\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Prove that ABDE is a cyclic quadrilateral\nD ˆBC = 90◦\n(∠in semi circle)\nand ˆE = 90◦\n(given)\n∴D ˆBC = ˆE\n∴ABDE is a cyclic quadrilateral\n(ext. ∠equals int. opp. ∠)\nExercise 8 – 4: Cyclic quadrilaterals\n1. Find the values of the unknown angles.\na)\nX\nY\nZ\nW\na\nb\n106◦\n87◦\nb)\nH\nI\nJ\nK\nL\n114◦\na\nc)\nU\nV\nW\nX\n57◦\na\n86◦\n352\n8.2.\nCircle geometry\n\n2. Prove that ABCD is a cyclic quadrilateral:\na) D\nC\n72◦\nB\nA\n32◦\nM\n40◦\nb) D\nC\n70◦\nB\nA\n35◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236R\n1b. 236S\n1c. 236T\n2a. 236V\n2b. 236W\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTangent line to a circle\nA tangent is a line that touches the circumference of a circle at only one place. The\nradius of a circle is perpendicular to the tangent at the point of contact.\nb\nO\n353\nChapter 8.\nEuclidean geometry\n\nTheorem: Two tangents drawn from the same point outside a circle\nSTATEMENT\nIf two tangents are drawn from the same point outside a circle, then they are equal in\nlength.\n(Reason: tangents from same point equal)\nGiven:\nCircle with centre O and tangents PA and PB, where A and B are the respective\npoints of contact for the two lines.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = BP\nPROOF\nIn △AOP and △BOP,\nO ˆAP = O ˆBP = 90◦\n(tangent ⊥radius)\nAO = BO\n(equal radii)\nOP = OP\n(common side)\n∴△AOP ≡△BOP\n(RHS)\n∴AP = BP\n354\n8.2.\nCircle geometry\n\nWorked example 6: Tangents from the same point outside a circle\nQUESTION\nIn the diagram below AE = 5 cm, AC = 8 cm and CE = 9 cm. Determine the values\nof a, b and c.\nA\nB\nC\nD\nE\nF\nAE = 5 cm\nAC = 8 cm\nCE = 9 cm\na\nb\nc\nb\nb\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for a, b and c\nAB = AF = a\n(tangents from A)\nEF = ED = c\n(tangents from E)\nCB = CD = b\n(tangents from C)\n∴AE = a + c = 5\nand AC = a + b = 8\nand CE = b + c = 9\nStep 3: Solve for the unknown variables using simultaneous equations\na + c = 5\n. . . (1)\na + b = 8\n. . . (2)\nb + c = 9\n. . . (3)\nSubtract equation (1) from equation (2) and then substitute into equation (3):\n(2) −(1)\nb −c = 8 −5\n= 3\n∴b = c + 3\nSubstitute into (3)\nc + 3 + c = 9\n2c = 6\nc = 3\n∴a = 2\nand b = 6\n355\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 5: Tangents to a circle\nFind the values of the unknown lengths.\n1.\nb\nG\nH\nI\nJ\nd\n5 cm\n8 cm\n2.\nb\nK\nL\nM\nN\nO\nP\ne\nLN = 7,5 cm\n2 cm\n6 cm\n3.\nb\nb\nR\nQ\nS\nf\n3 cm\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236X\n2. 236Y\n3. 236Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Tangent-chord theorem\nConsider the diagrams given below:\nDiagram 1\nDiagram 2\nDiagram 3\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\n1. Measure the following angles with a protractor and complete the table:\nDiagram 1\nDiagram 2\nDiagram 3\nA ˆBC =\nˆD =\nˆE =\n2. Use your results to complete the following: the angle between a tangent to a\ncircle and a chord is . . . . . . to the angle in the alternate segment.\n356\n8.2.\nCircle geometry\n\nTheorem: Tangent-chord theorem\nSTATEMENT\nThe angle between a tangent to a circle and a chord drawn at the point of contact, is\nequal to the angle which the chord subtends in the alternate segment.\n(Reason: tan. chord theorem)\nGiven:\nCircle with centre O and tangent SR touching the circle at B. Chord AB subtends ˆP1\nand ˆQ1.\nb\nO\nA\nB\nP\n1\n1\nQ\nT\n1\nS\nR\nRequired to prove:\n1. A ˆBR = A ˆPB\n2. A ˆBS = A ˆQB\nPROOF\nDraw diameter BT and join T to A.\nLet A ˆTB = T1.\nA ˆBS + A ˆBT = 90◦\n(tangent ⊥radius)\nB ˆAT = 90◦\n(∠in semi circle)\n∴A ˆBT + T1 = 90◦\n(∠sum of △BAT)\n∴A ˆBS = T1\nbut Q1 = T1\n(∠s in same segment)\n∴Q1 = A ˆBS\nA ˆBS + A ˆBR = 180◦\n(∠s on str. line)\nˆQ1 + ˆP1 = 180◦\n(opp. ∠s cyclic quad. supp.)\n∴A ˆBS + A ˆBR = Q1 + P1\nand A ˆBS = Q1\n∴A ˆBR = P1\n357\nChapter 8.\nEuclidean geometry\n\nWorked example 7: Tangent-chord theorem\nQUESTION\nDetermine the values of h and s.\nP\nO\nQ\nS\nR\nh + 20◦s\n4h\n4h −70◦\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for h\nO ˆQS = S ˆRQ\n(tangent chord theorem)\nh + 20◦= 4h −70◦\n90◦= 3h\n∴h = 30◦\nStep 3: Solve for s\nP ˆQR = Q ˆSR\n(tangent chord theorem)\ns = 4h\n= 4(30◦)\n= 120◦\n358\n8.2.\nCircle geometry\n\nExercise 8 – 6: Tangent-chord theorem\n1. Find the values of the unknown letters, stating reasons.\nQ\nR\nS\nO\nP\na\nb\n33◦\na)\nO\nP\nQ\nR\nS\nc\nd\n72◦\nb)\nO\nP\nQ\nR\nS\ng\nf\n38◦\n47◦\nc)\nR\nP\nO\nQ\nl\n1\n1\n66◦\nd)\nO\nP\nQ\nR\nS\ni\nj\nk\n39◦\n101◦\ne)\nO\nR\nQ\nS\nT\nm\nn\no\n34◦\nf)\nO\n•\nP\nR\nQ\nS\nT\np\nq\nr\n52◦\ng)\n359\nChapter 8.\nEuclidean geometry\n\n2. O is the centre of the circle and SPT is a tangent, with OP ⊥ST. Determine\na, b and c, giving reasons.\nO•\nS\nT\nP\nM\nN\na\nb\nc\n64◦\n3.\nP\nL\nA\nB\nC\n1 2\n3\n1\n2\nD\nGiven AB = AC, AP ∥BC and ˆA2 = ˆB2. Prove:\na) PAL is a tangent to the circle ABC.\nb) AB is a tangent to the circle ADP.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2372\n1b. 2373\n1c. 2374\n1d. 2375\n1e. 2376\n1f. 2377\n1g. 2378\n2. 2379\n3. 237B\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nConverse: tangent-chord theorem\nIf a line drawn through the end point of a chord forms an angle equal to the angle\nsubtended by the chord in the alternate segment, then the line is a tangent to the\ncircle.\n(Reason: ∠between line and chord = ∠in alt. seg. )\n360\n8.2.\nCircle geometry\n\nWorked example 8: Applying the theorems\nQUESTION\nA\nD\nB\nC\nO\nE\nF\nBD is a tangent to the circle with centre O, with BO ⊥AD.\nProve that:\n1. CFOE is a cyclic quadrilateral\n2. FB = BC\n3. ∠A ˆOC = 2B ˆFC\n4. Will DC be a tangent to the circle passing through C, F, O and E? Motivate your\nanswer.\nSOLUTION\nStep 1: Prove CFOE is a cyclic quadrilateral by showing opposite angles are supple-\nmentary\nBO ⊥OD\n(given)\n∴F ˆOE = 90◦\nF ˆCE = 90◦\n(∠in semi circle)\n∴CFOE is a cyclic quad.\n(opp. ∠s suppl.)\nStep 2: Prove BFC is an isosceles triangle\nTo show that FB = BC we first prove △BFC is an isosceles triangle by showing that\nB ˆFC = B ˆCF.\nB ˆCF = C ˆEO\n(tangent-chord)\nC ˆEO = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴B ˆFC = B ˆCF\n∴FB = BC\n(△BFC isosceles)\n361\nChapter 8.\nEuclidean geometry\n\nStep 3: Prove A ˆOC = 2B ˆFC\nA ˆOC = 2A ˆEC\n(∠at centre = 2∠at circum.)\nand A ˆEC = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴A ˆOC = 2B ˆFC\nStep 4: Determine if DC is a tangent to the circle through C, F, O and E\nProof by contradiction.\nLet us assume that DC is a tangent to the circle passing through the points C, F, O\nand E:\n∴D ˆCE = C ˆOE\n(tangent-chord)\nAnd using the circle with centre O and tangent BD we have that:\nD ˆCE = C ˆAE\n(tangent-chord)\nbut C ˆAE = 1\n2C ˆOE\n(∠at centre = 2∠at circum.)\n∴D ˆCE ̸= C ˆOE\nTherefore our assumption is not correct and we can conclude that DC is not a tangent\nto the circle passing through the points C, F, O and E.\nWorked example 9: Applying the theorems\nQUESTION\nA\nB\nC\nD\nE\nF\nG\nH\nFD is drawn parallel to the tangent CB\n362\n8.2.\nCircle geometry\n\nProve that:\n1. FADE is a cyclic quadrilateral\n2. F ˆEA = ˆB\nSOLUTION\nStep 1: Prove FADE is a cyclic quadrilateral using angles in the same segment\nF ˆDC = D ˆCB\n(alt. ∠s FD ∥CB)\nand D ˆCB = C ˆAE\n(tangent-chord)\n∴F ˆDC = C ˆAE\n∴FADE is a cyclic quad.\n(∠s in same seg.)\nStep 2: Prove F ˆEA = ˆB\nF ˆDA = ˆB\n(corresp. ∠s FD ∥CB)\nand F ˆEA = F ˆDA\n(∠s same seg. cyclic quad. FADE)\n∴F ˆEA = ˆB\n8.3\nSummary\nEMBJC\nSee presentation: 237C at www.everythingmaths.co.za\n• Arc An arc is a portion of the circumference of a circle.\n• Chord - a straight line joining the ends of an arc.\n• Circumference - perimeter or boundary line of a circle.\n• Radius (r) - any straight line from the centre of the circle to a point on the cir-\ncumference.\n• Diameter - a special chord that passes through the centre of the circle. A diame-\nter is the length of a straight line segment from one point on the circumference to\nanother point on the circumference, that passes through the centre of the circle.\n• Segment A segment is a part of the circle that is cut off by a chord. A chord\ndivides a circle into two segments.\n• Tangent - a straight line that makes contact with a circle at only one point on the\ncircumference.\n• A tangent line is perpendicular to the radius, drawn at the point of contact with\nthe circle.\n363\nChapter 8.\nEuclidean geometry\n\nb O\nM\nA\nB\n• If O is the centre and OM ⊥AB, then AM =\nMB.\n• If O is the centre and AM\n= MB, then\nA ˆ\nMO = B ˆ\nMO = 90◦.\n• If AM = MB and OM ⊥AB, then ⇒MO\npasses through centre O.\nb\n2x\nx\n2y\ny\nx\nIf an arc subtends an angle at the centre of a cir-\ncle and at the circumference, then the angle at the\ncentre is twice the size of the angle at the circum-\nference.\nb\nb\nAngles at the circumference subtended by the same\narc (or arcs of equal length) are equal.\nA\nB\nC\nD\n1\n2\nE\nThe four sides of a cyclic quadrilateral ABCD are\nchords of the circle with centre O.\n• ˆA + ˆC = 180◦(opp. ∠s supp.)\n• ˆB + ˆD = 180◦(opp. ∠s supp.)\n• E ˆBC = ˆD (ext. ∠cyclic quad.)\n• ˆA1 = ˆA2 = ˆC (vert. opp. ∠, ext. ∠cyclic\nquad.)\nA\nB\nC\nD\nProving a quadrilateral is cyclic: If ˆA + ˆC = 180◦or\nˆB+ ˆD = 180◦, then ABCD is a cyclic quadrilateral.\n364\n8.3.\nSummary\n\nA\nB\nC\nD\n1\n1\nIf ˆA1 = ˆC or ˆD1 = ˆB, then ABCD is a cyclic\nquadrilateral.\nA\nB\nC\nD\nIf ˆA = ˆB or ˆC = ˆD, then ABCD is a cyclic quadri-\nlateral.\nb\nA\nB\nO\nT\nIf AT and BT are tangents to circle O, then\n• OA ⊥AT (tangent ⊥radius)\n• OB ⊥BT (tangent ⊥radius)\n• TA = TB (tangents from same point equal)\nA\nB\nT\nD\nC\nx\ny\nx\ny\n• If DC is a tangent, then D ˆTA = T ˆBA and\nC ˆTB = T ˆAB\n• If D ˆTA = T ˆBA or C ˆTB = T ˆAB, then DC is\na tangent touching at T\n365\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 7: End of chapter exercises\n1.\nO\n•\nA\nB\nC\nD\nE\nF\n×\n×\nx\nAOC is a diameter of the circle with centre O. F is the mid-point of chord EC.\nB ˆOC = C ˆOD and ˆB = x. Express the following angles in terms of x, stating\nreasons:\na) ˆA\nb) C ˆOD\nc) ˆD\n2.\nM•\nD\nE\nF\nG\n1 2\n1\n2\n1\n2\n1 2\nD, E, F and G are points on circle with centre M.\nˆF1 = 7◦and ˆD2 = 51◦.\nDetermine the sizes of the following angles, stating reasons:\na)\nˆ\nM1\nb) ˆD1\nc) ˆF2\nd) ˆG\ne) ˆE1\n366\n8.3.\nSummary\n\n3.\nM\n•\nO•\nD\nA\nB\nC\n1 2\nO is a point on the circle with centre M. O is also the centre of a second circle.\nDA cuts the smaller circle at C and ˆD1 = x. Express the following angles in\nterms of x, stating reasons:\na) ˆD2\nb) O ˆAB\nc) O ˆBA\nd) A ˆOB\ne) ˆC\n4.\nO•\nA\nB\nC\nM\nO is the centre of the circle with radius 5 cm and chord BC = 8 cm. Calculate\nthe lengths of:\na) OM\nb) AM\nc) AB\n5.\nO•\nA\nB\nC\n70◦\nx\nAO ∥CB in circle with centre O. A ˆOB = 70◦and O ˆAC = x. Calculate the\nvalue of x, giving reasons.\n367\nChapter 8.\nEuclidean geometry\n\n6.\nO\n•\nP\nQ\nR\nS\nT\nx\nPQ is a diameter of the circle with centre O. SQ bisects P ˆQR and P ˆQS = x.\na) Write down two other angles that are also equal to x.\nb) Calculate P ˆOS in terms of x, giving reasons.\nc) Prove that OS is a perpendicular bisector of PR.\n7.\nO•\nA\nB\nC\nD\n35◦\nB ˆOD is a diameter of the circle with centre O. AB = AD and O ˆCD = 35◦.\nCalculate the value of the following angles, giving reasons:\na) O ˆDC\nb) C ˆOD\nc) C ˆBD\nd) B ˆAD\ne) A ˆDB\n8.\nO\n•\nR\nP\nT\nQ\nx\ny\nQP in the circle with centre O is protracted to T so that PR = PT. Express y in\nterms of x.\n368\n8.3.\nSummary\n\n9.\nO•\nA\nB\nC\nD\nE\nP\nF\nO is the centre of the circle with diameter AB. CD ⊥AB at P and chord DE\ncuts AB at F. Prove that:\na) C ˆBP = D ˆPB\nb) C ˆED = 2C ˆBA\nc) A ˆBD = 1\n2C ˆOA\n10.\nO\n•\nP\nQ\nR\nx\nS\nIn the circle with centre O, OR ⊥QP, PQ = 30 mm and RS = 9 mm. Deter-\nmine the length of OQ.\n11.\nM •\nP\nQ\nR\nS\nT\nP, Q, R and S are points on the circle with centre M. PS and QR are extended\nand meet at T. PQ = PR and P ˆQR = 70◦.\na) Determine, stating reasons, three more angles equal to 70◦.\nb) If Q ˆPS = 80◦, calculate S ˆRT, S ˆTR and P ˆQS.\nc) Explain why PQ is a tangent to the circle QST at point Q.\nd) Determine P ˆ\nMQ.\n369\nChapter 8.\nEuclidean geometry\n\n12.\nO\n•\nA\nP\nQ\nC\nB\nPOQ is a diameter of the circle with centre O. QP is protruded to A and AC is\na tangent to the circle. BA ⊥AQ and BCQ is a straight line. Prove:\na) P ˆCQ = B ˆAP\nb) BAPC is a cyclic quadrilateral\nc) AB = AC\n13.\nO•\nT\nC\nA\nB\nx\nTA and TB are tangents to the circle with centre O. C is a point on the circum-\nference and A ˆTB = x. Express the following in terms of x, giving reasons:\na) A ˆBT\nb) O ˆBA\nc) ˆC\n14.\nO•\nA\nB\nC\nE\nD\nAOB is a diameter of the circle\nAECB with centre O. OE ∥BC\nand cuts AC at D.\na) Prove AD = DC\nb) Show that A ˆBC is bisected\nby EB\nc) If O ˆEB = x, express B ˆAC\nin terms of x\nd) Calculate the radius of the\ncircle if AC = 10 cm and\nDE = 1 cm\n370\n8.3.\nSummary\n\n15.\nV\nQ\nS\nR\nP\nT\nW\nx\ny\nPQ and RS are chords of the circle and PQ ∥RS. The tangent to the circle at\nQ meets RS protruded at T. The tangent at S meets QT at V . QS and PR are\ndrawn.\nLet T ˆQS = x and Q ˆRP = y. Prove that:\na) T ˆV S = 2Q ˆRS\nb) QV SW is a cyclic quadrilateral\nc) Q ˆPS + ˆT = P ˆRT\nd) W is the centre of the circle\n16.\nF\nD\nB\nC\nE\nA\nK\nT\n1\n2\n1\n2\n1\n2\n3\n4\nThe two circles shown intersect at points F and D. BFT is a tangent to the\nsmaller circle at F. Straight line AFE is drawn such that DF = EF. CDE is a\nstraight line and chord AC and BF cut at K. Prove that:\na) BT ∥CE\nb) BCEF is a parallelogram\nc) AC = BF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237D\n2. 237F\n3. 237G\n4. 237H\n5. 237J\n6. 237K\n7. 237M\n8. 237N\n9. 237P\n10. 237Q\n11. 237R\n12. 237S\n13. 237T\n14. 237V\n15. 237W\n16. 237X\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n371\nChapter 8.\nEuclidean geometry\n\n\nCHAPTER\n9\nFinance, growth and decay\n9.1\nRevision\n374\n9.2\nSimple and compound depreciation\n377\n9.3\nTimelines\n388\n9.4\nNominal and effective interest rates\n394\n9.5\nSummary\n398\n\n9\nFinance, growth and decay\n9.1\nRevision\nEMBJD\nSimple interest is the interest calculated only on the initial amount invested, the prin-\ncipal amount. Compound interest is the interest earned on the principal amount and\non its accumulated interest. This means that interest is being earned on interest. The\naccumulated amount is the final amount; the sum of the principal amount and the\namount of interest earned.\nFormula for simple interest:\nA = P(1 + in)\nFormula for compound interest:\nA = P(1 + i)n\nwhere\nA = accumulated amount\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nWorked example 1: Simple and compound interest\nQUESTION\nSam wants to invest R 3450 for 5 years. Wise Bank offers a savings account which pays\nsimple interest at a rate of 12,5% per annum, and Grand Bank offers a savings account\npaying compound interest at a rate of 10,4% per annum. Which bank account would\ngive Sam the greatest accumulated balance at the end of the 5 year period?\nSOLUTION\nStep 1: Calculation using the simple interest formula\nWrite down the known variables and the simple interest formula\nP = 3450\ni = 0,125\nn = 5\nA = P(1 + in)\nSubstitute the values to determine the accumulated amount for the Wise Bank savings\n374\n9.1.\nRevision\n\naccount.\nA = 3450(1 + 0,125 × 5)\n= R 5606,25\nStep 2: Calculation using the compound interest formula\nWrite down the known variables and the compound interest formula.\nP = 3450\ni = 0,104\nn = 5\nA = P(1 + i)n\nSubstitute the values to determine the accumulated amount for the Grand Bank savings\naccount.\nA = 3450(1 + 0,104)5\n= R 5658,02\nStep 3: Write the final answer\nThe Grand Bank savings account would give Sam the highest accumulated balance at\nthe end of the 5 year period.\nWorked example 2: Finding i\nQUESTION\nBongani decides to put R 30 000 in an investment account. What compound interest\nrate must the investment account achieve for Bongani to double his money in 6 years?\nGive your answer correct to one decimal place.\nSOLUTION\nStep 1: Write down the known variables and the compound interest formula\nA = 60 000\nP = 30 000\nn = 6\nA = P(1 + i)n\n375\nChapter 9.\nFinance, growth and decay\n\nStep 2: Substitute the values and solve for i\n60 000 = 30 000(1 + i)6\n60 000\n30 000 = (1 + i)6\n2 = (1 + i)6\n6√\n2 = 1 + i\n6√\n2 −1 = i\n∴i = 0,122 . . .\nStep 3: Write the final answer and comment\nWe round up to a rate of 12,3% p.a. to make sure that Bongani doubles his invest-\nment.\nExercise 9 – 1: Revision\n1. Determine the value of an investment of R 10 000 at 12,1% p.a. simple interest\nfor 3 years.\n2. Calculate the value of R 8000 invested at 8,6% p.a. compound interest for 4\nyears.\n3. Calculate how much interest John will earn if he invests R 2000 for 4 years at:\na) 6,7% p.a. simple interest\nb) 5,4% p.a. compound interest\n4. The value of an investment grows from R 2200 to R 3850 in 8 years. Determine\nthe simple interest rate at which it was invested.\n5. James had R 12 000 and invested it for 5 years. If the value of his investment is\nR 15 600, what compound interest rate did it earn?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237Y\n2. 237Z\n3. 2382\n4. 2383\n5. 2384\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n376\n9.1.\nRevision\n\n9.2\nSimple and compound depreciation\nEMBJF\nAs soon as a new car leaves the dealership, its value decreases and it is considered\n“second-hand”.\nVehicles, equipment, machinery and other similar assets, all lose\nvalue over time as a result of usage and age. This loss in value is called deprecia-\ntion. Assets that have a relatively long useful lifetime, such as machines, trucks, farm-\ning equipment etc., depreciate slower than assets like office equipment, computers,\nfurniture etc. which need to be replaced more often and therefore depreciate more\nquickly.\nDepreciation is used to calculate the value of a company’s assets, which determines\nhow much tax a company must pay. Companies can take depreciation into account as\nan expense, and thereby reduce their taxable income. A lower taxable income means\nthat the company will pay less income tax to SARS (South African Revenue Service).\nWe can calculate two different kinds of depreciation: simple decay and compound\ndecay. Decay is also a term used to describe a reduction or decline in value. Simple\ndecay is also called straight-line depreciation and compound decay can also be re-\nferred to as reducing-balance depreciation. In the straight-line method the value of the\nasset is reduced by a constant amount each year, which is calculated on the principal\namount. In reducing-balance depreciation we calculate the depreciation on the re-\nduced value of the asset. This means that the value of an asset decreases by a different\namount each year.\nInvestigation: Simple and compound depreciation\n1. Mr. Sontange buys an Opel Fiesta for R 72 000. He expects that the value of the\ncar will depreciate by R 6000 every year. He draws up a table to calculate the\ndepreciated value of his Opel Fiesta.\nComplete Mr. Sontange’s table of values for the 7 year period:\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 6000\nR 66 000\n2\nR 66 000\nR 6000\n3\n4\n5\n6\n7\n2. His son, David, does not agree that the value of the car will reduce by the same\namount each year. David thinks that the car will depreciate by 10% every year.\nComplete David’s table of values:\n377\nChapter 9.\nFinance, growth and decay\n\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 7200\nR 64 800\n2\nR 64 800\nR 6480\n3\n4\n5\n6\n7\n3. Compare and discuss the results of the two different tables.\n4. Consider the graph below, which represents Mr. Sontange’s table of values:\n10 000\n20 000\n30 000\n40 000\n50 000\n60 000\n70 000\n80 000\n1\n2\n3\n4\n5\n6\n7\n8\n0\nTime (years)\nValue (Rands)\na) Draw a similar graph using David’s table of values.\nb) Interpret the two graphs and discuss the differences between them.\nc) Explain how the graphs can be used to determine the total depreciation in\neach case.\nd)\ni. Draw two new graphs by plotting the maximum value of each bar.\nii. Join the points with a line to show the general trend.\niii. Is it mathematically correct to join these points? Explain your answer.\n378\n9.2.\nSimple and compound depreciation\n\nSimple depreciation\nEMBJG\nWorked example 3: Straight-line depreciation\nQUESTION\nA new smartphone costs R 6000 and depreciates at 22% p.a. on a straight-line basis.\nDetermine the value of the smartphone at the end of each year over a 4 year period.\nSOLUTION\nStep 1: Calculate depreciation amount\nDepreciation = 6000 × 22\n100\n= 1320\nTherefore the smartphone depreciates by R 1320 every year.\nStep 2: Complete a table of values\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 6000\nR 1320\nR 4680\n2\nR 4680\nR 1320\nR 3360\n3\nR 3360\nR 1320\nR 2040\n4\nR 2040\nR 1320\nR 720\nWe notice that\nTotal depreciation = P × i × n\nwhere\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nTherefore the depreciated value of the asset (also called the book value) can be calcu-\nlated as:\nA = P(1 −in)\nNote the similarity to the simple interest formula A = P(1 + in). Interest increases the\nvalue of the principal amount, whereas with simple decay, depreciation reduces the\nvalue of the principal amount.\nImportant: to get an accurate answer do all calculations in one step on your calculator.\nDo not round off answers in your calculations until the final answer. In the worked\nexamples in this chapter, we use dots to show that the answer has not been rounded\noff. We always round the final answer to two decimal places (cents).\n379\nChapter 9.\nFinance, growth and decay\n\nWorked example 4: Straight-line depreciation method\nQUESTION\nA car is valued at R 240 000. If it depreciates at 15% p.a. using straight-line deprecia-\ntion, calculate the value of the car after 5 years.\nSOLUTION\nStep 1: Write down the known variables and the simple decay formula\nP = 240 000\ni = 0,15\nn = 5\nA = P(1 −in)\nStep 2: Substitute the values and solve for A\nA = 240 000(1 −0,15 × 5)\n= 240 000(0,25)\n= 60 000\nStep 3: Write the final answer\nAt the end of 5 years, the car is worth R 60 000.\nWorked example 5: Simple decay\nQUESTION\nA small business buys a photocopier for R 12 000. For the tax return the owner depre-\nciates this asset over 3 years using a straight-line depreciation method. What amount\nwill he fill in on his tax form at the end of each year?\nSOLUTION\nStep 1: Write down the known variables\nThe owner of the business wants the photocopier to have a book value of R 0 after 3\nyears.\nA = 0\nP = 12 000\nn = 3\n380\n9.2.\nSimple and compound depreciation\n\nTherefore we can calculate the annual depreciation as\nDepreciation = P\nn\n= 12 000\n3\n= R 4000\nStep 2: Determine the book value at the end of each year\nBook value end of first year = 12 000 −4000\n= R 8000\nBook value end of second year = 8000 −4000\n= R 4000\nBook value end of third year = 4000 −4000\n= R 0\nExercise 9 – 2: Simple decay\n1. A business buys a truck for R 560 000. Over a period of 10 years the value of\nthe truck depreciates to R 0 using the straight-line method. What is the value of\nthe truck after 8 years?\n2. Harry wants to buy his grandpa’s donkey for R 800. His grandpa is quite pleased\nwith the offer, seeing that it only depreciated at a rate of 3% per year using the\nstraight-line method. Grandpa bought the donkey 5 years ago. What did grandpa\npay for the donkey then?\n3. Seven years ago, Rocco’s drum kit cost him R 12 500. It has now been valued at\nR 2300. What rate of simple depreciation does this represent?\n4. Fiona buys a DStv satellite dish for R 3000. Due to weathering, its value depre-\nciates simply at 15% per annum. After how long will the satellite dish have a\nbook value of zero?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2385\n2. 2386\n3. 2387\n4. 2388\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n381\nChapter 9.\nFinance, growth and decay\n\nCompound depreciation\nEMBJH\nWorked example 6: Reducing-balance depreciation\nQUESTION\nA second-hand farm tractor worth R 60 000 has a limited useful life of 5 years and\ndepreciates at 20% p.a. on a reducing-balance basis. Determine the value of the\ntractor at the end of each year over the 5 year period.\nSOLUTION\nStep 1: Write down the known variables\nP = 60 000\ni = 0,2\nn = 5\nWhen we calculate depreciation using the reducing-balance method:\n1. the depreciation amount changes for each year.\n2. the depreciation amount gets smaller each year.\n3. the book value at the end of a year becomes the principal amount for the next\nyear.\n4. the asset will always have some value (the book value will never equal zero).\nStep 2: Complete a table of values\nYear\nBook value\nDepreciation\nValue at end of\nyear\n1\nR 60 000\n60 000 × 0,2 = 12 000\nR 48 000\n2\nR 48 000\n48 000 × 0,2 = 9600\nR 38 400\n3\nR 38 400\n38 400 × 0,2 = 7680\nR 30 720\n4\nR 30 720\n30 720 × 0,2 = 6144\nR 24 576\n5\nR 24 576\n24 576 × 0,2 = 4915,20\nR 19 660,80\n382\n9.2.\nSimple and compound depreciation\n\nNotice in the example above that we could also write the book value at the end of\neach year as:\nBook value end of first year\n= 60 000(1 −0,2)\nBook value end of second year = 48 000(1 −0,2) = 60 000(1 −0,2)2\nBook value end of third year\n= 38 400(1 −0,2) = 60 000(1 −0,2)3\nBook value end of fourth year = 30 720(1 −0,2) = 60 000(1 −0,2)4\nBook value end of fifth year\n= 24 576(1 −0,2) = 60 000(1 −0,2)5\nUsing the formula for simple decay and the observed pattern in the calculation above,\nwe obtain the following formula for compound decay:\nA = P(1 −i)n\nwhere\nA = book value or depreciated value\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nAgain, notice the similarity to the compound interest formula A = P(1 + i)n.\nWorked example 7: Reducing-balance depreciation\nQUESTION\nThe number of pelicans at the Berg river mouth is decreasing at a compound rate of\n12% p.a. If there are currently 3200 pelicans in the wetlands of the Berg river mouth,\nwhat will the population be in 5 years?\nSOLUTION\nStep 1: Write down the known variables and the compound decay formula\nP = 3200\ni = 0,12\nn = 5\nA = P(1 −i)n\nStep 2: Substitute the values and solve for A\nA = 3200(1 −0,12)5\n= 3200(0,88)5\n= 1688,7421 . . .\nStep 3: Write the final answer\nIn 5 years, the pelican population will be approximately 1689.\n383\nChapter 9.\nFinance, growth and decay\n\nWorked example 8: Compound decay\nQUESTION\n1. A school buys a minibus for R 950 000, which depreciates at 13,5% per annum.\nDetermine the value of the minibus after 3 years if the depreciation is calculated:\na) on a straight-line basis.\nb) on a reducing-balance basis.\n2. Which is the better option?\nSOLUTION\nStep 1: Write down known variables\nP = 950 000\ni = 0,135\nn = 3\nStep 2: Use the simple decay formula and solve for A\nA = 950 000(1 −3 × 0,135)\n= 950 000(0,865)\n= 565 250\n∴A = R 565 250\nStep 3: Use the compound decay formula and solve for A\nA = 950 000(1 −0,135)3\n= 950 000(0,865)3\n= 614 853,89\n∴A = R 614 853,89\nStep 4: Interpret the answers\nAfter a period of 3 years, the value of the minibus calculated on the straight-line\nmethod is less than the value of the minibus calculated on the reducing-balance\nmethod. The value of the minibus depreciated less on the reducing-balance basis\nbecause the amount of depreciation is calculated on a smaller amount every year,\nwhereas the straight-line method is based on the full value of the minibus every year.\n384\n9.2.\nSimple and compound depreciation\n\nWorked example 9: Compound depreciation\nQUESTION\nFarmer Jack bought a tractor and it has depreciated by 20% p.a. on a reducing-balance\nbasis. If the current value of the tractor is R 52 429, calculate how much Farmer Jack\npaid for his tractor if he bought it 7 years ago.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 52 429\ni = 0,2\nn = 7\nA = P(1 −i)n\nStep 2: Substitute the values and solve for P\n52 429 = P(1 −0,2)7\n= P(0,8)7\n∴P = 52 429\n(0,8)7\n= 250 000,95 . . .\nStep 3: Write the final answer\n7 years ago, Farmer Jack paid R 250 000 for his tractor.\nExercise 9 – 3: Compound depreciation\n1. Jwayelani buys a truck for R 89 000 and depreciates it by 9% p.a. using the\ncompound depreciation method. What is the value of the truck after 14 years?\n2. The number of cormorants at the Amanzimtoti river mouth is decreasing at a\ncompound rate of 8% p.a. If there are now 10 000 cormorants, how many will\nthere be in 18 years’ time?\n3. On January 1, 2008 the value of my Kia Sorento is R 320 000. Each year after\nthat, the car’s value will decrease 20% of the previous year’s value. What is the\nvalue of the car on January 1, 2012?\n385\nChapter 9.\nFinance, growth and decay\n\n4. The population of Bonduel decreases at a reducing-balance rate of 9,5% per\nannum as people migrate to the cities. Calculate the decrease in population over\na period of 5 years if the initial population was 2 178 000.\n5. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses\n3% of its water each day. How much does it weigh after a month of 31 days?\n6. Richard bought a car 15 years ago and it depreciated by 17% p.a. on a com-\npound depreciation basis. How much did he pay for the car if it is now worth\nR 5256?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2389\n2. 238B\n3. 238C\n4. 238D\n5. 238F\n6. 238G\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFinding i\nEMBJJ\nWorked example 10: Finding i for simple decay\nQUESTION\nAfter 4 years, the value of a computer is halved. Assuming simple decay, at what\nannual rate did it depreciate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and simple decay formula\nLet the value of the computer be x, therefore:\nA = x\n2\nP = x\nn = 4\nA = P(1 −in)\nStep 2: Substitute the values and solve for i\n386\n9.2.\nSimple and compound depreciation\n\nx\n2 = x(1 −3i)\n1\n2 = 1 −3i\n∴3i = 1 −1\n2\n∴i = 0,1667\nStep 3: Write the final answer\nThe computer depreciated at a rate of 16,67% p.a.\nWorked example 11: Finding i for compound decay\nQUESTION\nCristina bought a fridge at the beginning of 2009 for R 8999 and sold it at the end\nof 2011 for R 4500. At what rate did the value of her fridge depreciate assuming a\nreducing-balance method? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 4500\nP = 8999\nn = 3\nA = P(1 −i)n\nStep 2: Substitute the values and solve for i\n4500 = 8999(1 −i)3\n4500\n8999 = (1 −i)3\n3\nr\n4500\n8999 = 1 −i\n∴i = 1 −\n3\nr\n4500\n8999\n= 0,206\nStep 3: Write the final answer\nCristina’s fridge depreciated at a rate of 20,6% p.a.\n387\nChapter 9.\nFinance, growth and decay\n\nExercise 9 – 4: Finding i\n1. A machine costs R 45 000 and has a scrap value of R 9000 after 10 years. Deter-\nmine the annual rate of depreciation if it is calculated on the reducing balance\nmethod.\n2. After 15 years, an aeroplane is worth 1\n6 of its original value. At what annual rate\nwas depreciation compounded?\n3. Mr. Mabula buys furniture for R 20 000. After 6 years he sells the furniture for\nR 9300. Calculate the annual compound rate of depreciation of the furniture.\n4. Ayanda bought a new car 7 years ago for double what it is worth today. At what\nyearly compound rate did her car depreciate?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238H\n2. 238J\n3. 238K\n4. 238M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.3\nTimelines\nEMBJK\nInterest can be compounded more than once a year. For example, an investment can\nbe compounded monthly or quarterly. Below is a table of compounding terms and\ntheir corresponding numeric value (p). When amounts are compounded more than\nonce per annum, we multiply the number of years by p and we also divide the interest\nrate by p.\nTerm\np\nyearly / annually\n1\nhalf-yearly / bi-annually\n2\nquarterly\n4\nmonthly\n12\nweekly\n52\ndaily\n365\nWorked example 12: Timelines\nQUESTION\nR 5500 is invested for a period of 4 years in a savings account. For the first year, the\ninvestment grows at a simple interest rate of 11% p.a. and then at a rate of 12,5%\np.a. compounded quarterly for the rest of the period. Determine the value of the\ninvestment at the end of the 4 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\n388\n9.3.\nTimelines\n\nT0\nT1\nT2\nT3\nT4\n11% p.a. simple interest\n12,5% p.a. compounded quarterly\nR 5500\nIn the timeline above, the intervals are given in years. For example, T0 is the start of\nthe investment, T1 is the end of the first year and T4 is the end of the fourth year.\nStep 2: Use the simple interest formula to calculate A at T1\nA = P(1 + in)\n= 5500(1 + 0,11)\n= R 6105\nStep 3: Use the compound interest formula to calculate A at T4\nThe investment is compounded quarterly, therefore:\nn = 3 × 4\n= 12\nand i = 0,125\n4\nAlso notice that the accumulated amount at the end of the first year becomes the\nprincipal amount at the beginning of the second year.\nA = P(1 + i)n\n= 6105\n\u0012\n1 + 0,125\n4\n\u001312\n= R 8831,88\nStep 4: Write the final answer\nThe value of the investment at the end of the 4 years is R 8831,88.\n389\nChapter 9.\nFinance, growth and decay\n\nWorked example 13: Timelines\nQUESTION\nR 150 000 is deposited in an investment account for a period of 6 years at an interest\nrate of 12% p.a. compounded half-yearly for the first 4 years and then 8,5% p.a.\ncompounded yearly for the rest of the period. A deposit of R 8000 is made into the\naccount after the first year and then another deposit of R 2000 is made 5 years after\nthe initial investment. Calculate the value of the investment at the end of the 6 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n8,5% p.a. compounded yearly\nR 15 000\nT5\nT6\n12% p.a. compounded half-yearly\n+R 8000\n+R 2000\nRemember to show when the additional deposits of R 8000 and R 2000 where made\ninto the account. It is very important to note that the interest rate changes at T4.\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nBetween T0 and T4:\nWe notice that interest for the first 4 years is compounded half-yearly, therefore:\nn1 = 4 × 2\n= 8\nand i1 = 0,12\n2\nBetween T4 and T6:\nn2 = 2\nand i2 = 0,085\nTherefore the total growth of the initial deposit over the 6 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\nStep 3: The deposit at T1\nBetween T1 and T4:\n390\n9.3.\nTimelines\n\nInterest on this deposit is compounded half-yearly for 3 years, therefore:\nn3 = 3 × 2\n= 6\nand i3 = 0,12\n2\nBetween T4 and T6:\nn4 = 2\nand i4 = 0,085\nTherefore the total growth of the deposit over the 5 years is:\nA = P(1 + i3)n3(1 + i4)n4\n= 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2\nStep 4: The deposit at T5\nAccumulate interest for only 1 year:\nA = P(1 + i)n\n= 2000(1 + 0,085)1\nStep 5: Determine the total calculation\nTo get as accurate an answer as possible, we do the the calculation on the calculator\nin one step. Using the memory and answer recall function on the calculator, we avoid\nrounding off until we get the final answer.\nA = 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\n+ 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2 + 2000(1 + 0,085)1\n= R 296 977,00\nStep 6: Write the final answer\nThe value of the investment at the end of the 6 years is R 296 977,00.\n391\nChapter 9.\nFinance, growth and decay\n\nWorked example 14: Timelines\nQUESTION\nR 60 000 is invested in an account which offers interest at 7% p.a.\ncompounded\nquarterly for the first 18 months. Thereafter the interest rate changes to 5% p.a. com-\npounded monthly. Three years after the initial investment, R 5000 is withdrawn from\nthe account. How much will be in the account at the end of 5 years?\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n5% p.a. compounded monthly\nR 60 000\nT5\n7% p.a. compounded quarterly\n−R 5000\nRemember to show when the withdrawal of R 5000 was taken out of the account. It is\nalso important to note that the interest rate changes after 18 months (T1 1\n2 ).\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nInterest for the first 1,5 years is compounded quarterly, therefore:\nn1 = 1,5 × 4\n= 6\nand i1 = 0,07\n4\nInterest for the remaining 3,5 years is compounded monthly, therefore:\nn2 = 3,5 × 12\n= 42\nand i2 = 0,05\n12\nTherefore the total growth of the initial deposit over the 5 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n392\n9.3.\nTimelines\n\nStep 3: The withdrawal at T3\nWe calculate the interest that the R 5000 would have earned if it had remained in the\naccount:\nn = 2 × 12\n= 24\nand i = 0,05\n12\nTherefore we have that:\nA = P(1 + i)n\n= 5000\n\u0012\n1 + 0,05\n12\n\u001324\nStep 4: Determine the total calculation\nWe subtract the withdrawal and the interest it would have earned from the accumu-\nlated amount at the end of the 5 years:\nA = 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n−5000\n\u0012\n1 + 0,05\n12\n\u001324\n= R 73 762,19\nStep 5: Write the final answer\nThe value of the investment at the end of the 5 years is R 73 762,19.\nExercise 9 – 5: Timelines\n1. After a 20-year period Josh’s lump sum investment matures to an amount of\nR 313 550. How much did he invest if his money earned interest at a rate of\n13,65% p.a. compounded half yearly for the first 10 years, 8,4% p.a. com-\npounded quarterly for the next five years and 7,2% p.a. compounded monthly\nfor the remaining period?\n2. Sindisiwe wants to buy a motorcycle. The cost of the motorcycle is R 55 000.\nIn 1998 Sindisiwe opened an account at Sutherland Bank with R 16 000. Then\nin 2003 she added R 2000 more into the account. In 2007 Sindisiwe made\nanother change: she took R 3500 from the account. If the account pays 6% p.a.\ncompounded half-yearly, will Sindisiwe have enough money in the account at\nthe end of 2012 to buy the motorcycle?\n3. A loan has to be returned in two equal semi-annual instalments. If the rate of\ninterest is 16% per annum, compounded semi-annually and each instalment is\nR 1458, find the sum borrowed.\n393\nChapter 9.\nFinance, growth and decay\n\n4. A man named Phillip invests R 10 000 into an account at North Bank at an\ninterest rate of 7,5% p.a. compounded monthly. After 5 years the bank changes\nthe interest rate to 8% p.a. compounded quarterly. How much money will\nPhillip have in his account 9 years after the original deposit?\n5. R 75 000 is invested in an account which offers interest at 11% p.a.\ncom-\npounded monthly for the first 24 months.\nThen the interest rate changes to\n7,7% p.a. compounded half-yearly. If R 9000 is withdrawn from the account\nafter one year and then a deposit of R 3000 is made three years after the initial\ninvestment, how much will be in the account at the end of 6 years?\n6. Christopher wants to buy a computer, but right now he doesn’t have enough\nmoney. A friend told Christopher that in 5 years the computer will cost R 9150.\nHe decides to start saving money today at Durban United Bank. Christopher\ndeposits R 5000 into a savings account with an interest rate of 7,95% p.a. com-\npounded monthly.\nThen after 18 months the bank changes the interest rate\nto 6,95% p.a. compounded weekly. After another 6 months, the interest rate\nchanges again to 7,92% p.a.\ncompounded two times per year.\nHow much\nmoney will Christopher have in the account after 5 years, and will he then have\nenough money to buy the computer?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238N\n2. 238P\n3. 238Q\n4. 238R\n5. 238S\n6. 238T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.4\nNominal and effective interest rates\nEMBJM\nWe have seen that although interest is quoted as a percentage per annum it can be\ncompounded more than once a year. We therefore need a way of comparing interest\nrates. For example, is an annual interest rate of 8% compounded quarterly higher or\nlower than an interest rate of 8% p.a. compounded yearly?\nInvestigation: Nominal and effective interest rates\n1. Calculate the accumulated amount at the end of one year if R 1000 is invested\nat 8% p.a. compound interest:\nA = P(1 + i)n\n= . . . . . .\n2. Calculate the value of R 1000 if it is invested for one year at 8% p.a. com-\npounded:\n394\n9.4.\nNominal and effective interest rates\n\nFrequency\nCalculation\nAccumulated\namount\nInterest\namount\nhalf-yearly\nA = 1000\n\u0010\n1 + 0,08\n2\n\u00111×2\nR 1081,60\nR 81,60\nquarterly\nmonthly\nweekly\ndaily\n3. Use your results from the table above to calculate the effective rate that the\ninvestment of R 1000 earns in one year:\nFrequency\nAccumulated\namount\nCalculation\nEffective\ninterest\nrate\nhalf-yearly\nR 1081,60\n1081,60 = 1000(1 + i)\n1081,60\n1000\n= 1 + i\n1081,60\n1000\n−1 = i\n∴i = 0,0816\ni = 8,16%\nquarterly\nmonthly\nweekly\ndaily\n4. If you wanted to borrow R 10 000 from the bank, would it be better to pay it\nback at an interest rate of 22% p.a. compounded quarterly or 22% compounded\nmonthly? Show your calculations.\nAn interest rate compounded more than once a year is called the nominal interest rate.\nIn the investigation above, we determined that the nominal interest rate of 8% p.a.\ncompounded half-yearly is actually an effective rate of 8,16% p.a.\nGiven a nominal interest rate i(m) compounded at a frequency of m times per year\nand the effective interest rate i, the accumulated amount calculated using both interest\nrates will be equal so we can write:\nP(1 + i) = P\n \n1 + i(m)\nm\n!m\n∴1 + i =\n \n1 + i(m)\nm\n!m\n395\nChapter 9.\nFinance, growth and decay\n\nWorked example 15: Nominal and effective interest rates\nQUESTION\nInterest on a credit card is quoted as 23% p.a. compounded monthly. What is the\neffective annual interest rate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down the known variables\nInterest is being added monthly, therefore:\nm = 12\ni(12) = 0,23\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i\n1 + i =\n\u0012\n1 + 0,23\n12\n\u001312\n∴i = 1 −\n\u0012\n1 + 0,23\n12\n\u001312\n= 25,59%\nStep 3: Write the final answer\nThe effective interest rate is 25,59% per annum.\nWorked example 16: Nominal and effective interest rates\nQUESTION\nDetermine the nominal interest rate compounded quarterly if the effective interest rate\nis 9% per annum (correct to two decimal places).\nSOLUTION\nStep 1: Write down the known variables\n396\n9.4.\nNominal and effective interest rates\n\nInterest is being added quarterly, therefore:\nm = 4\ni = 0,09\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i(m)\n1 + 0,09 =\n \n1 + i(4)\n4\n!4\n4p\n1,09 = 1 + i(4)\n4\n4p\n1,09 −1 = i(4)\n4\n4\n\u0010\n4p\n1,09 −1\n\u0011\n= i(4)\n∴i(4) = 8,71%\nStep 3: Write the final answer\nThe nominal interest rate is 8,71% p.a. compounded quarterly.\nExercise 9 – 6: Nominal and effect interest rates\n1. Determine the effective annual interest rate if the nominal interest rate is:\na) 12% p.a. compounded quarterly.\nb) 14,5% p.a. compounded weekly.\nc) 20% p.a. compounded daily.\n2. Consider the following:\n• 16,8% p.a. compounded annually.\n• 16,4% p.a. compounded monthly.\n• 16,5% p.a. compounded quarterly.\na) Determine the effective annual interest rate of each of the nominal rates\nlisted above.\nb) Which is the best interest rate for an investment?\nc) Which is the best interest rate for a loan?\n397\nChapter 9.\nFinance, growth and decay\n\n3. Calculate the effective annual interest rate equivalent to a nominal interest rate\nof 8,75% p.a. compounded monthly.\n4. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every\nfour months on her investment of R 85 000.\nCalculate the effective rate per\nannum.\n5. Determine which of the following would be the better agreement for paying back\na student loan:\na) 9,1% p.a. compounded quarterly.\nb) 9% p.a. compounded monthly.\nc) 9,3% p.a. compounded half-yearly.\n6. Miranda invests R 8000 for 5 years for her son’s study fund. Determine how\nmuch money she will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 6% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 238V\n1b. 238W\n1c. 238X\n2. 238Y\n3. 238Z\n4. 2392\n5. 2393\n6. 2394\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.5\nSummary\nEMBJN\nSee presentation: 2395 at www.everythingmaths.co.za\n• Simple interest: A = P(1 + in)\n• Compound interest: A = P(1 + i)n\n• Simple depreciation: A = P(1 −in)\n• Compound depreciation: A = P(1 −i)n\n• Nominal and effective annual interest rates: 1 + i =\n\u0010\n1 + i(m)\nm\n\u0011m\n398\n9.5.\nSummary\n\nExercise 9 – 7: End of chapter exercises\n1. Thabang buys a Mercedes worth R 385 000 in 2007. What will the value of the\nMercedes be at the end of 2013 if:\na) the car depreciates at 6% p.a. straight-line depreciation.\nb) the car depreciates at 6% p.a. reducing-balance depreciation.\n2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R 8900.\nThe interest rate is quoted as 11% per annum based on simple interest. Calculate\nthe required monthly payment for this contract.\n3. A computer is purchased for R 16 000. It depreciates at 15% per annum.\na) Determine the book value of the computer after 3 years if depreciation is\ncalculated according to the straight-line method.\nb) Find the rate according to the reducing-balance method that would yield,\nafter 3 years, the same book value as calculated in the previous question.\n4. Maggie invests R 12 500 for 5 years at 12% per annum compounded monthly\nfor the first 2 years and 14% per annum compounded semi-annually for the next\n3 years. How much will Maggie receive in total after 5 years?\n5. Tintin invests R 120 000. He is quoted a nominal interest rate of 7,2% per an-\nnum compounded monthly.\na) Calculate the effective rate per annum (correct to two decimal places).\nb) Use the effective rate to calculate the value of Tintin’s investment if he\ninvested the money for 3 years.\nc) Suppose Tintin invests his money for a total period of 4 years, but after 18\nmonths makes a withdrawal of R 20 000, how much will he receive at the\nend of the 4 years?\n6. Ntombi opens accounts at a number of clothing stores and spends freely. She\ngets herself into terrible debt and she cannot pay off her accounts. She owes\nFashion World R 5000 and the shop agrees to let her pay the bill at a nominal\ninterest rate of 24% compounded monthly.\na) How much money will she owe Fashion World after two years?\nb) What is the effective rate of interest that Fashion World is charging her?\n7. John invests R 30 000 in the bank for a period of 18 months. Calculate how\nmuch money he will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 8% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\ndaily\n399\nChapter 9.\nFinance, growth and decay\n\n8. Convert an effective annual interest rate of 11,6% p.a. to a nominal interest rate\ncompounded:\na) half-yearly\nb) quarterly\nc) monthly\n9. Joseph must sell his plot on the West Coast and he needs to get R 300 000 on the\nsale of the land. If the estate agent charges him 7% commission on the selling\nprice, what must the buyer pay for the plot?\n10. Mrs. Brown retired and received a lump sum of R 200 000. She deposited the\nmoney in a fixed deposit savings account for 6 years. At the end of the 6 years\nthe value of the investment was R 265 000. If the interest on her investment was\ncompounded monthly, determine:\na) the nominal interest rate per annum\nb) the effective annual interest rate\n11. R 145 000 is invested in an account which offers interest at 9% p.a.\ncom-\npounded half-yearly for the first 2 years. Then the interest rate changes to 4%\np.a. compounded quarterly. Four years after the initial investment, R 20 000 is\nwithdrawn. 6 years after the initial investment, a deposit of R 15 000 is made.\nDetermine the balance of the account at the end of 8 years.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2396\n2. 2397\n3. 2398\n4. 2399\n5. 239B\n6. 239C\n7. 239D\n8. 239F\n9. 239G\n10. 239H\n11. 239J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n400\n9.5.\nSummary\n\nCHAPTER\n10\nProbability\n10.1\nRevision\n402\n10.2\nDependent and independent events\n411\n10.3\nMore Venn diagrams\n419\n10.4\nTree diagrams\n426\n10.5\nContingency tables\n431\n10.6\nSummary\n435\n\n10\nProbability\n10.1\nRevision\nEMBJP\nTerminology\nEMBJQ\nOutcome: a single observation of an uncertain or random process (called an experi-\nment). For example, when you accidentally drop a book, it might fall on its cover, on\nits back or on its side. Each of these options is a possible outcome.\nSample space of an experiment: the set of all possible outcomes of the experiment. For\nexample, the sample space when you roll a single 6-sided die is the set {1; 2; 3; 4; 5; 6}.\nFor a given experiment, there is exactly one sample space. The sample space is de-\nnoted by the letter S.\nEvent: a set of outcomes of an experiment. For example, during radioactive decay of\n1 gramme of uranium-234, one possible event is that the number of alpha-particles\nemitted during 1 microsecond is between 225 and 235.\nProbability of an event: a real number between 0 and 1 that describes how likely it\nis that the event will occur. A probability of 0 means the outcome of the experiment\nwill never be in the event set. A probability of 1 means the outcome of the experiment\nwill always be in the event set. When all possible outcomes of an experiment have\nequal chance of occurring, the probability of an event is the number of outcomes in\nthe event set as a fraction of the number of outcomes in the sample space.\nRelative frequency of an event: the number of times that the event occurs during\nexperimental trials, divided by the total number of trials conducted. For example, if\nwe flip a coin 10 times and it landed on heads 3 times, then the relative frequency of\nthe heads event is 3\n10 = 0,3.\nUnion of events: the set of all outcomes that occur in at least one of the events. For\n2 events called A and B, we write the union as “A or B”. Another way of writing the\nunion is using set notation: A ∪B.\nIntersection of events: the set of all outcomes that occur in all of the events. For 2\nevents called A and B, we write the intersection as “A and B”. Another way of writing\nthe intersection is using set notation: A ∩B.\nMutually exclusive events: events with no outcomes in common, that is (A and B) =\n∅. Mutually exclusive events can never occur simultaneously. For example the event\nthat a number is even and the event that the same number is odd are mutually exclu-\nsive, since a number can never be both even and odd.\nComplementary events: two mutually exclusive events that together contain all the\noutcomes in the sample space. For an event called A, we write the complement as\n“not A”. Another way of writing the complement is as A′.\nSee video: 239K at www.everythingmaths.co.za\n402\n10.1.\nRevision\n\nIdentities\nEMBJR\nThe addition rule (also called the sum rule) for any 2 events, A and B is\nP(A or B) = P(A) + P(B) −P(A and B)\nThis rule relates the probabilities of 2 events with the probabilities of their union and\nintersection.\nThe addition rule for 2 mutually exclusive events is\nP(A or B) = P(A) + P(B)\nThis rule is a special case of the previous rule. Because the events are mutually exclu-\nsive, P(A and B) = 0.\nThe complementary rule is\nP(not A) = 1 −P(A)\nThis rule is a special case of the previous rule. Since A and (not A) are mutually\nexclusive, P(A or (not A)) = 1.\nSee video: 239M at www.everythingmaths.co.za\nWorked example 1: Events\nQUESTION\nYou take all the hearts from a deck of cards. You then select a random card from the set\nof hearts. What is the sample space? What is the probability of each of the following\nevents?\n1. The card is the ace of hearts.\n2. The card has a prime number on it.\n3. The card has a letter of the alphabet on it.\nSOLUTION\nStep 1: Write down the sample space\nSince we are considering only one suit from the deck of cards (the hearts), we need to\nwrite down only the letters and numbers on the cards. Therefore the sample space is\nS = {A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K}\nStep 2: Write down the event sets\n• ace of hearts: {A}\n• prime number: {2; 3; 5; 7}\n• letter of alphabet: {A; J; Q; K}\n403\nChapter 10.\nProbability\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are 13 elements in the\nsample space. So the probability of each event is\n• ace of hearts:\n1\n13\n• prime number:\n4\n13\n• letter of alphabet:\n4\n13\nWorked example 2: Events\nQUESTION\nYou roll two 6-sided dice. Let E be the event that the total number of dots on the dice\nis 10. Let F be the event that at least one die is a 3.\n1. Write down the event sets for E and F.\n2. Determine the probabilities for E and F.\n3. Are E and F mutually exclusive? Why or why not?\nSOLUTION\nStep 1: Write down the sample space\nThe sample space of a single 6-sided die is just {1; 2; 3; 4; 5; 6}. To get the sample\nspace of two 6-sided dice, we have to take every possible pair of numbers from 1 to 6.\nS =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(1; 1)\n(1; 2)\n(1; 3)\n(1; 4)\n(1; 5)\n(1; 6)\n(2; 1)\n(2; 2)\n(2; 3)\n(2; 4)\n(2; 5)\n(2; 6)\n(3; 1)\n(3; 2)\n(3; 3)\n(3; 4)\n(3; 5)\n(3; 6)\n(4; 1)\n(4; 2)\n(4; 3)\n(4; 4)\n(4; 5)\n(4; 6)\n(5; 1)\n(5; 2)\n(5; 3)\n(5; 4)\n(5; 5)\n(5; 6)\n(6; 1)\n(6; 2)\n(6; 3)\n(6; 4)\n(6; 5)\n(6; 6)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nStep 2: Write down the events\nFor E the dice have to add to 10.\nE = {(4; 6); (5; 5); (6; 4)}\nFor F at least one die has to be 3.\nF = {(1; 3); (3; 1); (2; 3); (3; 2); (3; 3); (4; 3); (3; 4); (5; 3); (3; 5); (6; 3); (3; 6)}\n404\n10.1.\nRevision\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are\n• 6 × 6 = 36 outcomes in the sample space, S;\n• 3 outcomes in event E; and\n• 11 outcomes in event F.\nTherefore\nP(E) = 3\n36 = 1\n12\nand\nP(F) = 11\n36\nStep 4: Are they mutually exclusive\nTo test whether two events are mutually exclusive, we have to test whether their in-\ntersection is empty. Since E has no outcomes that contain a 3 on one of the dice,\nthe intersection of E and F is empty: (E and F) = ∅. This means that the events are\nmutually exclusive.\nSee video: 239N at www.everythingmaths.co.za\nExercise 10 – 1: Revision\n1. A bag contains r red balls, b blue balls and y yellow balls. What is the probability\nthat a ball drawn from the bag at random is yellow?\n2. A packet has yellow and pink sweets. The probability of taking out a pink sweet\nis 7\n12. What is the probability of taking out a yellow sweet?\n3. You flip a coin 4 times. What is the probability that you get 2 heads and 2 tails?\nWrite down the sample space and the event set to determine the probability of\nthis event.\n4. In a class of 37 children, 15 children walk to school, 20 children have pets at\nhome and 12 children who have a pet at home also walk to school. How many\nchildren walk to school and do not have a pet at home?\n5. You roll two 6-sided dice and are interested in the following two events:\n• A: the sum of the dice equals 8\n• B: at least one of the dice shows a 1\nShow that these events are mutually exclusive.\n405\nChapter 10.\nProbability\n\n6. You ask a friend to think of a number from 1 to 100. You then ask her the\nfollowing questions:\n• Is the number even?\n• Is the number divisible by 7?\nHow many possible numbers are less than 80 if she answered “yes” to both\nquestions?\n7. In a group of 42 pupils, all but 3 had a packet of chips or a Fanta or both. If 23\nhad a packet of chips and 7 of these also had a Fanta, what is the probability that\none pupil chosen at random has:\na) both chips and Fanta\nb) only Fanta\n8. Tamara has 18 loose socks in a drawer. Eight of these are orange and two are\npink. Calculate the probability that the first sock taken out at random is:\na) orange\nb) not orange\nc) pink\nd) not pink\ne) orange or pink\nf) neither orange nor pink\n9. A box contains coloured blocks. The number of blocks of each colour is given\nin the following table.\nColour\nPurple\nOrange\nWhite\nPink\nNumber of blocks\n24\n32\n41\n19\nA block is selected randomly. What is the probability that the block will be:\na) purple\nb) purple or white\nc) pink and orange\nd) not orange?\n10. The surface of a soccer ball is made up of 32 faces. 12 faces are regular pen-\ntagons, each with a surface area of about 37 cm2. The other 20 faces are regular\nhexagons, each with a surface area of about 56 cm2.\nYou roll the soccer ball. What is the probability that it stops with a pentagon\ntouching the ground?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 239P\n2. 239Q\n3. 239R\n4. 239S\n5. 239T\n6. 239V\n7. 239W\n8. 239X\n9. 239Y\n10. 239Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n406\n10.1.\nRevision\n\nVenn diagrams\nEMBJS\nA Venn diagram is used to show how events are related to one another.\nA Venn\ndiagram can be very helpful when doing calculations with probabilities. In a Venn\ndiagram each event is represented by a shape, often a circle or a rectangle. The region\ninside the shape represents the outcomes included in the event and the region outside\nthe shape represents the outcomes that are not in the event.\nS\nA\nB\nA and B\nA Venn diagram representing a sample space, S, as a square; and two events, A and\nB, as circles. The intersection of the two circles contains outcomes that are in both A\nand B.\nVenn diagrams can be used in slightly different ways and it is important to notice the\ndifferences between them. The following 3 examples show how a Venn diagram is\nused to represent\n• the outcomes included in each event;\n• the number of outcomes in each event; and\n• the probability of each event.\nWorked example 3: Venn diagrams with outcomes\nQUESTION\nChoose a number between 1 and 20. Draw a Venn diagram to answer the following\nquestions.\n1. What is the probability that the number is a multiple of 3?\n2. What is the probability that the number is a multiple of 5?\n3. What is the probability that the number is a multiple of 3 or 5?\n4. What is the probability that the number is a multiple of 3 and 5?\nSOLUTION\nStep 1: Draw a Venn diagram\nThe Venn diagram should show the sample space of all numbers from 1 to 20. It should\nalso show an event set that contains all the multiples of 3, let A = {3; 6; 9; 12; 15; 18},\n407\nChapter 10.\nProbability\n\nand another event set that contains all the multiples of 5, let B = {5; 10; 15; 20}. Note\nthat there is one shared outcome between these two events, namely 15.\n3\n18\n15\n12\n9\n6\n10\n20\n5\n1\n2\n4\n7\n8\n11\n13\n14\n16\n17\n19\nStep 2: Compute probabilities\nThe probability of an event is the number of outcomes in the event set divided by the\nnumber of outcomes in the sample space. There are 20 outcomes in the sample space.\n1. Since there are 6 outcomes in the multiples of 3 event set, the probability of a\nmultiple of 3 is P(A) = 6\n20 = 3\n10.\n2. Since there are 4 outcomes in the multiples of 5 event set, the probability of a\nmultiple of 5 is P(B) = 4\n20 = 1\n5.\n3. The event that the number is a multiple of 3 or 5 is the union of the above two\nevent sets. There are 9 elements in the union of the event sets, so the probability\nis 9\n20.\n4. The event that the number is a multiple of 3 and 5 is the intersection of the\ntwo event sets. There is 1 element in the intersection of the event sets, so the\nprobability is 1\n20.\nWorked example 4: Venn diagrams with counts\nQUESTION\nIn a group of 50 learners, 35 take Mathematics and 30 take History, while 12 take\nneither of the two subjects. Draw a Venn diagram representing this information. If a\nlearner is chosen at random from this group, what is the probability that he takes both\nMathematics and History?\nSOLUTION\nStep 1: Draw outline of Venn diagram\nThere are 2 events in this question, namely\n• M: that a learner takes Mathematics; and\n• H: that a learner takes History.\n408\n10.1.\nRevision\n\nWe need to do some calculations before drawing the full Venn diagram, but with the\ninformation above we can already draw the outline.\nS\nM\nH\nStep 2: Write down sizes of the event sets, their union and intersection\nWe are told that 12 learners take neither of the two subjects. Graphically we can\nrepresent this as:\nS\nM\nH\n12\nSince there are 50 elements in the sample space, we can see from this figure that there\nare 50 −12 = 38 elements in (M or H). So far we know\n• n(M) = 35\n• n(H) = 30\n• n(M or H) = 38\nFrom the addition rule,\nn(M or H) = n(M) + n(H) −n(M and H)\n∴n(M and H) = 35 + 30 −38\n= 27\nStep 3: Draw the final Venn diagram\nS\nM\nH\n12\n27\n8\n3\n409\nChapter 10.\nProbability\n\nWorked example 5: Venn diagrams with probabilities\nQUESTION\nDraw a Venn diagram to represent the same information as in the previous example,\nexcept showing the probabilities of the different events, rather than the counts.\nIf a learner is chosen at random from this group, what is the probability that she takes\nboth Mathematics and History?\nSOLUTION\nStep 1: Use counts to compute probabilities\nSince there are 50 elements (learners) in the sample space, we can compute the prob-\nability of any event by dividing the size of the event set by 50. This gives the following\nprobabilities:\n• P(M) = 35\n50 = 7\n10\n• P(H) = 30\n50 = 3\n5\n• P(M or H) = 38\n50 = 19\n25\n• P(M and H) = 27\n50\nStep 2: Draw the Venn diagram\nNext we replace each count from the Venn diagram in the previous example with a\nprobability.\nS\nM\nH\n6\n25\n27\n50\n4\n25\n3\n50\nStep 3: Find the answer\nThe probability that a random learner will take both Mathematics and History is\nP(M and H) = 27\n50.\nSee video: 23B2 at www.everythingmaths.co.za\n410\n10.1.\nRevision\n\nExercise 10 – 2: Venn diagram revision\n1. Given the following information:\n• P(A) = 0,3\n• P(B and A) = 0,2\n• P(B) = 0,7\nFirst draw a Venn diagram to represent this information. Then compute the value\nof P(B and (not A)).\n2. You are given the following information:\n• P(A) = 0,5\n• P(A and B) = 0,2\n• P(not B) = 0,6\nDraw a Venn diagram to represent this information and determine P(A or B).\n3. A study was undertaken to see how many people in Port Elizabeth owned either\na Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60%\nowned a Volkswagen. What percentage of people owned neither car?\n4. Let S denote the set of whole numbers from 1 to 15, X denote the set of even\nnumbers from 1 to 15 and Y denote the set of prime numbers from 1 to 15.\nDraw a Venn diagram depicting S, X and Y .\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B3\n2. 23B4\n3. 23B5\n4. 23B6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.2\nDependent and independent events\nEMBJT\nSometimes the presence or absence of one event tells us something about other events.\nWe call events dependent if knowing whether one of them happened tells us some-\nthing about whether the others happened. Independent events give us no information\nabout one another; the probability of one event occurring does not affect the probabil-\nity of the other events occurring.\nDEFINITION: Independent events\nTwo events, A and B are independent if and only if\nP(A and B) = P(A) × P(B)\nAt first it might not be clear why we should call events that satisfy the equation above\nindependent. We will explore this further using a number of examples.\n411\nChapter 10.\nProbability\n\nInvestigation: Independence\nRoll a single 6-sided die and consider the following two events:\n• E: you get an even number\n• T: you get a number that is divisible by three\nNow answer the following questions:\n• What is the probability of E?\n• What is the probability of getting an even number if you are told that the number\nwas also divisible by three?\n• Does knowing that the number was divisible by three change the probability that\nthe number was even?\nAre the events E and T dependent or independent according to the definition (hint:\ncompute the probabilities in the definition of independence)?\nSee video: 23B7 at www.everythingmaths.co.za\nSo, why do we call it independence when P(A and B) = P(A) × P(B)? For two\nevents, A and B, independence means that knowing the outcome of B does not affect\nthe probability of A.\nConsider the following Venn diagram.\nS\nA\nB\nA and B\nThe probability of A is the ratio between the number of outcomes in A and the number\nof outcomes in the sample space, S.\nP(A) = n(A)\nn(S)\n412\n10.2.\nDependent and independent events\n\nNow, let’s say that we know that event B happened. How does this affect the proba-\nbility of A? Here is how the Venn diagram changes:\nS\nA\nB\nA and B\nA lot of the possible outcomes (all of the outcomes outside B) are now out of the pic-\nture, because we know that they did not happen. Now the probability of A happening,\ngiven that we know that B happened, is the ratio between the size of the region where\nA is present (A and B) and the size of all possible events (B).\nP(A if we know B) = n(A and B)\nn(B)\nIf P(A) = P(A if we know B) we call them independent, because knowing B does\nnot change the probability of A.\nWith some algebra, we can prove that this statement of independence is the same\nas the definition of independence that we saw at the beginning of this section. For\nindependent events\nP(A and B) = P(A) × P(B)\nThis is equivalent to\nP(A) = P(A and B) ÷ P(B)\n= n(A and B)\nn(S)\n÷ n(B)\nn(S)\n= n(A and B)\nn(B)\n= P(A if we know B)\nThat is why we call events independent!\n(For enrichment only):\nThe ratio\nP(A and B)\nP(B)\nis called a conditional probability and written using the notation P(A | B). This\nnotation is read as “the probability of A given B.”\nIf (and only if) A and B are independent: P(A | B) = P(A) and P(B | A) = P(B).\nTry to prove this using the definition of independence.\n413\nChapter 10.\nProbability\n\nWorked example 6: Independent and dependent events\nQUESTION\nA bag contains 5 red and 5 blue balls. We remove a random ball from the bag, record\nits colour and put it back into the bag. We then remove another random ball from the\nbag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Probability of a red ball first\nSince there are a total of 10 balls, of which 5 are red, the probability of getting a red\nball is\nP(first ball red) = 5\n10 = 1\n2\nStep 2: Probability of a blue ball second\nThe problem states that the first ball is placed back into the bag before we take the\nsecond ball. This means that when we draw the second ball, there are again a total of\n10 balls in the bag, of which 5 are blue. Therefore the probability of drawing a blue\nball is\nP(second ball blue) = 5\n10 = 1\n2\nStep 3: Probability of red first and blue second\nWhen drawing two balls from the bag, there are 4 possibilities. We can get\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nWe want to know the probability of the second outcome, where we have to get a red\nball first. Since there are 5 red balls and 10 balls in total, there are\n5\n10 ways to get a\nred ball first. Now we put the first ball back, so there are again 5 red balls and 5 blue\nballs in the bag. Therefore there are\n5\n10 ways to get a blue ball second if the first ball\nwas red. This means that there are\n5\n10 × 5\n10 = 25\n100\n414\n10.2.\nDependent and independent events\n\nways to get a red ball first and a blue ball second. So, the probability of getting a red\nball first and a blue ball second is 1\n4.\nStep 4: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 1\n4\nSince 1\n4 = 1\n2 × 1\n2, the events are independent.\nSee video: 23B8 at www.everythingmaths.co.za\nWorked example 7: Independent and dependent events\nQUESTION\nIn the previous example, we picked a random ball and put it back into the bag before\ncontinuing. This is called sampling with replacement. In this example, we will follow\nthe same process, except that we will not put the first ball back into the bag. This is\ncalled sampling without replacement.\nSo, from a bag with 5 red and 5 blue balls, we remove a random ball and record its\ncolour. Then, without putting back the first ball, we remove another random ball from\nthe bag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Count the number of outcomes\nWe will look directly at the number of possible ways in which we can get the 4 possible\noutcomes when removing 2 balls. In the previous example, we saw that the 4 possible\noutcomes are\n415\nChapter 10.\nProbability\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nFor the first outcome, we have to get a red ball first. Since there are 5 red balls and\n10 balls in total, there are\n5\n10 ways to get a red ball first. After we have taken out a red\nball, there are now 4 red balls and 5 blue balls left. Therefore there are 4\n9 ways to get\na red ball second if the first ball was also red. This means that there are\n5\n10 × 4\n9 = 20\n90\nways to get a red ball first and a red ball second. The probability of the first outcome\nis 2\n9.\nFor the second outcome, we have to get a red ball first. As in the first outcome, there\nare\n5\n10 ways to get a red ball first; and there are now 4 red balls and 5 blue balls left.\nTherefore there are 5\n9 ways to get a blue ball second if the first ball was red. This means\nthat there are\n5\n10 × 5\n9 = 25\n90\nways to get a red ball first and a blue ball second. The probability of the second\noutcome is 5\n18.\nWe can compute the probabilities of the third and fourth outcomes in the same way as\nthe first two, but there is an easier way. Notice that there are only 2 types of ball and\nthat there are exactly equal numbers of them at the start. This means that the problem\nis completely symmetric in red and blue. We can use this symmetry to compute the\nprobabilities of the other two outcomes.\nIn the third outcome, the first ball is blue and the second ball is red. Because of\nsymmetry this outcome must have the same probability as the second outcome (when\nthe first ball is red and the second ball is blue). Therefore the probability of the third\noutcome is 5\n18.\nIn the fourth outcome, the first and second balls are both blue. From symmetry, this\noutcome must have the same probability as the first outcome (when both balls are red).\nTherefore the probability of the fourth outcome is 2\n9.\nTo summarise, these are the possible outcomes and their probabilities:\n• first ball red and second ball red: 2\n9;\n• first ball red and second ball blue:\n5\n18;\n• first ball blue and second ball red:\n5\n18;\n• first ball blue and second ball blue: 2\n9.\nStep 2: Probability of a red ball first\nTo determine the probability of getting a red ball on the first draw, we look at all of the\noutcomes that contain a red ball first. These are\n416\n10.2.\nDependent and independent events\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball.\nThe probability of the first outcome is 2\n9 and the probability of the second outcome is\n5\n18. By adding these two probabilities, we see that the probability of getting a red ball\nfirst is\nP(first ball red) = 2\n9 + 5\n18 = 1\n2\nThis is the same as in the previous exercise, which should not be too surprising since\nthe probability of the first ball being red is not affected by whether or not we put it\nback into the bag before drawing the second ball.\nStep 3: Probability of a blue ball second\nTo determine the probability of getting a blue ball on the second draw, we look at all\nof the outcomes that contain a blue ball second. These are\n• a red ball and then a blue ball;\n• a blue ball and then another blue ball.\nThe probability of the first outcome is 5\n18 and the probability of the second outcome is\n2\n9. By adding these two probabilities, we see that the probability of getting a blue ball\nsecond is\nP(second ball blue) = 5\n18 + 2\n9 = 1\n2\nThis is also the same as in the previous exercise! You might find it surprising that the\nprobability of the second ball is not affected by whether or not we replace the first ball.\nThe reason why this probability is still 1\n2 is that we are computing the probability that\nthe second ball is blue without knowing the colour of the first ball. Because there are\nonly two equal possibilities for the second ball (red and blue) and because we don’t\nknow whether the first ball is red or blue, there is an equal chance that the second ball\nwill be one colour or the other.\nStep 4: Probability of red first and blue second\nWe have already calculated the probability that the first ball is red and the second ball\nis blue. It is 5\n18.\nStep 5: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 5\n18\nSince 5\n18 ̸= 1\n2 × 1\n2, the events are dependent.\n417\nChapter 10.\nProbability\n\nWARNING!\nJust because two events are mutually exclusive does not necessarily mean that they\nare independent. To test whether events are mutually exclusive, always check that\nP(A and B) = 0. To test whether events are independent, always check that P(A and B) =\nP(A) × P(B). See the exercises below for examples of events that are mutually ex-\nclusive and independent in different combinations.\nExercise 10 – 3: Dependent and independent events\n1. Use the following Venn diagram to determine whether events X and Y are\na) mutually exclusive or not mutually exclusive;\nb) dependent or independent.\nS\nX\nY\n11\n7\n3\n14\n2. Of the 30 learners in a class 17 have black hair, 11 have brown hair and 2 have\nred hair. A learner is selected from the class at random.\na) What is the probability that the learner has black hair?\nb) What is the probability that the learner has brown hair?\nc) Are these two events mutually exclusive?\nd) Are these two events independent?\n3. P(M) = 0,45; P(N) = 0,3 and P(M or N) = 0,615. Are the events M and N\nmutually exclusive, independent or neither mutually exclusive nor independent?\n4. (For enrichment)\nProve that if event A and event B are mutually exclusive with P(A) ̸= 0 and\nP(B) ̸= 0, then A and B are always dependent.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B9\n2. 23BB\n3. 23BC\n4. 23BD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n418\n10.2.\nDependent and independent events\n\n10.3\nMore Venn diagrams\nEMBJV\nIn the rest of this chapter we will look at tools and techniques for working with proba-\nbility problems.\nWhen working with more complex problems, we can have three or more events that\nintersect in various ways.\nTo solve these problems, we usually want to count the\nnumber (or percentage) of outcomes in an event, or a combination of events. Venn\ndiagrams are a useful tool for recording and visualising the counts.\nInvestigation: Venn diagram for 3 events\nThe diagram below shows a general Venn diagram for 3 events.\nS\nA\nB\nC\nWrite down the sets corresponding to each of the three coloured regions and also\nto the shaded region. Remember that the intersections between circles represent the\nintersections between the different events.\nWhat is the event for\n• the red region;\n• the green region;\n• the blue region; and\n• the shaded region?\n419\nChapter 10.\nProbability\n\nWorked example 8: Venn diagram for 3 events\nQUESTION\nDraw a Venn diagram that shows the following sample space and events:\n• S: all the integers from 1 to 30\n• P: prime numbers\n• M: multiples of 3\n• F: factors of 30\nSOLUTION\nStep 1: Write down the sample space and event sets\nThe sample space contains all the positive integers up to 30.\nS = {1; 2; 3; . . . ; 30}\nThe prime numbers between 1 and 30 are\nP = {2; 3; 5; 7; 11; 13; 17; 19; 23; 29}\nThe multiples of 3 between 1 and 30 are\nM = {3; 6; 9; 12; 15; 18; 21; 24; 27; 30}\nThe factors of 30 are\nF = {1; 2; 3; 5; 6; 10; 15; 30}\nStep 2: Draw the outline of the Venn diagram\nThere are 3 events, namely P, M and F, and the sample space, S. Put this information\non a Venn diagram:\nS\nP\nM\nF\n420\n10.3.\nMore Venn diagrams\n\nStep 3: Place the outcomes in the appropriate event sets\nS\nP\nM\nF\n3\n2\n5\n6\n15\n30\n1\n10\n7\n11\n13\n17\n19\n23\n29\n9\n12\n21\n24\n18\n27\n4\n8\n14\n16\n20\n22\n25\n26\n28\nWorked example 9: Venn diagram for 3 events\nQUESTION\nAt Dawnview High there are 400 Grade 11 learners. 270 do Computer Science, 300\ndo English and 50 do Business studies. All those doing Computer Science do English,\n20 take Computer Science and Business studies and 35 take English and Business\nstudies. Using a Venn diagram, calculate the probability that a pupil drawn at random\nwill take:\n1. English, but not Business studies or Computer Science\n2. English but not Business studies\n3. English or Business studies but not Computer Science\n4. English or Business studies\nSOLUTION\nStep 1: Draw the outline of the Venn diagram\nWe need to be careful with this problem. In the question statement we are told that all\nthe learners who do Computer Science also do English. This means that the circle for\nComputer Science on the Venn diagram needs to be inside the circle for English.\n421\nChapter 10.\nProbability\n\nS\nE\nC\nB\nStep 2: Fill in the counts on the Venn diagram\nS\nE\nC\nB\n20\n250\n15\n15\n15\n85\nStep 3: Compute probabilities\nTo find the number of learners taking English, but not Business studies or Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 15 and there are a total of 400 learners in the grade. There-\nfore the probability that a learner will take English but not Business studies or Computer\nScience is\n15\n400 = 3\n80.\nTo find the number of learners taking English but not Business studies, we need to look\nat this region of the Venn diagram:\n422\n10.3.\nMore Venn diagrams\n\nThe count in this region is 265. Therefore the probability that a learner will take\nEnglish but not Business studies is 265\n400 = 53\n80.\nTo find the number of learners taking English or Business studies but not Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 45. Therefore the probability that a learner will take English\nor Business studies but not Computer Science is\n45\n400 = 9\n80.\nTo find the number of learners taking English or Business studies, we need to look at\nthis region of the Venn diagram:\nThe count in this region is 315. Therefore the probability that a learner will take\nEnglish or Business studies is 315\n400 = 63\n80.\n423\nChapter 10.\nProbability\n\nThere are some words that tell you which part of the Venn diagram should be filled in.\nThe following table summarises the most important ones:\nWords\nSymbols\nVenn diagram\n“all”\nA and B and C / A ∩B ∩C\n“none”\n“at least one”\nA or B or C / A ∪B ∪C\n“both A and B”\nA and B / A ∩B\n“A or B”\nA or B / A ∪B\nExercise 10 – 4: Venn diagrams\n1. Use the Venn diagram below to answer the following questions. Also given:\nn(S) = 120.\nS\nF\n8\n10\nG\n24\n15\nH\n14\n7\n2\na) Compute P(F).\nb) Compute P(G or H).\nc) Compute P(F and G).\nd) Are F and G dependent or independent?\n424\n10.3.\nMore Venn diagrams\n\n2. The Venn diagram below shows the probabilities of 3 events. Complete the Venn\ndiagram using the additional information provided.\nS\nZ\n1\n25\nY\n17\n100\nX\n17\n100\n3\n20\n• P(Z and (not Y )) =\n31\n100\n• P(Y and X) =\n23\n100\n• P(Y ) =\n39\n100\nAfter completing the Venn diagram, compute the following:\nP (Z and not (X or Y ))\n3. There are 79 Grade 10 learners at school. All of these take some combination of\nMaths, Geography and History. The number who take Geography is 41; those\nwho take History is 36; and 30 take Maths. The number who take Maths and\nHistory is 16; the number who take Geography and History is 6, and there are 8\nwho take Maths only and 16 who take History only.\na) Draw a Venn diagram to illustrate all this information.\nb) How many learners take Maths and Geography but not History?\nc) How many learners take Geography only?\nd) How many learners take all three subjects?\n4. Draw a Venn diagram with 3 mutually exclusive events. Use the diagram to\nshow that for 3 mutually exclusive events, A, B and C, the following is true:\nP(A or B or C) = P(A) + P(B) + P(C)\nThis is the addition rule for 3 mutually exclusive events.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BF\n2. 23BG\n3. 23BH\n4. 23BJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n425\nChapter 10.\nProbability\n\n10.4\nTree diagrams\nEMBJW\nTree diagrams are useful for organising and visualising the different possible outcomes\nof a sequence of events. For each possible outcome of the first event, we draw a line\nwhere we write down the probability of that outcome and the state of the world if that\noutcome happened. Then, for each possible outcome of the second event we do the\nsame thing.\nBelow is an example of a simple tree diagram, showing the possible outcomes of\nrolling a 6-sided die.\n1\n1\n6\n2\n1\n6\n3\n1\n6\n4\n1\n6\n5\n1\n6\n6\n1\n6\noutcomes\nprobabilities\nNote that each outcome (the numbers 1 to 6) is shown at the end of a line; and that\nthe probability of each outcome (all 1\n6 in this case) is shown shown on a line. The\nprobabilities have to add up to 1 in order to cover all of the possible outcomes. In the\nexamples below, we will see how to draw tree diagrams with multiple events and how\nto compute probabilities using the diagrams.\nEarlier in this chapter you learned about dependent and independent events. Tree\ndiagrams are very helpful for analysing dependent events. A tree diagram allows you\nto show how each possible outcome of one event affects the probabilities of the other\nevents.\nTree diagrams are not so useful for independent events since we can just multiply the\nprobabilities of separate events to get the probability of the combined event. Remem-\nber that for independent events:\nP(A and B) = P(A) × P(B)\nSo if you already know that events are independent, it is usually easier to solve a\nproblem without using tree diagrams. But if you are uncertain about whether events\nare independent or if you know that they are not, you should use a tree diagram.\nWorked example 10: Drawing a tree diagram\nQUESTION\nIf it rains on a given day, the probability that it rains the next day is 1\n3. If it does not rain\non a given day, the probability that it rains the next day is 1\n6. The probability that it will\nrain tomorrow is 1\n5. What is the probability that it will rain the day after tomorrow?\nDraw a tree diagram of all the possibilities to determine the answer.\nSOLUTION\nStep 1: Draw the first level of the tree diagram\nBefore we can determine what happens on the day after tomorrow, we first have to\ndetermine what might happen tomorrow. We are told that there is a 1\n5 probability that\n426\n10.4.\nTree diagrams\n\nit will rain tomorrow. Here is how to represent this information using a tree diagram:\n1\n5\nrain\n4\n5\nno rain\ntoday:\ntomorrow:\nStep 2: Draw the second level of the tree diagram\nWe are also told that if it does rain on one day, there is a 1\n3 probability that it will also\nrain on the following day. On the other hand, if it does not rain on one day, there is\nonly a 1\n6 probability that it will also rain on the following day. Using this information\nwe complete the tree diagram:\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nStep 3: Compute the probability\nWe are asked what the probability is that it will rain the day after tomorrow. On the\ntree diagram above we can see that there are 2 situations where it rains on the day\nafter tomorrow. They are marked in red below.\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nTo get the probability for the first situation (that it rains tomorrow and the day after\ntomorrow) we have to multiply the probabilies along the first red line.\nP(rain tomorrow and rain day after tomorrow)\n=1\n5 × 1\n3\n= 1\n15\n427\nChapter 10.\nProbability\n\nTo get the probability for the second situation (that it does not rain tomorrow, but it\ndoes rain the day after tomorrow) we have to multiply the probabilies along the second\nred line.\nP(not rain tomorrow and rain day after tomorrow)\n=4\n5 × 1\n6\n= 2\n15\nTherefore the total probability that it will rain the day after tomorrow is the sum of the\nprobabilities along the two red paths, namely\n1\n15 + 2\n15 = 1\n5\nWorked example 11: Drawing a tree diagram\nQUESTION\nYou play the following game. You flip a coin. If it comes up tails, you get 2 points\nand your turn ends. If it comes up heads, you get only 1 point, but you can flip the\ncoin again. If you flip the coin multiple times in one turn, you add up the points. You\ncan flip the coin at most 3 times in one turn. What is the probability that you will get\nexactly 3 points in one turn? Draw a tree diagram to visualise the different possibilities.\nSOLUTION\nStep 1: Write down the events and their symbols\nEach coin toss has on of two possible outcomes, namely heads (H) and tails (T). Each\noutcome has a probability of 1\n2. We are asked to count the number of points, so we\nwill also indicate how many points we have for each outcome.\nStep 2: Draw the first level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\nThis tree diagram shows the possible outcomes after 1 flip of the coin. Remember that\nwe can have up to 3 flips, so the diagram is not complete yet. If the coin comes up\nheads, we flip the coin again. If the coin comes up tails, we stop.\n428\n10.4.\nTree diagrams\n\nStep 3: Draw the second and third level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\n1\n2\nH\n2 pts\n1\n2\nT\n3 pts\n1\n2\nH\n3 pts\n1\n2\nT\n4 pts\nIn this tree diagram you can see that we add up the points we get with each coin flip.\nAfter three coin flips, the game is over.\nStep 4: Find the relevant outcomes and compute the probability\nWe are interested in getting exactly 3 points during the game. To find these outcomes\nwe look only at the tips of the tree. We end with exactly 3 points when the coin flips\nare\n• (H; T) with probability 1\n2 × 1\n2 = 1\n4;\n• (H; H; H) with probability 1\n2 × 1\n2 × 1\n2 = 1\n8.\nNotice that we compute the probability of an outcome by multiplying all the probabil-\nities along the path from the start of the tree to the tip where the outcome is. We add\nthe above two probabilites to obtain the final probability of getting exactly 3 points as\n1\n4 + 1\n8 = 3\n8.\nWorked example 12: Drawing a tree diagram\nQUESTION\nA person takes part in a medical trial that tests the effect of a medicine on a disease.\nHalf the people are given medicine and the other half are given a sugar pill, which has\nno effect on the disease. The medicine has a 60% chance of curing someone. But,\npeople who do not get the medicine still have a 10% chance of getting well. There are\n50 people in the trial and they all have the disease. Talwar takes part in the trial, but\nwe do not know whether he got the medicine or the sugar pill. Draw a tree diagram\nof all the possible cases. What is the probability that Talwar gets cured?\nSOLUTION\nStep 1: Summarise the information in the problem\nThere are two uncertain events in this problem. Each person either receives medicine\n(probability 1\n2) or a sugar pill (probability 1\n2). Each person also gets cured (probability\n429\nChapter 10.\nProbability\n\n3\n5 with medicine and\n1\n10 without) or stays ill (probability 2\n5 with medicine and\n9\n10\nwithout).\nStep 2: Draw the tree diagram\n1\n2\nmedicine\n1\n2\nsugar pill\n3\n5\ncured\n2\n5\nnot cured\n1\n10\ncured\n9\n10\nnot cured\nIn the first level of the tree diagram we show that Talwar either gets the medicine or\nthe sugar pill. The second level of the tree diagram shows whether Talwar is cured or\nnot, depending on which one of the pills he got.\nStep 3: Compute the required probability\nWe multiply the probabilites along each path in the tree diagram that leads to Talwer\nbeing cured:\n1\n2 × 3\n5 = 3\n10\n1\n2 × 1\n10 = 1\n20\nWe then add these probabilites to get the final answer. The probability that Talwar is\ncured is 7\n20.\nExercise 10 – 5: Tree diagrams\n1. You roll a die twice and add up the dots to get a score. Draw a tree diagram to\nrepresent this experiment. What is the probability that your score is a multiple\nof 5?\n2. What is the probability of throwing at least one five in four rolls of a regular\n6-sided die? Hint: do not show all possible outcomes of each roll of the die. We\nare interested in whether the outcome is 5 or not 5 only.\n3. You flip one coin 4 times.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\n430\n10.4.\nTree diagrams\n\n4. You flip 4 different coins at the same time.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BK\n2. 23BM\n3. 23BN\n4. 23BP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.5\nContingency tables\nEMBJX\nA contingency table is another tool for keeping a record of the counts or percentages\nin a probability problem. Contingency tables are especially helpful for figuring out\nwhether events are dependent or independent.\nWe will be studying two-way contingency tables, where we count the number of out-\ncomes for 2 events and their complements, making 4 events in total. A two-way contin-\ngency table always shows the counts for the 4 possible combinations of events, as well\nas the totals for each event and its complement. We can use a contingency table to\ncompute the probabilities of various events by computing the ratios between counts,\nand to determine whether the events are dependent or independent. The example\nbelow shows a two-way contingency table, representing the outcome of a medical\nstudy.\nWorked example 13: Contingency tables\nQUESTION\nA medical trial into the effectiveness of a new medication was carried out. 120 females\nand 90 males took part in the trial. Out of those people, 50 females and 30 males\nresponded positively to the medication. Given below is a contingency table with the\ngiven information filled in.\nFemale\nMale\nTotals\nPositive\n50\n30\nNegative\nTotals\n120\n90\n1. What is the probability that the medicine gives a positive result for females?\n2. What is the probability that the medicine gives a negative result for males?\n3. Was the medication’s success independent of gender? Explain.\n431\nChapter 10.\nProbability\n\nSOLUTION\nStep 1: Complete the contingency table\nThe best place to start is always to complete the contingency table. Because the each\ncolumn has to sum up to its total, we can work out the number of females and males\nwho responded negatively to the medication. Then we can add each row to get the\ntotals on the right hand side of the table.\nFemale\nMale\nTotals\nPositive\n50\n30\n80\nNegative\n70\n60\n130\nTotals\n120\n90\n210\nStep 2: Compute the required probabilities\nThe way the first question is phrased, we need to determine the probability that a\nperson responds positively if she is female. This means that we do not include males\nin this calculation. So, the probability that the medicine gives a positive result for\nfemales is the ratio between the number of females who got a positive response and\nthe total number of females.\nP(positive if female) = n(positive and female)\nn(female)\n= 50\n120\n= 5\n12\nSimilarly, the probability that the medicine gives a negative result for males is:\nP(negative if male) = n(negative and male)\nn(male)\n= 60\n90\n= 2\n3\nStep 3: Independence\nWe need to determine whether the effect of the medicine and the gender of a par-\nticipant are dependent or independent. According to the definition, two events are\nindependent if and only if\nP(A and B) = P(A) × P(B)\nWe will look at the events that a participant is female and that the participant re-\nsponded positively to the trial.\nP(female) =\nn(female)\nn(total trials)\n= 120\n210\n= 4\n7\n432\n10.5.\nContingency tables\n\nP(positive) =\nn(positive)\nn(total trials)\n= 80\n210\n= 8\n21\nP(female and positive) = n(female and positive)\nn(total trials)\n= 50\n210\n= 5\n21\nFrom these probabilities we can see that\nP(female and positive) ̸= P(female) × P(positive)\nand therefore the gender of a participant and the outcome of a trial are dependent\nevents.\nWorked example 14: Contingency tables\nQUESTION\nUse the contingency table below to answer the following questions.\nGrade 11\nGrade 12\nTotals\nHas cellphone\n59\n50\n109\nNo cellphone\n6\n3\n9\nTotals\n65\n53\n118\n1. What is the probability that a learner from Grade 11 has a cellphone?\n2. What is the probability that a learner who does not have a cellphone is from\nGrade 11.\n3. Are the grade of a learner and whether he has a cellphone or not independent\nevents? Explain your answer.\nSOLUTION\n1. There are 65 learners in Grade 11 and 59 of them have a cellphone. Therefore\nthe probability that a learner from Grade 11 has a cellphone is 59\n65.\n2. There are 9 learners who do not have a cellphone and 6 of them are in Grade\n11. Therefore the probability that a learner who does not have a cellphone is\nfrom from Grade 11 is 6\n9 = 2\n3.\n433\nChapter 10.\nProbability\n\n3. To test for independence, we will consider whether a learner is in Grade 11 and\nwhether a learner has a cellphone. The probability that a learner is in Grade 11\nis\n65\n118. The probability that a learner has a cellphone is 109\n118. The probability that\na learner is in Grade 11 and has a cellphone is\n59\n118 = 1\n2. Since 1\n2 ̸=\n65\n118 × 109\n118\nthe grade of a learner and whether he has a cellphone are dependent.\nExercise 10 – 6: Contingency tables\n1. Use the contingency table below to answer the following questions.\nBrown eyes\nNot brown eyes\nTotals\nBlack hair\n50\n30\n80\nRed hair\n70\n80\n150\nTotals\n120\n110\n230\na) What is the probability that someone with black hair has brown eyes?\nb) What is the probability that someone has black hair?\nc) What is the probability that someone has brown eyes?\nd) Are having black hair and having brown eyes dependent or independent\nevents?\n2. Given the following contingency table, identify the events and determine\nwhether they are dependent or independent.\nLocation A\nLocation B\nTotals\nBuses left late\n15\n40\n55\nBuses left on time\n25\n20\n45\nTotals\n40\n60\n100\n3. You are given the following information.\n• Events A and B are independent.\n• P(not A) = 0,3.\n• P(B) = 0,4.\nComplete the contingency table below.\nA\nnot A\nTotals\nB\nnot B\nTotals\n50\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BQ\n2. 23BR\n3. 23BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n434\n10.5.\nContingency tables\n\n10.6\nSummary\nEMBJY\nSee presentation: 23BT at www.everythingmaths.co.za\n• Terminology:\n– Outcome: a single observation of an experiment.\n– Sample space of an experiment: the set of all possible outcomes of the\nexperiment.\n– Event: a set of outcomes of an experiment.\n– Probability of an event: a real number between 0 and 1 that describes how\nlikely it is that the event will occur.\n– Relative frequency of an event: the number of times that the event occurs\nduring experimental trials, divided by the total number of trials conducted.\n– Union of events: the set of all outcomes that occur in at least one of the\nevents, written as “A or B”.\n– Intersection of events: the set of all outcomes that occur in all of the events,\nwritten as “A and B”.\n– Mutually exclusive events: events with no outcomes in common, that is\n(A and B) = ∅.\n– Complementary events: two mutually exclusive events that together con-\ntain all the outcomes in the sample space. We write the complement as\n“not A”.\n– Independent events: two events where knowing the outcome of one event\ndoes not affect the probability of the other event. Events are independent if\nand only if P(A and B) = P(A) × P(B).\n• Identities:\n– The addition rule: P(A or B) = P(A) + P(B) −P(A and B)\n– The addition rule for 2 mutually exclusive events: P(A or B) = P(A) +\nP(B)\n– The complementary rule: P(not A) = 1 −P(A)\n• A Venn diagram is a visual tool used to show how events overlap. Each region\nin a Venn diagram represents an event and could contain either the outcomes in\nthe event, the number of outcomes in the event or the probability of the event.\n• A tree diagram is a visual tool that helps with computing probabilities for depen-\ndent events. The outcomes of each event are shown along with the probability\nof each outcome. For each event that depends on a previous event, we go one\nlevel deeper into the tree. To compute the probability of some combination of\noutcomes, we\n– find all the paths that contain the outcome of interest;\n– multiply the probabilities along each path;\n– add the probabilities between different paths.\n• A 2-way contingency table is a tool for organising data, especially when we want\nto determine whether two events, each with only two outcomes, are dependent\nor independent. The counts for each possible combination of outcomes are\nentered into the table, along with the totals of each row and column.\n435\nChapter 10.\nProbability\n\nExercise 10 – 7: End of chapter exercises\n1. Jane invested in the stock market. The probability that she will not lose all her\nmoney is 0,32. What is the probability that she will lose all her money? Explain.\n2. If D and F are mutually exclusive events, with P(not D)\n=\n0,3 and\nP(D or F) = 0,94, find P(F).\n3. A car sales person has pink, lime-green and purple models of car A and purple,\norange and multicolour models of car B. One dark night a thief steals a car.\na) What is the experiment and sample space?\nb) What is the probability of stealing either a model of A or a model of B?\nc) What is the probability of stealing both a model of A and a model of B?\n4. The probability of event X is 0,43 and the probability of event Y is 0,24. The\nprobability of both occurring together is 0,10. What is the probability that X or\nY will occur?\n5. P(H) = 0,62; P(J) = 0,39 and P(H and J) = 0,31. Calculate:\na) P(H′)\nb) P(H or J)\nc) P(H′ or J′)\nd) P(H′ or J)\ne) P(H′ and J′)\n6. The last ten letters of the alphabet are placed in a hat and people are asked to\npick one of them. Event D is picking a vowel, event E is picking a consonant\nand event F is picking one of the last four letters. Draw a Venn diagram showing\nthe outcomes in the sample space and the different events. Then calculate the\nfollowing probabilities:\na) P(not F)\nb) P(F or D)\nc) P(neither E nor F)\nd) P(D and E)\ne) P(E and F)\nf) P(E and D′)\n7. Thobeka compares three neighbourhoods (we’ll call them A, B and C) to see\nwhere the best place is to live. She interviews 80 people and asks them whether\nthey like each of the neighbourhoods, or not.\n• 40 people like neighbourhood A.\n• 35 people like neighbourhood B.\n• 40 people like neighbourhood C.\n• 21 people like both neighbourhoods A and C.\n• 18 people like both neighbourhoods B and C.\n• 68 people like at least one neighbourhood.\n• 7 people like all three neighbourhoods.\n436\n10.6.\nSummary\n\na) Use this information to draw a Venn diagram.\nb) How many people like none of the neighbourhoods?\nc) How many people like neighbourhoods A and B, but not C?\nd) What is the probability that a randomly chosen person from the survey likes\nat least one of the neighbourhoods?\n8. Let G and H be two events in a sample space.\nSuppose that P(G) = 0,4;\nP(H) = h; and P(G or H) = 0,7.\na) For what value of h are G and H mutually exclusive?\nb) For what value of h are G and H independent?\n9. The following tree diagram represents points scored by two teams in a soccer\ngame. At each level in the tree, the points are shown as (points for Team 1;\npoints for Team 2).\n0,75\n(3; 0)\n0,25\n(2; 1)\n0,5\n(2; 1)\n0,5\n(1; 2)\n0,65\n(2; 0)\n0,35\n(1; 1)\n0,4\n(1; 1)\n0,6\n(0; 2)\n0,52\n(1; 0)\n0,48\n(0; 1)\n(0; 0)\nUse this diagram to determine the probability that:\na) Team 1 will win\nb) The game will be a draw\nc) The game will end with an even number of total points\n10. A bag contains 10 orange balls and 7 black balls. You draw 3 balls from the bag\nwithout replacement. What is the probability that you will end up with exactly\n2 orange balls? Represent this experiment using a tree diagram.\n11. Complete the following contingency table and determine whether the events are\ndependent or independent.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\nDid not like living there\n140\n340\nTotals\n230\n500\n12. Summarise the following information about a medical trial with 2 types of multi-\nvitamin in a contingency table and determine whether the events are dependent\nor independent.\n• 960 people took part in the medical trial.\n• 540 people used multivitamin A for a month and 400 of those people\nshowed an improvement in their health.\n437\nChapter 10.\nProbability\n\n• 300 people showed an improvement in health when using multivitamin B\nfor a month.\nIf the events are independent, it means that the two multivitamins have the same\neffect on people. If the events are dependent, it means that one multivitamin is\nbetter than the other. Which multivitamin is better than the other, or are the both\nequally effective?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BV\n2. 23BW\n3. 23BX\n4. 23BY\n5. 23BZ\n6. 23C2\n7. 23C3\n8. 23C4\n9. 23C5\n10. 23C6\n11. 23C7\n12. 23C8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n438\n10.6.\nSummary\n\nCHAPTER\n11\nStatistics\n11.1\nRevision\n440\n11.2\nHistograms\n444\n11.3\nOgives\n451\n11.4\nVariance and standard deviation\n455\n11.5\nSymmetric and skewed data\n461\n11.6\nIdentification of outliers\n464\n11.7\nSummary\n467\n\n11\nStatistics\n11.1\nRevision\nEMBJZ\nMeasures of central tendency\nEMBK2\nThe mean and median of a data set both give an indication where the centre of the\ndata distribution is located. The mean, or average, is calculated as\nx =\nPn\ni=1 xi\nn\nwhere the xi are the data and n is the number of data. We read x as “x bar”.\nThe median is the middle value of an ordered data set. To find the median, we first\nsort the data and then pick out the value in the middle of the sorted list. If the middle\nis in between two values, the median is the average of those two values.\nSee video: 23C9 at www.everythingmaths.co.za\nWorked example 1: Computing measures of central tendency\nQUESTION\nCompute the mean and median of the following data set:\n72,5 ; 92,6 ; 15,6 ; 53,0 ; 86,4 ; 89,9 ; 90,9 ; 21,7 ; 46,0 ; 4,1 ; 51,7 ; 2,2\nSOLUTION\nStep 1: Compute the mean\nUsing the formula for the mean, we first compute the sum of the values and then divide\nby the number of values.\nx = 626,6\n12\n≈52,22\nStep 2: Compute the median\nTo find the median, we first have to sort the data:\n2,2 ; 4,1 ; 15,6 ; 21,7 ; 46,0 ; 51,7 ; 53,0 ; 72,5 ; 86,4 ; 89,9 ; 90,9 ; 92,6\nSince there are an even number of values, the median will lie between two values.\nIn this case, the two values in the middle are 51,7 and 53,0. Therefore the median is\n52,35.\n440\n11.1.\nRevision\n\nMeasures of dispersion\nEMBK3\nMeasures of dispersion tell us how spread out a data set is. If a measure of dispersion\nis small, the data are clustered in a small region. If a measure of dispersion is large,\nthe data are spread out over a large region.\nThe range is the difference between the maximum and minimum values in the data\nset.\nThe inter-quartile range is the difference between the first and third quartiles of the\ndata set. The quartiles are computed in a similar way to the median. The median is\nhalfway into the ordered data set and is sometimes also called the second quartile.\nThe first quartile is one quarter of the way into the ordered data set; whereas the third\nquartile is three quarters of the way into the ordered data set.\nSee video: 23CB at www.everythingmaths.co.za\nWorked example 2: Range and inter-quartile range\nQUESTION\nDetermine the range and the inter-quartile range of the following data set.\n14 ; 17 ; 45 ; 20 ; 19 ; 36 ; 7 ; 30 ; 8\nSOLUTION\nStep 1: Sort the values in the data set\nTo determine the range we need to find the minimum and maximum values in the\ndata set. To determine the inter-quartile range we need to compute the first and third\nquartiles of the data set. For both of these requirements, it is easier to order the data\nset first.\nThe sorted data set is\n7 ; 8 ; 14 ; 17 ; 19 ; 20 ; 30 ; 36 ; 45\nStep 2: Find the minimum, maximum and range\nThe minimum value is the first value in the ordered data set, namely 7. The maximum\nis the last value in the ordered data set, namely 45. The range is the difference between\nthe minimum and maximum: 45 −7 = 38.\nStep 3: Find the quartiles and inter-quartile range\nThe diagram below shows how we find the quartiles one quarter, one half and three\nquarters of the way into the ordered list of values.\n441\nChapter 11.\nStatistics\n\n7\n8\n14\n17\n19\n20\n30\n36\n45\n0\n1\n4\n1\n2\n3\n4\n1\nFrom this diagram we can see that the first quartile is at a value of 14, the second\nquartile (median) is at a value of 19 and the third quartile is at a value of 30.\nThe inter-quartile range is the difference between the first and third quartiles. The\nfirst quartile is 14 and the third quartile is 30. Therefore the inter-quartile range is\n30 −14 = 16.\nFive number summary\nEMBK4\nThe five number summary combines a measure of central tendency, namely the me-\ndian, with measures of dispersion, namely the range and the inter-quartile range. This\ngives a good overview of the overall data distribution. More precisely, the five number\nsummary is written in the following order:\n• minimum;\n• first quartile;\n• median;\n• third quartile;\n• maximum.\nThe five number summary is often presented visually using a box and whisker diagram.\nA box and whisker diagram is shown below, with the positions of the five relevant\nnumbers labelled. Note that this diagram is drawn vertically, but that it may also be\ndrawn horizontally.\nmaximum\nupper quartile\nmedian\nlower quartile\nminimum\ninter-quartile range\ndata range\nSee video: 23CC at www.everythingmaths.co.za\n442\n11.1.\nRevision\n\nWorked example 3: Five number summary\nQUESTION\nDraw a box and whisker diagram for the following data set:\n1,25 ; 1,5 ; 2,5 ; 2,5 ; 3,1 ; 3,2 ; 4,1 ; 4,25 ; 4,75 ; 4,8 ; 4,95 ; 5,1\nSOLUTION\nStep 1: Determine the minimum and maximum\nSince the data set is already ordered, we can read off the minimum as the first value\n(1,25) and the maximum as the last value (5,1).\nStep 2: Determine the quartiles\nThere are 12 values in the data set.\n1,25 1,5\n2,5\n2,5\n3,1\n3,2\n4,1 4,25 4,75 4,8 4,95 5,1\n0\n1\n4\n1\n2\n3\n4\n1\nUsing the figure above we can see that the median is between the sixth and seventh\nvalues, making it.\n3,2 + 4,1\n2\n= 3,65\nThe first quartile lies between the third and fourth values, making it\nQ1 = 2,5 + 2,5\n2\n= 2,5\nThe third quartile lies between the ninth and tenth values, making it\nQ3 = 4,75 + 4,8\n2\n= 4,775\nStep 3: Draw the box and whisker diagram\nWe now have the five number summary as (1,25; 2,5; 3,65; 4,775; 5,1). The box and\nwhisker diagram representing the five number summary is given below.\n1,25\n2,5\n3,65\n4,775 5,1\n443\nChapter 11.\nStatistics\n\nExercise 11 – 1: Revision\n1. For each of the following data sets, compute the mean and all the quartiles.\nRound your answers to one decimal place.\na) −3,4 ; −3,1 ; −6,1 ; −1,5 ; −7,8 ; −3,4 ; −2,7 ; −6,2\nb) −6 ; −99 ; 90 ; 81 ; 13 ; −85 ; −60 ; 65 ; −49\nc) 7 ; 45 ; 11 ; 3 ; 9 ; 35 ; 31 ; 7 ; 16 ; 40 ; 12 ; 6\n2. Use the following box and whisker diagram to determine the range and inter-\nquartile range of the data.\n−5,52\n−2,41−1,53\n0,10\n4,08\n3. Draw the box and whisker diagram for the following data.\n0,2 ; −0,2 ; −2,7 ; 2,9 ; −0,2 ; −4,2 ; −1,8 ; 0,4 ; −1,7 ; −2,5 ; 2,7 ; 0,8 ; −0,5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23CD\n1b. 23CF\n1c. 23CG\n2. 23CH\n3. 23CJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.2\nHistograms\nEMBK5\nA histogram is a graphical representation of how many times different, mutually exclu-\nsive events are observed in an experiment. To interpret a histogram, we find the events\non the x-axis and the counts on the y-axis. Each event has a rectangle that shows what\nits count (or frequency) is.\nSee video: 23CK at www.everythingmaths.co.za\nWorked example 4: Reading histograms\nQUESTION\nUse the following histogram to determine the events that were recorded and the rela-\ntive frequency of each event. Summarise your answer in a table.\n444\n11.2.\nHistograms\n\n0\n2\n4\n6\n8\n10\nnot yet\nin school\nin primary\nschool\nin high\nschool\nSOLUTION\nStep 1: Determine the events\nThe events are shown on the x-axis. In this example we have “not yet in school”, “in\nprimary school” and “in high school”.\nStep 2: Read off the count for each event\nThe counts are shown on the y-axis and the height of each rectangle shows the fre-\nquency for each event.\n• not yet in school: 2\n• in primary school: 5\n• in high school: 9\nStep 3: Calculate relative frequency\nThe relative frequency of an event in an experiment is the number of times that the\nevent occurred divided by the total number of times that the experiment was com-\npleted. In this example we add up the frequencies for all the events to get a total\nfrequency of 16. Therefore the relative frequencies are:\n• not yet in school:\n2\n16 = 1\n8\n• in primary school:\n5\n16\n• in high school:\n9\n16\nStep 4: Summarise\nEvent\nCount\nRelative frequency\nnot yet in school\n2\n1\n8\nin primary school\n5\n5\n16\nin high school\n9\n9\n16\n445\nChapter 11.\nStatistics\n\nTo draw a histogram of a data set containing numbers, the numbers first have to be\ngrouped.\nEach group is defined by an interval.\nWe then count how many times\nnumbers from each group appear in the data set and draw a histogram using the counts.\nWorked example 5: Draw a histogram\nQUESTION\nThe following data represent the heights of 16 adults in centimetres.\n162 ; 168 ; 177 ; 147 ; 189 ; 171 ; 173 ; 168\n178 ; 184 ; 165 ; 173 ; 179 ; 166 ; 168 ; 165\nDivide the data into 5 equal length intervals between 140 cm and 190 cm and draw a\nhistogram.\nSOLUTION\nStep 1: Determine intervals\nTo have 5 intervals of the same length between 140 and 190, we need and interval\nlength of 10. Therefore the intervals are (140; 150]; (150; 160]; (160; 170]; (170; 180];\nand (180; 190].\nStep 2: Count data\nThe following table summarises the number of data values in each of the intervals.\nInterval\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\n(180; 190]\nCount\n1\n0\n7\n6\n2\nStep 3: Draw the histogram\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\n446\n11.2.\nHistograms\n\nFrequency polygons\nEMBK6\nA frequency polygon is sometimes used to represent the same information as in a his-\ntogram. A frequency polygon is drawn by using line segments to connect the middle of\nthe top of each bar in the histogram. This means that the frequency polygon connects\nthe coordinates at the centre of each interval and the count in each interval.\nWorked example 6: Drawing a frequency polygon\nQUESTION\nUse the histogram from the previous example to draw a frequency polygon of the same\ndata.\nSOLUTION\nStep 1: Draw the histogram\nWe already know that the histogram looks like this:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\nStep 2: Connect the tops of the rectangles\nWhen we draw line segments between the tops of the rectangles in the histogram, we\nget the following picture:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\n447\nChapter 11.\nStatistics\n\nStep 3: Draw final frequency polygon\nFinally, we remove the histogram to show only the frequency polygon.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\nFrequency polygons are particularly useful for comparing two data sets. Comparing\ntwo histograms would be more difficult since we would have to draw the rectangles of\nthe two data sets on top of each other. Because frequency polygons are just lines, they\ndo not pose the same problem.\nWorked example 7: Drawing frequency polygons\nQUESTION\nHere is another data set of heights, this time of Grade 11 learners.\n132 ; 132 ; 156 ; 147 ; 162 ; 168 ; 152 ; 174\n141 ; 136 ; 161 ; 148 ; 140 ; 174 ; 174 ; 162\nDraw the frequency polygon for this data set using the same interval length as in the\nprevious example. Then compare the two frequency polygons on one graph to see the\ndifferences between the distributions.\nSOLUTION\nStep 1: Frequency table\nWe first create the table of counts for the new data set.\nInterval\n(130; 140]\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\nCount\n4\n3\n2\n4\n3\n448\n11.2.\nHistograms\n\nStep 2: Draw histogram and frequency polygon\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\nStep 3: Compare frequency polygons\nWe draw the two frequency polygons on the same axes. The red line indicates the\ndistribution over heights for adults and the blue line, for Grade 11 learners.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\n190\nFrom this plot we can easily see that the heights for Grade 11 learners are distributed\nmore towards the left (shorter) than adults. The learner heights also seem to be more\nevenly distributed between 130 and 180 cm, whereas the adult heights are mostly\nbetween 160 and 180 cm.\n449\nChapter 11.\nStatistics\n\nExercise 11 – 2: Histograms\n1. Use the histogram below to answer the following questions.\nThe histogram\nshows the number of people born around the world each year. The ticks on\nthe x-axis are located at the start of each year.\npeople (millions)\nyear\n79\n80\n81\n82\n83\n84\n85\n86\n87\n1994 1995 1996 1997 1998 1999 2000 2001\na) How many people were born between the beginning of 1994 and the be-\nginning of 1996?\nb) Is the number people in the world population increasing or decreasing?\n(Ignore the rate at which people are dying for this question.)\nc) How many more people were born in 1994 than in 1997?\n2. In a traffic survey, a random sample of 50 motorists were asked the distance (d)\nthey drove to work daily. The results of the survey are shown in the table below.\nDraw a histogram to represent the data.\nd\n0 < d ≤10\n10 < d ≤20\n20 < d ≤30\n30 < d ≤40\n40 < d ≤50\nf\n9\n19\n15\n5\n4\n3. Below is data for the prevalence of HIV in South Africa. HIV prevalence refers to\nthe percentage of people between the ages of 15 and 49 who are infected with\nHIV.\nyear\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nprevalence (%)\n17,7\n18,0\n18,1\n18,1\n18,1\n18,0\n17,9\n17,9\nDraw a frequency polygon of this data set.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CM\n2. 23CN\n3. 23CP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n450\n11.2.\nHistograms\n\n11.3\nOgives\nEMBK7\nCumulative histograms, also known as ogives, are graphs that can be used to deter-\nmine how many data values lie above or below a particular value in a data set. The\ncumulative frequency is calculated from a frequency table, by adding each frequency\nto the total of the frequencies of all data values before it in the data set. The last value\nfor the cumulative frequency will always be equal to the total number of data values,\nsince all frequencies will already have been added to the previous total.\nAn ogive is drawn by\n• plotting the beginning of the first interval at a y-value of zero;\n• plotting the end of every interval at the y-value equal to the cumulative count for\nthat interval; and\n• connecting the points on the plot with straight lines.\nIn this way, the end of the final interval will always be at the total number of data since\nwe will have added up across all intervals.\nWorked example 8: Cumulative frequencies and ogives\nQUESTION\nDetermine the cumulative frequencies of the following grouped data and complete the\ntable below. Use the table to draw an ogive of the data.\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n20 < n ≤30\n7\n30 < n ≤40\n12\n40 < n ≤50\n10\n50 < n ≤60\n6\nSOLUTION\nStep 1: Compute cumulative frequencies\nTo determine the cumulative frequency, we add up the frequencies going down the\ntable. The first cumulative frequency is just the same as the frequency, because we are\nadding it to zero. The final cumulative frequency is always equal to the sum of all the\nfrequencies. This gives the following table:\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n5\n20 < n ≤30\n7\n12\n30 < n ≤40\n12\n24\n40 < n ≤50\n10\n34\n50 < n ≤60\n6\n40\n451\nChapter 11.\nStatistics\n\nStep 2: Plot the ogive\nThe first coordinate in the plot always starts at a y-value of 0 because we always start\nfrom a count of zero. So, the first coordinate is at (10; 0) — at the beginning of the\nfirst interval. The second coordinate is at the end of the first interval (which is also the\nbeginning of the second interval) and at the first cumulative count, so (20; 5). The third\ncoordinate is at the end of the second interval and at the second cumulative count,\nnamely (30; 12), and so on.\nComputing all the coordinates and connecting them with straight lines gives the fol-\nlowing ogive.\nn\n0\n10\n20\n30\n40\n10\n20\n30\n40\n50\n60\n•\n•\n•\n•\n•\n•\nOgives do look similar to frequency polygons, which we saw earlier. The most impor-\ntant difference between them is that an ogive is a plot of cumulative values, whereas\na frequency polygon is a plot of the values themselves. So, to get from a frequency\npolygon to an ogive, we would add up the counts as we move from left to right in the\ngraph.\nOgives are useful for determining the median, percentiles and five number summary\nof data. Remember that the median is simply the value in the middle when we order\nthe data. A quartile is simply a quarter of the way from the beginning or the end of an\nordered data set. With an ogive we already know how many data values are above or\nbelow a certain point, so it is easy to find the middle or a quarter of the data set.\nWorked example 9: Ogives and the five number summary\nQUESTION\nUse the following ogive to compute the five number summary of the data. Remember\nthat the five number summary consists of the minimum, all the quartiles (including the\nmedian) and the maximum.\n452\n11.3.\nOgives\n\ncount\nvalue\n0\n10\n20\n30\n40\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\nSOLUTION\nStep 1: Find the minimum and maximum\nThe minimum value in the data set is 1 since this is where the ogive starts on the\nhorizontal axis. The maximum value in the data set is 10 since this is where the ogive\nstops on the horizontal axis.\nStep 2: Find the quartiles\nThe quartiles are the values that are 1\n4, 1\n2 and 3\n4 of the way into the ordered data set.\nHere the counts go up to 40, so we can find the quartiles by looking at the values\ncorresponding to counts of 10, 20 and 30. On the ogive a count of\n• 10 corresponds to a value of 3 (first quartile);\n• 20 corresponds to a value of 7 (second quartile); and\n• 30 corresponds to a value of 8 (third quartile).\nStep 3: Write down the five number summary\nThe five number summary is (1; 3; 7; 8; 10). The box-and-whisker plot of this data set\nis given below.\n1\n3\n7\n8\n10\n453\nChapter 11.\nStatistics\n\nExercise 11 – 3: Ogives\n1. Use the ogive to answer the questions below. Note that marks are given as a\npercentage.\nnumber of students\nmark\n0\n10\n20\n30\n40\n50\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n•\n•\n•\n•\n•\n•\n•\n•\n•\na) How many students got between 50% and 70%?\nb) How many students got at least 70%?\nc) Compute the average mark for this class, rounded to the nearest integer.\n2. Draw the histogram corresponding to this ogive.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n−25\n−15\n−5\n5\n15\n25\n•\n•\n•\n•\n•\n•\n3. The following data set lists the ages of 24 people.\n2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53; 38\n4; 28; 5; 73; 79; 17; 15; 5; 34; 37; 45; 56\nUse the data to answer the following questions.\na) Using an interval width of 8 construct a cumulative frequency plot.\nb) How many are below 30?\nc) How many are below 60?\nd) Giving an explanation state below what value the bottom 50% of the ages\nfall.\ne) Below what value do the bottom 40% fall?\nf) Construct a frequency polygon.\n454\n11.3.\nOgives\n\n4. The weights of bags of sand in grams is given below (rounded to the nearest\ntenth):\n50,1; 40,4; 48,5; 29,4; 50,2; 55,3; 58,1; 35,3; 54,2; 43,5\n60,1; 43,9; 45,3; 49,2; 36,6; 31,5; 63,1; 49,3; 43,4; 54,1\na) Decide on an interval width and state what you observe about your choice.\nb) Give your lowest interval.\nc) Give your highest interval.\nd) Construct a cumulative frequency graph and a frequency polygon.\ne) Below what value do 53% of the cases fall?\nf) Below what value of 60% of the cases fall?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CQ\n2. 23CR\n3. 23CS\n4. 23CT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.4\nVariance and standard deviation\nEMBK8\nMeasures of central tendency (mean, median and mode) provide information on the\ndata values at the centre of the data set. Measures of dispersion (quartiles, percentiles,\nranges) provide information on the spread of the data around the centre. In this section\nwe will look at two more measures of dispersion called the variance and the standard\ndeviation.\nSee video: 23CV at www.everythingmaths.co.za\nVariance\nEMBK9\nDEFINITION: Variance\nLet a population consist of n elements, {x1; x2; . . . ; xn}. Write the mean of the data as\nx.\nThe variance of the data is the average squared distance between the mean and each\ndata value.\nσ2 =\nPn\ni=1 (xi −x)2\nn\nNOTE:\nThe variance is written as σ2. It might seem strange that it is written in squared form,\nbut you will see why soon when we discuss the standard deviation.\n455\nChapter 11.\nStatistics\n\nThe variance has the following properties.\n• It is never negative since every term in the variance sum is squared and therefore\neither positive or zero.\n• It has squared units. For example, the variance of a set of heights measured in\ncentimetres will be given in centimeters squared. Since the population variance\nis squared, it is not directly comparable with the mean or the data themselves. In\nthe next section we will describe a different measure of dispersion, the standard\ndeviation, which has the same units as the data.\nWorked example 10: Variance\nQUESTION\nYou flip a coin 100 times and it lands on heads 44 times. You then use the same\ncoin and do another 100 flips. This time in lands on heads 49 times. You repeat this\nexperiment a total of 10 times and get the following results for the number of heads.\n{44; 49; 52; 62; 53; 48; 54; 49; 46; 51}\nCompute the mean and variance of this data set.\nSOLUTION\nStep 1: Compute the mean\nThe formula for the mean is\nx =\nPn\ni=1 xi\nn\nIn this case, we sum the data and divide by 10 to get x = 50,8.\nStep 2: Compute the variance\nThe formula for the variance is\nσ2 =\nPn\ni=1 (xi −x)2\nn\nWe first subtract the mean from each datum and then square the result.\nxi\n44\n49\n52\n62\n53\n48\n54\n49\n46\n51\nxi −x\n−6,8\n−1,8\n1,2\n11,2\n2,2\n−2,8\n3,2\n−1,8\n−4,8\n0,2\n(xi −x)2\n46,24\n3,24\n1,44\n125,44 4,84\n7,84\n10,24\n3,24\n23,04\n0,04\nThe variance is the sum of the last row in this table divided by 10, so σ2 = 22,56.\n456\n11.4.\nVariance and standard deviation\n\nStandard deviation\nEMBKB\nSince the variance is a squared quantity, it cannot be directly compared to the data val-\nues or the mean value of a data set. It is therefore more useful to have a quantity which\nis the square root of the variance. This quantity is known as the standard deviation.\nDEFINITION: Standard deviation\nLet a population consist of n elements, {x1; x2; . . . ; xn}, with a mean of x. The stan-\ndard deviation of the data is\nσ =\nsPn\ni=1 (xi −x)2\nn\nIn statistics, the standard deviation is a very common measure of dispersion. Standard\ndeviation measures how spread out the values in a data set are around the mean. More\nprecisely, it is a measure of the average distance between the values of the data in the\nset and the mean. If the data values are all similar, then the standard deviation will be\nlow (closer to zero). If the data values are highly variable, then the standard variation\nis high (further from zero).\nThe standard deviation is always a positive number and is always measured in the\nsame units as the original data. For example, if the data are distance measurements in\nkilogrammes, the standard deviation will also be measured in kilogrammes.\nThe mean and the standard deviation of a set of data are usually reported together. In\na certain sense, the standard deviation is a natural measure of dispersion if the centre\nof the data is taken as the mean.\nInvestigation: Tabulating results\nIt is often useful to set your data out in a table so that you can apply the for-\nmulae easily.\nComplete the table below to calculate the standard deviation of\n{57; 53; 58; 65; 48; 50; 66; 51}.\n• Firstly, remember to calculate the mean, x.\n• Complete the following table.\nindex: i\ndatum: xi\ndeviation: xi −x\ndeviation\nsquared: (xi −x)2\n1\n57\n2\n53\n3\n58\n4\n65\n5\n48\n6\n50\n7\n66\n8\n51\nP xi = . . .\nP(xi −x) = . . .\nP(xi −x)2 = . . .\n• The sum of the deviations is always zero. Why is this? Find out.\n• Calculate the variance using the completed table.\n• Then calculate the standard deviation.\n457\nChapter 11.\nStatistics\n\nWorked example 11: Variance and standard deviation\nQUESTION\nWhat is the variance and standard deviation of the possibilities associated with rolling\na fair die?\nSOLUTION\nStep 1: Determine all the possible outcomes\nWhen rolling a fair die, the sample space consists of 6 outcomes. The data set is\ntherefore x = {1; 2; 3; 4; 5; 6} and n = 6.\nStep 2: Calculate the mean\nThe mean is:\nx = 1\n6 (1 + 2 + 3 + 4 + 5 + 6)\n= 3,5\nStep 3: Calculate the variance\nThe variance is:\nσ2 =\nP (x −x)2\nn\n= 1\n6 (6,25 + 2,25 + 0,25 + 0,25 + 2,25 + 6,25)\n= 2,917\nStep 4: Calculate the standard deviation\nThe standard deviation is:\nσ =\np\n2,917\n= 1,708\nSee video: 23CW at www.everythingmaths.co.za\n458\n11.4.\nVariance and standard deviation\n\nInterpretation and application\nEMBKC\nA large standard deviation indicates that the data values are far from the mean and a\nsmall standard deviation indicates that they are clustered closely around the mean.\nFor example, consider the following three data sets:\n{65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\n{85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\n{43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nEach of these data sets has the same mean, namely 67. However, they have different\nstandard deviations, namely 8,97, 17,75 and 21,23. The following figures show plots\nof the data sets with the mean and standard deviation indicated on each. You can see\nhow the standard deviation is larger when the data are more spread out.\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 8,97\ndata:\n{xi} = {65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\nmean:\nx = 67\nstandard deviation:\nσ ≈8,97\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 17,75\ndata:\n{xi} = {85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\nmean:\nx = 67\nstandard deviation:\nσ ≈17,75\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 21,23\ndata:\n{xi} = {43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nmean:\nx = 67\nstandard deviation:\nσ ≈21,23\nThe standard deviation may also be thought of as a measure of uncertainty. In the phys-\nical sciences, for example, the reported standard deviation of a group of repeated mea-\nsurements represents the precision of those measurements. When deciding whether\n459\nChapter 11.\nStatistics\n\nmeasurements agree with a theoretical prediction, the standard deviation of those mea-\nsurements is very important: if the mean of the measurements is too far away from the\nprediction (with the distance measured in standard deviations), then we consider the\nmeasurements as contradicting the prediction. This makes sense since they fall outside\nthe range of values that could reasonably be expected to occur if the prediction were\ncorrect.\nExercise 11 – 4: Variance and standard deviation\n1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban.\nThe data, in rands per litre, are given below.\nCape Town\n3,96\n3,76\n4,00\n3,91\n3,69\n3,72\nDurban\n3,97\n3,81\n3,52\n4,08\n3,88\n3,68\na) Find the mean price in each city and then state which city has the lowest\nmean.\nb) Find the standard deviation of each city’s prices.\nc) Which city has the more consistently priced petrol? Give reasons for your\nanswer.\n2. Compute the mean and variance of the following set of values.\n150 ; 300 ; 250 ; 270 ; 130 ; 80 ; 700 ; 500 ; 200 ; 220 ; 110 ; 320 ; 420 ; 140\n3. Compute the mean and variance of the following set of values.\n−6,9 ; −17,3 ; 18,1 ; 1,5 ; 8,1 ; 9,6 ; −13,1 ; −14,0 ; 10,5 ; −14,8 ; −6,5 ; 1,4\n4. The times for 8 athletes who ran a 100 m sprint on the same track are shown\nbelow. All times are in seconds.\n10,2 ; 10,8 ; 10,9 ; 10,3 ; 10,2 ; 10,4 ; 10,1 ; 10,4\na) Calculate the mean time.\nb) Calculate the standard deviation for the data.\nc) How many of the athletes’ times are more than one standard deviation away\nfrom the mean?\n5. The following data set has a mean of 14,7 and a variance of 10,01.\n18 ; 11 ; 12 ; a ; 16 ; 11 ; 19 ; 14 ; b ; 13\nCompute the values of a and b.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CX\n2. 23CY\n3. 23CZ\n4. 23D2\n5. 23D3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n460\n11.4.\nVariance and standard deviation\n\n11.5\nSymmetric and skewed data\nEMBKD\nWe are now going to classify data sets into 3 categories that describe the shape of the\ndata distribution: symmetric, left skewed, right skewed. We can use this classification\nfor any data set, but here we will look only at distributions with one peak. Most of\nthe data distributions that you have seen so far have only one peak, so the plots in this\nsection should look familiar.\nDistributions with one peak are called unimodal distributions.\nUnimodal literally\nmeans having one mode. (Remember that a mode is a maximum in the distribution.)\nSymmetric distributions\nEMBKF\nA symmetric distribution is one where the left and right hand sides of the distribution\nare roughly equally balanced around the mean. The histogram below shows a typical\nsymmetric distribution.\nmean ≈median\nbalanced left and right tails\nFor symmetric distributions, the mean is approximately equal to the median. The tails\nof the distribution are the parts to the left and to the right, away from the mean. The\ntail is the part where the counts in the histogram become smaller. For a symmetric\ndistribution, the left and right tails are equally balanced, meaning that they have about\nthe same length.\nThe figure below shows the box and whisker diagram for a typical symmetric data set.\nmedian halfway\nbetween\nfirst and third quartiles\nAnother property of a symmetric distribution is that its median (second quartile) lies\nin the middle of its first and third quartiles. Note that the whiskers of the plot (the\nminimum and maximum) do not have to be equally far away from the median. In the\nnext section on outliers, you will see that the minimum and maximum values do not\nnecessarily match the rest of the data distribution well.\n461\nChapter 11.\nStatistics\n\nSkewed\nEMBKG\nA distribution that is skewed right (also known as positively skewed) is shown below.\nmean\nmedian\nmean > median\nlong right tail\nshort left tail\nNow the picture is not symmetric around the mean anymore.\nFor a right skewed\ndistribution, the mean is typically greater than the median. Also notice that the tail of\nthe distribution on the right hand (positive) side is longer than on the left hand side.\nmedian closer to first quartile\nFrom the box and whisker diagram we can also see that the median is closer to the first\nquartile than the third quartile. The fact that the right hand side tail of the distribution\nis longer than the left can also be seen.\nA distribution that is skewed left has exactly the opposite characteristics of one that is\nskewed right:\n• the mean is typically less than the median;\n• the tail of the distribution is longer on the left hand side than on the right hand\nside; and\n• the median is closer to the third quartile than to the first quartile.\nThe table below summarises the different categories visually.\nSymmetric\nSkewed right (positive)\nSkewed left (negative)\n462\n11.5.\nSymmetric and skewed data\n\nExercise 11 – 5: Symmetric and skewed data\n1. Is the following data set symmetric, skewed right or skewed left? Motivate your\nanswer.\n27 ; 28 ; 30 ; 32 ; 34 ; 38 ; 41 ; 42 ; 43 ; 44 ; 46 ; 53 ; 56 ; 62\n2. State whether each of the following data sets are symmetric, skewed right or\nskewed left.\na) A data set with this histogram:\nb) A data set with this box and whisker plot:\nc) A data set with this frequency polygon:\n• • • • • • • • •\n•\n•\n•\n• • • •\nd) The following data set:\n11,2 ; 5 ; 9,4 ; 14,9 ; 4,4 ; 18,8 ; −0,4 ; 10,5 ; 8,3 ; 17,8\n3. Two data sets have the same range and interquartile range, but one is skewed\nright and the other is skewed left. Sketch the box and whisker plot for each of\nthese data sets. Then, invent data (6 points in each data set) that matches the\ndescriptions of the two data sets.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23D4\n2a. 23D5\n2b. 23D6\n2c. 23D7\n2d. 23D8\n3. 23D9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n463\nChapter 11.\nStatistics\n\n11.6\nIdentification of outliers\nEMBKH\nAn outlier in a data set is a value that is far away from the rest of the values in the\ndata set. In a box and whisker diagram, outliers are usually close to the whiskers of\nthe diagram. This is because the centre of the diagram represents the data between\nthe first and third quartiles, which is where 50% of the data lie, while the whiskers\nrepresent the extremes — the minimum and maximum — of the data.\nWorked example 12: Identifying outliers\nQUESTION\nFind the outliers in the following data set by drawing a box and whisker diagram and\nlocating the data values on the diagram.\n0,5 ; 1 ; 1,1 ; 1,4 ; 2,4 ; 2,8 ; 3,5 ; 5,1 ; 5,2 ; 6 ; 6,5 ; 9,5\nSOLUTION\nStep 1: Determine the five number summary\nThe minimum of the data set is 0,5. The maximum of the data set is 9,5. Since there\nare 12 values in the data set, the median lies between the sixth and seventh values,\nmaking it equal to 2,8+3,5\n2\n= 3,15. The first quartile lies between the third and fourth\nvalues, making it equal to 1,1+1,4\n2\n= 1,25. The third quartile lies between the ninth\nand tenth values, making it equal to 5,2+6\n2\n= 5,6.\nStep 2: Draw the box and whisker diagram\n0,5 1,25\n3,15\n5,6\n9,5\n• •• •\n• •\n•\n••\n• •\n•\nIn the figure above, each value in the data set is shown with a black dot.\nStep 3: Find the outliers\nFrom the diagram we can see that most of the values are between 1 and 6. The only\nvalue that is very far away from this range is the maximum at 9,5. Therefore 9,5 is the\nonly outlier in the data set.\nYou should also be able to identify outliers in plots of two variables. A scatter plot\nis a graph that shows the relationship between two random variables. We call these\ndata bivariate (literally meaning two variables) and we plot the data for two different\nvariables on one set of axes. The following example shows what a typical scatter plot\nlooks like. For Grade 11 you do not need to learn how to draw these 2-dimensional\n464\n11.6.\nIdentification of outliers\n\nscatter plots, but you should be able to identify outliers on them. As before, an outlier\nis a value that is far removed from the main distribution of data.\nWorked example 13: Scatter plot\nQUESTION\nWe have a data set that relates the heights and weights of a number of people. The\nheight is the first variable and its value is plotted along the horizontal axis. The weight\nis the second variable and its value is plotted along the vertical axis. The data values\nare shown on the plot below. Identify any outliers on the scatter plot.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\nSOLUTION\nWe inspect the plot visually and notice that there are two points that lie far away from\nthe main data distribution. These two points are circled in the plot below.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\n465\nChapter 11.\nStatistics\n\nExercise 11 – 6: Outliers\n1. For each of the following data sets, draw a box and whisker diagram and deter-\nmine whether there are any outliers in the data.\na) 30 ; 21,4 ; 39,4 ; 33,4 ; 21,1 ; 29,3 ; 32,8 ; 31,6 ; 36 ;\n27,9 ; 27,3 ; 29,4 ; 29,1 ; 38,6 ; 33,8 ; 29,1 ; 37,1\nb) 198 ; 166 ; 175 ; 147 ; 125 ; 194 ; 119 ; 170 ; 142 ; 148\nc) 7,1 ; 9,6 ; 6,3 ; −5,9 ; 0,7 ; −0,1 ; 4,4 ; −11,7 ; 10 ; 2,3 ; −3,7 ; 5,8 ; −1,4\n; 1,7 ; −0,7\n2. A class’s results for a test were recorded along with the amount of time spent\nstudying for it. The results are given below. Identify any outliers in the data.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23DB\n1b. 23DC\n1c. 23DD\n2. 23DF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n466\n11.6.\nIdentification of outliers\n\n11.7\nSummary\nEMBKJ\nSee presentation: 23DG at www.everythingmaths.co.za\n• Histograms visualise how many times different events occurred. Each rectangle\nin a histogram represents one event and the height of the rectangle is relative to\nthe number of times that the event occurred.\n• Frequency polygons represent the same information as histograms, but using\nlines and points rather than rectangles. A frequency polygon connects the mid-\ndle of the top edge of each rectangle in a histogram.\n• Ogives (also known as cumulative histograms) show the total number of times\nthat a value or anything less than that value appears in the data set. To draw an\nogive you need to add up all the counts in a histogram from left to right.\n– The first count in an ogive is always zero.\n– The last count in an ogive is always the sum of all the counts in the data\nset.\n• The variance and standard deviation are measures of dispersion.\n– The standard deviation is the square root of the variance.\n– Variance: σ2 = 1\nn\nPn\ni=1(xi −x)2\n– Standard deviation: σ =\nq\n1\nn\nPn\ni=1(xi −x)2\n– The standard deviation is measured in the same units as the mean and the\ndata, but the variance is not. The variance is measured in the square of the\ndata units.\n• In a symmetric distribution\n– the mean is approximately equal to the median; and\n– the tails of the distribution are balanced.\n• In a right (positively) skewed distribution\n– the mean is greater than the median;\n– the tail on the right hand side is longer than the tail on the left hand side;\nand\n– the median is closer to the first quartile than the third quartile.\n• In a left (negatively) skewed distribution\n– the mean is less than the median;\n– the tail on the left hand side is longer than the tail on the right hand side;\nand\n– the median is closer to the third quartile than the first quartile.\n• An outlier is a value that is far away from the rest of the data.\n467\nChapter 11.\nStatistics\n\nExercise 11 – 7: End of chapter exercises\n1. Draw a histogram, frequency polygon and ogive of the following data set. To\ncount the data, use intervals with a width of 1, starting from 0.\n0,4 ; 3,1 ; 1,1 ; 2,8 ; 1,5 ; 1,3 ; 2,8 ; 3,1 ; 1,8 ; 1,3 ;\n2,6 ; 3,7 ; 3,3 ; 5,7 ; 3,7 ; 7,4 ; 4,6 ; 2,4 ; 3,5 ; 5,3\n2. Draw a box and whisker diagram of the following data set and explain whether\nit is symmetric, skewed right or skewed left.\n−4,1 ; −1,1 ; −1 ; −1,2 ; −1,5 ; −3,2 ; −4 ; −1,9 ; −4 ;\n−0,8 ; −3,3 ; −4,5 ; −2,5 ; −4,4 ; −4,6 ; −4,4 ; −3,3\n3. Eight children’s sweet consumption and sleeping habits were recorded. The data\nare given in the following table and scatter plot.\nNumber of sweets\nper week\n15\n12\n5\n3\n18\n23\n11\n4\nAverage sleeping\ntime (hours per day)\n4\n4,5\n8\n8,5\n3\n2\n5\n8\n5\n10\n15\n20\n25\nnumber of sweets\n1\n2\n3\n4\n5\n6\n7\n8\n9\nsleeping time (hours per day)\na) What is the mean and standard deviation of the number of sweets eaten per\nday?\nb) What is the mean and standard deviation of the number of hours slept per\nday?\nc) Make a list of all the outliers in the data set.\n4. The monthly incomes of eight teachers are as follows:\nR 10 050;\nR 14 300;\nR 9800;\nR 15 000;\nR 12 140;\nR 13 800;\nR 11 990;\nR 12 900.\na) What is the mean and standard deviation of their incomes?\nb) How many of the salaries are less than one standard deviation away from\nthe mean?\nc) If each teacher gets a bonus of R 500 added to their pay what is the new\nmean and standard deviation?\nd) If each teacher gets a bonus of 10% on their salary what is the new mean\nand standard deviation?\ne) Determine for both of the above, how many salaries are less than one stan-\ndard deviation away from the mean.\n468\n11.7.\nSummary\n\nf) Using the above information work out which bonus is more beneficial fi-\nnancially for the teachers.\n5. The weights of a random sample of boys in Grade 11 were recorded. The cumu-\nlative frequency graph (ogive) below represents the recorded weights.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100 110 120\n0\n10\n20\n30\n40\n50\n60\nWeight (in kilogrammes)\nCumulative frequency\nCumulative frequency curve showing weight of boys\na) How many of the boys weighed between 90 and 100 kilogrammes?\nb) Estimate the median weight of the boys.\nc) If there were 250 boys in Grade 11, estimate how many of them would\nweigh less than 80 kilogrammes?\n6. Three sets of 12 learners each had their test scores recorded. The test was out of\n50. Use the given data to answer the following questions.\nSet A\nSet B\nSet C\n25\n32\n43\n47\n34\n47\n15\n35\n16\n17\n32\n43\n16\n25\n38\n26\n16\n44\n24\n38\n42\n27\n47\n50\n22\n43\n50\n24\n29\n44\n12\n18\n43\n31\n25\n42\na) For each of the sets calculate the mean and the five number summary.\nb) Make box and whisker plots of the three data sets on the same set of axes.\nc) State, with reasons, whether each of the three data sets are symmetric or\nskewed (either right or left).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DH\n2. 23DJ\n3. 23DK\n4. 23DM\n5. 23DN\n6. 23DP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n469\nChapter 11.\nStatistics\n\n\nCHAPTER\n12\nLinear programming\n12.1\nIntroduction\n472\n\n12\nLinear programming\n12.1\nIntroduction\nEMBKK\nIn everyday life people are interested in knowing the most efficient way of carrying out\na task or achieving a goal. For example, a farmer wants to know how many hectares to\nplant during a season in order to maximise the yield (produce), a stock broker wants to\nknow how much to invest in stocks in order to maximise profit, an entrepreneur wants\nto know how many people to employ to minimise expenditure. These are optimisation\nproblems; we want to to determine either the maximum or the minimum in a specific\nsituation.\nTo describe this mathematically, we assign variables to represent the different factors\nthat influence the situation. Optimisation means finding the combination of variables\nthat gives the best result.\nSee video: 23DQ at www.everythingmaths.co.za\nWorked example 1: Mountees and Roadees\nQUESTION\nInvestigate the following situation and use your knowledge of mathematics to solve the\nproblem:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make the maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nStep 2: Organise the information given\nWrite down a summary of the information given in the problem so that we consider\n472\n12.1.\nIntroduction\n\nall the different components in the situation.\nmaximum number for M\n= 5\nmaximum number for R\n= 3\nnumber of technicians needed for M = 1\nnumber of technicians needed for R = 2\ntotal number of technicians\n= 8\nprofit per M\n= 800\nprofit per R\n= 2400\nStep 3: Draw up a table\nUse the summary to draw up a table of all the possible combinations of the number of\nMountees and Roadees that can be manufactured per day:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n(4; 3)\n5\n(5; 0)\n(5; 1)\n(5; 2)\n(5; 3)\nNote that there are 24 possible combinations.\nStep 4: Consider the limitation of the number of technicians\nIt takes 1 technician to assemble a Mountee and 2 technicians to assemble a Roadee.\nThere are a total of 8 technicians in the assembly department, therefore we can write\nthat 1(M) + 2(R) ≤8.\nWith this limitation, we are able to eliminate some of the combinations in the table\nwhere M + 2R > 8:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n\b\b\b\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n\b\b\b\n(4; 3)\n5\n(5; 0)\n(5; 1)\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nThese combinations have been excluded as possible answers. For example, (5; 3) gives\n5 + 2(3) = 11 technicians.\n473\nChapter 12.\nLinear programming\n\nStep 5: Consider the profit on the bicycles\nWe can express the profit (P) per day as: P = 800(M)+2400(R). Notice that a higher\nprofit is made on a Roadee.\nBy substituting the different combinations for M and R, we can find the values that\ngive the maximum profit:\nFor (5; 0)\nP = 800(5) + 2400(0)\n= R 4000\nFor (3; 1)\nP = 800(3) + 2400(1)\n= R 4800\nM\nR\n0\n1\n2\n3\n0\n(0; 0) ⇒R 0\n(0; 1) ⇒R 2400\n(0; 2) ⇒R 4800\n(0; 3) ⇒R 7200\n1\n(1; 0) ⇒R 800\n(1; 1) ⇒R 3200\n(1; 2) ⇒R 5600\n(1; 3) ⇒R 8000\n2\n(2; 0) ⇒R 1600\n(2; 1) ⇒R 4000\n(2; 2) ⇒R 6400\n(2; 3) ⇒R 8800\n3\n(3; 0) ⇒R 2400\n(3; 1) ⇒R 4800\n(3; 2) ⇒R 7200\n\b\b\b\n(3; 3)\n4\n(4; 0) ⇒R 3200\n(4; 1) ⇒R 5600\n(4; 2) ⇒R 8000\n\b\b\b\n(4; 3)\n5\n(5; 0) ⇒R 4000\n(5; 1) ⇒R 6400\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nStep 6: Write the final answer\nTherefore the maximum profit of R 8800 is obtained if 2 Mountees and 3 Roadees are\nmanufactured per day.\nExercise 12 – 1: Optimisation\n1. Furniture store opening special:\nAs part of their opening special, a furniture store has promised to give away at\nleast 40 prizes with a total value of at least R 4000. They intend to give away\nkettles and toasters. They decide there will be at least 10 units of each prize. A\nkettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the\ncompany. Calculate how much this combination of kettles and toasters will cost.\nUse a suitable strategy to organise the information and solve the problem.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n474\n12.1.\nIntroduction\n\nOptimisation using graphs\nA more efficient way to solve optimisation problems is using graphs.\nWe write the limitations in the situation, called constraints, as inequalities. Some con-\nstraints can be modelled by an equation, which needs to be maximised or minimized.\nWe sketch the inequalities and indicate the region above or below the line that is to be\nconsidered in determining the solution. This method of solving optimisation problems\nis called linear programming.\nSee video: 23DS at www.everythingmaths.co.za\nWorked example 2: Optimisation using graphs\nQUESTION\nConsider again the example of Mr. Hunter who manufactures Mountees and Roadees:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make a maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nNotice that the values of M and R are limited to positive integers; Mr. Hunter cannot\nsell negative numbers of bikes nor can he sell a fraction of a bike.\nStep 2: Organise the information\nWe can write these constraints as inequalities:\nnumber of Mountees: 0 ≤M ≤5\nnumber of Roadees: 0 ≤R ≤3\ntotal number of technicians: M + 2R ≤8\nWe also know that P = 800M + 2400R. This is called the objective function, some-\ntimes also referred to as the search line, because the objective (goal) is to determine\nthe maximum value of P.\n475\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nWe represent the number of Mountees manufactured daily on the horizontal axis and\nthe number of Roadees manufactured daily on the vertical axis. Since M and R are\npositive integers, we only use the first quadrant of the Cartesian plane. Note that the\ngraph only includes the integer values of M between 0 and 5 and R between 0 and 3.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nFor the number of technicians in the assembly department M + 2R ≤8. If we make\nR (represented on the y-axis) the subject of the inequality we get R ≤−1\n2M + 4.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR = −1\n2M + 4\nThe arrows indicate the region in which the solution will lie, where R ≤−1\n2M + 4.\nThis area is called the feasible region.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR ≤−1\n2M + 4\nA\n476\n12.1.\nIntroduction\n\nWe substitute the possible combinations into the profit equation P = 800M + 2400R,\nand find the combination that gives the maximum profit.\nAt A(2; 3) :\nP = 800(2) + 2400(3)\n= R 8800\nStep 4: Write the final answer\nTherefore the maximum profit is obtained if 2 Mountees and 3 Roadees are manufac-\ntured per day.\nSee video: 23DT at www.everythingmaths.co.za\nWorked example 3: Optimisation using graphs\nQUESTION\nSolve the “furniture store opening special” problem using graphs:\nAs part of their opening special, a furniture store has promised to give away at least 40\nprizes. They intend to give away kettles and toasters. They decide there will be at least\n10 units of each prize. A kettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the company.\nCalculate how much this combination of kettles and toasters will cost.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nkettles be k and the number of toasters be t, with k, t ∈Z.\nStep 2: Organise the information\nWe can write the given information as inequalities:\nnumber of kettles: k ≥10\nnumber of toasters: t ≥10\ntotal number of prizes: k + t ≥40\nWe make t the subject of the inequality:\nt ≥−k + 40\n477\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nRepresent the constraints on a set of axes:\nKettles (k)\nToasters (t)\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nt ≥−k + 40\nt ≥10\nk ≥10\nWe shade the feasible region as shown in the diagram. Remember that in this situation\nonly the points with integer coordinates inside or on the border of the feasible region\nare possible solutions. The combination giving the minimum cost will lie towards or\non the lower border of the feasible region, which gives us many points to consider. To\nfind the optimum value of C, we use the graph of the objective function\nC = 120k + 100t\nTo draw the line, we make t the subject of the formula\nt = −6\n5k + C\n100\nWe see that the gradient of the objective function is −6\n5, but we do not know the exact\nvalue of the t-intercept ( C\n100). To find the minimum value of C, we need to determine\nthe position of the objective function where it first touches the feasible region and also\ngives the lowest t-intercept.\nKettles (k)\nToasters (t)\nA\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nWe indicate the gradient of the objective function on the graph (the green search\nline). Keeping the gradient the same, we “slide” the objective function towards the\nlower border of the feasible region and find that it touches the feasible region at point\n478\n12.1.\nIntroduction\n\nA(10; 30). This optimum position of the objective function is indicated on the graph\nby the dotted line passing through point A.\nWe substitute the coordinates of A into the cost equation C = 120k + 100t:\nAt A(10; 30) :\nC = 120(10) + 100(30)\n= R 4200\nThe minimum cost can also be determined graphically by reading off the coordinates\nof the t-intercept of the objective function in the optimum position:\ntint = 42\n∴C\n100 = 42\n∴C = R 4200\nStep 4: Write the final answer\nTherefore the minimum cost to the company is R 4200 with 10 kettles and 30 toast-\ners.\nExercise 12 – 2: Optimisation\n1. You are given a test consisting of two sections. The first section is on algebra and\nthe second section is on geometry. You are not allowed to answer more than 10\nquestions from any section, but you have to answer at least 4 algebra questions.\nThe time allowed is not more than 30 minutes. An algebra problem will take 2\nminutes and a geometry problem will take 3 minutes to solve.\nLet x be the number of algebra questions and y be the number of geometry\nquestions.\na) Formulate the equations and inequalities that satisfy the above constraints.\nb) The algebra questions carry 5 marks each and the geometry questions carry\n10 marks each. If T is the total marks, write down an expression for T.\n2. A local clinic wants to produce a guide to healthy living. The clinic intends to\nproduce the guide in two formats: a short video and a printed book. The clinic\nneeds to decide how many of each format to produce for sale. Estimates show\nthat no more than 10 000 copies of both items together will be sold. At least\n4000 copies of the video and at least 2000 copies of the book could be sold,\nalthough sales of the book are not expected to exceed 4000 copies. Let x be the\nnumber of videos sold, and y the number of printed books sold.\na) Write down the constraint inequalities that can be deduced from the given\ninformation.\nb) Represent these inequalities graphically and indicate the feasible region\nclearly.\n479\nChapter 12.\nLinear programming\n\nc) The clinic is seeking to maximise the income, I, earned from the sales of\nthe two products. Each video will sell for R 50 and each book for R 30.\nWrite down the objective function for the income.\nd) What maximum income will be generated by the two guides?\n3. A certain motorcycle manufacturer produces two basic models, the Super X and\nthe Super Y. These motorcycles are sold to dealers at a profit of R 20 000 per\nSuper X and R 10 000 per Super Y. A Super X requires 150 hours for assembly,\n50 hours for painting and finishing and 10 hours for checking and testing. The\nSuper Y requires 60 hours for assembly, 40 hours for painting and finishing and\n20 hours for checking and testing. The total number of hours available per month\nis: 30 000 in the assembly department, 13 000 in the painting and finishing\ndepartment and 5000 in the checking and testing department.\nThe above information is summarised by the following table:\nDepartment\nHours for\nSuper X\nHours for\nSuper Y\nHours available\nper month\nAssembly\n150\n60\n30 000\nPainting and\nfinishing\n50\n40\n13 000\nChecking and testing\n10\n20\n5000\nLet x be the number of Super X and y be the number of Super Y models manu-\nfactured per month.\na) Write down the set of constraint inequalities.\nb) Use graph paper to represent the set of constraint inequalities.\nc) Shade the feasible region on the graph paper.\nd) Write down the profit generated in terms of x and y.\ne) How many motorcycles of each model must be produced in order to max-\nimise the monthly profit?\nf) What is the maximum monthly profit?\n4. A group of students plan to sell x hamburgers and y chicken burgers at a rugby\nmatch. They have meat for at most 300 hamburgers and at most 400 chicken\nburgers. Each burger of both types is sold in a packet. There are 500 packets\navailable. The demand is likely to be such that the number of chicken burgers\nsold is at least half the number of hamburgers sold.\na) Write the constraint inequalities and draw a graph of the feasible region.\nb) A profit of R 3 is made on each hamburger sold and R 2 on each chicken\nburger sold. Write the equation which represents the total profit P in terms\nof x and y.\nc) The objective is to maximise profit. How many of each type of burger\nshould be sold?\n5. Fashion-Cards is a small company that makes two types of cards, type X and type\nY. With the available labour and material, the company can make at most 150\ncards of type X and at most 120 cards of type Y per week. Altogether they cannot\nmake more than 200 cards per week.\n480\n12.1.\nIntroduction\n\nThere is an order for at least 40 type X cards and 10 type Y cards per week.\nFashion-Cards makes a profit of R 5 for each type X card sold and R 10 for each\ntype Y card.\nLet the number of type X cards manufactured per week be x and the number of\ntype Y cards manufactured per week be y.\na) One of the constraint inequalities which represents the restrictions above is\n0 ≤x ≤150. Write the other constraint inequalities.\nb) Represent the constraints graphically and shade the feasible region.\nc) Write the equation that represents the profit P (the objective function), in\nterms of x and y.\nd) Calculate the maximum weekly profit.\n6. To meet the requirements of a specialised diet a meal is prepared by mixing\ntwo types of cereal, Vuka and Molo. The mixture must contain x packets of\nVuka cereal and y packets of Molo cereal. The meal requires at least 15 g of\nprotein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains\n4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains\n3 g of protein and 24 g of carbohydrates. There are at most 5 packets of cereal\navailable. The feasible region is shaded on the attached graph paper.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\nNumber of packets of Vuka\nNumber of packets of Molo\na) Write down the constraint inequalities.\nb) If Vuka cereal costs R 6 per packet and Molo cereal also costs R 6 per\npacket, use the graph to determine how many packets of each cereal must\nbe used so that the total cost for the mixture is a minimum.\nc) Use the graph to determine how many packets of each cereal must be used\nso that the total cost for the mixture is a maximum (give all possibilities).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DV\n2. 23DW\n3. 23DX\n4. 23DY\n5. 23DZ\n6. 23F2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n481\nChapter 12.\nLinear programming\n\n\nSolutions to exercises\n1\nExponents and surds\nExercise 1 – 1: The number system\n1. R; Q′\n2. R; Q\n3. R; Q\n4. R; Q\n5. R; Q; Z; N0\n6. R′Q′\n7. R; Q\n8. R; Q′\n9. R′\n10. R; Q′\n11. R; Q\n12. R; Q; Z\n13. R; Q\n14. R; Q′\n15. R; Q\n16. R; Q; Z\nExercise 1 – 2: Laws of exponents\n1. 43a+3\n2. 72\n3. 9p10\n4. k2x−2\n5. 52z−2 + 5z\n6. 1\n7. x10\n8.\nb2\na2\n9.\n1\nm+n\n10. 2pts\n11.\n1\na\n12. k\n13. 2a+1\n14. h4\n15.\na4b6\nc6d2\n16. 4\n17.\nm2n2\n2\n18. 400\n19.\n1\ny7\n20. 8\n21. 26a+2\n22. 2pt\n23. 81q2sy8a+2\nExercise 1 – 3: Rational exponents and surds\n1.\na) 7\nb)\n1\n6\nc)\n1\n3√\n36\nd) −4\n3\ne) 8x3\n2.\na) s\n1\n6\nb) 16m4\nc)\n3\n2 m2\nd) 8\n3. x\n31\n16\n483\nChapter 12.\nLinear programming\n\nExercise 1 – 4: Simplification of surds\n1.\na) 4\nb) ab4c2\nc) 2\nd) xy4\n2.\na)\nab\nb−a\nb) −\n\u0010\na\n1\n2 + b\n1\n2\n\u0011\nExercise 1 – 5: Rationalising the denominator\n1. 2\n√\n5\n2.\n√\n6\n2\n3.\n√\n6\n4.\n3\n√\n5 + 3\n4\n5.\nx√y\ny\n6.\n√\n6 +\n√\n14\n2\n7.\n3p −4√p\np\n8.\n√\nt −2\n9.\n1−√m\n1−m\n10.\n√\nab\nExercise 1 – 6: Solving surd equations\n1. x = 4\n2. p = 3\n3. y = 1\n4. t = 3\n5. z = 9 or z = 1\n4\n6. x = 8 or x = −27\n7. n = −1\n4\n8. d = 3 or d = −5\n9. y = 1 or y = 81\n10. f = 5\nExercise 1 – 7: Applications of exponentials\n1. 9,7%\n2. 4 254 691\n3. 7\n4. 26 893\n484\n12.1.\nIntroduction\n\nExercise 1 – 8: End of chapter exercises\n1.\na)\n1\n4\nb) 4 1\n4\n2.\na) x4\nb) s\nc) m\n25\n3\nd) m\n8\n3\ne) −m\n8\n3\nf) 81y\n16\n3\n3.\na)\n3b\n45\n2\n(a12c\n5\n2\nb) 3a3b2\nc) a24b12\nd) x\n7\n2\ne) x\n4\n3 b\n5\n3\n4.\n1\nx\n1\n16\n5. x −2\n6.\n10√x + 10\nx −1\n7.\n3√x + 2x√x\n2x\n8.\na) 6\n√\n2\nb) 7\n√\n5\nc) 2\nd)\n1\n4\n√\n2\ne) 2\nf)\n16\n√\n15\n5\n9.\na) 6 + 4\n√\n2\nb) 6 + 5\n√\n2\nc) 4+2\n√\n2+2\n√\n3+2\n√\n6\n10.\na) 55\nb) 1\n11. 15\n√\n2x3\n12.\na) 1 + 2\n√\n5\n5\nb)\n2y + y√y −4√y −8\ny −4\nc) 2√x + 2\n√\n10\n13.\n3\n2\n15. 3\n16. −\n√\n288\n17.\na) 4\nb) −1\n3\nc) 3\nd) No solution\ne) x = 1\n8 or x = −8\n18.\nb) x = 1\n2\nEquations and inequalities\nExercise 2 – 1: Solution by factorisation\n1. t = 0 or t = −2\n2. y = −1\n3. s = ±5\n4. y = 3 or y = 2\n5. y = 4 or y = −9\n6. p = −2\n7. y = −3 or y = −8\n8. y = 6 or y = 7\n9. x = −7 or x = −2\n10. y = 4k or y = k\n11. y = 9 or y = −9\n12. y = ±\n√\n5\n13. h = ±6\n14. y = ±\n√\n14\n15. p = −2\n16. y = ±6\n√\n2\n17. f = 5\n2 or f = −3\n18. x = 1\n4\n19. y = 1\n7\n20. x ∈R, x ̸= ±3\n21. y = −13 or y = −1\n22. t = 3\n2 or t = −2\n23. m = −6\n24. t = 0 or t = 3\nExercise 2 – 2: Solution by completing the square\n1.\na) x = −5 −3\n√\n3 or x = −5 + 3\n√\n3\nb) x = −1 or x = −3\nc) p = −4 ±\n√\n21\nd) x = −3 ±\n√\n7\ne) No real solution\nf) t = −8 ± 3\n√\n6\ng) x = −1 ±\nq\n5\n3\nh) z = −4 ±\n√\n22\ni) z = 11\n2 or z = 0\nj) z = 5 or z = −1\n2. k = −3 ± √9 −a\n3. y = −q±√\nq2−4pr\n2p\n485\nChapter 12.\nLinear programming\n\nExercise 2 – 3: Solution by the quadratic formula\n1. t = 1 or t = −4\n3\n2. x = 5+\n√\n37\n2\nor t = 5−\n√\n37\n2\n3. No real solution\n4. p = 1\n2 or p = −1\n5. No real solution\n6. t = −3+\n√\n69\n10\nor t = −3−\n√\n69\n10\n7. t = 2 ±\n√\n2\n8. k = 7+\n√\n373\n18\nor k = 7−\n√\n373\n18\n9. f = 1\n2 or f = −2\n10. No real solution\nExercise 2 – 4:\n1. x = −1, x = −4, x = −2 and x = −3\n2. x = 1, x = 4 and x = −2\n3. x = −7, x = 4, x = −1 and x = −2\n4. x = −4, x = 3, x = −3 and x = 2\n5. x = 8±\n√\n40\n4\n6. x = −5, x = 3, x = −1 +\n√\n10 and\nx = −1 −\n√\n10\nExercise 2 – 5: Finding the equation\n1. x2 −x −6 = 0\n2. x2 −16 = 0\n3. 2x2 −5x −3 = 0\n4. k = 3 and x = 3\n4\n5. p = 5 and x = −1\nExercise 2 – 6: Mixed exercises\n1. y = 1\n8 or y = −8\n3\n2. x = 3\n2 or x = −7\n2\n3. t = 2\n3 or t = 2\n4. y = 1 or y = −1\n5. m = 1 or m = 4\n6. y = ± 5\n7\n7. w = 3\n2 or w = 4\n8. y = 6\n5 or y = 1\n4\n9. n = 8\n3 or n = −9\n8\n10. y = −8\n3 or y = 3\n2\n11. x = −1\n2 or x = 3\n12. y = −5\n2 or y = −5\n9\n13. y = 4\n5 or y = 1\n5\n14. g = −1\n4 or g = 1\n15. y = 2 or y = −5\n9\n16. p = 3\n7 or p = −1\n5\n17. y = −2\n9 or y = −1\n18. y = 9\n2 or y = 9\n7\n486\n12.1.\nIntroduction\n\nExercise 2 – 7: From past papers\n1.\na) Real, unequal and rational\nb) Real and equal\nc) Real, unequal and irrational\nd) Real, unequal and rational\ne) Real, unequal and irrational\nf) Non-real\ng) Real, unequal and rational\nh) Real, unequal and irrational\ni) Non-real\nj) Real and equal\n2.\nb) real and unequal\nc) k = −6 ± 2\n√\n6\n4.\na) k = 6\nb) k = 1\n3\n5.\na) k = 4 or k = 1\nb) k = 0 or k = 5\n6.\na) all real values of a, b and p\nb) a = b and p = 0\nExercise 2 – 8: Solving quadratic inequalities\n1.\na) −3 < x < 4\nb) x < −4\n3 or when x > 1\nc) no real solutions\nd) −1 < t < 3\ne) All real values of s.\nf) All real values of x.\ng) x ≤−1\n4 or x ≥0\ni) x < 3 or x > 6 with x ̸= 3\nj) −2 ≤x ≤2 and x > 7 with x ̸= 7\nk) x > 0 with x ̸= 0\n2.\na) x < −3 or x > 3\nb) −\n√\n5 ≤x ≤\n√\n5\nc) no solution\nd) All real values of x\nExercise 2 – 9: Solving simultaneous equations\n1.\na) (0; 5) and (2; 3)\nb) x = 3 ±\n√\n2 and y = 2 ±\n√\n2\nc) (−1; 0) and ( 1\n4 ; 5\n8 )\nd) b = 2 ±\n√\n88\n6\nand a = 11 ±\n√\n88\n6\ne) (−3; −20) and (2; 0)\nf) x = 6 ±\n√\n264\n2\nand y = 70 ±\n√\n264\n2\n2.\na) (−3; 8) and (2; 3)\nb) (−4; 14) and (3; 7)\nc) (3; 4) and (4; 3)\nExercise 2 – 10:\n1. b = 2 m, l = 4 m\n2. 187\n3. t = 10,5 s\n4. t = 5d; 105 minutes; 1,4 km\n5. 24 A; 70 W; 12 A\n487\nChapter 12.\nLinear programming\n\nExercise 2 – 11: End of chapter exercises\n1. x = 1,62 or x = −0,62\n2. x = ±4 or x = −1\n3. y = 0 or y = ±1\n4. x = ±2\n5.\na) x = 7 or x = 2\nb) x = 2,3 or x = −1,3\nc) x = 1,65 or x = −3,65\nd) x = 0 or x = −3\n6. x =\n√\n16+p2−2\n2\n7. a = 3; b = 10 and c = −8\n8. p = ±16\n9. x2 + 2x −15\n10. Undefined:b = −2 Zero:b = 2 or b = 3\n13. a ≥4\n14. x = −3\n2 or x = 1\n15.\na) x < 3 or x ≥7:\nb) x < 1 or x > 5:\nc) 3 < x < 7:\nd) x < −1 or x > 3\ne) 0,5 < x < 2,5\nf) x ≤−3 or 0 < x ≤5\n2\ng) x < 2\n3\nh) −1 ≤x < 0 or x ≥3\ni) −4 ≤x ≤1\nj) 2 1\n2 ≤x < 3\n16.\na) x = ±\n√\n3 and y = ±2\n√\n3\nb) a = −3 and b = −1 or a = 12 and b = 4\nc) x = −5 and y = 0 or x = 2 and y = 14\nd) p = 5\n3 and q = 2\n9 or p = −1 and q = −2\n3\ne) b = 3±\n√\n5\n2\nand a = 7±3\n√\n5\n2\nf) b = −10±\n√\n140\n4\nand a = −12±\n√\n140\n2\ng) x = 3,4 and y = 5,4 or x = 3 and y = 5\nh) b = −1,4 and a = 23,6 or\nb = 3 and a = 6\n17.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\nx\n0\ny\nb\nb\nb)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n1\n2\n3\n4\n5\n6\n−1\n−2\nx\n0\ny\nb\nb\n18. 35 m\n20.\na) y = −5\n4 or y = −9\nb) x = −9\n4 or x = 1\nc) p = −8\n7 or p = −4\n3\nd) y = −1\n4 or y = 1\n2\ne) y = −2\n9 or y = −1\nf) y = 7\n3 or y = −1\n2\ng) y = 9\n4 or y = −9\n4\nh) y = 8\n3 or y = −6\ni) y = 9\n5 or y = −7\nj) x = ±4\nk) y = ±7\n21. k = 76 and 4\n9\n22. x = 3 or x = −2 and y = 1±√−7\n2\n23. x = 4 or x = −1\n24. y = 3\n2 , y = 1\n2 and p = 9\n2 , p = 7\n2\n25.\n69\n4\n26. 7\n27.\n2±\n√\n12\n2\n28. t = 1\n2 , t = 1 or t = 3±\n√\n33\n4\n488\n12.1.\nIntroduction\n\n3\nNumber patterns\nExercise 3 – 1: Linear sequences\n1. −19; −35; −51\n2.\na) −19\nb) T2 = 15; T4 = 33\n3.\na) Tn = 10 + 3n; T10 = 40; T15 = 55;\nT30 = 100\nb) Tn = 12 + 6n; T10 = 72; T15 = 102;\nT30 = 192\nc) Tn = −5 −5n; T10 = −55; T15 = −80;\nT30 = −155\n4. T9 = 36\n5.\na) 44; 66; 121\nExercise 3 – 2: Quadratic sequences\n1.\na) 10\nb) 2\nc) 2\nd) −2\ne) 2\nf) −4\ng) 4\nh) −2\ni) 6a\nj) 6\nk) 2t\n2.\na) T4 = 53\nb) T2 = 30\nc) T1 = 17\nd) T2 = −3\ne) T4 = 63\nf) T1 = 2\n3.\na) 3; 9; 17; 27\nb) −6; −9; −14; −21\nc) 1; 8; 21; 40\nd) 0; −5; −14; −27\nExercise 3 – 3: Quadratic sequences\n1.\na) 1\nb) 2\nc) 4\nd) 8\ne) −2\n2. 12; 30; 58; 96; 144\n3. T9 = 379\n4. n = 4\n5.\na) T5 = 84; T6 = 111\nb) Tn = 2n2 + 5n + 9\n489\nChapter 12.\nLinear programming\n\nExercise 3 – 4: End of chapter exercises\n1. −4; 9; 16; 25; 36\n2.\na) Quadratic sequence\nb) Quadratic sequence\nc) Quadratic sequence\nd) Quadratic sequence\ne) Quadratic sequence\nf) Quadratic sequence\ng) Linear sequence\nh) Linear sequence\ni) Quadratic sequence\nj) Quadratic sequence\nk) Quadratic sequence\nl) Linear sequence\nm) Quadratic sequence\n3. x = 31\n4. n = 11\n5. T11 = 363\n6. n = 9\n7. T5 = 114\n8. n = 8\n9.\na) T5 = 19;\nTn = 4n −1;\nT10 = 39\nb) T5 = −3;\nTn = 22 −5n;\nT10 = −28\nc) T5 = 2 1\n2 ; Tn = 1\n2 n;\nT10 = 5\nd) T5 = a + 4b;\nTn = a −b + bn;\nT10 = a + 9b\ne) T5 = −7;\nTn = 3 −2n;\nT10 = −17\n10.\na) Tn = n2 + 3;\nT100 = 10 003\nb) Tn = 6n −4;\nT100 = 596\nc) Tn = 2n2 + 5;\nT100 = 20 005\nd) Tn = 3n2 + 2;\nT100 = 30 002\n11.\na) 2; 5; 8; 11; 14\nb) Constant difference,\nd = 3\nc) Yes\n12.\na) Tn = 4n −19\nb) n = 48\n13.\na) Incorrect\nb) Correct\n14.\nc) Linear\n15.\nb) Linear\nd) Quadratic\ne) Tn = 1\n2 n2 + 3\n2 n + 1\nf) T21 = 253\ng) 31 cm\n16.\na) −1\nb) 7\n17.\nb) 2\nc) Tn = n2 −n\nd) 210\ne) 25\n18. 4; 14; 34; 64; 104; 154\n4\nAnalytical geometry\nExercise 4 – 1: Revision\n1.\na) 2\n√\n26units\nb) 7 units\nc) x + 1units\n2. p = 6 or p = 2\n3.\na) −1\n2\nb) 3\n5. 2\n6.\na) (1; 2)\nb)\n\u0000 −1\n2 ; −1\n2\n\u0001\n7. B(4; 2)\n8.\na) y = −4x + 3 and\ny = −4x + 19\nc) AD =\n√\n17units and\nBC =\n√\n17units\nd) y = 4\n3 x −7\n3\ne) Parallelogram (one\nopposite side equal\nand parallel)\n9. N(0; 3)\n10.\na) PQ =\n√\n20 and\nSR =\n√\n20\nb) M( 3\n2 ; 1)\nd) PS: y = −2\n5 x −1\n5\nand SR: y = 1\n2 x −2\ne) No\nf) Parallelogram\nExercise 4 – 2: The two-point form of the straight line equation\n1. y = 2\n3 x + 5\n2. y = −3x + 1\n4\n3. y = x + 3\n4. y = 2x −1\n5. y = −5\n6. y = 3\n4 x + 3\n7. y = −x + (s + t)\n8. y = 5x + 2\n9. y = q\npx −q\n490\n12.1.\nIntroduction\n\nExercise 4 – 3: Gradient–point form of a straight line equation\n1. y = 2\n3 x + 4\n2. y = −x −2\n3. y = −1\n3 x\n4. y = 11\n5. y = −2x + 7\n6. x = −3\n2\n7. y = −4\n5 x + 1\n8. x = 4\n9. y = 3ax + b\nExercise 4 – 4: The gradient–intercept form of a straight line equation\n1. y = 2x + 3\n2. y = 4x −4\n3. y = −x −1\n4. y = −3\n7 x\n5. y = 1\n2 x −1\n5\n6. y = 2x −2\n7. y = −3\n2\n8. y = 3x + 4\n9. y = −5x\nExercise 4 – 5: Angle of inclination\n1.\na) 1,7\nb) −1\nc) 0\nd) 1,4\ne) Undefined\nf) 1\ng) −0,8\nh) 0\ni) 3,7\n2.\na) 36,8◦\nb) 26,6◦\nc) 45◦\nd) Horizontal line\ne) 18,4◦\nf) Vertical line\ng) 71,6◦\nh) 30◦\nExercise 4 – 6: Inclination of a straight line\n1.\na) 38,7◦\nb) 135◦\nc) 80◦\nd) 80◦\ne) 102,5◦\nf) 45◦\ng) 56,3◦\nh) 63,4◦\ni) 161,6◦\nj) Gradient undefined\n2. 85,2◦\n3. 90◦\n4. 81,8◦\nExercise 4 – 7: Parallel lines\n1.\na) Parallel\nb) Parallel\nc) Parallel\nd) Not parallel\ne) Parallel\nf) Parallel\n2. y = −2x −3\n3. y = 3x\n4. y = 3\n2 x + 1\n5. y = −7\n10 x −1\n491\nChapter 12.\nLinear programming\n\nExercise 4 – 8: Perpendicular lines\n1.\na) Perpendicular\nb) Not perpendicular\nc) Perpendicular\nd) Perpendicular\ne) Perpendicular\nf) Not perpendicular\ng) Not perpendicular\n2. y = 1\n2 x −3\n3. y = −5x + 3\n4. y = −x + 2\n5. x = −2\nExercise 4 – 9: End of chapter exercises\n1.\na) y = 1\n2 x + 7\n2\nb) y = −x + 4\nc) y = 1\n2 x + 4\nd) y = 2x + 4\ne) y = 3x\n2.\na) θ = 63,4◦\nb) θ = 18,4◦\nc) θ = 36,9◦\nd) θ = 146,3◦\ne) θ = 161,6◦\n3.\na) y = −2x + 7\nb)\n\u0000 7\n2 ; 0\n\u0001\nc) θ = 116,6◦\nd) m = 1\n2\ne) Q ˆPR = 90◦\nf) y = −2x\ng)\n\u0000 1\n2 ; −3\n2\n\u0001\nh) y = −2x −1\n2\n4.\na)\n1\n2\n3\n4\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\nb\nb\nb\ny\nx\nA(−3; 5)\nB(−7; −4)\nC(2; 0)\nD(x; y)\nb) D (6; 9)\n5.\na) (−1; −2)\nb) (8; 3)\nc) x = −1\nd) MN = 5 units\ne) M ˆ\nNP = 21,8◦\nf) y = 5\n2 x + 11\n2\n6.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nA(−2; 3)\nB(2; 4)\nC(3; 0)\ny\nx\n0\nb\nb\nb\nc) y = 1\n4 x + 7\n2\nd) D(−1; −1)\ne) E\n\u0000 5\n2 ; 2\n\u0001\n7.\na) y = 3\n2 x + 2\nb) T ˆSV = 49,6◦\n8.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nb\nb\nb\nF(−1; 3)\nH(4; 4)\nG(2; 1)\ny\nx\n0\nc) y = −5x + 11\nd) Yes\ne) y = 3\n2 x + 9\n2\n9.\na)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nb\nb\nb\nA(−1; 5)\nB(5; −3)\nC(0; −6)\nx\ny\nM\nN\n492\n12.1.\nIntroduction\n\n5\nFunctions\nExercise 5 – 1: Revision\n1.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nc)\n1\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nd)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nExercise 5 – 2: Domain and range\n1. {x : x ∈R} ; {y : y ≥−1, y ∈R}\n2. {x : x ∈R} ; {y : y ≤4, y ∈R}\n3. {x : x ∈R} ; {y : y ≥0, y ∈R}\n4. {x : x ∈R} ; {y : y ≤0, y ∈R}\n5. {x : x ∈R} ; {y : y ≤2, y ∈R}\nExercise 5 – 3: Intercepts\n1. (0; 15) and (−5; 0); (−3; 0)\n2. (0; 16) and (4; 0)\n3. (0; −3) and (1; 0); (3; 0)\n4. (0; 35) and (−7\n2 ; 0); (−5\n2 ; 0)\n5. (0; 37) and no x-intercepts\n6. (0; −4) and\n(−0,85; 0); (−2,35; 0)\nExercise 5 – 4: Turning points\n1. (3; −1)\n2. (2; 1)\n3. (−2; −1)\n4. (−1\n2 ; 1\n2 )\n5. (1; 21)\n6. (−1; −6)\nExercise 5 – 5: Axis of symmetry\n1.\na) Axis of symmetry:\nx = 5\n4\nb) Axis of symmetry:\nx = 2\nc) Axis of symmetry:\nx = 2\n2. y = ax2 + q\n493\nChapter 12.\nLinear programming\n\nExercise 5 – 6: Sketching parabolas\n1.\na)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n1\n2\n3\n4\n5\n6\n−1\ny\nx\n0\nIntercepts: (−1; 0), (5; 0), (0; 5)\nTurning point: (2; 9)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≤9, y ∈R}\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−1; 0), (0; 2)\nTurning point: (−1; 0)\nAxes of symmetry: x = −1\nDomain: {x : x ∈R}\nRange: {y : y ≥0, y ∈R}\nc)\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−0,87; 0), (1,54; 0), (0; −4)\nTurning point: (0,33; −4,33)\nAxes of symmetry: x = −0,33\nDomain: {x : x ∈R}\nRange: {y : y ≥4,33, y ∈R}\nd)\n1\n2\n3\n4\n5\n6\n−1\n1\n2\n3\n4\n−1\ny\nx\n0\nIntercepts: (0; 13) Turning point: (2; 1)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≥1, y ∈R}\n3.\na)\ny\nx\n0\nb)\ny\nx\n0\nc)\ny\nx\n0\nd)\ny\nx\n0\ne)\ny\nx\n0\nf)\ny\nx\n0\n4.\na) yshifted = 2x2 + 16x + 32\nb) yshifted = −x2 −2x\nc) yshifted = 3x2 −16x + 22\n494\n12.1.\nIntroduction\n\nExercise 5 – 7: Finding the equation\n1. y = −3(x + 1)2 + 6 or y = −3x2 −6x + 3\n2. y = 1\n2 x2 −5\n2 x\n3. y = 2\n3 (x + 2)2\n4. y = −x2 + 3x + 4\nExercise 5 – 8:\n1.\na) 11\n2.\na)\n1\n2\n3\n4\n1\n2\n−1\n−2\nf(x)\nx\n0\nA(1; 3)\nb\nb) 6\nc) y = 6x −3\n3.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n−1\n−2\ng(x)\nx\n0\nb) 1\nc) 4\nd) 0\nExercise 5 – 10: Domain and range\n1. {x : x ∈R, x ̸= 0} ; {y : y ∈R, y ̸= 1}\n2. {x : x ∈R, x ̸= 8} ; {y : y ∈R, y ̸= 4}\n3. {x : x ∈R, x ̸= −1} ; {y : y ∈R, y ̸= −3}\n4. {x : x ∈R, x ̸= 5} ; {y : y ∈R, y ̸= 3}\n5. {x : x ∈R, x ̸= −2} ; {y : y ∈R, y ̸= 2}\nExercise 5 – 11: Intercepts\n1. (0; −1 3\n4 ) and\n\u0000−3 1\n2 ; 0\n\u0001\n2.\n\u0000 5\n2 ; 0\n\u0001\n3. (0; 1) and\n\u0000 1\n3 ; 0\n\u0001\n4.\n\u00000; 3\n2\n\u0001\nand\n\u0000 1\n3 ; 0\n\u0001\n5. (0; 2) and (8; 0)\nExercise 5 – 12: Asymptotes\n1. y = −2 and x = −4\n2. y = 0 and x = 0\n3. y = 1 and x = 2\n4. y = −8 and x = 0\n5. y = 0 and x = 2\n495\nChapter 12.\nLinear programming\n\nExercise 5 – 13: Axes of symmetry\n1.\na) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (0; 1); y1 = x + 1 and\ny2 = −x + 1\nb) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (−1; 0); y1 = x + 1 and\ny2 = −x −1\nc) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (1; −1); y1 = x −2 and\ny2 = −x\n2. k(x) =\n5\nx+1 + 2\nExercise 5 – 14: Sketching graphs\n1.\na) Asymptotes: x = 0; y = 2\nIntercepts:\n\u0000−1\n2 ; 0\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= 0}\nRange: {y : y ∈R, y ̸= 2}\nb) Asymptotes: x = −4; y = −2\nIntercepts:\n\u0000−3 1\n2 ; 0\n\u0001\nand\n\u00000; −1 3\n4\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x −6\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\nc) Asymptotes: x = −1; y = 3\nIntercepts:\n\u0000−2\n3 ; 0\n\u0001\nand (0; 2)\nAxes of symmetry: y = x + 4 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 3}\nd) Asymptotes: x = −2 1\n2 ; y = −2\nIntercepts: (0; 0)\nAxes of symmetry: y = x −4 1\n2 and\ny = −x + 1\n2\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\ne) Asymptotes: x = 8; y = 4\nIntercepts: (6; 0) and (0; 3)\nAxes of symmetry: y = x −4 and\ny = −x + 12\nDomain: {x : x ∈R, x ̸= 8}\nRange: {y : y ∈R, y ̸= 4}\n2. y =\n1\nx+2 −1\n3. y = −4\nx + 2\n4.\na)\nb) Average gradient = 1\nc) Average gradient = 12\nExercise 5 – 16: Domain and range\n1. {x : x ∈R} ; {y : y > 0, y ∈R}\n2. {x : x ∈R} ; {y : y < 1, y ∈R}\n3. {x : x ∈R} ; {y : y > −3, y ∈R}\n4. {x : x ∈R} ; {y : y > n, y ∈R}\n5. {x : x ∈R} ; {y : y > 2, y ∈R}\nExercise 5 – 17: Intercepts\n1. (0; −6) and (2; 0)\n2. (0; −17 1\n3 ) and (3; 0)\n3. (0; −20) and (−1; 0)\n4. (0; 15\n16 ) and (−2; 0)\n496\n12.1.\nIntroduction\n\nExercise 5 – 18: Asymptote\n1. y = 0\n2. y = 1\n3. y = −2\n3\n4. y = −2\n5. y = −2\nExercise 5 – 19: Mixed exercises\n1.\nb)\ni. y = 3\nx + 3\nii. y =\n3\nx−3\niii. y = −3\nx\niv. y = 3\nx −1\n4\nv. y = 3\nx + 4\nvi. y =\n3\nx+2 −1\n2.\na) M(−2; 2)\nb) g(x) = −4\nx\nc) f(x) = 2(x + 1)2\nd) −2 < x < 0\ne) Range: {y : y ∈R, y ≥0}\n3.\na) For k(x) :\nIntercepts:\n(−2; 0), (1; 0) and (0; −4)\nTurning point:\n\u0000−1\n2 ; −4 1\n2\n\u0001\nAsymptote:\nnone\nFor h(x) :\nIntercepts:\n(1,41; 0)\nTurning point:\nnone\nAsymptote:\ny = 0\n6.\na) f(x) = −3\n4 (x −2)2 + 3 ;\nAxes of symmetry: x = 2 ;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≤3}\nb) g(x) = 1\n4 x2 −2;\nAxes of symmetry: x = 0;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≥−2}; h(x) = 2\nx ;\nAxes of symmetry: y = x\nDomain: {x : x ∈R, x < 0};\nRange: {y : y ∈R, y < 0};\nc) k(x) =\n\u0000 1\n2\n\u0001x + 1\n2 ;\nDomain: {x : x ∈R};\nRange:\n\b\ny : y ∈R, y > 1\n2\n\t\n7.\nb) p = 9\nc) Average gradient = −2 8\n9\nd) y =\n\u0000 1\n3\n\u0001x+2 −2\n8.\na) f(x) = 2x −3\n2 and g(x) = −1\n4 x −1\n2\nb) h(x) = −\n3\nx+2 + 1\n9.\na) AO = 2 units OB = 5 units\nOC = 10 units DE = 12,25 units\nb) DE = 12 1\n4\nc) h(x) = −2x + 10\nd) {x : x ∈R, x < −2 and x > 5}\ne) {x : x ∈R, 0 ≤x ≤5}\nf) 5,25 units\n497\nChapter 12.\nLinear programming\n\nExercise 5 – 20: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\nPeriod: = 360◦\nAmplitude: = 1\nDomain: = [0◦; 360◦]\nRange: = [−1; 1]\nx-intercepts: = (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: = (0◦; 0)\nMax. turning point: = (90◦; 1)\nMin. turning point: = (270◦; −1)\n2.\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny2 = −2 sin θ\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 2\nDomain: [0◦; 360◦]\nRange: [−2; 2]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nMax. turning point: (270◦; 2)\nMin. turning point: (90◦; −2)\n3.\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny3 = sin θ + 1\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [0; 2]\nx-intercepts: (270◦; 0)\ny-intercepts: (0◦; 1)\nMax. turning point: (90◦; 2)\nMin. turning point: (270◦; 0)\n4.\n1\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny4 = 1\n2 sin θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (90◦; 1\n2 )\nMin. turning point: (270◦; −3\n2 )\nExercise 5 – 21: Sine functions of the form y = sin kθ\n2.\na) k = 2\nb) k = −3\n4\n498\n12.1.\nIntroduction\n\nExercise 5 – 23: The sine function\n1.\na)\n1\n2\n−1\n−2\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 2 sin( θ\n2 )\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 1\n2 sin(θ −45◦)\nc)\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(θ + 90◦) + 1\nd)\n1\n−1\n60◦\n120◦\n180◦\n−60◦\n−120◦\n−180◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(−3θ\n2 )\ne)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(30◦−θ)\n2. a = 2; p = 90◦∴y = 2 sin(θ + 90◦) and\ny = 2 cos θ\nExercise 5 – 24: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\n2.\n1\n2\n3\n−1\n−2\n−3\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny2 = −3 cos θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−3; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; −3)\n3.\n1\n2\n3\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny3 = cos θ + 2\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [1; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; 1)\n4.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny4 = 1\n2 cos θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (0◦; −1\n2 ); (360◦; −1\n2 )\nMin. turning point: (180◦; −3\n2 )\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n2.\na) k = 3\n2\nb) k = 2\n3\n499\nChapter 12.\nLinear programming\n\nExercise 5 – 27: The cosine function\n1.\na)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nθ\n0◦\ny\ny = cos θ\ny = cos(θ + 15◦)\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny = cos θ\nf(θ) = 1\n3 cos(θ −60◦)\nc)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ny = −2 cos θ\nd)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(30◦−θ)\ne)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ng(θ) = 1 + cos(θ −90◦)\nf)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(2θ + 60◦)\n2.\na) a = −1\nb) p = −180◦\nc) cos(θ −180◦) = −cos θ\nExercise 5 – 28: Revision\n1.\n1\n2\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1\n2 )\nAsymptotes: 90◦; 270◦\n2.\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: 90◦; 270◦\n3.\n1\n2\n3\n4\n5\n−1\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (116,6◦; 0); (296,6◦; 0)\ny-intercepts: (0◦; 2)\nAsymptotes: 90◦; 270◦\n4.\n1\n2\n−1\n−2\n−3\n−4\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1)\nAsymptotes: 90◦; 270◦\n500\n12.1.\nIntroduction\n\nExercise 5 – 29: Tangent functions of the form y = tan kθ\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦] Range: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 240◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −120◦; 120◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n4.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 270◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; 135◦\n501\nChapter 12.\nLinear programming\n\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−225◦; 0); (−45◦; 0); (135◦; 0);\n(315◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−330◦; 0); (−150◦; 0); (30◦; 0);\n(210◦; 0)\ny-intercepts: (0◦; −0,58)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−240◦; 0); (−60◦; 0); (120◦; 0);\n(300◦; 0)\ny-intercepts: (0◦; 1.73)\nAsymptotes: −330◦; −150◦; 30◦; 210◦\nExercise 5 – 31: The tangent function\n1.\na)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n−45◦\n−90◦\nθ\ny\nb)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\nθ\ny\nc)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nd)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\ny\n2. a = −1; k = 1\n2\n502\n12.1.\nIntroduction\n\nExercise 5 – 32: Mixed exercises\n1.\na) f(θ) = 3\n2 sin 2θ and g(θ) = −3\n2 tan θ\nb) f(θ) = −2 sin θ and\ng(θ) = 2 cos(θ + 360◦\nc) y = 3 tan θ\n2\nd) y = y = 2 cos θ + 2\n2.\na)\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\nθ\n0\ng\nf\ny\nb\nb\nb\nb\n(90◦; 2)\n(270◦; −2)\nb) 360◦\nc) 1\nd) At θ = 180◦\n3.\na) a = 2, b = −1 and c = 240◦\nb) 180◦\nc) θ = 60◦; 300◦\nd) y = −tan(θ −45◦)\n4.\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny1\ny2\ny\nb\nb\nExercise 5 – 33: End of chapter exercises\n2. a = −2; k = −1\n4.\na) y = x + 22 + 2\nb) y = x −12 + 5\n5. (−1; 0)\n6.\ny\nx\n0\n4\n−4\n4\n−4\ny =\n2\nx−3 −1\n7. y =\n1\n(x−1) + 2\n9.\na) a = −1\nb) f(−15) = 0,99997\nc) x = −1\nd) h(x) = −2(x−2) + 1\n10.\na) a = 256\nb) f(x) = 256\n\u0000 3\n4\n\u0001x\nc) f(13) = 6,08\n11.\na)\n1\n−1\n90◦\n180◦\n−90◦\n−180◦\nθ\n0\ny\nb)\n1\n−1\n90◦\n180◦\nθ\n0\ny\nd)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\n0\ny\ne)\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny\nf)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\n503\nChapter 12.\nLinear programming\n\n6\nTrigonometry\nExercise 6 – 1: Revision\n1.\na) True\nb) True\nc) False\nd) True\n2.\na) 50,2◦\nb) 40,5◦\nc) 26,6◦\nd) 109,8◦\ne) No solution\nf) 17,7◦\ng) 69,4◦\n3.\na) 17,3 cm\nb) 10 cm\nc) 64,8◦\n4.\na) 10 cm\nb) 5,2 cm and 19,3 cm\nc) 50 cm2\n5.\na) 2\nb) 0\nc) −1 1\n2\nd) 1\ne) 1\n6.\na) 60◦\nb)\n1\n2\nc) 1\n7. No\nExercise 6 – 2: Trigonometric identities\n1.\na) cos α\nb) tan2 θ\nc) cos2 θ\nd) 0\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1.\na)\n√\n3\n3\nb)\n1\n8\nc) 1\n2.\na)\n1−cos2 θ\ncos θ\nb) −1\n3.\na) 2t\nb) −1\nt\nExercise 6 – 4: Using reduction formula\n1.\na) −tan θ\nb) 1\nc) 1\n2. −cos β\n3.\na)\n1\n√\n3\nb) 2\nc) 2\nd) −3\n2\ne) −4\n√\n3\n5\n5.\na) −t\nb) 1 −t2\nc) ±\nt\n√\n1−t2\nExercise 6 – 5: Co-functions\n1.\na) cos θ\nb)\n3\n2\n2.\na) p\nb)\np\n1 −p2\nc) −\np\n√\n1−p2\nd) p\n504\n12.1.\nIntroduction\n\nExercise 6 – 6: Reduction formulae\n1.\na) sin2 θ\nb) cos2 θ\nc)\ni. 1\nii. tan2 θ\n2.\na) sin 17◦\nb) cos 33◦\nc) tan 68◦\nd) −cos 33◦\n3.\na)\n√\n3\nb)\n√\n3\n2\nc)\n1\n4\nd) 1\nExercise 6 – 7: Solving trigonometric equations\n1.\na) α = 60◦; 300◦\nb) α = 220,5◦; 319,5◦\nc) α = 79,2◦; 259,2◦\nd) α = 200,1◦; 339,9◦\ne) α = 36,9◦; 143,1◦\nf) α = 109,7◦; 289,7◦\n2.\na) θ = −323,1◦; −216,9◦; 36,9◦; 143,1◦\nb) θ = −221,4◦; −138,6◦; 138,6◦; 221,4◦\nc) θ = −278,5◦; −98,5◦; 81,5◦; 261,5◦\nd) θ = −90◦; 270◦\ne) θ = −293,6◦; −66,4◦; 66,4◦; 293,6◦\nExercise 6 – 8: General solution\n1.\na) θ = −128,36◦; −101,64◦; 51,64◦\nb) θ = −80,45◦; −9,54◦; 99,55◦; 170,46◦\nc) θ = −53,27◦; 126,73◦\nd) α = 0◦\ne) θ = −180◦; 0◦; 180◦\nf) θ = −180◦; 180◦\ng) θ = 84◦\nh) θ = −120◦; 120◦\ni) θ = −60◦; −30◦; 120◦; 150◦\n2.\na) θ = −20◦+ n . 360◦\nb) α = 30◦+ n . 120◦\nc) β = 10,25◦+ n . 45◦or\nβ = 55,25◦+ n . 45◦\nd) α = 70◦+ n . 360◦or\nα = 340◦+ n . 360◦\ne) θ = 140◦+ n . 240◦or\nθ = 220◦+ n . 240◦\nf) β = 15◦+ n . 180◦\nExercise 6 – 9: Solving trigonometric equations\n1.\na) θ = 45◦+ k . 180◦or\nθ = 135◦+ k . 180◦\nb) α = 50◦+ k . 360◦or\nα = 110◦+ k . 360◦\nc) θ = 60◦+ k . 720◦or\nθ = 660◦+ k . 720◦\nd) β = 146,6◦+ k . 180◦\ne) θ = 110,27◦+ k . 360◦or\nθ = 249,73◦+ k . 360◦\nf) α = 210◦+ k . 360◦or\nα = 330◦+ k . 360◦\ng) β = 23,3◦+ k . 120◦\nh) θ = 122◦+ k . 180◦\ni) α = 21◦+ k . 180◦or\nα = 39,5◦+ k . 90◦\nj) β = 22,5◦+ k . 90◦\n2. θ = 0◦, 180◦, 210◦, 330◦or 360◦\n3.\na) θ = 120◦+ k . 360◦or\nθ = 240◦+ k . 360◦\nb) θ = 0◦+ k . 180◦or\nθ = 146,3◦+ k . 180◦\nc) α = 36,9◦+ k . 360◦or\nα = 143,1◦+ k . 360◦or\nα = 216,9◦+ k . 360◦or\nα = 323,1◦+ k . 360◦\nd) β = 15◦+ k . 120◦or\nβ = 75◦+ k . 120◦\ne) α = 48,4◦+ k . 180◦\nf) θ = 63,4◦+ k . 180◦or\nθ = 116,6◦+ k . 180◦\ng) θ = 54,8◦+ k . 180◦or\nθ = 95,25◦+ k . 180◦\n4. β = −70,5◦or β = 109,5◦\n505\nChapter 12.\nLinear programming\n\nExercise 6 – 10: The area rule\n1.\na)\nP\nQ\nR\n30◦\n10\n7\nArea △PQR = 17,5 square units\nb)\nP\nQ\nR\n110◦\n9\n8\nArea △PQR = 33,8 square units\n2. Area △XY Z = 645,6 square units\n3. Area = 106,5 square units\n4.\nˆC = 72,2◦or ˆC = 107,8◦\nExercise 6 – 11: Sine rule\n1.\na)\nˆP = 92◦, q = 6,6, p = 7,4\nb)\nˆL = 87◦, l = 1,3, k = 0,89\nc)\nˆB = 76,8◦, b = 94,3, c = 91,3\nd)\nˆY = 84◦, y = 60, z = 38,8\n2.\nˆB = 32◦, AB = 23, BC = 39\n3. ST = 78,1 km\n4. m = 26,2\n5. BC = 3,2\nExercise 6 – 12: The cosine rule\n1.\na) a = 8,5, ˆC = 83,9◦, ˆB = 26,1◦\nb)\nˆR = 120◦, ˆS = 32,2◦, ˆT = 27,8◦\nc)\nˆ\nM = 27,7◦, ˆL = 40,5◦, ˆ\nK = 111,8◦\nd) h = 19,1, ˆJ = 18,2◦, ˆ\nK = 31,8◦\ne)\nˆD = 34◦, ˆE = 44,4◦, ˆF = 101,6◦\n2.\na) x = 4,4 km\nb) y = 63,5 cm\n3.\na)\nˆ\nK = 117,3◦\nb)\nˆQ = 78,5◦\nExercise 6 – 13: Area, sine and cosine rule\n1.\na) 7,78 km\nb) 6 km\n2. XZ = 1,73 km, XY = 0,87 km\n3.\na) 1053 km\nb) 4,42◦\n4. DC = x sin a sin(b+c)\nsin(a+c) sin b\n5.\nb) 438,5 km\n6. 9,38 m2\n7. DC = x sin α\nsin β\n506\n12.1.\nIntroduction\n\nExercise 6 – 14: End of chapter exercises\n1. sin2 A\n2. 1 1\n4\n3. cos α\n4. 3\n7.\na) −1\nb) θ = 135◦or θ = 315◦\n8.\na)\nb\nx\ny\n0\nθ\n(−12; −5)\nb) −5\n13 and 12\n13\nc) θ = 202,62◦\n9.\na) a = 1 and b = −\n√\n3\nb) −\n√\n3\n2\n10.\na) x = 50,9◦or x = 309,1◦\nb) x = 127,3◦or x = 307,3◦\nc) x = 26,6◦; 153,4◦206,6◦or 333,4◦\n11.\na) x = 55◦+ k . 360◦or\nx = 175◦+ k . 360◦\nb) x = 180◦+ k . 360◦\n12.\na) x = 28,6◦+ k . 180◦or\nx = 61,4◦+ k . 180◦\nb)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nb\nb\nb\nb\nθ\n0◦\ny\ny = sin 2α\nc) 28,6◦; 61,4◦; 208,6◦; 241,4◦\n13.\na) A ˆGN = α −β\nb)\nˆ\nA = 90◦−α\nd) H = 5 m\n14.\na) AC = 9,43 m\nb) AD = 6,2 m\nc) Area = 49,25 m2\nd) Area = 49,23 m2\n7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "4.1" }, { "title": "Equation of a line", "content": "", "chapter_id": "4.2" }, { "title": "Inclination of a line", "content": "", "chapter_id": "4.3" }, { "title": "Parallel lines", "content": "", "chapter_id": "4.4" }, { "title": "Perpendicular lines", "content": "", "chapter_id": "4.5" }, { "title": "Summary", "content": "6.6\nSummary\n301\n\n6\nTrigonometry\n6.1\nRevision\nEMBHG\nTrigonometric ratios\nb\nb\nb\ny\nx\nP(x; y)\nQ(−x; y)\nO\nα\nβ\nr\nr\nWe plot the points P(x; y) and Q(−x; y) in the Cartesian plane and measure the angles\nfrom the positive x-axis to the terminal arms (OP and OQ).\nP(x; y) lies in the first quadrant with P ˆOX = α and Q(−x; y) lies in the second\nquadrant with Q ˆOX = β.\nUsing the theorem of Pythagoras we have that\nOP 2 = x2 + y2\nAnd OQ2 = (−x)2 + y2\n= x2 + y2\n∴OP = OQ\nLet OP = OQ = r.\nTrigonometric ratios\nsin α = y\nr\ncos α = x\nr\ntan α = y\nx\nIn the second quadrant we notice that −x < 0\nsin β = y\nr\ncos β = −x\nr\ntan β = −y\nx\n240\n6.1.\nRevision\n\nSimilarily, in the third and fourth quadrants the sign of the trigonometric ratios depends\non the signs of x and y:\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nSpecial angles\n30◦\n60◦\n1\n√\n3\n2\n45◦\n45◦\n1\n1\n√\n2\nθ\n0◦\n30◦\n45◦\n60◦\n90◦\ncos θ\n1\n√\n3\n2\n1\n√\n2\n1\n2\n0\nsin θ\n0\n1\n2\n1\n√\n2\n√\n3\n2\n1\ntan θ\n0\n1\n√\n3\n1\n√\n3\nundef\nSee video: 22XJ at www.everythingmaths.co.za\n241\nChapter 6.\nTrigonometry\n\nSolving equations\nWorked example 1: Solving equations\nQUESTION\nDetermine the values of a and b in the right-angled triangle TUW (correct to one\ndecimal place):\nT\nW\nU\n47◦\nb\n30\na\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of a\nsin θ = opposite side\nhypotenuse\nsin 47◦= 30\na\na =\n30\nsin 47◦\n∴a = 41,0\nStep 3: Determine the value of b\nAlways try to use the information that is given for calculations and not answers that you\nhave worked out in case you have made an error. For example, avoid using a = 41,0\nto determine the value of b.\ntan θ = opposite side\nadjacent side\ntan 47◦= 30\nb\nb =\n30\ntan 47◦\n∴b = 28,0\nStep 4: Write the final answer\na = 41,0 units and b = 28,0 units.\n242\n6.1.\nRevision\n\nFinding an angle\nWorked example 2: Finding an angle\nQUESTION\nCalculate the value of θ in the right-angled triangle MNP (correct to one decimal\nplace):\nP\nN\nM\n41\nθ\n24\nSOLUTION\nStep 1: Identify the opposite and adjacent sides and the hypotenuse\nStep 2: Determine the value of θ\ntan θ = opposite side\nadjacent side\ntan θ = 41\n24\n∴θ = tan−1\n\u001241\n24\n\u0013\nθ = 59,7◦\n243\nChapter 6.\nTrigonometry\n\nWorked example 3: Finding an angle\nQUESTION\nGiven 2 sin θ\n2 = cos 43◦, for θ ∈[0◦; 90◦], determine the value of θ (correct to one\ndecimal place).\nSOLUTION\nStep 1: Simplify the equation\nAvoiding rounding off in calculations until you have determined the final answer. In\nthe calculation below, the dots indicate that the number has not been rounded so that\nthe answer is as accurate as possible.\n2 sin θ\n2 = cos 43◦\nsin θ\n2 = cos 43◦\n2\nθ\n2 = sin−1(0,365 . . .)\nθ = 2(21,449 . . .)\n∴θ = 42,9◦\nTwo-dimensional problems\nWorked example 4: Flying a kite\nQUESTION\nThelma flies a kite on a 22 m piece of string and the height of the kite above the\nground is 20,4 m. Determine the angle of inclination of the string (correct to one\ndecimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the opposite and adjacent sides and the hy-\npotenuse\nLet the angle of inclination of the string be θ.\n244\n6.1.\nRevision\n\nKite\nThelma\n20,4\nθ\n22\nStep 2: Use an appropriate trigonometric ratio to find θ\nsin θ = opposite side\nhypotenuse\n= 20,4\n22\nθ = sin−1(0,927 . . .)\n∴θ = 68,0◦\nExercise 6 – 1: Revision\n1. If p = 49◦and q = 32◦, use a calculator to determine whether the following\nstatements are true of false:\na) sin p + 3 sin p = 4 sin p\nb) sin q\ncos q = tan q\nc) cos(p −q) = cos p −cos q\nd) sin(2p) = 2 sin p cos p\n2. Determine the following angles (correct to one decimal place):\na) cos α = 0,64\nb) sin θ + 2 = 2,65\nc) 1\n2 cos 2β = 0,3\nd) tan θ\n3 = sin 48◦\ne) cos 3p = 1,03\nf) 2 sin 3β + 1 = 2,6\ng) sin θ\ncos θ = 42\n3\n3. In △ABC, A ˆCB = 30◦, AC = 20 cm and BC = 22 cm. The perpendicular\nline from A intersects BC at T.\n245\nChapter 6.\nTrigonometry\n\nDetermine:\nA\nC\nB\nT\n20 cm\n22 cm\n30◦\na) the length TC\nb) the length AT\nc) the angle B ˆAT\n4. A rhombus has a perimeter of 40 cm and one of the internal angles is 30◦.\na) Determine the length of the sides.\nb) Determine the lengths of the diagonals.\nc) Calculate the area of the rhombus.\n5. Simplify the following without using a calculator:\na) 2 sin 45◦× 2 cos 45◦\nb) cos2 30◦−sin2 60◦\nc) sin 60◦cos 30◦−cos 60◦sin 30◦−tan 45◦\nd) 4 sin 60◦cos 30◦−2 tan 45◦+ tan 60◦−2 sin 60◦\ne) sin 60◦×\n√\n2 tan 45◦+ 1 −sin 30◦\n6. Given the diagram below.\nb\nx\ny\n0\nB(2; 2\n√\n3)\nβ\nDetermine the following without using a calculator:\na) β\nb) cos β\nc) cos2 β + sin2 β\n7. The 10 m ladder of a fire truck leans against the wall of a burning building at an\nangle of 60◦. The height of an open window is 9 m from the ground. Will the\nladder reach the window?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22XK\n2a. 22XM\n2b. 22XN\n2c. 22XP\n2d. 22XQ\n2e. 22XR\n2f. 22XS\n2g. 22XT\n3. 22XV\n4. 22XW\n5a. 22XX\n5b. 22XY\n5c. 22XZ\n5d. 22Y2\n5e. 22Y3\n6. 22Y4\n7. 22Y5\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n246\n6.1.\nRevision\n\n6.2\nTrigonometric identities\nEMBHH\nAn identity is a mathematical statement that equates one quantity with another. Trigono-\nmetric identities allow us to simplify a given expression so that it contains sine and co-\nsine ratios only. This enables us to solve equations and also to prove other identities.\nQuotient identity\nInvestigation: Quotient identity\n1. Complete the table without using a calculator, leaving your answer in surd form\nwhere applicable:\nθ = 45◦\nθ\n3\n5\nx\ny\n(3; 2)\nθ\nb\nsin θ\ncos θ\nsin θ\ncos θ\ntan θ\n2. Examine the last two rows of the table and make a conjecture.\n3. Are there any values of θ for which your conjecture would not be true? Explain\nyour answer.\nWe know that tan θ is defined as:\ntan θ = opposite side\nadjacent side\nUsing the diagram below and the theorem of Pythagoras, we can write the tangent\nfunction in terms of x, y and r:\nx\ny\n(x; y)\nθ\nb\nO\n247\nChapter 6.\nTrigonometry\n\ntan θ = y\nx\n= y\nx × r\nr\n= y\nr × r\nx\n= y\nr ÷ x\nr\n= sin θ ÷ cos θ\n= sin θ\ncos θ\nThis is the quotient identity:\ntan θ = sin θ\ncos θ\nNotice that tan θ is undefined if cos θ = 0, therefore θ ̸= k × 90◦, where k is an odd\ninteger.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\ntan θ\nSquare identity\nInvestigation: Square identity\n1. Use a calculator to complete the following table:\nsin2 80◦+ cos2 80◦=\ncos2 23◦+ sin2 23◦=\nsin 50◦+ cos 50◦=\nsin2 67◦−cos2 67◦=\nsin2 67◦+ cos2 67◦=\n2. What do you notice? Make a conjecture.\n3. Draw a sketch and prove your conjecture in general terms, using x, y and r.\n248\n6.2.\nTrigonometric identities\n\nx\ny\n(x; y)\nα\nb\nO\nr\nUsing the theorem of Pythagoras, we can write the sine and cosine functions in terms\nof x, y and r:\nsin2 θ + cos2 θ =\n\u0010y\nr\n\u00112\n+\n\u0010x\nr\n\u00112\n= y2\nr2 + x2\nr2\n= y2 + x2\nr2\n= r2\nr2\n= 1\nThis is the square identity:\nsin2 θ + cos2 θ = 1\nOther forms of the square identity\nComplete the following:\n1. sin2 θ = 1 −. . . . . .\n2. cos θ = ±√. . . . . .\n3. sin2 θ = (1 + . . . . . .)(1 −. . . . . .)\n4. cos2 θ −1 = . . . . . .\nHere are some useful tips for proving identities:\n• Change all trigonometric ratios to sine and cosine.\n• Choose one side of the equation to simplify and show that it is equal to the other\nside.\n• Usually it is better to choose the more complicated side to simplify.\n• Sometimes we need to simplify both sides of the equation to show that they are\nequal.\n• A square root sign often indicates that we need to use the square identity.\n• We can also add to the expression to make simplifying easier:\n– replace 1 with sin2 θ + cos2 θ.\n– multiply by 1 in the form of a suitable fraction, for example 1 + sin θ\n1 + sin θ.\n249\nChapter 6.\nTrigonometry\n\nSee video: 22Y6 at www.everythingmaths.co.za\nWorked example 5: Trigonometric identities\nQUESTION\nSimplify the following:\n1. tan2 θ × cos2 θ\n2.\n1\ncos2 θ −tan2 θ\nSOLUTION\nStep 1: Write the expression in terms of sine and cosine only\nWe use the square and quotient identities to write the given expression in terms of sine\nand cosine and then simplify as far as possible.\n1.\ntan2 θ × cos2 θ =\n\u0012 sin θ\ncos θ\n\u00132\n× cos2 θ\n= sin2 θ\ncos2 θ × cos2 θ\n= sin2 θ\n2.\n1\ncos2 θ −tan2 θ =\n1\ncos2 θ −\n\u0012 sin θ\ncos θ\n\u00132\n=\n1\ncos2 θ −sin2 θ\ncos2 θ\n= 1 −sin2 θ\ncos2 θ\n= cos2 θ\ncos2 θ\n= 1\n250\n6.2.\nTrigonometric identities\n\nWorked example 6: Trigonometric identities\nQUESTION\nProve: 1 −sin α\ncos α\n=\ncos α\n1 + sin α\nSOLUTION\nStep 1: Note restrictions\nWhen working with fractions, we must be careful that the denominator does not equal\n0. Therefore cos θ ̸= 0 for the fraction on the left-hand side and sin θ + 1 ̸= 0 for the\nfraction on the right-hand side.\nStep 2: Simplify the left-hand side\nThis is not an equation that needs to be solved. We are required to show that one side\nof the equation is equal to the other. We can choose either of the two sides to simplify.\nLHS = 1 −sin α\ncos α\n= 1 −sin α\ncos α\n× 1 + sin α\n1 + sin α\nNotice that we have not changed the equation — this is the same as multiplying by 1\nsince the numerator and the denominator are the same.\nStep 3: Determine the lowest common denominator and simplify\nLHS =\n1 −sin2 α\ncos α(1 + sin α)\n=\ncos2 α\ncos α(1 + sin α)\n=\ncos α\n1 + sin α\n= RHS\n251\nChapter 6.\nTrigonometry\n\nExercise 6 – 2: Trigonometric identities\n1. Reduce the following to one trigonometric ratio:\na) sin α\ntan α\nb) cos2 θ tan2 θ + tan2 θ sin2 θ\nc) 1 −sin θ cos θ tan θ\nd)\n\u00121 −cos2 β\ncos2 β\n\u0013\n−tan2 β\n2. Prove the following identities and state restrictions where appropriate:\na) 1 + sin θ\ncos θ\n=\ncos θ\n1 −sin θ\nb) sin2α + (cos α −tan α) (cos α + tan α) = 1 −tan2α\nc)\n1\ncos θ −cos θtan2θ\n1\n= cos θ\nd)\n2 sin θ cos θ\nsin θ + cos θ = sin θ + cos θ −\n1\nsin θ + cos θ\ne)\n\u0012cos β\nsin β + tan β\n\u0013\ncos β =\n1\nsin β\nf)\n1\n1 + sin θ +\n1\n1 −sin θ = d\n2 tan θ\nsin θ cos θ\ng) (1 + tan2 α) cos α\n(1 −tan α)\n=\n1\ncos α −sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Y7\n1b. 22Y8\n1c. 22Y9\n1d. 22YB\n2a. 22YC\n2b. 22YD\n2c. 22YF\n2d. 22YG\n2e. 22YH\n2f. 22YJ\n2g. 22YK\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n252\n6.2.\nTrigonometric identities\n\n6.3\nReduction formula\nEMBHJ\nAny trigonometric function whose argument is 90◦± θ; 180◦± θ and 360◦± θ can be\nwritten simply in terms of θ.\nDeriving reduction formulae\nEMBHK\nInvestigation: Reduction formulae for function values of 180◦± θ\n1. Function values of 180◦−θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the y-axis, determine the coordi-\nnates of P ′.\nb) Write down values for sin θ, cos θ and tan θ.\nc) Use the coordinates for P ′ to determine sin(180◦−θ), cos(180◦−θ),\ntan(180◦−θ).\nd) From your results determine a relationship between the trigonometric func-\ntion values of (180◦−θ) and θ.\n253\nChapter 6.\nTrigonometry\n\n2. Function values of 180◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n180◦+ θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the origin (the two points are sym-\nmetrical about both the x-axis and the y-axis), determine the coordinates of\nP ′.\nb) Use the coordinates for P ′ to determine sin(180◦+ θ), cos(180◦+ θ) and\ntan(180◦+ θ).\nc) From your results determine a relationship between the trigonometric func-\ntion values of (180◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(180◦−θ) = . . . . . .\nb) cos(180◦−θ) = . . . . . .\nc) tan(180◦−θ) = . . . . . .\nd) sin(180◦+ θ) = . . . . . .\ne) cos(180◦+ θ) = . . . . . .\nf) tan(180◦+ θ) = . . . . . .\n254\n6.3.\nReduction formula\n\nWorked example 7: Reduction formulae for function values of 180◦± θ\nQUESTION\nWrite the following as a single trigonometric ratio:\nsin 163◦\ncos 197◦+ tan 17◦+ cos(180◦−θ) × tan(180◦+ θ)\nSOLUTION\nStep 1: Use reduction formulae to write the trigonometric function values in terms\nof acute angles and θ\n= sin(180◦−17◦)\ncos(180◦+ 17◦) + tan 17◦+ (−cos θ) × tan θ\nStep 2: Simplify\n=\nsin 17◦\n−cos 17◦+ tan 17◦−cos θ × sin θ\ncos θ\n= −tan 17◦+ tan 17◦−sin θ\n= −sin θ\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1. Determine the value of the following expressions without using a calculator:\na) tan 150◦sin 30◦−cos 210◦\nb) (1 + cos 120◦)(1 −sin2 240◦)\nc) cos2 140◦+ sin2 220◦\n2. Write the following in terms of a single trigonometric ratio:\na) tan(180◦−θ) × sin(180◦+ θ)\nb) tan(180◦+ θ) cos(180◦−θ)\nsin(180◦−θ)\n255\nChapter 6.\nTrigonometry\n\n3. If t = tan 40◦, express the following in terms of t:\na) tan 140◦+ 3 tan 220◦\nb) cos 220◦\nsin 140◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YM\n1b. 22YN\n1c. 22YP\n2a. 22YQ\n2b. 22YR\n3. 22YS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of (360◦± θ) and (−θ)\n1. Function values of (360◦−θ) and (−θ)\nIn the Cartesian plane we measure angles from the positive x-axis to the terminal\narm, which means that an anti-clockwise rotation gives a positive angle. We can\ntherefore measure negative angles by rotating in a clockwise direction.\nFor an acute angle θ, we know that −θ will lie in the fourth quadrant.\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\n360◦−θ\nP ′\n2\n2\na) If points P and P ′ are symmetrical about the x-axis (y = 0), determine the\ncoordinates of P ′.\nb) Use the coordinates of P ′ to determine sin(360◦−θ), cos(360◦−θ) and\ntan(360◦−θ).\nc) Use the coordinates of P ′ to determine sin(−θ), cos(−θ) and tan(−θ).\nd) From your results determine a relationship between the function values of\n(360◦−θ) and −θ.\n256\n6.3.\nReduction formula\n\ne) Complete the following reduction formulae:\ni. sin(360◦−θ) = . . . . . .\nii. cos(360◦−θ) = . . . . . .\niii. tan(360◦−θ) = . . . . . .\niv. sin(−θ) = . . . . . .\nv. cos(−θ) = . . . . . .\nvi. tan(−θ) = . . . . . .\n2. Function values of 360◦+ θ\nWe can also have an angle that is larger than 360◦. The angle completes a\nrevolution of 360◦and then continues to give an angle of θ.\nComplete the following reduction formulae:\na) sin(360◦+ θ) = . . . . . .\nb) cos(360◦+ θ) = . . . . . .\nc) tan(360◦+ θ) = . . . . . .\nFrom working with functions, we know that the graph of y = sin θ has a period of\n360◦. Therefore, one complete wave of a sine graph is the same as one complete\nrevolution for sin θ in the Cartesian plane.\n0\n1\n−1\n90◦\n180◦\n270◦\n360◦\n1st\n2nd\n3rd\n4th\npositive\npositive\nnegative\nnegative\n0◦/360◦\n90◦\n180◦\n270◦\n2nd\npos.\nneg.\n3rd\nneg.\n4th\n1st\npos.\nWe can also have multiple revolutions. The periodicity of the trigonometric graphs\nshows this clearly. A complete sine or cosine curve is completed in 360◦.\ny = cos θ\ny = sin θ\nθ\ny\n257\nChapter 6.\nTrigonometry\n\nIf k is any integer, then\nsin(k . 360◦+ θ) = sin θ\ncos(k . 360◦+ θ) = cos θ\ntan(k . 360◦+ θ) = tan θ\nWorked example 8: Reduction formulae for function values of 360◦± θ\nQUESTION\nIf f = tan 67◦, express the following in terms of f\nsin 293◦\ncos 427◦+ tan(−67◦) + tan 1147◦\nSOLUTION\nStep 1: Using reduction formula\n= sin(360◦−67◦)\ncos(360◦+ 67◦) −tan(67◦) + tan (3(360◦) + 67◦)\n= −sin 67◦\ncos 67◦−tan 67◦+ tan 67◦\n= −tan 67◦\n= −f\nWorked example 9: Using reduction formula\nQUESTION\nEvaluate without using a calculator:\ntan2 210◦−(1 + cos 120◦) sin2 405◦\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and special angles\n258\n6.3.\nReduction formula\n\n= tan2(180◦+ 30◦) −(1 + cos(180◦−60◦)) sin2(360◦+ 45◦)\n= tan2 30◦−(1 + (−cos 60◦)) sin2 45◦\n=\n\u0012 1\n√\n3\n\u00132\n−\n\u0012\n1 −1\n2\n\u0013 \u0012 1\n√\n2\n\u00132\n= 1\n3 −\n\u00121\n2\n\u0013 \u00121\n2\n\u0013\n= 1\n3 −1\n4\n= 1\n12\nExercise 6 – 4: Using reduction formula\n1. Simplify the following:\na) tan(180◦−θ) sin(360◦+ θ)\ncos(180◦+ θ) tan(360◦−θ)\nb) cos2(360◦+ θ) + cos(180◦+ θ) tan(360◦−θ) sin(360◦+ θ)\nc)\nsin(360◦+ α) tan(180◦+ α)\ncos(360◦−α) tan2(360◦+ α)\n2. Write the following in terms of cos β:\ncos(360◦−β) cos(−β) −1\nsin(360◦+ β) tan(360◦−β)\n3. Simplify the following without using a calculator:\na)\ncos 300◦tan 150◦\nsin 225◦cos(−45◦)\nb) 3 tan 405◦+ 2 tan 330◦cos 750◦\nc) cos 315◦cos 405◦+ sin 45◦sin 135◦\nsin 750◦\nd) tan 150◦cos 390◦−2 sin 510◦\ne) 2 sin 120◦+ 3 cos 765◦−2 sin 240◦−3 cos 45◦\n5 sin 300◦+ 3 tan 225◦−6 cos 60◦\n4. Given 90◦< α < 180◦, use a sketch to help explain why:\na) sin(−α) = −sin α\nb) cos(−α) = −cos α\n259\nChapter 6.\nTrigonometry\n\n5. If t = sin 43◦, express the following in terms of t:\na) sin 317◦\nb) cos2 403◦\nc) tan(−43◦)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22YT\n1b. 22YV\n1c. 22YW\n2. 22YX\n3a. 22YY\n3b. 22YZ\n3c. 22Z2\n3d. 22Z3\n3e. 22Z4\n4. 22Z5\n5. 22Z6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Reduction formulae for function values of 90◦± θ\nIn any right-angled triangle, the two acute angles are complements of each other, ˆA +\nˆC = 90◦\nA\nB\nC\nb\na\nc\nComplete the following:\nIn △ABC\nsin ˆC = c\nb = cos . . .\ncos ˆC = a\nb = sin . . .\nComplementary angles are positive acute angles that add up to 90◦. For example 20◦\nand 70◦are complementary angles.\n260\n6.3.\nReduction formula\n\nIn the figure P(\n√\n3; 1) and P ′ lie on a circle with radius 2. OP makes an angle of\nθ = 30◦with the x-axis.\nb\nP\nO\nx\ny\nθ\nb\nθ\nP ′\n2\n2\n90◦−θ\n1. Function values of 90◦−θ\na) If points P and P ′ are symmetrical about the line y = x, determine the\ncoordinates of P ′.\nb) Use the coordinates for P ′ to determine sin(90◦−θ) and cos(90◦−θ).\nc) From your results determine a relationship between the function values of\n(90◦−θ) and θ.\n2. Function values of 90◦+ θ\nIn the figure P(\n√\n3; 1) and P ′ lie on the circle with radius 2. OP makes an angle\nθ = 30◦with the x-axis.\nθ\nb\nP\nO\nx\ny\nθ\nb\nP ′\n90◦+ θ\n2\n2\n261\nChapter 6.\nTrigonometry\n\na) If point P is rotated through 90◦to get point P ′, determine the coordinates\nof P ′.\nb) Use the coordinates for P ′ to determine sin(90◦+ θ) and cos(90◦+ θ).\nc) From your results determine a relationship between the function values of\n(90◦+ θ) and θ.\n3. Complete the following reduction formulae:\na) sin(90◦−θ) = . . . . . .\nb) cos(90◦−θ) = . . . . . .\nc) sin(90◦+ θ) = . . . . . .\nd) cos(90◦+ θ) = . . . . . .\nSine and cosine are known as co-functions. Two functions are called co-functions if\nf (A) = g (B) whenever A + B = 90◦(that is, A and B are complementary angles).\nThe function value of an angle is equal to the co-function of its complement.\nThus for sine and cosine we have\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nThe sine and cosine graphs illustrate this clearly: the two graphs are identical except\nthat they have a 90◦phase difference.\nθ\ny\ny = cos θ\ny = sin θ\n262\n6.3.\nReduction formula\n\nWorked example 10: Using the co-function rule\nQUESTION\nWrite each of the following in terms of sin 40◦:\n1. cos 50◦\n2. sin 320◦\n3. cos 230◦\n4. cos 130◦\nSOLUTION\n1. cos 50◦= sin(90◦−50◦) = sin 40◦\n2. sin 320◦= sin(360◦−40◦) = −sin 40◦\n3. cos 230◦= cos(180◦+ 50◦) = −cos 50◦= −cos(90◦−40◦) = −sin 40◦\n4. cos 130◦= cos(90◦+ 40◦) = −sin 40◦\nFunction values of θ −90◦\nWe can write sin(θ −90◦) as\nsin(θ −90◦) = sin [−(90◦−θ)]\n= −sin(90◦−θ)\n= −cos θ\nsimilarly, we can show that cos (θ −90◦) = sin θ\nTherefore, sin (θ −90◦) = −cos θ and cos (θ −90◦) = sin θ.\nWorked example 11: Co-functions\nQUESTION\nExpress the following in terms of t if t = sin θ:\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\nSOLUTION\nStep 1: Simplify the expression using reduction formulae and co-functions\n263\nChapter 6.\nTrigonometry\n\nUse the CAST diagram to check in which quadrants the trigonometric ratios are positive\nand negative.\ncos(θ −90◦) cos(720◦+ θ) tan(θ −360◦)\nsin2(θ + 360◦) cos(θ + 90◦)\n=cos[−(90◦−θ)] cos[2(360◦) + θ] tan[−(360◦−θ)]\nsin2(360◦+ θ) cos(90◦+ θ)\n=sin θ cos θ tan θ\nsin2 θ(−sin θ)\n= −cos θ\n\u0000 sin θ\ncos θ\n\u0001\nsin2 θ\n= −\n1\nsin θ\n= −1\nt\nExercise 6 – 5: Co-functions\n1. Simplify the following:\na) cos(90◦+ θ) sin(θ + 90◦)\nsin(−θ)\nb) 2 sin(90◦−x) + sin(90◦+ x)\nsin(90◦−x) + cos(180◦+ x)\n2. Given cos 36◦= p, express the following in terms on p:\na) sin 54◦\nb) sin 36◦\nc) tan 126◦\nd) cos 324◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22Z7\n1b. 22Z8\n2. 22Z9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n264\n6.3.\nReduction formula\n\nReduction formulae and co-functions:\n1. The reduction formulae hold for any angle θ. For convenience, we assume θ is\nan acute angle (0◦< θ < 90◦).\n2. When determining function values of (180◦±θ), (360◦±θ) and (−θ) the function\ndoes not change.\n3. When determining function values of (90◦±θ) and (θ±90◦) the function changes\nto its co-function.\nsecond quadrant (180◦−θ) or (90◦+ θ)\nfirst quadrant (θ) or (90◦−θ)\nsin(180◦−θ) = + sin θ\nall trig functions are positive\ncos(180◦−θ) = −cos θ\nsin(360◦+ θ) = sin θ\ntan(180◦−θ) = −tan θ\ncos(360◦+ θ) = cos θ\nsin(90◦+ θ) = + cos θ\ntan(360◦+ θ) = tan θ\ncos(90◦+ θ) = −sin θ\nsin(90◦−θ) = cos θ\ncos(90◦−θ) = sin θ\nthird quadrant (180◦+ θ)\nfourth quadrant (360◦−θ)\nsin(180◦+ θ) = −sin θ\nsin(360◦−θ) = −sin θ\ncos(180◦+ θ) = −cos θ\ncos(360◦−θ) = + cos θ\ntan(180◦+ θ) = + tan θ\ntan(360◦−θ) = −tan θ\nExercise 6 – 6: Reduction formulae\n1. Write A and B as a single trigonometric ratio:\na) A = sin(360◦−θ) cos(180◦−θ) tan(360◦+ θ)\nb) B = cos(360◦+ θ) cos(−θ) sin(−θ)\ncos(90◦+ θ)\nc) Hence, determine:\ni. A + B = . . .\nii.\nA\nB = . . .\n2. Write the following as a function of an acute angle:\na) sin 163◦\nb) cos 327◦\nc) tan 248◦\nd) cos(−213◦)\n3. Determine the value of the following, without using a calculator:\na) sin(−30◦)\ntan(150◦) + cos 330◦\nb) tan 300◦cos 120◦\nc) (1 −cos 30◦)(1 −cos 210◦)\nd) cos 780◦−(sin 315◦)(cos 405◦)\n4. Prove that the following identity is true and state any restrictions:\nsin(180◦+ α) tan(360◦+ α) cos α\ncos(90◦−α)\n= sin α\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 22ZB\n2a. 22ZC\n2b. 22ZD\n2c. 22ZF\n2d. 22ZG\n3a. 22ZH\n3b. 22ZJ\n3c. 22ZK\n3d. 22ZM\n4. 22ZN\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n265\nChapter 6.\nTrigonometry\n\n6.4\nTrigonometric equations\nEMBHM\nSolving trigonometric equations requires that we find the value of the angles that satisfy\nthe equation. If a specific interval for the solution is given, then we need only find the\nvalue of the angles within the given interval that satisfy the equation. If no interval is\ngiven, then we need to find the general solution. The periodic nature of trigonometric\nfunctions means that there are many values that satisfy a given equation, as shown in\nthe diagram below.\n1\n−1\n90◦180◦270◦360◦\n−90◦\n−180◦\n−270◦\n−360◦\nb\nb\nb\nb\nθ\n0\ny\ny = 0,5\ny = sin θ\nWorked example 12: Solving trigonometric equations\nQUESTION\nSolve for θ (correct to one decimal place), given tan θ = 5 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to solve for θ\ntan θ = 5\n∴θ = tan−1 5\n= 78,7◦\nThis value of θ is an acute angle which lies in the first quadrant and is called the\nreference angle.\nStep 2: Use the CAST diagram to determine in which quadrants tan θ is positive\nThe CAST diagram indicates that tan θ is positive in the first and third quadrants, there-\nfore we must determine the value of θ such that 180◦< θ < 270◦.\nUsing reduction formulae, we know that tan(180◦+ θ) = tan θ\nθ = 180◦+ 78,7◦\n∴θ = 258,7◦\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 78,7◦or θ = 258,7◦.\n266\n6.4.\nTrigonometric equations\n\nWorked example 13: Solving trigonometric equations\nQUESTION\nSolve for α (correct to one decimal place), given cos α = −0,7 and θ ∈[0◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we do not include the negative sign. The reference\nangle must be an acute angle in the first quadrant, where all the trigonometric functions\nare positive.\nref ∠= cos−1 0,7\n= 45,6◦\nStep 2: Use the CAST diagram to determine in which quadrants cos α is negative\nThe CAST diagram indicates that cos α is negative in the second and third quadrants,\ntherefore we must determine the value of α such that 90◦< α < 270◦.\nUsing reduction formulae, we know that cos(180◦−α) = −cos α and cos(180◦+α) =\n−cos α\nIn the second quadrant:\nα = 180◦−45,6◦\n= 134,4◦\nIn the third quadrant:\nα = 180◦+ 45,6◦\n= 225,6◦\nNote: the reference angle (45,6◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nα = 134,4◦or α = 225,6◦.\n267\nChapter 6.\nTrigonometry\n\nWorked example 14: Solving trigonometric equations\nQUESTION\nSolve for β (correct to one decimal place), given sin β = −0,5 and β ∈[−360◦; 360◦].\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nTo determine the reference angle, we use a positive value.\nref ∠= sin−1 0,5\n= 30◦\nStep 2: Use the CAST diagram to determine in which quadrants sin β is negative\nThe CAST diagram indicates that sin β is negative in the third and fourth quadrants.\nWe also need to find the values of β such that −360◦≤β ≤360◦.\nUsing reduction formulae, we know that sin(180◦+β) = −sin β and sin(360◦−β) =\n−sin β\nIn the third quadrant:\nβ = 180◦+ 30◦\n= 210◦\nor β = −180◦+ 30◦\n= −150◦\nIn the fourth quadrant:\nβ = 360◦−30◦\n= 330◦\nor β = 0◦−30◦\n= −30◦\nNotice: the reference angle (30◦) does not form part of the solution.\nStep 3: Use a calculator to check that the solution satisfies the original equation\nStep 4: Write the final answer\nβ = −150◦, −30◦, 210◦or 330◦.\n268\n6.4.\nTrigonometric equations\n\nExercise 6 – 7: Solving trigonometric equations\n1. Determine the values of α for α ∈[0◦; 360◦] if:\na) 4 cos α = 2\nb) sin α + 3,65 = 3\nc) tan α = 51\n4\nd) cos α + 0,939 = 0\ne) 5 sin α = 3\nf)\n1\n2 tan α = −1,4\n2. Determine the values of θ for θ ∈[−360◦; 360◦] if:\na) sin θ = 0,6\nb) cos θ + 3\n4 = 0\nc) 3 tan θ = 20\nd) sin θ = cos 180◦\ne) 2 cos θ = 4\n5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 22ZP\n1b. 22ZQ\n1c. 22ZR\n1d. 22ZS\n1e. 22ZT\n1f. 22ZV\n2a. 22ZW\n2b. 22ZX\n2c. 22ZY\n2d. 22ZZ\n2e. 2322\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe general solution\nEMBHN\nIn the previous worked example, the solution was restricted to a certain interval. How-\never, the periodicity of the trigonometric functions means that there are an infinite\nnumber of positive and negative angles that satisfy an equation. If we do not restrict\nthe solution, then we need to determine the general solution to the equation. We know\nthat the sine and cosine functions have a period of 360◦and the tangent function has\na period of 180◦.\nMethod for finding the general solution:\n1. Determine the reference angle (use a positive value).\n2. Use the CAST diagram to determine where the function is positive or negative\n(depending on the given equation).\n3. Find the angles in the interval [0◦; 360◦] that satisfy the equation and add multi-\nples of the period to each answer.\n4. Check answers using a calculator.\n269\nChapter 6.\nTrigonometry\n\nWorked example 15: Finding the general solution\nQUESTION\nDetermine the general solution for sin θ = 0,3 (correct to one decimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nsin θ = 0,3\n∴ref ∠= sin−1 0,3\n= 17,5◦\nStep 2: Use CAST diagram to determine in which quadrants sin θ is positive\nThe CAST diagram indicates that sin θ is positive in the first and second quadrants.\nUsing reduction formulae, we know that sin(180◦−θ) = sin θ.\nIn the first quadrant:\nθ = 17,5◦\n∴θ = 17,5◦+ k . 360◦\nIn the second quadrant:\nθ = 180◦−17,5◦\n∴θ = 162,5◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 4:\nθ = 17,5◦+ 4(360)◦\n∴θ = 1457,5◦\nAnd sin 1457,5◦= 0,3007 . . .\nThis solution is correct.\nSimilarly, if we let k = −2:\nθ = 162,5◦−2(360)◦\n∴θ = −557,5◦\nAnd sin(−557,5◦) = 0,3007 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 17,5◦+ k . 360◦or θ = 162,5◦+ k . 360◦.\n270\n6.4.\nTrigonometric equations\n\nWorked example 16: Finding the general solution\nQUESTION\nDetermine the general solution for cos 2θ = −0,6427 (give answers correct to one\ndecimal place).\nSOLUTION\nStep 1: Use a calculator to find the reference angle\nref ∠= sin−1 0,6427\n= 50,0◦\nStep 2: Use CAST diagram to determine in which quadrants cos θ is negative\nThe CAST diagram shows that cos θ is negative in the second and third quadrants.\nTherefore we use the reduction formulae cos(180◦−θ) = −cos θ and cos(180◦+θ) =\n−cos θ.\nIn the second quadrant:\n2θ = 180◦−50◦+ k . 360◦\n= 130◦+ k . 360◦\n∴θ = 65◦+ k . 180◦\nIn the third quadrant:\n2θ = 180◦+ 50◦+ k . 360◦\n= 230◦+ k . 360◦\n∴θ = 115◦+ k . 180◦\nwhere k ∈Z.\nRemember: also divide the period (360◦) by the coefficient of θ.\nStep 3: Check that the solution satisfies the original equation\nWe can select random values of k to check that the answers satisfy the original equa-\ntion.\nLet k = 2:\nθ = 65◦+ 2(180◦)\n∴θ = 425◦\nAnd cos 2(425)◦= −0,6427 . . .\nThis solution is correct.\n271\nChapter 6.\nTrigonometry\n\nSimilarly, if we let k = −5:\nθ = 115◦−5(180◦)\n∴θ = −785◦\nAnd cos 2(−785◦) = −0,6427 . . .\nThis solution is also correct.\nStep 4: Write the final answer\nθ = 65◦+ k . 180◦or θ = 115◦+ k . 180◦.\nWorked example 17: Finding the general solution\nQUESTION\nDetermine the general solution for tan(2α −10◦) = 2,5 such that −180◦≤α ≤180◦\n(give answers correct to one decimal place).\nSOLUTION\nStep 1: Make a substitution\nTo solve this equation, it can be useful to make a substitution: let x = 2α −10◦.\ntan(x) = 2,5\nStep 2: Use a calculator to find the reference angle\ntan x = 2,5\n∴ref ∠= tan−1 2,5\n= 68,2◦\nStep 3: Use CAST diagram to determine in which quadrants the tangent function is\npositive\nWe see that tan x is positive in the first and third quadrants, so we use the reduction\nformula tan(180◦+ x) = tan x. It is also important to remember that the period of the\ntangent function is 180◦.\n272\n6.4.\nTrigonometric equations\n\nIn the first quadrant:\nx = 68,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 68,2◦+ k . 180◦\n2α = 78,2◦+ k . 180◦\n∴α = 39,1◦+ k . 90◦\nIn the third quadrant:\nx = 180◦+ 68,2◦+ k . 180◦\n= 248,2◦+ k . 180◦\nSubstitute x = 2α −10◦\n2α −10◦= 248,2◦+ k . 180◦\n2α = 258,2◦+ k . 180◦\n∴α = 129,1◦+ k . 90◦\nwhere k ∈Z.\nRemember: to divide the period (180◦) by the coefficient of α.\nStep 4: Find the answers within the given interval\nSubstitute suitable values of k to determine the values of α that lie within the interval\n(−180◦≤α ≤180◦).\nI: α = 39,1◦+ k . 90◦\nIII: α = 129,1◦+ k . 90◦\nk = 0\n39,1◦\n129,1◦\nk = 1\n129,1◦\n219,1◦\n(outside)\nk = 2\n219,1◦\n(outside)\nk = −1\n−50,9◦\n39,1◦\nk = −2\n−140,9◦\n−50,9◦\nk = −3\n−230,9◦\n(outside)\n−140,9◦\nk = −4\n−230,9◦\n(outside)\nNotice how some of the values repeat. This is because of the periodic nature of the\ntangent function. Therefore we need only determine the solution:\nα = 39,1◦+ k . 90◦\nfor k ∈Z.\nStep 5: Write the final answer\nα = −140,9◦; −50,9◦; 39,1◦or 129,1◦.\n273\nChapter 6.\nTrigonometry\n\nWorked example 18: Finding the general solution using co-functions\nQUESTION\nDetermine the general solution for sin(θ −20◦) = cos 2θ.\nSOLUTION\nStep 1: Use co-functions to simplify the equation\nsin(θ −20◦) = cos 2θ\n= sin(90◦−2θ)\n∴θ −20◦= 90◦−2θ + k . 360◦,\nk ∈Z\n3θ = 110◦+ k . 360◦\n∴θ = 36,7◦+ k . 120◦\nStep 2: Use the CAST diagram to determine the correct quadrants\nSince the original equation equates a sine and cosine function, we need to work in the\nquadrant where both functions are positive or in the quadrant where both functions\nare negative so that the equation holds true. We therefore determine the solution using\nthe first and third quadrants.\nIn the first quadrant: θ = 36,7◦+ k . 120◦.\nIn the third quadrant:\n3θ = 180◦+ 110◦+ k . 360◦\n= 290◦+ k . 360◦\n∴θ = 96,6◦+ k . 120◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 36,7◦+ k . 120◦or θ = 96,6◦+ k . 120◦\n274\n6.4.\nTrigonometric equations\n\nExercise 6 – 8: General solution\n1.\n• Find the general solution for each equation.\n• Hence, find all the solutions in the interval [−180◦; 180◦].\na) cos(θ + 25◦) = 0,231\nb) sin 2α = −0,327\nc) 2 tan β = −2,68\nd) cos α = 1\ne) 4 sin θ = 0\nf) cos θ = −1\ng) tan θ\n2 = 0,9\nh) 4 cos θ + 3 = 1\ni) sin 2θ = −\n√\n3\n2\n2. Find the general solution for each equation.\na) cos(θ + 20◦) = 0\nb) sin 3α = −1\nc) tan 4β = 0,866\nd) cos(α −25◦) = 0,707\ne) 2 sin 3θ\n2 = −1\nf) 5 tan(β + 15◦) =\n5\n√\n3\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2323\n1b. 2324\n1c. 2325\n1d. 2326\n1e. 2327\n1f. 2328\n1g. 2329\n1h. 232B\n1i. 232C\n2a. 232D\n2b. 232F\n2c. 232G\n2d. 232H\n2e. 232J\n2f. 232K\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nSolving quadratic trigonometric equations\nWe can use our knowledge of algebraic equations to solve quadratic trigonometric\nequations.\nWorked example 19: Quadratic trigonometric equations\nQUESTION\nFind the general solution of 4 sin2 θ = 3.\nSOLUTION\nStep 1: Simplify the equation and determine the reference angle\n4 sin2 θ = 3\nsin2 θ = 3\n4\n∴sin θ = ±\nr\n3\n4\n= ±\n√\n3\n2\n∴ref ∠= 60◦\n275\nChapter 6.\nTrigonometry\n\nStep 2: Determine in which quadrants the sine function is positive and negative\nThe CAST diagram shows that sin θ is positive in the first and second quadrants and\nnegative in the third and fourth quadrants.\nPositive in the first and second quadrants:\nθ = 60◦+ k . 360◦\nor θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nNegative in the third and fourth quadrants:\nθ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor θ = 360◦−60◦+ k . 360◦\n= 300◦+ k . 360◦\nwhere k ∈Z.\nStep 3: Check that the solution satisfies the original equation\nStep 4: Write the final answer\nθ = 60◦+ k . 360◦or 120◦+ k . 360◦or 240◦+ k . 360◦or 300◦+ k . 360◦\nWorked example 20: Quadratic trigonometric equations\nQUESTION\nFind θ if 2 cos2 θ −cos θ −1 = 0 for θ ∈[−180◦; 180◦].\nSOLUTION\nStep 1: Factorise the equation\n2 cos2 θ −cos θ −1 = 0\n(2 cos θ + 1)(cos θ −1) = 0\n∴2 cos θ + 1 = 0 or cos θ −1 = 0\n276\n6.4.\nTrigonometric equations\n\nStep 2: Simplify the equations and solve for θ\n2 cos θ + 1 = 0\n2 cos θ = −1\ncos θ = −1\n2\n∴ref ∠= 60◦\nII quadrant: θ = 180◦−60◦+ k . 360◦\n= 120◦+ k . 360◦\nIII quadrant: θ = 180◦+ 60◦+ k . 360◦\n= 240◦+ k . 360◦\nor\ncos θ −1 = 0\ncos θ = 1\n∴ref ∠= 0◦\nII and IV quadrants: θ = k . 360◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of θ that lie within the the given interval θ ∈[−180◦; 180◦] by\nsubstituting suitable values of k.\nIf k = −1,\nθ = 240◦+ k . 360◦\n= 240◦−(360◦)\n= −120◦\nIf k = 0,\nθ = 120◦+ k . 360◦\n= 120◦+ 0(360◦)\n= 120◦\nIf k = 1,\nθ = k . 360◦\n= 0(360◦)\n= 0◦\n277\nChapter 6.\nTrigonometry\n\nStep 4: Alternative method: substitution\nWe can simplify the given equation by letting y = cos θ and then factorising as:\n2y2 −y −1 = 0\n(2y + 1)(y −1) = 0\n∴y = −1\n2 or y = 1\nWe substitute y = cos θ back into these two equations and solve for θ.\nStep 5: Write the final answer\nθ = −120◦; 0◦; 120◦\nWorked example 21: Quadratic trigonometric equations\nQUESTION\nFind α if 2 sin2 α −sin α cos α = 0 for α ∈[0◦; 360◦].\nSOLUTION\nStep 1: Factorise the equation by taking out a common factor\n2 sin2 α −sin α cos α = 0\nsin α(2 sin α −cos α) = 0\n∴sin α = 0 or 2 sin α −cos α = 0\nStep 2: Simplify the equations and solve for α\nsin α = 0\n∴ref ∠= 0◦\n∴α = 0◦+ k . 360◦\nor α = 180◦+ k . 360◦\nand since 360◦= 2 × 180◦\nwe therefore have α = k . 180◦\n278\n6.4.\nTrigonometric equations\n\nor\n2 sin α −cos α = 0\n2 sin α = cos α\nTo simplify further, we divide both sides of the equation by cos α.\n2 sin α\ncos α = cos α\ncos α\n(cos α ̸= 0)\n2 tan α = 1\ntan α = 1\n2\n∴ref ∠= 26,6◦\n∴α = 26,6◦+ k . 180◦\nwhere k ∈Z.\nStep 3: Substitute suitable values of k\nDetermine the values of α that lie within the the given interval α ∈[0◦; 360◦] by\nsubstituting suitable values of k.\nIf k = 0:\nα = 0◦\nor α = 26,6◦\nIf k = 1:\nα = 180◦\nor α = 26,6◦+ 180◦\n= 206,6◦\nIf k = 2:\nα = 360◦\nStep 4: Write the final answer\nα = 0◦; 26,6◦; 180◦; 206,6◦; 360◦\n279\nChapter 6.\nTrigonometry\n\nExercise 6 – 9: Solving trigonometric equations\n1. Find the general solution for each of the following equations:\na) cos 2θ = 0\nb) sin(α + 10◦) =\n√\n3\n2\nc) 2 cos θ\n2 −\n√\n3 = 0\nd)\n1\n2 tan(β −30◦) = −1\ne) 5 cos θ = tan 300◦\nf) 3 sin α = −1,5\ng) sin 2β = cos(β + 20◦)\nh) 0,5 tan θ + 2,5 = 1,7\ni) sin(3α −10◦) = sin(α + 32◦)\nj) sin 2β = cos 2β\n2. Find θ if sin2 θ + 1\n2 sin θ = 0 for θ ∈[0◦; 360◦].\n3. Determine the general solution for each of the following:\na) 2 cos2 θ −3 cos θ = 2\nb) 3 tan2 θ + 2 tan θ = 0\nc) cos2 α = 0,64\nd) sin(4β + 35◦) = cos(10◦−β)\ne) sin(α + 15◦) = 2 cos(α + 15◦)\nf) sin2 θ −4 cos2 θ = 0\ng) cos(2θ + 30◦)\n2\n+ 0,38 = 0\n4. Find β if 1\n3 tan β = cos 200◦for β ∈[−180◦; 180◦].\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 232M\n1b. 232N\n1c. 232P\n1d. 232Q\n1e. 232R\n1f. 232S\n1g. 232T\n1h. 232V\n1i. 232W\n1j. 232X\n2. 232Y\n3a. 232Z\n3b. 2332\n3c. 2333\n3d. 2334\n3e. 2335\n3f. 2336\n3g. 2337\n4. 2338\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n6.5\nArea, sine, and cosine rules\nEMBHP\nThere are three identities relating to the trigonometric functions that make working\nwith triangles easier:\n1. the area rule\n2. the sine rule\n3. the cosine rule\n280\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nEMBHQ\nInvestigation: The area rule\n1. Consider △ABC:\nB\nA\nC\n10\n54◦\n7\nComplete the following:\na) Area △ABC = 1\n2 × . . . × AC\nb) sin ˆB = . . . and AC = . . . × . . .\nc) Therefore area △ABC = . . . × . . . × . . . × . . .\n2. Consider △A′B′C′:\nB′\nA′\nC′\n10\n54◦\n7\nComplete the following:\na) How is △A′B′C′ different from △ABC?\nb) Calculate area △A′B′C′.\n3. Use your results to write a general formula for determining the area of △PQR:\nQ\nP\nR\nr\np\nq\n281\nChapter 6.\nTrigonometry\n\nFor any △ABC with AB = c, BC = a and AC = b, we can construct a perpendicular\nheight (h) from vertex A to the line BC:\nB\nA\nC\nc\na\nb\nh\nIn △ABC:\nsin ˆB = h\nc\n∴h = c sin ˆB\nAnd we know that\nArea △ABC = 1\n2 × a × h\n= 1\n2 × a × c sin ˆB\n∴Area △ABC = 1\n2ac sin ˆB\nAlternatively, we could write that\nsin ˆC = h\nb\n∴h = b sin ˆC\nAnd then we would have that\nArea △ABC = 1\n2 × a × h\n= 1\n2ab sin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nThe area rule\nIn any △ABC:\nArea △ABC = 1\n2bc sin ˆA\n= 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n282\n6.5.\nArea, sine, and cosine rules\n\nWorked example 22: The area rule\nQUESTION\nFind the area of △ABC (correct to two decimal places):\nA\n7\nB\nC\n50◦\nSOLUTION\nStep 1: Use the given information to determine unknown angles and sides\nAB = AC = 7\n(given)\n∴ˆB = ˆC = 50◦\n(∠s opp. equal sides)\nAnd ˆA = 180◦−50◦−50◦\n(∠s sum of △ABC)\n∴ˆA = 80◦\nStep 2: Use the area rule to calculate the area of △ABC\nNotice that we do not know the length of side a and must therefore choose the form\nof the area rule that does not include this side of the triangle.\nIn △ABC:\nArea = 1\n2bc sin ˆA\n= 1\n2(7)(7) sin 80◦\n= 24,13\nStep 3: Write the final answer\nArea of △ABC = 24,13 square units.\n283\nChapter 6.\nTrigonometry\n\nWorked example 23: The area rule\nQUESTION\nShow that the area of △DEF = 1\n2df sin ˆE.\nD\nF\nE\nH\ne\nd\nf\nh\n1\n2\nSOLUTION\nStep 1: Construct a perpendicular height h\nDraw DH such that DH ⊥EF and let DH = h, D ˆEF = ˆE1 and D ˆEH = ˆE2.\nIn △DHE:\nsin ˆE2 = h\nf\nh = f sin(180◦−ˆE1)\n(∠s on str. line)\n= f sin ˆE1\nStep 2: Use the area rule to calculate the area of △DEF\nIn △DEF:\nArea = 1\n2d × h\n= 1\n2df sin ˆE1\n284\n6.5.\nArea, sine, and cosine rules\n\nThe area rule\nIn any △PQR:\nP\nP\nQ\nQ\nR\nR\nq\nq\nr\nr\np\np\nArea △PQR = 1\n2qr sin ˆP\n= 1\n2pr sin ˆQ\n= 1\n2pq sin ˆR\nThe area rule states that the area of any triangle is equal to half the product of the\nlengths of the two sides of the triangle multiplied by the sine of the angle included by\nthe two sides.\nExercise 6 – 10: The area rule\n1. Draw a sketch and calculate the area of △PQR given:\na) ˆQ = 30◦; r = 10 and p = 7\nb) ˆR = 110◦; p = 8 and q = 9\n2. Find the area of △XY Z given XZ = 52 cm, XY = 29 cm and ˆX = 58,9◦.\n3. Determine the area of a parallelogram in which two adjacent sides are 10 cm\nand 13 cm and the angle between them is 55◦.\n4. If the area of △ABC is 5000 m2 with a = 150 m and b = 70 m, what are the two\npossible sizes of ˆC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2339\n1b. 233B\n2. 233C\n3. 233D\n4. 233F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n285\nChapter 6.\nTrigonometry\n\nThe sine rule\nEMBHR\nSo far we have only applied the trigonometric ratios to right-angled triangles. We now\nexpand the application of the trigonometric ratios to triangles that do not have a right\nangle:\nInvestigation: The sine rule\nIn △ABC, AC = 15, BC = 11 and ˆA = 48◦. Find ˆB.\nA\nC\nB\nb = 15\nF\na = 11\n48◦\n1. Method 1: using the sine ratio\na) Draw a sketch of △ABC.\nb) Construct CF ⊥AB.\nc) In △CBF:\nCF\n. . . = sin ˆB\n∴CF = . . . × sin ˆB\nd) In △CAF:\nCF\n15 = . . .\n∴CF = 15 × . . .\ne) Therefore we have that:\nCF = 15 × . . .\nand CF = . . . × sin ˆB\n∴15 × . . . = . . . × sin ˆB\n∴sin ˆB = . . . . . . . . .\n∴ˆB = . . .\n2. Method 2: using the area rule\n286\n6.5.\nArea, sine, and cosine rules\n\na) In △ABC:\nArea △ABC = 1\n2AB × AC × . . .\n= 1\n2AB × . . . × . . .\nb) And we also know that\nArea △ABC = 1\n2AB × . . . × sin ˆB\nc) We can equate these two equations and solve for ˆB:\n1\n2AB × . . . × sin ˆB = 1\n2AB × . . . × . . .\n∴. . . × sin ˆB = . . . × . . .\n∴sin ˆB = . . . × . . .\n∴ˆB = . . .\n3. Use your results to write a general formula for the sine rule given △PQR:\nP\nQ\nR\nq\nr\np\nFor any triangle ABC with AB = c, BC = a and AC = b, we can construct a perpen-\ndicular height (h) at F:\nA\nC\nB\nb\nF\na\nh\nc\nMethod 1: using the sine ratio\nIn △ABF:\nsin ˆB = h\nc\n∴h = c sin ˆB\n287\nChapter 6.\nTrigonometry\n\nIn △ACF:\nsin ˆC = h\nb\n∴h = b sin ˆC\nWe can equate the two equations\nc sin ˆB = b sin ˆC\n∴sin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that:\nsin ˆA\na\n= sin ˆC\nc\nor\na\nsin ˆA\n=\nc\nsin ˆC\nSimilarly, by constructing a perpendicular height from vertex B to the line AC, we can\nalso show that area △ABC = 1\n2bc sin ˆA.\nMethod 2: using the area rule\nIn △ABC:\nArea △ABC = 1\n2ac sin ˆB\n= 1\n2ab sin ˆC\n∴1\n2ac sin ˆB = 1\n2ab sin ˆC\nc sin ˆB = b sin ˆC\nsin ˆB\nb\n= sin ˆC\nc\nor\nb\nsin ˆB\n=\nc\nsin ˆC\n288\n6.5.\nArea, sine, and cosine rules\n\nThe sine rule\nIn any △ABC:\nA\nC\nB\nb\na\nc\nsin ˆA\na\n= sin ˆB\nb\n= sin ˆC\nc\na\nsin ˆA\n=\nb\nsin ˆB\n=\nc\nsin ˆC\nSee video: 233G at www.everythingmaths.co.za\nWorked example 24: The sine rule\nQUESTION\nGiven △TRS with S ˆTR = 55◦, TR = 30 and R ˆST = 40◦, determine RS, ST and\nT ˆRS.\nSOLUTION\nStep 1: Draw a sketch\nLet RS = t, ST = r and TR = s.\nR\nS\nT\n55◦\n30\n40◦\nStep 2: Find T ˆRS using angles in a triangle\nT ˆRS + R ˆST + S ˆTR = 180◦\n(∠s sum of △TRS)\n∴T ˆRS = 180◦−40◦−55◦\n= 85◦\n289\nChapter 6.\nTrigonometry\n\nStep 3: Determine t and r using the sine rule\nt\nsin ˆT\n=\ns\nsin ˆS\nt\nsin 55◦=\n30\nsin 40◦\n∴t =\n30\nsin 40◦× sin 55◦\n= 38,2\nr\nsin ˆR\n=\ns\nsin ˆS\nr\nsin 85◦=\n30\nsin 40◦\n∴r =\n30\nsin 40◦× sin 85◦\n= 46,5\nWorked example 25: The sine rule\nQUESTION\nProve the sine rule for △MNP with MS ⊥NP.\nM\nP\nN\nS\nn\nm\np\nh\n1\n2\nSOLUTION\nStep 1: Use the sine ratio to express the angles in the triangle in terms of the length\nof the sides\nIn △MSN:\nsin ˆN2 = h\np\n∴h = p sin ˆN2\nand ˆN2 = 180◦−ˆN1\n∠s on str. line\n∴h = p sin(180◦−ˆN1)\n= p sin ˆN1\n290\n6.5.\nArea, sine, and cosine rules\n\nIn △MSP:\nsin ˆP = h\nn\n∴h = n sin ˆP\nStep 2: Equate the two equations to derive the sine rule\np sin ˆN1 = n sin ˆP\n∴sin ˆN1\nn\n= sin ˆP\np\nor\nn\nsin ˆN1\n=\np\nsin ˆP\nThe ambiguous case\nIf two sides and an interior angle of a triangle are given, and the side opposite the given\nangle is the shorter of the two sides, then we can draw two different triangles (△NMP\nand △NMP ′), both having the given dimensions. We call this the ambiguous case\nbecause there are two ways of interpreting the given information and it is not certain\nwhich is the required solution.\nM\nP ′\np\nN\nP\nn\nn\nWorked example 26: The ambiguous case\nQUESTION\nIn △ABC, AB = 82, BC = 65 and ˆA = 50◦. Draw △ABC and find ˆC (correct to\none decimal place).\nSOLUTION\nStep 1: Draw a sketch and identify the ambiguous case\nWe notice that for the given dimensions of △ABC, the side BC opposite ˆA is shorter\nthan AB. This means that we can draw two different triangles with the given dimen-\nsions.\nA\nB\nC\n50◦\n65\n82\nA′\nC′\nB′\n65\n82\n50◦\n291\nChapter 6.\nTrigonometry\n\nStep 2: Solve for unknown angle using the sine rule\nIn △ABC:\nsin ˆA\nBC = sin ˆC\nAB\nsin 50◦\n65\n= sin ˆC\n82\n∴sin 50◦\n65\n× 82 = sin ˆC\n∴ˆC = 75,1◦\nIn △A′B′C′:\nWe know that sin(180 −ˆC) = sin ˆC, which means we can also have the solution\nˆC′ = 180◦−75,1◦\n= 104,9◦\nBoth solutions are correct.\nWorked example 27: Lighthouses\nQUESTION\nThere is a coastline with two lighthouses, one on either side of a beach. The two\nlighthouses are 0,67 km apart and one is exactly due east of the other. The lighthouses\ntell how close a boat is by taking bearings to the boat (a bearing is an angle measured\nclockwise from north). These bearings are shown on the diagram below.\nCalculate how far the boat is from each lighthouse.\nˆA = 127◦\nˆB = 255◦\nC\nSOLUTION\nWe see that the two lighthouses and the boat form a triangle. Since we know the\ndistance between the lighthouses and we have two angles we can use trigonometry\n292\n6.5.\nArea, sine, and cosine rules\n\nto find the remaining two sides of the triangle, the distance of the boat from the two\nlighthouses.\nb\nA\nb B\nb\nC\n15◦\n37◦\n128◦\n0,67 km\nWe need to determine the lengths of the two sides AC and BC. We can use the sine\nrule to find the missing lengths.\nBC\nsin ˆA\n= AB\nsin ˆC\nBC = AB . sin ˆA\nsin ˆC\n= (0,67 km) sin 37◦\nsin 128◦\n= 0,51 km\nAC\nsin ˆB\n= AB\nsin ˆC\nAC = AB . sin ˆB\nsin ˆC\n= (0,67 km) sin 15◦\nsin 128◦\n= 0,22 km\nExercise 6 – 11: Sine rule\n1. Find all the unknown sides and angles of the following triangles:\na) △PQR in which ˆQ = 64◦; ˆR = 24◦and r = 3\nb) △KLM in which ˆK = 43◦; ˆ\nM = 50◦and m = 1\nc) △ABC in which ˆA = 32,7◦; ˆC = 70,5◦and a = 52,3\nd) △XY Z in which ˆX = 56◦; ˆZ = 40◦and x = 50\n2. In △ABC, ˆA = 116◦; ˆC = 32◦and AC = 23 m. Find the lengths of the sides\nAB and BC.\n3. In △RST, ˆR = 19◦; ˆS = 30◦and RT = 120 km. Find the length of the side\nST.\n4. In △KMS, ˆK = 20◦; ˆ\nM = 100◦and s = 23 cm. Find the length of the side m.\n293\nChapter 6.\nTrigonometry\n\n5. In △ABD, ˆB = 90◦, AB = 10 cm and A ˆDB = 40◦. In △BCD, ˆC = 106◦and\nC ˆDB = 15◦. Determine BC.\nA\nB\nD\nC\n10\n106◦\n15◦\n40◦\n6. In △ABC, ˆA = 33◦, AC = 21 mm and AB = 17 mm. Can you determine BC?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233H\n1b. 233J\n1c. 233K\n1d. 233M\n2. 233N\n3. 233P\n4. 233Q\n5. 233R\n6. 233S\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nThe cosine rule\nEMBHS\nInvestigation: The cosine rule\nIf a triangle is given with two sides and the included angle known, then we can not\nsolve for the remaining unknown sides and angles using the sine rule. We therefore\ninvestigate the cosine rule:\nIn △ABC, AB = 21, AC = 17 and ˆA = 33◦. Find ˆB.\nA\nC\nB\nH\n21\nc\n17\n33◦\n1. Determine CB:\na) Construct CH ⊥AB.\nb) Let AH = c and therefore HB = . . .\n294\n6.5.\nArea, sine, and cosine rules\n\nc) Applying the theorem of Pythagoras in the right-angled triangles:\nIn△CHB:\nCB2 = BH2 + CH2\n= (. . .)2 + CH2\n= 212 −(2)(21)c + c2 + CH2 . . . . . . (1)\nIn △CHA:\nCA2 = c2 + CH2\n172 = c2 + CH2 . . . . . . (2)\nSubstitute equation (2) into equation (1):\nCB2 = 212 −(2)(21)c + 172\nNow c is the only remaining unknown. In △CHA:\nc\n17 = cos 33◦\n∴c = 17 cos 33◦\nTherefore we have that\nCB2 = 212 −(2)(21)c + 172\n= 212 −(2)(21)(17 cos 33◦) + 172\n= 212 + 172 −(2)(21)(17) cos 33◦\n= 131,189 . . .\n∴CB = 11,5\n2. Use your results to write a general formula for the cosine rule given △PQR:\nP\nQ\nR\nq\nr\np\nThe cosine rule relates the length of a side of a triangle to the angle opposite it and the\nlengths of the other two sides.\n295\nChapter 6.\nTrigonometry\n\nConsider △ABC with CD ⊥AB:\nb\nD\nb\nA\nb\nB\nbC\nh\na\nb\nc\nc −d\nd\nIn △DCB: a2 = (c −d)2 + h2 from the theorem of Pythagoras.\nIn △ACD: b2 = d2 + h2 from the theorem of Pythagoras.\nSince h2 is common to both equations we can write:\na2 = (c −d)2 + h2\n∴h2 = a2 −(c −d)2\nAnd b2 = d2 + h2\n∴h2 = b2 −d2\n∴b2 −d2 = a2 −(c −d)2\na2 = b2 + (c2 −2cd + d2) −d2\n= b2 + c2 −2cd\nIn order to eliminate d we look at △ACD, where we have: cos ˆA = d\nb. So, d = b cos ˆA.\nSubstituting back we get: a2 = b2 + c2 −2bc cos ˆA.\nThe cosine rule\nIn any △ABC:\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\na2 = b2 + c2 −2bc cos ˆA\nb2 = a2 + c2 −2ac cos ˆB\nc2 = a2 + b2 −2ab cos ˆC\nSee video: 233T at www.everythingmaths.co.za\n296\n6.5.\nArea, sine, and cosine rules\n\nWorked example 28: The cosine rule\nQUESTION\nDetermine the length of QR.\nP\nR\nQ\n13 cm\n4 cm\n70◦\nSOLUTION\nStep 1: Use the cosine rule to solve for the unknown side\nQR2 = PR2 + QP 2 −2(PR)(QP) cos ˆP\n= 42 + 132 −2(4)(13) cos 70◦\n= 149,42 . . .\n∴QR = 12,2\nStep 2: Write the final answer\nQR = 12,2 cm\nWorked example 29: The cosine rule\nQUESTION\nDetermine ˆA.\n5\n7\n8\nA\nB\nC\nSOLUTION\nApplying the cosine rule:\na2 = b2 + c2 −2bc cos ˆA\n∴cos ˆA = b2 + c2 −a2\n2bc\n= 82 + 52 −72\n2 . 8 . 5\n= 0,5\n∴ˆA = 60◦\n297\nChapter 6.\nTrigonometry\n\nIt is very important:\n• not to round off before the final answer as this will affect accuracy;\n• to take the square root;\n• to remember to give units where applicable.\nHow to determine which rule to use:\n1. Area rule:\n• if no perpendicular height is given\n2. Sine rule:\n• if no right angle is given\n• if two sides and an angle are given (not the included angle)\n• if two angles and a side are given\n3. Cosine rule:\n• if no right angle is given\n• if two sides and the included angle are given\n• if three sides are given\nExercise 6 – 12: The cosine rule\n1. Solve the following triangles (that is, find all unknown sides and angles):\na) △ABC in which ˆA = 70◦; b = 4 and c = 9\nb) △RST in which RS = 14; ST = 26 and RT = 16\nc) △KLM in which KL = 5; LM = 10 and KM = 7\nd) △JHK in which ˆH = 130◦; JH = 13 and HK = 8\ne) △DEF in which d = 4; e = 5 and f = 7\n2. Find the length of the third side of the △XY Z where:\na) ˆX = 71,4◦; y = 3,42 km and z = 4,03 km\nb) x = 103,2 cm; ˆY = 20,8◦and z = 44,59 cm\n3. Determine the largest angle in:\na) △JHK in which JH = 6; HK = 4 and JK = 3\nb) △PQR where p = 50; q = 70 and r = 60\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 233V\n1b. 233W\n1c. 233X\n1d. 233Y\n1e. 233Z\n2a. 2342\n2b. 2343\n3a. 2344\n3b. 2345\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n298\n6.5.\nArea, sine, and cosine rules\n\nSee video: 2346 at www.everythingmaths.co.za\nExercise 6 – 13: Area, sine and cosine rule\n1. Q is a ship at a point 10 km due south of another ship P. R is a lighthouse on\nthe coast such that ˆP = ˆQ = 50◦.\n10 km\nP\nQ\nR\n50◦\n50◦\nDetermine:\na) the distance QR\nb) the shortest distance from the lighthouse to the line joining the two ships\n(PQ).\n2. WXY Z is a trapezium, WX ∥Y Z with WX = 3 m; Y Z = 1,5 m; ˆZ = 120◦\nand ˆW = 30◦.\nDetermine the distances XZ and XY .\n1,5 m\n3 m\n30◦\n120◦\nW\nX\nY\nZ\n3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has\nbeen flying 3◦off course. At this point the plane is 500 km from Johannesburg.\nThe direct distance between Cape Town and Johannesburg airports is 1552 km.\nDetermine, to the nearest km:\na) The distance the plane has to travel to get to Cape Town and hence the\nextra distance that the plane has had to travel due to the pilot’s error.\nb) The correction, to one hundredth of a degree, to the plane’s heading (or\ndirection).\n4. ABCD is a trapezium (meaning that AB ∥CD). AB = x; B ˆAD = a; B ˆCD = b\nand B ˆDC = c.\nFind an expression for the length of CD in terms of x, a, b and c.\nA\nB\nC\nD\na\nb\nc\nx\n299\nChapter 6.\nTrigonometry\n\n5. A surveyor is trying to determine the distance between points X and Z. However\nthe distance cannot be determined directly as a ridge lies between the two points.\nFrom a point Y which is equidistant from X and Z, he measures the angle X ˆY Z.\nY\nX\nZ\nx\nθ\na) If XY = x and X ˆY Z = θ, show that XZ = x\np\n2(1 −cos θ).\nb) Calculate XZ (to the nearest kilometre) if x = 240 km and θ = 132◦.\n6. Find the area of WXY Z (to two decimal places):\nW\nX\nY\nZ\n120◦\n3\n4\n3,5\n7. Find the area of the shaded triangle in terms of x, α, β, θ and φ:\nA\nB\nC\nD\nE\nx\nα\nβ\nθ\nφ\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2347\n2. 2348\n3. 2349\n4. 234B\n5. 234C\n6. 234D\n7. 234F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n300\n6.5.\nArea, sine, and cosine rules\n\n6.6\nSummary\nEMBHT\nSee presentation: 234G at www.everythingmaths.co.za\nsquare identity\nquotient identity\ncos2 θ + sin2 θ = 1\ntan θ = sin θ\ncos θ\nA\nA\nB\nB\nC\nC\nb\nb\nc\nc\na\na\nnegative angles\nperiodicity identities\nco-function identities\nsin(−θ) = −sin θ\nsin(θ ± 360◦) = sin θ\nsin(90◦−θ) = cos θ\ncos(−θ) = cos θ\ncos(θ ± 360◦) = cos θ\ncos(90◦−θ) = sin θ\nsine rule\narea rule\ncosine rule\nsin A\na\n= sin B\nb\n= sin C\nc\narea △ABC = 1\n2bc sin A\na2 = b2 + c2 −2bc cos A\na\nsin A =\nb\nsin B =\nc\nsin C\narea △ABC = 1\n2ac sin B\nb2 = a2 + c2 −2ac cos B\narea △ABC = 1\n2ab sin C\nc2 = a2 + b2 −2ab cos C\nQuadrant I\nQuadrant II\nQuadrant III\nQuadrant IV\nA\nall\nS\nsin θ\nT\ntan θ\nC\ncos θ\nx\ny\n0◦\n360◦\n90◦\n180◦\n270◦\n0\nGeneral solution:\n301\nChapter 6.\nTrigonometry\n\n1.\nIf sin θ = x\nθ = sin−1 x + k . 360◦\nor θ =\n\u0000180◦−sin−1 x\n\u0001\n+ k . 360◦\n2.\nIf cos θ = x\nθ = cos−1 x + k . 360◦\nor θ =\n\u0000360◦−cos−1 x\n\u0001\n+ k . 360◦\n3.\nIf tan θ = x\nθ = tan−1 x + k . 180◦\nfor k ∈Z.\nHow to determine which rule to use:\n1. Area rule:\n• no perpendicular height is given\n2. Sine rule:\n• no right angle is given\n• two sides and an angle are given (not the included angle)\n• two angles and a side are given\n3. Cosine rule:\n• no right angle is given\n• two sides and the included angle angle are given\n• three sides are given\nExercise 6 – 14: End of chapter exercises\n1. Write the following as a single trigonometric ratio:\ncos(90◦−A) sin 20◦\nsin(180◦−A) cos 70◦+ cos(180◦+ A) sin(90◦+ A)\n2. Determine the value of the following expression without using a calculator:\nsin 240◦cos 210◦−tan2 225◦cos 300◦cos 180◦\n302\n6.6.\nSummary\n\n3. Simplify:\nsin(180◦+ θ) sin(θ + 360◦)\nsin(−θ) tan(θ −360◦)\n4. Without the use of a calculator, evaluate:\n3 sin 55◦sin2 325◦\ncos(−145◦)\n−3 cos 395◦sin 125◦\n5. Prove the following identities:\na)\n1\n(cos x −1)(cos x + 1) =\n−1\ntan2 x cos2 x\nb) (1 −tan α) cos α = sin(90 + α) + cos(90 + α)\n6.\na) Prove: tan y +\n1\ntan y =\n1\ncos2 y tan y\nb) For which values of y ∈[0◦; 360◦] is the identity above undefined?\n7.\na) Simplify: sin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\nb) Hence, solve the equation\nsin(180◦+ θ) tan(360◦−θ)\nsin(−θ) tan(180◦+ θ)\n= tan θ\nfor θ ∈[0◦; 360◦].\n8. Given 12 tan θ = 5 and θ > 90◦.\na) Draw a sketch.\nb) Determine without using a calculator sin θ and cos(180◦+ θ).\nc) Use a calculator to find θ (correct to two decimal places).\n9.\nθ\nP(a; b)\n2\nx\ny\nO\nb\nIn the figure, P is a point on the Cartesian plane such that OP = 2 units and\nθ = 300◦. Without the use of a calculator, determine:\na) the values of a and b\nb) the value of sin(180◦−θ)\n10. Solve for x with x ∈[−180◦; 180◦] (correct to one decimal place):\na) 2 sin x\n2 = 0,86\n303\nChapter 6.\nTrigonometry\n\nb) tan(x + 10◦) = cos 202,6◦\nc) cos2 x −4 sin2 x = 0\n11. Find the general solution for the following equations:\na)\n1\n2 sin(x −25◦) = 0,25\nb) sin2 x + 2 cos x = −2\n12. Given the equation: sin 2α = 0,84\na) Find the general solution of the equation.\nb) Illustrate how this equation could be solved graphically for α ∈[0◦; 360◦].\nc) Write down the solutions for sin 2α = 0,84 for α ∈[0◦; 360◦].\n13.\nA\nT\nG\nN\nH\nn\nα\nβ\nA is the highest point of a vertical tower AT. At point N on the tower, n metres\nfrom the top of the tower, a bird has made its nest. The angle of inclination from\nG to point A is α and the angle of inclination from G to point N is β.\na) Express A ˆGN in terms of α and β.\nb) Express ˆA in terms of α and/or β.\nc) Show that the height of the nest from the ground (H) can be determined by\nthe formula\nH = n cos α sin β\nsin(α −β)\nd) Calculate the height of the nest H if n = 10 m, α = 68◦and β = 40◦(give\nyour answer correct to the nearest metre).\n304\n6.6.\nSummary\n\n14.\nA\nD\nB\nC\n11\n8\n5\nMr. Collins wants to pave his trapezium-shaped backyard, ABCD. AB ∥DC\nand ˆB = 90◦. DC = 11 m, AB = 8 m and BC = 5 m.\na) Calculate the length of the diagonal AC.\nb) Calculate the length of the side AD.\nc) Calculate the area of the patio using geometry.\nd) Calculate the area of the patio using trigonometry.\n15.\nA\nC\nB\n2t\nF\nt\nn\nn\n2n\nα\nIn △ABC, AC = 2A, AF = BF, A ˆFB = α and FC = 2AF. Prove that\ncos α = 1\n4.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 234H\n2. 234J\n3. 234K\n4. 234M\n5a. 234N\n5b. 234P\n6. 234Q\n7. 234R\n8. 234S\n9. 234T\n10a. 234V\n10b. 234W\n10c. 234X\n11a. 234Y\n11b. 234Z\n12. 2352\n13. 2353\n14. 2354\n15. 2355\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n305\nChapter 6.\nTrigonometry\n\n\nCHAPTER\n7\nMeasurement\n7.1\nArea of a polygon\n308\n7.2\nRight prisms and cylinders\n311\n7.3\nRight pyramids, right cones and spheres\n318\n7.4\nMultiplying a dimension by a constant factor\n322\n7.5\nSummary\n326\n\n7\nMeasurement\nThis chapter is a revision of perimeters and areas of two dimensional objects and\nvolumes of three dimensional objects. We also examine different combinations of\ngeometric objects and calculate areas and volumes in a variety of real-life contexts.\nSee video: 2356 at www.everythingmaths.co.za\n7.1\nArea of a polygon\nEMBHV\nSquare\ns\ns\nArea = s2\nRectangle\nh\nb\nArea = b × h\nTriangle\nh\nb\nArea = 1\n2b × h\nSee video: 2357 at www.everythingmaths.co.za\nTrapezium\nh\nb\na\nArea = 1\n2 (a + b) × h\nParallelogram\nh\nb\nArea = b × h\nCircle\nb r\nArea = πr2\n(Circumference = 2πr)\nSee video: 2358 at www.everythingmaths.co.za\n308\n7.1.\nArea of a polygon\n\nWorked example 1: Finding the area of a polygon\nQUESTION\nABCD is a parallelogram with DC = 15 cm, h = 8 cm and BF = 9 cm.\nA\nB\nC\nD\nH\n9 cm\n15 cm\nh\nF\nCalculate:\n1. the area of ABCD\n2. the perimeter of ABCD\nSOLUTION\nStep 1: Determine the area\nThe area of a parallelogram ABCD = base × height:\nArea = 15 × 8\n= 120 cm2\nStep 2: Determine the perimeter\nThe perimeter of a parallelogram ABCD = 2DC + 2BC.\nTo find the length of BC, we use AF ⊥BC and the theorem of Pythagoras.\nIn △ABF:\nAF 2 = AB2 −BF 2\n= 152 −92\n= 144\n∴AF = 12 cm\nAreaABCD = BC × AF\n120 = BC × 12\n∴BC = 10 cm\n∴PerimeterABCD = 2(15) + 2(10)\n= 50 cm\n309\nChapter 7.\nMeasurement\n\nExercise 7 – 1: Area of a polygon\n1. Vuyo and Banele are having a competition to see who can build the best kite\nusing balsa wood (a lightweight wood) and paper. Vuyo decides to make his kite\nwith one diagonal 1 m long and the other diagonal 60 cm long. The intersection\nof the two diagonals cuts the longer diagonal in the ratio 1 : 3.\nBanele also uses diagonals of length 60 cm and 1 m, but he designs his kite to\nbe rhombus-shaped.\na) Draw a sketch of Vuyo’s kite and write down all the known measurements.\nb) Determine how much balsa wood Vuyo will need to build the outside frame\nof the kite (give answer correct to the nearest cm).\nc) Calculate how much paper he will need to cover the frame of the kite.\nd) Draw a sketch of Banele’s kite and write down all the known measure-\nments.\ne) Determine how much wood and paper Banele will need for his kite.\nf) Compare the two designs and comment on the similarities and differences.\nWhich do you think is the better design? Motivate your answer.\n2. O is the centre of the bigger semi-circle with a radius of 10 units. Two smaller\nsemi-circles are inscribed into the bigger one, as shown on the diagram. Calcu-\nlate the following (in terms of π):\nO\nb\nb\na) The area of the shaded figure.\nb) The perimeter enclosing the shaded area.\n3. Karen’s engineering textbook is 30 cm long and 20 cm wide. She notices that\nthe dimensions of her desk are in the same proportion as the dimensions of her\ntextbook.\na) If the desk is 90 cm wide, calculate the area of the top of the desk.\nb) Karen uses some cardboard to cover each corner of her desk with an isosce-\nles triangle, as shown in the diagram:\n150 mm\n150 mm\ndesk\nCalculate the new perimeter and area of the visible part of the top of her\ndesk.\n310\n7.1.\nArea of a polygon\n\nc) Use this new area to calculate the dimensions of a square desk with the\nsame desk top area.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2359\n2. 235B\n3. 235C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.2\nRight prisms and cylinders\nEMBHW\nA right prism is a geometric solid that has a polygon as its base and vertical sides\nperpendicular to the base. The base and top surface are the same shape and size. It is\ncalled a “right” prism because the angles between the base and sides are right angles.\nA triangular prism has a triangle as its base, a rectangular prism has a rectangle as its\nbase, and a cube is a rectangular prism with all its sides of equal length. A cylinder is\nanother type of right prism which has a circle as its base. Examples of right prisms are\ngiven below: a rectangular prism, a cube, a triangular prism and a cylinder.\nSurface area of prisms and cylinders\nEMBHX\nSurface area is the total area of the exposed or outer surfaces of a prism. This is easier\nto understand if we imagine the prism to be a cardboard box that we can unfold. A\nsolid that is unfolded like this is called a net. When a prism is unfolded into a net, we\ncan clearly see each of its faces. In order to calculate the surface area of the prism, we\ncan then simply calculate the area of each face, and add them all together.\nFor example, when a triangular prism is unfolded into a net, we can see that it has\ntwo faces that are triangles and three faces that are rectangles. To calculate the surface\narea of the prism, we find the area of each triangle and each rectangle, and add them\ntogether.\nIn the case of a cylinder the top and bottom faces are circles and the curved surface\nflattens into a rectangle with a length that is equal to the circumference of the circular\nbase. To calculate the surface area we therefore find the area of the two circles and the\nrectangle and add them together.\n311\nChapter 7.\nMeasurement\n\nBelow are examples of right prisms that have been unfolded into nets. A rectangular\nprism unfolded into a net is made up of six rectangles.\nA cube unfolded into a net is made up of six identical squares.\nA triangular prism unfolded into a net is made up of two triangles and three rectangles.\nThe sum of the lengths of the rectangles is equal to the perimeter of the triangles.\nA cylinder unfolded into a net is made up of two identical circles and a rectangle with\nlength equal to the circumference of the circles.\n312\n7.2.\nRight prisms and cylinders\n\nWorked example 2: Calculating surface area\nQUESTION\nA box of chocolates has the following dimensions:\nlength = 25 cm\nwidth = 20 cm\nheight = 4 cm\n25 cm\n20 cm\n4 cm\nAnd a cylindrical tin of biscuits has the following dimensions:\ndiameter = 20 cm\nheight = 20 cm\nb\n20 cm\n20 cm\n1. Calculate the area of the wrapping paper needed to cover the entire box (assume\nno overlapping at the corners).\n2. Determine if this same sheet of wrapping paper would be enough to cover the\ntin of biscuits.\nSOLUTION\nStep 1: Determine the area of the rectangular box\nSurface area = 2 × (25 × 20) + 2 × (20 × 4) + 2 × (25 × 4)\n= 1360 cm2\n313\nChapter 7.\nMeasurement\n\nStep 2: Determine the area of the cylindrical tin\nThe radius of the cylinder = 20\n2 = 10 cm.\nSurface area = 2 × π(10)2 + 2π(10)(20)\n= 1885 cm2\nStep 3: Write the final answer\nNo, the area of the sheet of wrapping paper used to cover the box is not big enough to\ncover the tin.\nExercise 7 – 2: Calculating surface area\n1. A popular chocolate container is an equilateral right triangular prism with sides\nof 34 mm. The box is 170 mm long. Calculate the surface area of the box (to the\nnearest square centimetre).\n34 mm\n34 mm\n34 mm\n170 mm\n2. Gordon buys a cylindrical water tank to catch rain water off his roof. He discov-\ners a full 2 ℓtin of green paint in his garage and decides to paint the tank (not the\nbase). If he uses 250 ml to cover 1 m2, will he have enough green paint to cover\nthe tank with one layer of paint?\nDimensions of the tank:\ndiameter = 1,1 m\nheight = 1,4 m\nb\n1,1 m\n1,4 m\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235D\n2. 235F\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n314\n7.2.\nRight prisms and cylinders\n\nVolume of prisms and cylinders\nEMBHY\nVolume, sometimes also called capacity, is the three dimensional space occupied by\nan object, or the contents of an object. It is measured in cubic units.\nThe volume of a right prism is simply calculated by multiplying the area of the base of\na solid by the height of the solid.\nRectangular\nprism\nl\nb\nh\nVolume = area of base × height\n= area of rectangle × height\n= l × b × h\nTriangular\nprism\nH\nb\nh\nVolume = area of base × height\n= area of triangle × height\n=\n\u00121\n2b × h\n\u0013\n× H\nCylinder\nh\nr\nVolume = area of base × height\n= area of circle × height\n= πr2 × h\nSee video: 235G at www.everythingmaths.co.za\n315\nChapter 7.\nMeasurement\n\nWorked example 3: Calculating volume\nQUESTION\nA rectangular glass vase with dimensions 28 cm × 18 cm × 8 cm is used for flower\narrangements. A florist uses a platic cylindrical jug to pour water into the glass vase.\nThe jug has a diameter of 142 mm and a height of 28 cm.\n28 cm\n18 cm\n8 cm\n142 mm\n28 cm\njug\nvase\n1. Will the plastic jug hold 5 ℓof water?\n2. Will a full jug of water be enough to fill the glass vase?\nSOLUTION\nStep 1: Determine the volume of the plastic jug\nThe diameter of the jug is 142 mm, therefore the radius =\n142\n2×10 = 7,1 cm.\nVolume of a cylinder = area of the base × height\nVolume of the jug = πr2 × h\n= π × (7,1)2 × 28\n= 4434 cm3\nAnd 1000 cm3 = 1 ℓ\n∴Volume of the jug = 4434\n1000\n= 4,434 ℓ\nNo, the capacity of the jug is not enough to hold 5 ℓof water.\n316\n7.2.\nRight prisms and cylinders\n\nStep 2: Determine the volume of the glass vase\nVolume of a rectangular prism = area of the base × height\nVolume of the vase = l × b × h\n= 28 × 18 × 8\n= 4032 cm3\n∴Volume of the vase = 4032\n1000\n= 4,032 ℓ\nYes, the volume of the jug is greater than the volume of the vase.\nExercise 7 – 3: Calculating volume\n1. The roof of Phumza’s house is the shape of a right-angled trapezium. A cylindri-\ncal water tank is positioned next to the house so that the rain on the roof runs\ninto the tank. The diameter of the tank is 140 cm and the height is 2,2 m.\n10 m\n8 m\n7,5 m\n2,2 m\n140 cm\na) Determine the area of the roof.\nb) Determine how many litres of water the tank can hold.\n2. The length of a side of a hexagonal sweet tin is 8 cm and its height is equal to\nhalf of the side length.\nA\nB\nC\nD\nE\nF\n8 cm\nh\na) Show that the interior angles are equal to 120◦.\n317\nChapter 7.\nMeasurement\n\nb) Determine the length of the line AE.\nc) Calculate the volume of the tin.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235H\n2. 235J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.3\nRight pyramids, right cones and spheres\nEMBHZ\nA pyramid is a geometric solid that has a polygon as its base and sides that converge\nat a point called the apex. In other words the sides are not perpendicular to the base.\nb\nThe triangular pyramid and square pyramid take their names from the shape of their\nbase. We call a pyramid a “right pyramid” if the line between the apex and the centre\nof the base is perpendicular to the base. Cones are similar to pyramids except that\ntheir bases are circles instead of polygons. Spheres are solids that are perfectly round\nand look the same from any direction.\nSurface area of pyramids, cones and spheres\nEMBJ2\nSquare\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n= b2 + 4\n\u0000 1\n2bhs\n\u0001\n= b (b + 2hs)\n318\n7.3.\nRight pyramids, right cones and spheres\n\nTriangular\npyramid\nhs\nb\nH\nSurface area = area of base +\narea of triangular sides\n=\n\u0000 1\n2b × hb\n\u0001\n+ 3\n\u0000 1\n2b × hs\n\u0001\n= 1\n2b (hb + 3hs)\nRight cone\nh\nr\nH\nSurface area = area of base +\narea of walls\n= πr2 + 1\n2 × 2πrh\n= πr (r + h)\nSphere\nb\nr\nSurface area = 4πr2\nVolume of pyramids, cones and spheres\nEMBJ3\nSquare\npyramid\nb\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × b2 × H\nTriangular\npyramid\nb\nh\nH\nVolume = 1\n3 × area of base ×\nheight of pyramid\n= 1\n3 × 1\n2bh × H\n319\nChapter 7.\nMeasurement\n\nRight cone\nr\nH\nVolume = 1\n3 × area of base ×\nheight of cone\n= 1\n3 × πr2 × H\nSphere\nb\nr\nVolume = 4\n3πr3\nSee video: 235K at www.everythingmaths.co.za\nWorked example 4: Finding surface area and volume\nQUESTION\nThe Southern African Large Telescope (SALT) is housed in a cylindrical building with\na domed roof in the shape of a hemisphere. The height of the building wall is 17 m\nand the diameter is 26 m.\n17 m\n26 m\n1. Calculate the total surface area of the building.\n2. Calculate the total volume of the building.\n320\n7.3.\nRight pyramids, right cones and spheres\n\nSOLUTION\nStep 1: Calculate the total surface area\nTotal surface area = area of the dome + area of the cylinder\nSurface area =\n\u00141\n2(4πr2)\n\u0015\n+ [2πr × h]\n= 1\n2(4π)(13)2 + 2π(13)(17)\n= 2450 m2\nStep 2: Calculate the total volume\nTotal volume = volume of the dome + volume of the cylinder\nVolume =\n\u00141\n2 ×\n\u00124\n3πr3\n\u0013\u0015\n+\n\u0002\nπr2h\n\u0003\n= 2\n3π(13)3 + π(11)2(13)\n= 9543 m3\nExercise 7 – 4: Finding surface area and volume\n1. An ice-cream cone has a diameter of 52,4 mm and a total height of 146 mm.\n52,4 mm\n146 mm\na) Calculate the surface area of the ice-cream and the cone.\nb) Calculate the total volume of the ice-cream and the cone.\nc) How many ice-cream cones can be made from a 5 ℓtub of ice-cream (as-\nsume the cone is completely filled with ice-cream)?\n321\nChapter 7.\nMeasurement\n\nd) Consider the net of the cone given below. R is the length from the tip of\nthe cone to its perimeter, P.\nP\nR\nb\nM\ni. Determine the value of R.\nii. Calculate the length of arc P.\niii. Determine the length of arc M.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.4\nMultiplying a dimension by a constant factor\nEMBJ4\nWhen one or more of the dimensions of a prism or cylinder is multiplied by a constant,\nthe surface area and volume will change. The new surface area and volume can be\ncalculated by using the formulae from the preceding section.\nIt is important to see a relationship between the change in dimensions and the resulting\nchange in surface area and volume. These relationships make it simpler to calculate\nthe new volume or surface area of an object when its dimensions are scaled up or\ndown.\nConsider a rectangular prism of dimensions l, b and h. Below we multiply one, two\nand three of its dimensions by a constant factor of 5 and calculate the new volume and\nsurface area.\n322\n7.4.\nMultiplying a dimension by a constant factor\n\nDimensions\nVolume\nSurface\nOriginal dimensions\nl\nb\nh\nV = l × b × h\n= lbh\nA\n= 2 [(l × h) + (l × b) + (b × h)]\n= 2 (lh + lb + bh)\nMultiply one\ndimension by 5\nl\nb\n5h\nV1 = l × b × 5h\n= 5 (lbh)\n= 5V\nA1\n= 2 [(l × 5h) + (l × b) + (b × 5h)]\n= 2 (5lh + lb + 5bh)\nMultiply two\ndimensions by 5\n5l\nb\n5h\nV = 5l × b × 5h\n= 5 . 5(lbh)\n= 52V\nA2\n= 2 [(5l × 5h) + (5l × b) + (b × 5h)]\n= 2 × 5(5lh + lb + bh)\nMultiply all three\ndimensions by 5\n5l\n5b\n5h\nV = 5l × 5b × 5h\n= 53(lbh)\n= 53V\nA3\n= 2 [(5l × 5h) + (5l × 5b) + (5b × 5h)]\n= 2 × (52lh + 52lb + 52bh)\n= 52 × 2(lh + lb + bh)\n= 52A\nMultiply all three\ndimensions by k\nkl\nkb\nkh\nV = kl × kb × kh\n= k3(lbh)\n= k3V\nAk\n= 2 [(kl × kh) + (kl × kb) + (kb × kh)]\n= 2 × (k2lh + k2lb + k2bh)\n= k2 × 2(lh + lb + bh)\n= k2A\n323\nChapter 7.\nMeasurement\n\nWorked example 5: The effects of k\nQUESTION\nThe Nash family wants to build a television room onto their house. The dad draws up\nthe plans for the new square room of length k metres. The mum looks at the plans and\ndecides that the area of the room needs to be doubled. To achieve this:\n• the mum suggests doubling the length of the sides of the room\n• the dad recommends adding 2 m to the length of the sides\n• the daughter suggests multiplying the length of the sides by a factor of\n√\n2\n• the son suggests doubling only the width of the room\nWho’s suggestion will double the area of the square room? Show all calculations.\nSOLUTION\nStep 1: Draw a sketch\nk\nk\n2k\n2k\nk + 2\nk + 2\n√\n2k\n√\n2k\n2k\nk\nArea O\nArea M\nArea D\nArea d\nArea s\nStep 2: Calculate and compare\nFirst calculate the area of the square room in the original plan:\nArea O = length × length\n= k2\nTherefore, double the area of the room would be 2k2.\n324\n7.4.\nMultiplying a dimension by a constant factor\n\nConsider the mum’s suggestion of doubling the length of the sides of the room:\nArea M = length × length\n= 2k × 2k\n= 4k2\nThis area would be 4 times the original area.\nThe dad suggests adding 2 m to the length of the sides of the room:\nArea D = length × length\n= (k + 2) × (k + 2)\n= k2 + 4k + 2\n̸= 2k2\nThis is not double the original area.\nThe daughter suggests multiplying the length of the sides by a factor of\n√\n2:\nArea d = length × length\n=\n√\n2k ×\n√\n2k\n= 2k2\nThe daughter’s suggestion would double the area of the room. Practically, the length\nof the room could be multiplied by\n√\n2 ≈1,41 which would given an area of 1,96 m2.\nThe son suggests doubling only the width of the room:\nArea s = length × length\n= 2k × k\n= 2k2\nThe son’s suggestion would double the area of the room, however the room would no\nlonger be a square.\nStep 3: Write the final answer\nThe daughter’s suggestion of multiplying the length of the sides of the room by a factor\nof\n√\n2 would keep the shape of the room a square and would double the area of the\nroom.\nExercise 7 – 5: The effects of k\n1. Complete the following sentences:\na) If one dimension of a cube is multiplied by a factor 1\n2, the volume of the\ncube . . .\nb) If two dimensions of a cube are multiplied by a factor 7, the volume of the\ncube . . .\n325\nChapter 7.\nMeasurement\n\nc) If three dimensions of a cube are multiplied by a factor 3, then:\ni. each side of the cube will . . .\nii. the outer surface area of the cube will . . .\niii. the volume of the cube will . . .\nd) If each side of a cube is halved, then:\ni. the outer surface area of the cube will . . .\nii. the volume of the cube will . . .\n2. The municipality intends building a swimming pool of volume W 3 cubic metres.\nHowever, they realise that it will be very expensive to fill the pool with water, so\nthey decide to make the pool smaller.\na) The length and breadth of the pool are reduced by a factor of\n7\n10. Express\nthe new volume in terms of W.\nb) The dimensions of the pool are reduced so that the volume of the pool\ndecreases by a factor of 0,8. Determine the new dimensions of the pool in\nterms of W (remember that the pool must be a cube).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235N\n2. 235P\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n7.5\nSummary\nEMBJ5\nSee presentation: 235Q at www.everythingmaths.co.za\n1. Area is the two dimensional space inside the boundary of a flat object.\n2. Area formulae:\n• square: s2\n• rectangle: b × h\n• triangle: 1\n2b × h\n• trapezium: 1\n2 (a + b) × h\n• parallelogram: b × h\n• circle: πr2\n3. Surface area is the total area of the exposed or outer surfaces of a prism.\n4. A net is the unfolded “plan” of a solid.\n5. Volume is the three dimensional space occupied by an object, or the contents\nof an object.\n• Volume of a rectangular prism: l × b × h\n326\n7.5.\nSummary\n\n• Volume of a triangular prism:\n\u0000 1\n2b × h\n\u0001\n× H\n• Volume of a square prism or cube: s3\n• Volume of a cylinder: πr2 × h\n6. A pyramid is a geometric solid that has a polygon as its base and sides that\nconverge at a point called the apex. The sides are not perpendicular to the base.\n7. Surface area formulae:\n• square pyramid: b (b + 2h)\n• triangular pyramid: 1\n2b (hb + 3hs)\n• right cone: πr (r + hs)\n• sphere: 4πr2\n8. Volume formulae:\n• square pyramid: 1\n3 × b2 × H\n• triangular pyramid: 1\n3 × 1\n2bh × H\n• right cone: 1\n3 × πr2 × H\n• sphere: 4\n3πr3\nExercise 7 – 6: End of chapter exercises\n1.\na) Describe this figure in terms of a prism.\nb) Draw a net of this figure.\n2. Which of the following is a net of a cube?\na)\nb)\nc)\nd)\ne)\n327\nChapter 7.\nMeasurement\n\n3. Name and draw the following figures:\na) A prism with the least number of sides.\nb) A pyramid with the least number of vertices.\nc) A right prism with a kite base.\n4.\na)\ni. Determine how much paper is needed to make a box of width 16 cm,\nheight 3 cm and length 20 cm (assume no overlapping at corners).\nii. Give a mathematical name for the shape of the box.\niii. Calculate the volume of the box.\nb) Determine how much paper is needed to make a cube with a capacity of\n1 ℓ.\nc) Compare the box and the cube. Which has the greater volume and which\nrequires the most paper to make?\n5. ABCD is a rhombus with sides of length 3\n2x millimetres. The diagonals intersect\nat O and length DO = x millimetres. Express the area of ABCD in terms of x.\nO\nB\nD\nx\nC\nA\n3\n2x\n6. The diagram shows a rectangular pyramid with a base of length 80 cm and\nbreadth 60 cm. The vertical height of the pyramid is 45 cm.\n60 cm\n80 cm\n45 cm\nb\nh\nH\na) Calculate the volume of the pyramid.\nb) Calculate H and h.\nc) Calculate the surface area of the pyramid.\n7. A group of children are playing soccer in a field. The soccer ball has a capacity\nof 5000 cc (cubic centimetres). A drain pipe in the corner of the field has a\ndiameter of 20 cm. Is it possible for the children to lose their ball down the pipe?\nShow your calculations.\n328\n7.5.\nSummary\n\n8. A litre of washing powder goes into a standard cubic container at the factory.\na) Determine the length of the sides of the container.\nb) Determine the dimensions of the cubic container required to hold double\nthe volume of washing powder.\n9. A cube has sides of length k units.\na) Describe the effect on the volume of the cube if the height is tripled.\nb) If all three dimensions of the cube are tripled, determine the effect on the\nouter surface area.\nc) If all three dimensions of the cube are tripled, determine the effect on the\nvolume.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 235R\n2. 235S\n3a. 235T\n3b. 235V\n3c. 235W\n4. 235X\n5. 235Y\n6. 235Z\n7. 2362\n8. 2363\n9. 2364\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n329\nChapter 7.\nMeasurement\n\n\nCHAPTER\n8\nEuclidean geometry\n8.1\nRevision\n332\n8.2\nCircle geometry\n333\n8.3\nSummary\n363\n\n8\nEuclidean geometry\n8.1\nRevision\nEMBJ6\nParallelogram\nEMBJ7\nA parallelogram is a quadrilateral with both pairs of opposite sides parallel.\nSummary of the properties of a parallelogram:\n• Both pairs of opposite sides are parallel.\n• Both pairs of opposite sides are equal in length.\n• Both pairs of opposite angles are equal.\n• Both diagonals bisect each other.\nb\nA\nb B\nb C\nb\nD\nb\n///\n///\n/\n/\nThe mid-point theorem\nEMBJ8\nThe line joining the mid-points of two sides of a triangle is parallel to the third side\nand equal to half the length of the third side.\nA\nB\nC\nD\nE\nGiven: AD = DB and AE = EC, we can conclude that DE ∥BC and DE = 1\n2BC.\n332\n8.1.\nRevision\n\n8.2\nCircle geometry\nEMBJ9\nTerminology\nThe following terms are regularly used when referring to circles:\n• Arc — a portion of the circumference of a circle.\n• Chord — a straight line joining the ends of an arc.\n• Circumference — the perimeter or boundary line of a circle.\n• Radius (r) — any straight line from the centre of the circle to a point on the\ncircumference.\n• Diameter — a special chord that passes through the centre of the circle. A di-\nameter is a straight line segment from one point on the circumference to another\npoint on the circumference that passes through the centre of the circle.\n• Segment — part of the circle that is cut off by a chord. A chord divides a circle\ninto two segments.\n• Tangent — a straight line that makes contact with a circle at only one point on\nthe circumference.\nb\nb\nA\nB\nO\nP\na\nr\nc\nchord\ntangent\ndiameter\nradius\nsegment\nSee video: 2365 at www.everythingmaths.co.za\nAxioms\nAn axiom is an established or accepted principle. For this section, the following are\naccepted as axioms.\n333\nChapter 8.\nEuclidean geometry\n\n1. The theorem of Pythagoras states that the square of the hypotenuse of a right-\nangled triangle is equal to the sum of the squares of the other two sides.\n(AC)2 = (AB)2 + (BC)2\nC\nB\nA\n(AC)2\n(AB)2\n(BC)2\n2. A tangent is perpendicular to the radius (OT ⊥ST), drawn at the point of contact\nwith the circle.\nT\nS\nb\nO\nTheorems\nEMBJB\nA theorem is a hypothesis (proposition) that can be shown to be true by accepted\nmathematical operations and arguments. A proof is the process of showing a theorem\nto be correct.\nThe converse of a theorem is the reverse of the hypothesis and the conclusion. For\nexample, given the theorem “if A, then B”, the converse is “if B, then A”.\n334\n8.2.\nCircle geometry\n\nTheorem: Perpendicular line from circle centre bisects chord\nSTATEMENT\nIf a line is drawn from the centre of a circle perpendicular to a chord, then it bisects\nthe chord.\n(Reason: ⊥from centre bisects chord)\nGiven:\nCircle with centre O and line OP perpendicular to chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = PB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA2 = OP 2 + AP 2\n(Pythagoras)\nOB2 = OP 2 + BP 2\n(Pythagoras)\nand\nOA = OB\n(equal radii)\n∴AP 2 = BP 2\n∴AP = BP\nTherefore OP bisects AB.\nAlternative proof:\nIn △OPA and in △OPB,\nO ˆPA = O ˆPB\n(given OP ⊥AB)\nOA = OB\n(equal radii)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(RHS)\n∴AP = PB\nTherefore OP bisects AB.\n335\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Line from circle centre to mid-point of\nchord is perpendicular\nSTATEMENT\nIf a line is drawn from the centre of a circle to the mid-point of a chord, then the line\nis perpendicular to the chord.\n(Reason: line from centre to mid-point ⊥)\nGiven:\nCircle with centre O and line OP to mid-point P on chord AB.\nb\nA\nB\nO\nP\nRequired to prove:\nOP ⊥AB\nPROOF\nDraw OA and OB.\nIn △OPA and in △OPB,\nOA = OB\n(equal radii)\nAP = PB\n(given)\nOP = OP\n(common side)\n∴△OPA ≡△OPB\n(SSS)\n∴O ˆPA = O ˆPB\nand O ˆPA + O ˆPB = 180◦\n(∠on str. line)\n∴O ˆPA = O ˆPB = 90◦\nTherefore OP ⊥AB.\nSee video: 2366 at www.everythingmaths.co.za\n336\n8.2.\nCircle geometry\n\nTheorem: Perpendicular bisector of chord passes through circle centre\nSTATEMENT\nIf the perpendicular bisector of a chord is drawn, then the line will pass through the\ncentre of the circle.\n(Reason: ⊥bisector through centre)\nGiven:\nCircle with mid-point P on chord AB.\nLine QP is drawn such that Q ˆPA = Q ˆPB = 90◦.\nLine RP is drawn such that R ˆPA = R ˆPB = 90◦.\nb\nb\nA\nB\nQ\nP\nR\nRequired to prove:\nCircle centre O lies on the line PR\nPROOF\nDraw lines QA and QB.\nDraw lines RA and RB.\nIn △QPA and in △QPB,\nAP = PB\n(given)\nQP = QP\n(common side)\nQ ˆPA = Q ˆPB = 90◦\n(given)\n∴△QPA ≡△QPB\n(SAS)\n∴QA = QB\nSimilarly it can be shown that in △RPA and in △RPB, RA = RB.\nWe conclude that all the points that are equidistant from A and B will lie on the\nline PR extended. Therefore the centre O, which is equidistant to all points on the\ncircumference, must also lie on the line PR.\n337\nChapter 8.\nEuclidean geometry\n\nWorked example 1: Perpendicular line from circle centre bisects chord\nQUESTION\nGiven OQ ⊥PR and PR = 8 units, determine the value of x.\nO\nx\n5\nP\nQ\nR\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nPQ = QR = 4\n(⊥from centre bisects chord)\nStep 2: Solve for x\nIn △OQP:\nPQ = 4\n(⊥from centre bisects chord)\nOP 2 = OQ2 + QP 2\n(Pythagoras)\n52 = x2 + 42\n∴x2 = 25 −16\nx2 = 9\nx = 3\nStep 3: Write the final answer\nx = 3 units.\n338\n8.2.\nCircle geometry\n\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. In the circle with centre O, OQ ⊥PR,\nOQ = 4 units and PR = 10. Determine\nx.\nO\n4\nP\nQ\nR\nx\n2. In the circle with centre O and radius\n= 10 units, OQ ⊥PR and PR = 8. De-\ntermine x.\nO\nx\n10\nP\nQ\nR\n3. In the circle with centre O, OQ ⊥PR,\nPR = 12 units and SQ = 2 units. Deter-\nmine x.\nO\nx\nP\nQ\nR\nS\n4. In the circle with centre O, OT ⊥SQ,\nOT ⊥PR, OP = 10 units, ST = 5 units\nand PU = 8 units. Determine TU.\nO\nV\n8\nP\nR\nU\n10\n5\nT\nQ\nS\n5. In the circle with centre O, OT ⊥QP,\nOS ⊥PR, OT = 5 units, PQ = 24 units\nand PR = 25 units. Determine OS = x.\nO\nx\nP\nS\n5\nT\nQ\nR\n25\n24\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2367\n2. 2368\n3. 2369\n4. 236B\n5. 236C\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n339\nChapter 8.\nEuclidean geometry\n\nInvestigation: Angles subtended by an arc at the centre and the circumference of\na circle\n1. Measure angles x and y in each of the following graphs:\nb\nx1\ny1\nb\nx2\ny2\nb\nx3\ny3\n2. Complete the table:\nx\ny\n3. Use your results to make a conjecture about the relationship between angles\nsubtended by an arc at the centre of a circle and angles at the circumference of\na circle.\n4. Now draw three of your own similar diagrams and measure the angles to check\nyour conjecture.\n340\n8.2.\nCircle geometry\n\nTheorem: Angle at the centre of a circle is twice the size of the angle at the cir-\ncumference\nSTATEMENT\nIf an arc subtends an angle at the centre of a circle and at the circumference, then the\nangle at the centre is twice the size of the angle at the circumference.\n(Reason: ∠at centre = 2∠at circum.)\nGiven:\nCircle with centre O, arc AB subtending A ˆOB at the centre of the circle, and A ˆPB at\nthe circumference.\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nb\nQ\nO\nP\nB\nA\nRequired to prove:\nA ˆOB = 2A ˆPB\nPROOF\nDraw PO extended to Q and let A ˆOQ = ˆO1 and B ˆOQ = ˆO2.\nˆO1 = A ˆPO + P ˆAO\n(ext. ∠△= sum int. opp. ∠s)\nand A ˆPO = P ˆAO\n(equal radii, isosceles △APO)\n∴ˆO1 = A ˆPO + A ˆPO\nˆO1 = 2A ˆPO\nSimilarly, we can also show that ˆO2 = 2B ˆPO.\nFor the first two diagrams shown above we have that:\nA ˆOB = ˆO1 + ˆO2\n= 2A ˆPO + 2B ˆPO\n= 2(A ˆPO + B ˆPO)\n∴A ˆOB = 2(A ˆPB)\nAnd for the last diagram:\nA ˆOB = ˆO2 −ˆO1\n= 2B ˆPO −2A ˆPO\n= 2(B ˆPO −A ˆPO)\n∴A ˆOB = 2(A ˆPB)\n341\nChapter 8.\nEuclidean geometry\n\nWorked example 2: Angle at the centre of circle is twice angle at circumference\nQUESTION\nGiven HK, the diameter of the circle passing through centre O.\nb\nJ\nH\nK\nO\na\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles and sides on\nthe diagram\nStep 2: Solve for a\nIn △HJK:\nH ˆOK = 180◦\n(∠on str. line)\n= 2a\n(∠at centre = 2∠at circum.)\n∴2a = 180◦\na = 180◦\n2\n= 90◦\nStep 3: Conclusion\nThe diameter of a circle subtends a right angle at the circumference (angles in a semi-\ncircle).\n342\n8.2.\nCircle geometry\n\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\nGiven O is the centre of the circle, determine the unknown angle in each of the fol-\nlowing diagrams:\n1.\nb\nJ\nH\nK\nO\nb\n45◦\n2.\nbO\nJ\nK\nH\n45◦\nc\n3.\nb\nO\nK\nJ\n100◦\nH\nd\n4.\nb\nO\nH\nJ\ne\nK\n35◦\n5.\nb\nO\nJ\nK\nH\n120◦\nf\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236D\n2. 236F\n3. 236G\n4. 236H\n5. 236J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n343\nChapter 8.\nEuclidean geometry\n\nInvestigation: Subtended angles in the same segment of a circle\n1. Measure angles a, b, c, d and e in the diagram below:\na\ne\nd\nb\nc\nP\nQ\n2. Choose any two points on the circumference of the circle and label them A and\nB.\n3. Draw AP and BP, and measure A ˆPB.\n4. Draw AQ and BQ, and measure A ˆQB.\n5. What do you observe? Make a conjecture about these types of angles.\nTheorem: Subtended angles in the same segment of a circle are equal\nSTATEMENT\nIf the angles subtended by a chord of the circle are on the same side of the chord, then\nthe angles are equal.\n(Reason: ∠s in same seg.)\nGiven:\nCircle with centre O, and points P and Q on the circumference of the circle. Arc AB\nsubtends A ˆPB and A ˆQB in the same segment of the circle.\n344\n8.2.\nCircle geometry\n\nbO\nA\nB\nP\nQ\nRequired to prove:\nA ˆPB = A ˆQB\nPROOF\nA ˆOB = 2A ˆPB\n(∠at centre = 2∠at circum.)\nA ˆOB = 2A ˆQB\n(∠at centre = 2∠at circum.)\n∴2A ˆPB = 2A ˆQB\nA ˆPB = A ˆQB\nEqual arcs subtend equal angles\nFrom the theorem above we can deduce that if angles at the circumference of a circle\nare subtended by arcs of equal length, then the angles are equal. In the figure below,\nnotice that if we were to move the two chords with equal length closer to each other,\nuntil they overlap, we would have the same situation as with the theorem above. This\nshows that the angles subtended by arcs of equal length are also equal.\nb\nb\n345\nChapter 8.\nEuclidean geometry\n\n(PROOF NOT FOR EXAMS) Converse: Concyclic points\nSTATEMENT\nIf a line segment subtends equal angles at two other points on the same side of the line\nsegment, then these four points are concyclic (lie on a circle).\nGiven:\nLine segment AB subtending equal angles at points P and Q on the same side of the\nline segment AB.\nA\nB\nR\nQ\nP\nRequired to prove:\nA, B, P and Q lie on a circle.\nPROOF\nProof by contradiction:\nPoints on the circumference of a circle: we know that there are only two possible\noptions regarding a given point — it either lies on circumference or it does not.\nWe will assume that point P does not lie on the circumference.\nWe draw a circle that cuts AP at R and passes through A, B and Q.\nA ˆQB = A ˆRB\n(∠s in same seg.)\nbut A ˆQB = A ˆPB\n(given)\n∴A ˆRB = A ˆPB\nbut A ˆRB = A ˆPB + R ˆBP\n(ext. ∠△= sum int. opp.)\n∴R ˆBP = 0◦\nTherefore the assumption that the circle does not pass through P must be false.\nWe can conclude that A, B, Q and P lie on a circle (A, B, Q and P are concyclic).\n346\n8.2.\nCircle geometry\n\nWorked example 3: Concyclic points\nQUESTION\nGiven FH ∥EI and E ˆIF = 15◦, determine the value of b.\nE\nF\nG\nH\nI\n15◦\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nH ˆFI = 15◦\n(alt. ∠, FH ∥EI)\nand b = H ˆFI\n(∠s in same seg.)\n∴b = 15◦\nExercise 8 – 3: Subtended angles in the same segment\n1. Find the values of the unknown angles.\na)\nA\nB\nC\nD\n21◦\na\nb)\nJ\nK\nL\nM\n24◦\nc\n102◦\nd\nc)\nN\nO\nP\nQ\n17◦\nd\n347\nChapter 8.\nEuclidean geometry\n\n2.\nR\nS\nT\nU\nV\n45◦\n35◦\n15◦\ne\na) Given T ˆV S = S ˆV R, deter-\nmine the value of e.\nb) Is TV a diameter of the cir-\ncle? Explain your answer.\n3.\nb\nW\nX\nY\nZ\nO\n35◦\nf\nT\n1\n2\nGiven circle with centre O, WT =\nTY and X ˆWT = 35◦. Determine\nf.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236K\n1b. 236M\n1c. 236N\n2. 236P\n3. 236Q\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nCyclic quadrilaterals\nCyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference\nof a circle (concyclic).\nInvestigation: Cyclic quadrilaterals\nConsider the diagrams given below:\nCircle 1\nCircle 2\nCircle 3\nA\nB\nC\nD\nA\nB\nC\nD\nA\nB\nC\nD\n348\n8.2.\nCircle geometry\n\n1. Complete the following:\nABCD is a cyclic quadrilateral because . . . . . .\n2. Complete the table:\nCircle 1\nCircle 2\nCircle 3\nˆA =\nˆB =\nˆC =\nˆD =\nˆA + ˆC =\nˆB + ˆD =\n3. Use your results to make a conjecture about the relationship between angles of\ncyclic quadrilaterals.\nTheorem: Opposite angles of a cyclic quadrilateral\nSTATEMENT\nThe opposite angles of a cyclic quadrilateral are supplementary.\n(Reason: opp. ∠s cyclic quad.)\nGiven:\nCircle with centre O with points A, B, P and Q on the circumference such that ABPQ\nis a cyclic quadrilateral.\nbO\nA\nB\nP\nQ\n1\n2\nRequired to prove:\nA ˆBP + A ˆQP = 180◦and Q ˆAB + Q ˆPB = 180◦\n349\nChapter 8.\nEuclidean geometry\n\nPROOF\nDraw AO and OP. Label ˆO1 and ˆO2.\nˆO1 = 2A ˆBP\n(∠at centre = 2∠at circum.)\nˆO2 = 2A ˆQP\n(∠at centre = 2∠at circum.)\nand ˆO1 + ˆO2 = 360◦\n(∠s around a point)\n∴2A ˆBP + 2A ˆQP = 360◦\nA ˆBP + A ˆQP = 180◦\nSimilarly, we can show that Q ˆAB + Q ˆPB = 180◦.\nConverse: interior opposite angles of a quadrilateral\nIf the interior opposite angles of a quadrilateral are supplementary, then the quadrilat-\neral is cyclic.\nExterior angle of a cyclic quadrilateral\nIf a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle.\nb\nb\nWorked example 4: Opposite angles of a cyclic quadrilateral\nQUESTION\nGiven the circle with centre O and cyclic quadrilateral PQRS. SQ is drawn and\nS ˆPQ = 34◦. Determine the values of a, b and c.\nbO\nP\nQ\nR\nS\na\nb\nc\n34◦\n350\n8.2.\nCircle geometry\n\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for b\nS ˆPQ + c = 180◦\n(opp. ∠s cyclic quad supp.)\n∴c = 180◦−34◦\n= 146◦\na = 90◦\n(∠in semi circle)\nIn △PSQ:\na + b + 34◦= 180◦\n(∠sum of △)\n∴b = 180◦−90◦−34◦\n= 56◦\nMethods for proving a quadrilateral is cyclic\nThere are three ways to prove that a quadrilateral is a cyclic quadrilateral:\nMethod of proof\nReason\nR\nQ\nS\nP\nIf ˆP + ˆR = 180◦or ˆS +\nˆQ = 180◦, then PQRS is\na cyclic quad.\nopp.\nint.\nangles\nsuppl.\nR\nQ\nS\nP\nIf ˆP = ˆQ or ˆS = ˆR, then\nPQRS is a cyclic quad.\nangles in the same\nseg.\nR\nQ\nS\nP\nT\nIf T ˆQR = ˆS, then PQRS\nis a cyclic quad.\next.\nangle equal to\nint. opp. angle\n351\nChapter 8.\nEuclidean geometry\n\nWorked example 5: Proving a quadrilateral is a cyclic quadrilateral\nQUESTION\nProve that ABDE is a cyclic quadrilateral.\nbO\nE\nC\nD\nA\nB\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Prove that ABDE is a cyclic quadrilateral\nD ˆBC = 90◦\n(∠in semi circle)\nand ˆE = 90◦\n(given)\n∴D ˆBC = ˆE\n∴ABDE is a cyclic quadrilateral\n(ext. ∠equals int. opp. ∠)\nExercise 8 – 4: Cyclic quadrilaterals\n1. Find the values of the unknown angles.\na)\nX\nY\nZ\nW\na\nb\n106◦\n87◦\nb)\nH\nI\nJ\nK\nL\n114◦\na\nc)\nU\nV\nW\nX\n57◦\na\n86◦\n352\n8.2.\nCircle geometry\n\n2. Prove that ABCD is a cyclic quadrilateral:\na) D\nC\n72◦\nB\nA\n32◦\nM\n40◦\nb) D\nC\n70◦\nB\nA\n35◦\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 236R\n1b. 236S\n1c. 236T\n2a. 236V\n2b. 236W\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nTangent line to a circle\nA tangent is a line that touches the circumference of a circle at only one place. The\nradius of a circle is perpendicular to the tangent at the point of contact.\nb\nO\n353\nChapter 8.\nEuclidean geometry\n\nTheorem: Two tangents drawn from the same point outside a circle\nSTATEMENT\nIf two tangents are drawn from the same point outside a circle, then they are equal in\nlength.\n(Reason: tangents from same point equal)\nGiven:\nCircle with centre O and tangents PA and PB, where A and B are the respective\npoints of contact for the two lines.\nb\nA\nB\nO\nP\nRequired to prove:\nAP = BP\nPROOF\nIn △AOP and △BOP,\nO ˆAP = O ˆBP = 90◦\n(tangent ⊥radius)\nAO = BO\n(equal radii)\nOP = OP\n(common side)\n∴△AOP ≡△BOP\n(RHS)\n∴AP = BP\n354\n8.2.\nCircle geometry\n\nWorked example 6: Tangents from the same point outside a circle\nQUESTION\nIn the diagram below AE = 5 cm, AC = 8 cm and CE = 9 cm. Determine the values\nof a, b and c.\nA\nB\nC\nD\nE\nF\nAE = 5 cm\nAC = 8 cm\nCE = 9 cm\na\nb\nc\nb\nb\nb\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for a, b and c\nAB = AF = a\n(tangents from A)\nEF = ED = c\n(tangents from E)\nCB = CD = b\n(tangents from C)\n∴AE = a + c = 5\nand AC = a + b = 8\nand CE = b + c = 9\nStep 3: Solve for the unknown variables using simultaneous equations\na + c = 5\n. . . (1)\na + b = 8\n. . . (2)\nb + c = 9\n. . . (3)\nSubtract equation (1) from equation (2) and then substitute into equation (3):\n(2) −(1)\nb −c = 8 −5\n= 3\n∴b = c + 3\nSubstitute into (3)\nc + 3 + c = 9\n2c = 6\nc = 3\n∴a = 2\nand b = 6\n355\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 5: Tangents to a circle\nFind the values of the unknown lengths.\n1.\nb\nG\nH\nI\nJ\nd\n5 cm\n8 cm\n2.\nb\nK\nL\nM\nN\nO\nP\ne\nLN = 7,5 cm\n2 cm\n6 cm\n3.\nb\nb\nR\nQ\nS\nf\n3 cm\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 236X\n2. 236Y\n3. 236Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nInvestigation: Tangent-chord theorem\nConsider the diagrams given below:\nDiagram 1\nDiagram 2\nDiagram 3\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\nD\nC\nE\nB\nA\n1. Measure the following angles with a protractor and complete the table:\nDiagram 1\nDiagram 2\nDiagram 3\nA ˆBC =\nˆD =\nˆE =\n2. Use your results to complete the following: the angle between a tangent to a\ncircle and a chord is . . . . . . to the angle in the alternate segment.\n356\n8.2.\nCircle geometry\n\nTheorem: Tangent-chord theorem\nSTATEMENT\nThe angle between a tangent to a circle and a chord drawn at the point of contact, is\nequal to the angle which the chord subtends in the alternate segment.\n(Reason: tan. chord theorem)\nGiven:\nCircle with centre O and tangent SR touching the circle at B. Chord AB subtends ˆP1\nand ˆQ1.\nb\nO\nA\nB\nP\n1\n1\nQ\nT\n1\nS\nR\nRequired to prove:\n1. A ˆBR = A ˆPB\n2. A ˆBS = A ˆQB\nPROOF\nDraw diameter BT and join T to A.\nLet A ˆTB = T1.\nA ˆBS + A ˆBT = 90◦\n(tangent ⊥radius)\nB ˆAT = 90◦\n(∠in semi circle)\n∴A ˆBT + T1 = 90◦\n(∠sum of △BAT)\n∴A ˆBS = T1\nbut Q1 = T1\n(∠s in same segment)\n∴Q1 = A ˆBS\nA ˆBS + A ˆBR = 180◦\n(∠s on str. line)\nˆQ1 + ˆP1 = 180◦\n(opp. ∠s cyclic quad. supp.)\n∴A ˆBS + A ˆBR = Q1 + P1\nand A ˆBS = Q1\n∴A ˆBR = P1\n357\nChapter 8.\nEuclidean geometry\n\nWorked example 7: Tangent-chord theorem\nQUESTION\nDetermine the values of h and s.\nP\nO\nQ\nS\nR\nh + 20◦s\n4h\n4h −70◦\nSOLUTION\nStep 1: Use theorems and the given information to find all equal angles on the dia-\ngram\nStep 2: Solve for h\nO ˆQS = S ˆRQ\n(tangent chord theorem)\nh + 20◦= 4h −70◦\n90◦= 3h\n∴h = 30◦\nStep 3: Solve for s\nP ˆQR = Q ˆSR\n(tangent chord theorem)\ns = 4h\n= 4(30◦)\n= 120◦\n358\n8.2.\nCircle geometry\n\nExercise 8 – 6: Tangent-chord theorem\n1. Find the values of the unknown letters, stating reasons.\nQ\nR\nS\nO\nP\na\nb\n33◦\na)\nO\nP\nQ\nR\nS\nc\nd\n72◦\nb)\nO\nP\nQ\nR\nS\ng\nf\n38◦\n47◦\nc)\nR\nP\nO\nQ\nl\n1\n1\n66◦\nd)\nO\nP\nQ\nR\nS\ni\nj\nk\n39◦\n101◦\ne)\nO\nR\nQ\nS\nT\nm\nn\no\n34◦\nf)\nO\n•\nP\nR\nQ\nS\nT\np\nq\nr\n52◦\ng)\n359\nChapter 8.\nEuclidean geometry\n\n2. O is the centre of the circle and SPT is a tangent, with OP ⊥ST. Determine\na, b and c, giving reasons.\nO•\nS\nT\nP\nM\nN\na\nb\nc\n64◦\n3.\nP\nL\nA\nB\nC\n1 2\n3\n1\n2\nD\nGiven AB = AC, AP ∥BC and ˆA2 = ˆB2. Prove:\na) PAL is a tangent to the circle ABC.\nb) AB is a tangent to the circle ADP.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 2372\n1b. 2373\n1c. 2374\n1d. 2375\n1e. 2376\n1f. 2377\n1g. 2378\n2. 2379\n3. 237B\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nConverse: tangent-chord theorem\nIf a line drawn through the end point of a chord forms an angle equal to the angle\nsubtended by the chord in the alternate segment, then the line is a tangent to the\ncircle.\n(Reason: ∠between line and chord = ∠in alt. seg. )\n360\n8.2.\nCircle geometry\n\nWorked example 8: Applying the theorems\nQUESTION\nA\nD\nB\nC\nO\nE\nF\nBD is a tangent to the circle with centre O, with BO ⊥AD.\nProve that:\n1. CFOE is a cyclic quadrilateral\n2. FB = BC\n3. ∠A ˆOC = 2B ˆFC\n4. Will DC be a tangent to the circle passing through C, F, O and E? Motivate your\nanswer.\nSOLUTION\nStep 1: Prove CFOE is a cyclic quadrilateral by showing opposite angles are supple-\nmentary\nBO ⊥OD\n(given)\n∴F ˆOE = 90◦\nF ˆCE = 90◦\n(∠in semi circle)\n∴CFOE is a cyclic quad.\n(opp. ∠s suppl.)\nStep 2: Prove BFC is an isosceles triangle\nTo show that FB = BC we first prove △BFC is an isosceles triangle by showing that\nB ˆFC = B ˆCF.\nB ˆCF = C ˆEO\n(tangent-chord)\nC ˆEO = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴B ˆFC = B ˆCF\n∴FB = BC\n(△BFC isosceles)\n361\nChapter 8.\nEuclidean geometry\n\nStep 3: Prove A ˆOC = 2B ˆFC\nA ˆOC = 2A ˆEC\n(∠at centre = 2∠at circum.)\nand A ˆEC = B ˆFC\n(ext. ∠cyclic quad. CFOE)\n∴A ˆOC = 2B ˆFC\nStep 4: Determine if DC is a tangent to the circle through C, F, O and E\nProof by contradiction.\nLet us assume that DC is a tangent to the circle passing through the points C, F, O\nand E:\n∴D ˆCE = C ˆOE\n(tangent-chord)\nAnd using the circle with centre O and tangent BD we have that:\nD ˆCE = C ˆAE\n(tangent-chord)\nbut C ˆAE = 1\n2C ˆOE\n(∠at centre = 2∠at circum.)\n∴D ˆCE ̸= C ˆOE\nTherefore our assumption is not correct and we can conclude that DC is not a tangent\nto the circle passing through the points C, F, O and E.\nWorked example 9: Applying the theorems\nQUESTION\nA\nB\nC\nD\nE\nF\nG\nH\nFD is drawn parallel to the tangent CB\n362\n8.2.\nCircle geometry\n\nProve that:\n1. FADE is a cyclic quadrilateral\n2. F ˆEA = ˆB\nSOLUTION\nStep 1: Prove FADE is a cyclic quadrilateral using angles in the same segment\nF ˆDC = D ˆCB\n(alt. ∠s FD ∥CB)\nand D ˆCB = C ˆAE\n(tangent-chord)\n∴F ˆDC = C ˆAE\n∴FADE is a cyclic quad.\n(∠s in same seg.)\nStep 2: Prove F ˆEA = ˆB\nF ˆDA = ˆB\n(corresp. ∠s FD ∥CB)\nand F ˆEA = F ˆDA\n(∠s same seg. cyclic quad. FADE)\n∴F ˆEA = ˆB\n8.3\nSummary\nEMBJC\nSee presentation: 237C at www.everythingmaths.co.za\n• Arc An arc is a portion of the circumference of a circle.\n• Chord - a straight line joining the ends of an arc.\n• Circumference - perimeter or boundary line of a circle.\n• Radius (r) - any straight line from the centre of the circle to a point on the cir-\ncumference.\n• Diameter - a special chord that passes through the centre of the circle. A diame-\nter is the length of a straight line segment from one point on the circumference to\nanother point on the circumference, that passes through the centre of the circle.\n• Segment A segment is a part of the circle that is cut off by a chord. A chord\ndivides a circle into two segments.\n• Tangent - a straight line that makes contact with a circle at only one point on the\ncircumference.\n• A tangent line is perpendicular to the radius, drawn at the point of contact with\nthe circle.\n363\nChapter 8.\nEuclidean geometry\n\nb O\nM\nA\nB\n• If O is the centre and OM ⊥AB, then AM =\nMB.\n• If O is the centre and AM\n= MB, then\nA ˆ\nMO = B ˆ\nMO = 90◦.\n• If AM = MB and OM ⊥AB, then ⇒MO\npasses through centre O.\nb\n2x\nx\n2y\ny\nx\nIf an arc subtends an angle at the centre of a cir-\ncle and at the circumference, then the angle at the\ncentre is twice the size of the angle at the circum-\nference.\nb\nb\nAngles at the circumference subtended by the same\narc (or arcs of equal length) are equal.\nA\nB\nC\nD\n1\n2\nE\nThe four sides of a cyclic quadrilateral ABCD are\nchords of the circle with centre O.\n• ˆA + ˆC = 180◦(opp. ∠s supp.)\n• ˆB + ˆD = 180◦(opp. ∠s supp.)\n• E ˆBC = ˆD (ext. ∠cyclic quad.)\n• ˆA1 = ˆA2 = ˆC (vert. opp. ∠, ext. ∠cyclic\nquad.)\nA\nB\nC\nD\nProving a quadrilateral is cyclic: If ˆA + ˆC = 180◦or\nˆB+ ˆD = 180◦, then ABCD is a cyclic quadrilateral.\n364\n8.3.\nSummary\n\nA\nB\nC\nD\n1\n1\nIf ˆA1 = ˆC or ˆD1 = ˆB, then ABCD is a cyclic\nquadrilateral.\nA\nB\nC\nD\nIf ˆA = ˆB or ˆC = ˆD, then ABCD is a cyclic quadri-\nlateral.\nb\nA\nB\nO\nT\nIf AT and BT are tangents to circle O, then\n• OA ⊥AT (tangent ⊥radius)\n• OB ⊥BT (tangent ⊥radius)\n• TA = TB (tangents from same point equal)\nA\nB\nT\nD\nC\nx\ny\nx\ny\n• If DC is a tangent, then D ˆTA = T ˆBA and\nC ˆTB = T ˆAB\n• If D ˆTA = T ˆBA or C ˆTB = T ˆAB, then DC is\na tangent touching at T\n365\nChapter 8.\nEuclidean geometry\n\nExercise 8 – 7: End of chapter exercises\n1.\nO\n•\nA\nB\nC\nD\nE\nF\n×\n×\nx\nAOC is a diameter of the circle with centre O. F is the mid-point of chord EC.\nB ˆOC = C ˆOD and ˆB = x. Express the following angles in terms of x, stating\nreasons:\na) ˆA\nb) C ˆOD\nc) ˆD\n2.\nM•\nD\nE\nF\nG\n1 2\n1\n2\n1\n2\n1 2\nD, E, F and G are points on circle with centre M.\nˆF1 = 7◦and ˆD2 = 51◦.\nDetermine the sizes of the following angles, stating reasons:\na)\nˆ\nM1\nb) ˆD1\nc) ˆF2\nd) ˆG\ne) ˆE1\n366\n8.3.\nSummary\n\n3.\nM\n•\nO•\nD\nA\nB\nC\n1 2\nO is a point on the circle with centre M. O is also the centre of a second circle.\nDA cuts the smaller circle at C and ˆD1 = x. Express the following angles in\nterms of x, stating reasons:\na) ˆD2\nb) O ˆAB\nc) O ˆBA\nd) A ˆOB\ne) ˆC\n4.\nO•\nA\nB\nC\nM\nO is the centre of the circle with radius 5 cm and chord BC = 8 cm. Calculate\nthe lengths of:\na) OM\nb) AM\nc) AB\n5.\nO•\nA\nB\nC\n70◦\nx\nAO ∥CB in circle with centre O. A ˆOB = 70◦and O ˆAC = x. Calculate the\nvalue of x, giving reasons.\n367\nChapter 8.\nEuclidean geometry\n\n6.\nO\n•\nP\nQ\nR\nS\nT\nx\nPQ is a diameter of the circle with centre O. SQ bisects P ˆQR and P ˆQS = x.\na) Write down two other angles that are also equal to x.\nb) Calculate P ˆOS in terms of x, giving reasons.\nc) Prove that OS is a perpendicular bisector of PR.\n7.\nO•\nA\nB\nC\nD\n35◦\nB ˆOD is a diameter of the circle with centre O. AB = AD and O ˆCD = 35◦.\nCalculate the value of the following angles, giving reasons:\na) O ˆDC\nb) C ˆOD\nc) C ˆBD\nd) B ˆAD\ne) A ˆDB\n8.\nO\n•\nR\nP\nT\nQ\nx\ny\nQP in the circle with centre O is protracted to T so that PR = PT. Express y in\nterms of x.\n368\n8.3.\nSummary\n\n9.\nO•\nA\nB\nC\nD\nE\nP\nF\nO is the centre of the circle with diameter AB. CD ⊥AB at P and chord DE\ncuts AB at F. Prove that:\na) C ˆBP = D ˆPB\nb) C ˆED = 2C ˆBA\nc) A ˆBD = 1\n2C ˆOA\n10.\nO\n•\nP\nQ\nR\nx\nS\nIn the circle with centre O, OR ⊥QP, PQ = 30 mm and RS = 9 mm. Deter-\nmine the length of OQ.\n11.\nM •\nP\nQ\nR\nS\nT\nP, Q, R and S are points on the circle with centre M. PS and QR are extended\nand meet at T. PQ = PR and P ˆQR = 70◦.\na) Determine, stating reasons, three more angles equal to 70◦.\nb) If Q ˆPS = 80◦, calculate S ˆRT, S ˆTR and P ˆQS.\nc) Explain why PQ is a tangent to the circle QST at point Q.\nd) Determine P ˆ\nMQ.\n369\nChapter 8.\nEuclidean geometry\n\n12.\nO\n•\nA\nP\nQ\nC\nB\nPOQ is a diameter of the circle with centre O. QP is protruded to A and AC is\na tangent to the circle. BA ⊥AQ and BCQ is a straight line. Prove:\na) P ˆCQ = B ˆAP\nb) BAPC is a cyclic quadrilateral\nc) AB = AC\n13.\nO•\nT\nC\nA\nB\nx\nTA and TB are tangents to the circle with centre O. C is a point on the circum-\nference and A ˆTB = x. Express the following in terms of x, giving reasons:\na) A ˆBT\nb) O ˆBA\nc) ˆC\n14.\nO•\nA\nB\nC\nE\nD\nAOB is a diameter of the circle\nAECB with centre O. OE ∥BC\nand cuts AC at D.\na) Prove AD = DC\nb) Show that A ˆBC is bisected\nby EB\nc) If O ˆEB = x, express B ˆAC\nin terms of x\nd) Calculate the radius of the\ncircle if AC = 10 cm and\nDE = 1 cm\n370\n8.3.\nSummary\n\n15.\nV\nQ\nS\nR\nP\nT\nW\nx\ny\nPQ and RS are chords of the circle and PQ ∥RS. The tangent to the circle at\nQ meets RS protruded at T. The tangent at S meets QT at V . QS and PR are\ndrawn.\nLet T ˆQS = x and Q ˆRP = y. Prove that:\na) T ˆV S = 2Q ˆRS\nb) QV SW is a cyclic quadrilateral\nc) Q ˆPS + ˆT = P ˆRT\nd) W is the centre of the circle\n16.\nF\nD\nB\nC\nE\nA\nK\nT\n1\n2\n1\n2\n1\n2\n3\n4\nThe two circles shown intersect at points F and D. BFT is a tangent to the\nsmaller circle at F. Straight line AFE is drawn such that DF = EF. CDE is a\nstraight line and chord AC and BF cut at K. Prove that:\na) BT ∥CE\nb) BCEF is a parallelogram\nc) AC = BF\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237D\n2. 237F\n3. 237G\n4. 237H\n5. 237J\n6. 237K\n7. 237M\n8. 237N\n9. 237P\n10. 237Q\n11. 237R\n12. 237S\n13. 237T\n14. 237V\n15. 237W\n16. 237X\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n371\nChapter 8.\nEuclidean geometry\n\n\nCHAPTER\n9\nFinance, growth and decay\n9.1\nRevision\n374\n9.2\nSimple and compound depreciation\n377\n9.3\nTimelines\n388\n9.4\nNominal and effective interest rates\n394\n9.5\nSummary\n398\n\n9\nFinance, growth and decay\n9.1\nRevision\nEMBJD\nSimple interest is the interest calculated only on the initial amount invested, the prin-\ncipal amount. Compound interest is the interest earned on the principal amount and\non its accumulated interest. This means that interest is being earned on interest. The\naccumulated amount is the final amount; the sum of the principal amount and the\namount of interest earned.\nFormula for simple interest:\nA = P(1 + in)\nFormula for compound interest:\nA = P(1 + i)n\nwhere\nA = accumulated amount\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nWorked example 1: Simple and compound interest\nQUESTION\nSam wants to invest R 3450 for 5 years. Wise Bank offers a savings account which pays\nsimple interest at a rate of 12,5% per annum, and Grand Bank offers a savings account\npaying compound interest at a rate of 10,4% per annum. Which bank account would\ngive Sam the greatest accumulated balance at the end of the 5 year period?\nSOLUTION\nStep 1: Calculation using the simple interest formula\nWrite down the known variables and the simple interest formula\nP = 3450\ni = 0,125\nn = 5\nA = P(1 + in)\nSubstitute the values to determine the accumulated amount for the Wise Bank savings\n374\n9.1.\nRevision\n\naccount.\nA = 3450(1 + 0,125 × 5)\n= R 5606,25\nStep 2: Calculation using the compound interest formula\nWrite down the known variables and the compound interest formula.\nP = 3450\ni = 0,104\nn = 5\nA = P(1 + i)n\nSubstitute the values to determine the accumulated amount for the Grand Bank savings\naccount.\nA = 3450(1 + 0,104)5\n= R 5658,02\nStep 3: Write the final answer\nThe Grand Bank savings account would give Sam the highest accumulated balance at\nthe end of the 5 year period.\nWorked example 2: Finding i\nQUESTION\nBongani decides to put R 30 000 in an investment account. What compound interest\nrate must the investment account achieve for Bongani to double his money in 6 years?\nGive your answer correct to one decimal place.\nSOLUTION\nStep 1: Write down the known variables and the compound interest formula\nA = 60 000\nP = 30 000\nn = 6\nA = P(1 + i)n\n375\nChapter 9.\nFinance, growth and decay\n\nStep 2: Substitute the values and solve for i\n60 000 = 30 000(1 + i)6\n60 000\n30 000 = (1 + i)6\n2 = (1 + i)6\n6√\n2 = 1 + i\n6√\n2 −1 = i\n∴i = 0,122 . . .\nStep 3: Write the final answer and comment\nWe round up to a rate of 12,3% p.a. to make sure that Bongani doubles his invest-\nment.\nExercise 9 – 1: Revision\n1. Determine the value of an investment of R 10 000 at 12,1% p.a. simple interest\nfor 3 years.\n2. Calculate the value of R 8000 invested at 8,6% p.a. compound interest for 4\nyears.\n3. Calculate how much interest John will earn if he invests R 2000 for 4 years at:\na) 6,7% p.a. simple interest\nb) 5,4% p.a. compound interest\n4. The value of an investment grows from R 2200 to R 3850 in 8 years. Determine\nthe simple interest rate at which it was invested.\n5. James had R 12 000 and invested it for 5 years. If the value of his investment is\nR 15 600, what compound interest rate did it earn?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 237Y\n2. 237Z\n3. 2382\n4. 2383\n5. 2384\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n376\n9.1.\nRevision\n\n9.2\nSimple and compound depreciation\nEMBJF\nAs soon as a new car leaves the dealership, its value decreases and it is considered\n“second-hand”.\nVehicles, equipment, machinery and other similar assets, all lose\nvalue over time as a result of usage and age. This loss in value is called deprecia-\ntion. Assets that have a relatively long useful lifetime, such as machines, trucks, farm-\ning equipment etc., depreciate slower than assets like office equipment, computers,\nfurniture etc. which need to be replaced more often and therefore depreciate more\nquickly.\nDepreciation is used to calculate the value of a company’s assets, which determines\nhow much tax a company must pay. Companies can take depreciation into account as\nan expense, and thereby reduce their taxable income. A lower taxable income means\nthat the company will pay less income tax to SARS (South African Revenue Service).\nWe can calculate two different kinds of depreciation: simple decay and compound\ndecay. Decay is also a term used to describe a reduction or decline in value. Simple\ndecay is also called straight-line depreciation and compound decay can also be re-\nferred to as reducing-balance depreciation. In the straight-line method the value of the\nasset is reduced by a constant amount each year, which is calculated on the principal\namount. In reducing-balance depreciation we calculate the depreciation on the re-\nduced value of the asset. This means that the value of an asset decreases by a different\namount each year.\nInvestigation: Simple and compound depreciation\n1. Mr. Sontange buys an Opel Fiesta for R 72 000. He expects that the value of the\ncar will depreciate by R 6000 every year. He draws up a table to calculate the\ndepreciated value of his Opel Fiesta.\nComplete Mr. Sontange’s table of values for the 7 year period:\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 6000\nR 66 000\n2\nR 66 000\nR 6000\n3\n4\n5\n6\n7\n2. His son, David, does not agree that the value of the car will reduce by the same\namount each year. David thinks that the car will depreciate by 10% every year.\nComplete David’s table of values:\n377\nChapter 9.\nFinance, growth and decay\n\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 72 000\nR 7200\nR 64 800\n2\nR 64 800\nR 6480\n3\n4\n5\n6\n7\n3. Compare and discuss the results of the two different tables.\n4. Consider the graph below, which represents Mr. Sontange’s table of values:\n10 000\n20 000\n30 000\n40 000\n50 000\n60 000\n70 000\n80 000\n1\n2\n3\n4\n5\n6\n7\n8\n0\nTime (years)\nValue (Rands)\na) Draw a similar graph using David’s table of values.\nb) Interpret the two graphs and discuss the differences between them.\nc) Explain how the graphs can be used to determine the total depreciation in\neach case.\nd)\ni. Draw two new graphs by plotting the maximum value of each bar.\nii. Join the points with a line to show the general trend.\niii. Is it mathematically correct to join these points? Explain your answer.\n378\n9.2.\nSimple and compound depreciation\n\nSimple depreciation\nEMBJG\nWorked example 3: Straight-line depreciation\nQUESTION\nA new smartphone costs R 6000 and depreciates at 22% p.a. on a straight-line basis.\nDetermine the value of the smartphone at the end of each year over a 4 year period.\nSOLUTION\nStep 1: Calculate depreciation amount\nDepreciation = 6000 × 22\n100\n= 1320\nTherefore the smartphone depreciates by R 1320 every year.\nStep 2: Complete a table of values\nYear\nValue at\nbeginning of\nyear\nDepreciation\namount\nValue at end of\nyear\n1\nR 6000\nR 1320\nR 4680\n2\nR 4680\nR 1320\nR 3360\n3\nR 3360\nR 1320\nR 2040\n4\nR 2040\nR 1320\nR 720\nWe notice that\nTotal depreciation = P × i × n\nwhere\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nTherefore the depreciated value of the asset (also called the book value) can be calcu-\nlated as:\nA = P(1 −in)\nNote the similarity to the simple interest formula A = P(1 + in). Interest increases the\nvalue of the principal amount, whereas with simple decay, depreciation reduces the\nvalue of the principal amount.\nImportant: to get an accurate answer do all calculations in one step on your calculator.\nDo not round off answers in your calculations until the final answer. In the worked\nexamples in this chapter, we use dots to show that the answer has not been rounded\noff. We always round the final answer to two decimal places (cents).\n379\nChapter 9.\nFinance, growth and decay\n\nWorked example 4: Straight-line depreciation method\nQUESTION\nA car is valued at R 240 000. If it depreciates at 15% p.a. using straight-line deprecia-\ntion, calculate the value of the car after 5 years.\nSOLUTION\nStep 1: Write down the known variables and the simple decay formula\nP = 240 000\ni = 0,15\nn = 5\nA = P(1 −in)\nStep 2: Substitute the values and solve for A\nA = 240 000(1 −0,15 × 5)\n= 240 000(0,25)\n= 60 000\nStep 3: Write the final answer\nAt the end of 5 years, the car is worth R 60 000.\nWorked example 5: Simple decay\nQUESTION\nA small business buys a photocopier for R 12 000. For the tax return the owner depre-\nciates this asset over 3 years using a straight-line depreciation method. What amount\nwill he fill in on his tax form at the end of each year?\nSOLUTION\nStep 1: Write down the known variables\nThe owner of the business wants the photocopier to have a book value of R 0 after 3\nyears.\nA = 0\nP = 12 000\nn = 3\n380\n9.2.\nSimple and compound depreciation\n\nTherefore we can calculate the annual depreciation as\nDepreciation = P\nn\n= 12 000\n3\n= R 4000\nStep 2: Determine the book value at the end of each year\nBook value end of first year = 12 000 −4000\n= R 8000\nBook value end of second year = 8000 −4000\n= R 4000\nBook value end of third year = 4000 −4000\n= R 0\nExercise 9 – 2: Simple decay\n1. A business buys a truck for R 560 000. Over a period of 10 years the value of\nthe truck depreciates to R 0 using the straight-line method. What is the value of\nthe truck after 8 years?\n2. Harry wants to buy his grandpa’s donkey for R 800. His grandpa is quite pleased\nwith the offer, seeing that it only depreciated at a rate of 3% per year using the\nstraight-line method. Grandpa bought the donkey 5 years ago. What did grandpa\npay for the donkey then?\n3. Seven years ago, Rocco’s drum kit cost him R 12 500. It has now been valued at\nR 2300. What rate of simple depreciation does this represent?\n4. Fiona buys a DStv satellite dish for R 3000. Due to weathering, its value depre-\nciates simply at 15% per annum. After how long will the satellite dish have a\nbook value of zero?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2385\n2. 2386\n3. 2387\n4. 2388\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n381\nChapter 9.\nFinance, growth and decay\n\nCompound depreciation\nEMBJH\nWorked example 6: Reducing-balance depreciation\nQUESTION\nA second-hand farm tractor worth R 60 000 has a limited useful life of 5 years and\ndepreciates at 20% p.a. on a reducing-balance basis. Determine the value of the\ntractor at the end of each year over the 5 year period.\nSOLUTION\nStep 1: Write down the known variables\nP = 60 000\ni = 0,2\nn = 5\nWhen we calculate depreciation using the reducing-balance method:\n1. the depreciation amount changes for each year.\n2. the depreciation amount gets smaller each year.\n3. the book value at the end of a year becomes the principal amount for the next\nyear.\n4. the asset will always have some value (the book value will never equal zero).\nStep 2: Complete a table of values\nYear\nBook value\nDepreciation\nValue at end of\nyear\n1\nR 60 000\n60 000 × 0,2 = 12 000\nR 48 000\n2\nR 48 000\n48 000 × 0,2 = 9600\nR 38 400\n3\nR 38 400\n38 400 × 0,2 = 7680\nR 30 720\n4\nR 30 720\n30 720 × 0,2 = 6144\nR 24 576\n5\nR 24 576\n24 576 × 0,2 = 4915,20\nR 19 660,80\n382\n9.2.\nSimple and compound depreciation\n\nNotice in the example above that we could also write the book value at the end of\neach year as:\nBook value end of first year\n= 60 000(1 −0,2)\nBook value end of second year = 48 000(1 −0,2) = 60 000(1 −0,2)2\nBook value end of third year\n= 38 400(1 −0,2) = 60 000(1 −0,2)3\nBook value end of fourth year = 30 720(1 −0,2) = 60 000(1 −0,2)4\nBook value end of fifth year\n= 24 576(1 −0,2) = 60 000(1 −0,2)5\nUsing the formula for simple decay and the observed pattern in the calculation above,\nwe obtain the following formula for compound decay:\nA = P(1 −i)n\nwhere\nA = book value or depreciated value\nP = principal amount\ni = interest rate written as a decimal\nn = time period in years\nAgain, notice the similarity to the compound interest formula A = P(1 + i)n.\nWorked example 7: Reducing-balance depreciation\nQUESTION\nThe number of pelicans at the Berg river mouth is decreasing at a compound rate of\n12% p.a. If there are currently 3200 pelicans in the wetlands of the Berg river mouth,\nwhat will the population be in 5 years?\nSOLUTION\nStep 1: Write down the known variables and the compound decay formula\nP = 3200\ni = 0,12\nn = 5\nA = P(1 −i)n\nStep 2: Substitute the values and solve for A\nA = 3200(1 −0,12)5\n= 3200(0,88)5\n= 1688,7421 . . .\nStep 3: Write the final answer\nIn 5 years, the pelican population will be approximately 1689.\n383\nChapter 9.\nFinance, growth and decay\n\nWorked example 8: Compound decay\nQUESTION\n1. A school buys a minibus for R 950 000, which depreciates at 13,5% per annum.\nDetermine the value of the minibus after 3 years if the depreciation is calculated:\na) on a straight-line basis.\nb) on a reducing-balance basis.\n2. Which is the better option?\nSOLUTION\nStep 1: Write down known variables\nP = 950 000\ni = 0,135\nn = 3\nStep 2: Use the simple decay formula and solve for A\nA = 950 000(1 −3 × 0,135)\n= 950 000(0,865)\n= 565 250\n∴A = R 565 250\nStep 3: Use the compound decay formula and solve for A\nA = 950 000(1 −0,135)3\n= 950 000(0,865)3\n= 614 853,89\n∴A = R 614 853,89\nStep 4: Interpret the answers\nAfter a period of 3 years, the value of the minibus calculated on the straight-line\nmethod is less than the value of the minibus calculated on the reducing-balance\nmethod. The value of the minibus depreciated less on the reducing-balance basis\nbecause the amount of depreciation is calculated on a smaller amount every year,\nwhereas the straight-line method is based on the full value of the minibus every year.\n384\n9.2.\nSimple and compound depreciation\n\nWorked example 9: Compound depreciation\nQUESTION\nFarmer Jack bought a tractor and it has depreciated by 20% p.a. on a reducing-balance\nbasis. If the current value of the tractor is R 52 429, calculate how much Farmer Jack\npaid for his tractor if he bought it 7 years ago.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 52 429\ni = 0,2\nn = 7\nA = P(1 −i)n\nStep 2: Substitute the values and solve for P\n52 429 = P(1 −0,2)7\n= P(0,8)7\n∴P = 52 429\n(0,8)7\n= 250 000,95 . . .\nStep 3: Write the final answer\n7 years ago, Farmer Jack paid R 250 000 for his tractor.\nExercise 9 – 3: Compound depreciation\n1. Jwayelani buys a truck for R 89 000 and depreciates it by 9% p.a. using the\ncompound depreciation method. What is the value of the truck after 14 years?\n2. The number of cormorants at the Amanzimtoti river mouth is decreasing at a\ncompound rate of 8% p.a. If there are now 10 000 cormorants, how many will\nthere be in 18 years’ time?\n3. On January 1, 2008 the value of my Kia Sorento is R 320 000. Each year after\nthat, the car’s value will decrease 20% of the previous year’s value. What is the\nvalue of the car on January 1, 2012?\n385\nChapter 9.\nFinance, growth and decay\n\n4. The population of Bonduel decreases at a reducing-balance rate of 9,5% per\nannum as people migrate to the cities. Calculate the decrease in population over\na period of 5 years if the initial population was 2 178 000.\n5. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses\n3% of its water each day. How much does it weigh after a month of 31 days?\n6. Richard bought a car 15 years ago and it depreciated by 17% p.a. on a com-\npound depreciation basis. How much did he pay for the car if it is now worth\nR 5256?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2389\n2. 238B\n3. 238C\n4. 238D\n5. 238F\n6. 238G\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\nFinding i\nEMBJJ\nWorked example 10: Finding i for simple decay\nQUESTION\nAfter 4 years, the value of a computer is halved. Assuming simple decay, at what\nannual rate did it depreciate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and simple decay formula\nLet the value of the computer be x, therefore:\nA = x\n2\nP = x\nn = 4\nA = P(1 −in)\nStep 2: Substitute the values and solve for i\n386\n9.2.\nSimple and compound depreciation\n\nx\n2 = x(1 −3i)\n1\n2 = 1 −3i\n∴3i = 1 −1\n2\n∴i = 0,1667\nStep 3: Write the final answer\nThe computer depreciated at a rate of 16,67% p.a.\nWorked example 11: Finding i for compound decay\nQUESTION\nCristina bought a fridge at the beginning of 2009 for R 8999 and sold it at the end\nof 2011 for R 4500. At what rate did the value of her fridge depreciate assuming a\nreducing-balance method? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down known variables and compound decay formula\nA = 4500\nP = 8999\nn = 3\nA = P(1 −i)n\nStep 2: Substitute the values and solve for i\n4500 = 8999(1 −i)3\n4500\n8999 = (1 −i)3\n3\nr\n4500\n8999 = 1 −i\n∴i = 1 −\n3\nr\n4500\n8999\n= 0,206\nStep 3: Write the final answer\nCristina’s fridge depreciated at a rate of 20,6% p.a.\n387\nChapter 9.\nFinance, growth and decay\n\nExercise 9 – 4: Finding i\n1. A machine costs R 45 000 and has a scrap value of R 9000 after 10 years. Deter-\nmine the annual rate of depreciation if it is calculated on the reducing balance\nmethod.\n2. After 15 years, an aeroplane is worth 1\n6 of its original value. At what annual rate\nwas depreciation compounded?\n3. Mr. Mabula buys furniture for R 20 000. After 6 years he sells the furniture for\nR 9300. Calculate the annual compound rate of depreciation of the furniture.\n4. Ayanda bought a new car 7 years ago for double what it is worth today. At what\nyearly compound rate did her car depreciate?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238H\n2. 238J\n3. 238K\n4. 238M\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.3\nTimelines\nEMBJK\nInterest can be compounded more than once a year. For example, an investment can\nbe compounded monthly or quarterly. Below is a table of compounding terms and\ntheir corresponding numeric value (p). When amounts are compounded more than\nonce per annum, we multiply the number of years by p and we also divide the interest\nrate by p.\nTerm\np\nyearly / annually\n1\nhalf-yearly / bi-annually\n2\nquarterly\n4\nmonthly\n12\nweekly\n52\ndaily\n365\nWorked example 12: Timelines\nQUESTION\nR 5500 is invested for a period of 4 years in a savings account. For the first year, the\ninvestment grows at a simple interest rate of 11% p.a. and then at a rate of 12,5%\np.a. compounded quarterly for the rest of the period. Determine the value of the\ninvestment at the end of the 4 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\n388\n9.3.\nTimelines\n\nT0\nT1\nT2\nT3\nT4\n11% p.a. simple interest\n12,5% p.a. compounded quarterly\nR 5500\nIn the timeline above, the intervals are given in years. For example, T0 is the start of\nthe investment, T1 is the end of the first year and T4 is the end of the fourth year.\nStep 2: Use the simple interest formula to calculate A at T1\nA = P(1 + in)\n= 5500(1 + 0,11)\n= R 6105\nStep 3: Use the compound interest formula to calculate A at T4\nThe investment is compounded quarterly, therefore:\nn = 3 × 4\n= 12\nand i = 0,125\n4\nAlso notice that the accumulated amount at the end of the first year becomes the\nprincipal amount at the beginning of the second year.\nA = P(1 + i)n\n= 6105\n\u0012\n1 + 0,125\n4\n\u001312\n= R 8831,88\nStep 4: Write the final answer\nThe value of the investment at the end of the 4 years is R 8831,88.\n389\nChapter 9.\nFinance, growth and decay\n\nWorked example 13: Timelines\nQUESTION\nR 150 000 is deposited in an investment account for a period of 6 years at an interest\nrate of 12% p.a. compounded half-yearly for the first 4 years and then 8,5% p.a.\ncompounded yearly for the rest of the period. A deposit of R 8000 is made into the\naccount after the first year and then another deposit of R 2000 is made 5 years after\nthe initial investment. Calculate the value of the investment at the end of the 6 years.\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n8,5% p.a. compounded yearly\nR 15 000\nT5\nT6\n12% p.a. compounded half-yearly\n+R 8000\n+R 2000\nRemember to show when the additional deposits of R 8000 and R 2000 where made\ninto the account. It is very important to note that the interest rate changes at T4.\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nBetween T0 and T4:\nWe notice that interest for the first 4 years is compounded half-yearly, therefore:\nn1 = 4 × 2\n= 8\nand i1 = 0,12\n2\nBetween T4 and T6:\nn2 = 2\nand i2 = 0,085\nTherefore the total growth of the initial deposit over the 6 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\nStep 3: The deposit at T1\nBetween T1 and T4:\n390\n9.3.\nTimelines\n\nInterest on this deposit is compounded half-yearly for 3 years, therefore:\nn3 = 3 × 2\n= 6\nand i3 = 0,12\n2\nBetween T4 and T6:\nn4 = 2\nand i4 = 0,085\nTherefore the total growth of the deposit over the 5 years is:\nA = P(1 + i3)n3(1 + i4)n4\n= 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2\nStep 4: The deposit at T5\nAccumulate interest for only 1 year:\nA = P(1 + i)n\n= 2000(1 + 0,085)1\nStep 5: Determine the total calculation\nTo get as accurate an answer as possible, we do the the calculation on the calculator\nin one step. Using the memory and answer recall function on the calculator, we avoid\nrounding off until we get the final answer.\nA = 150 000\n\u0012\n1 + 0,12\n2\n\u00138\n(1 + 0,085)2\n+ 8000\n\u0012\n1 + 0,12\n2\n\u00136\n(1 + 0,085)2 + 2000(1 + 0,085)1\n= R 296 977,00\nStep 6: Write the final answer\nThe value of the investment at the end of the 6 years is R 296 977,00.\n391\nChapter 9.\nFinance, growth and decay\n\nWorked example 14: Timelines\nQUESTION\nR 60 000 is invested in an account which offers interest at 7% p.a.\ncompounded\nquarterly for the first 18 months. Thereafter the interest rate changes to 5% p.a. com-\npounded monthly. Three years after the initial investment, R 5000 is withdrawn from\nthe account. How much will be in the account at the end of 5 years?\nSOLUTION\nStep 1: Draw a timeline and write down known variables\nT0\nT1\nT2\nT3\nT4\n5% p.a. compounded monthly\nR 60 000\nT5\n7% p.a. compounded quarterly\n−R 5000\nRemember to show when the withdrawal of R 5000 was taken out of the account. It is\nalso important to note that the interest rate changes after 18 months (T1 1\n2 ).\nWe break this question down into parts and consider each amount separately.\nStep 2: The initial deposit at T0\nInterest for the first 1,5 years is compounded quarterly, therefore:\nn1 = 1,5 × 4\n= 6\nand i1 = 0,07\n4\nInterest for the remaining 3,5 years is compounded monthly, therefore:\nn2 = 3,5 × 12\n= 42\nand i2 = 0,05\n12\nTherefore the total growth of the initial deposit over the 5 years is:\nA = P(1 + i1)n1(1 + i2)n2\n= 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n392\n9.3.\nTimelines\n\nStep 3: The withdrawal at T3\nWe calculate the interest that the R 5000 would have earned if it had remained in the\naccount:\nn = 2 × 12\n= 24\nand i = 0,05\n12\nTherefore we have that:\nA = P(1 + i)n\n= 5000\n\u0012\n1 + 0,05\n12\n\u001324\nStep 4: Determine the total calculation\nWe subtract the withdrawal and the interest it would have earned from the accumu-\nlated amount at the end of the 5 years:\nA = 60 000\n\u0012\n1 + 0,07\n4\n\u00136 \u0012\n1 + 0,05\n12\n\u001342\n−5000\n\u0012\n1 + 0,05\n12\n\u001324\n= R 73 762,19\nStep 5: Write the final answer\nThe value of the investment at the end of the 5 years is R 73 762,19.\nExercise 9 – 5: Timelines\n1. After a 20-year period Josh’s lump sum investment matures to an amount of\nR 313 550. How much did he invest if his money earned interest at a rate of\n13,65% p.a. compounded half yearly for the first 10 years, 8,4% p.a. com-\npounded quarterly for the next five years and 7,2% p.a. compounded monthly\nfor the remaining period?\n2. Sindisiwe wants to buy a motorcycle. The cost of the motorcycle is R 55 000.\nIn 1998 Sindisiwe opened an account at Sutherland Bank with R 16 000. Then\nin 2003 she added R 2000 more into the account. In 2007 Sindisiwe made\nanother change: she took R 3500 from the account. If the account pays 6% p.a.\ncompounded half-yearly, will Sindisiwe have enough money in the account at\nthe end of 2012 to buy the motorcycle?\n3. A loan has to be returned in two equal semi-annual instalments. If the rate of\ninterest is 16% per annum, compounded semi-annually and each instalment is\nR 1458, find the sum borrowed.\n393\nChapter 9.\nFinance, growth and decay\n\n4. A man named Phillip invests R 10 000 into an account at North Bank at an\ninterest rate of 7,5% p.a. compounded monthly. After 5 years the bank changes\nthe interest rate to 8% p.a. compounded quarterly. How much money will\nPhillip have in his account 9 years after the original deposit?\n5. R 75 000 is invested in an account which offers interest at 11% p.a.\ncom-\npounded monthly for the first 24 months.\nThen the interest rate changes to\n7,7% p.a. compounded half-yearly. If R 9000 is withdrawn from the account\nafter one year and then a deposit of R 3000 is made three years after the initial\ninvestment, how much will be in the account at the end of 6 years?\n6. Christopher wants to buy a computer, but right now he doesn’t have enough\nmoney. A friend told Christopher that in 5 years the computer will cost R 9150.\nHe decides to start saving money today at Durban United Bank. Christopher\ndeposits R 5000 into a savings account with an interest rate of 7,95% p.a. com-\npounded monthly.\nThen after 18 months the bank changes the interest rate\nto 6,95% p.a. compounded weekly. After another 6 months, the interest rate\nchanges again to 7,92% p.a.\ncompounded two times per year.\nHow much\nmoney will Christopher have in the account after 5 years, and will he then have\nenough money to buy the computer?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 238N\n2. 238P\n3. 238Q\n4. 238R\n5. 238S\n6. 238T\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.4\nNominal and effective interest rates\nEMBJM\nWe have seen that although interest is quoted as a percentage per annum it can be\ncompounded more than once a year. We therefore need a way of comparing interest\nrates. For example, is an annual interest rate of 8% compounded quarterly higher or\nlower than an interest rate of 8% p.a. compounded yearly?\nInvestigation: Nominal and effective interest rates\n1. Calculate the accumulated amount at the end of one year if R 1000 is invested\nat 8% p.a. compound interest:\nA = P(1 + i)n\n= . . . . . .\n2. Calculate the value of R 1000 if it is invested for one year at 8% p.a. com-\npounded:\n394\n9.4.\nNominal and effective interest rates\n\nFrequency\nCalculation\nAccumulated\namount\nInterest\namount\nhalf-yearly\nA = 1000\n\u0010\n1 + 0,08\n2\n\u00111×2\nR 1081,60\nR 81,60\nquarterly\nmonthly\nweekly\ndaily\n3. Use your results from the table above to calculate the effective rate that the\ninvestment of R 1000 earns in one year:\nFrequency\nAccumulated\namount\nCalculation\nEffective\ninterest\nrate\nhalf-yearly\nR 1081,60\n1081,60 = 1000(1 + i)\n1081,60\n1000\n= 1 + i\n1081,60\n1000\n−1 = i\n∴i = 0,0816\ni = 8,16%\nquarterly\nmonthly\nweekly\ndaily\n4. If you wanted to borrow R 10 000 from the bank, would it be better to pay it\nback at an interest rate of 22% p.a. compounded quarterly or 22% compounded\nmonthly? Show your calculations.\nAn interest rate compounded more than once a year is called the nominal interest rate.\nIn the investigation above, we determined that the nominal interest rate of 8% p.a.\ncompounded half-yearly is actually an effective rate of 8,16% p.a.\nGiven a nominal interest rate i(m) compounded at a frequency of m times per year\nand the effective interest rate i, the accumulated amount calculated using both interest\nrates will be equal so we can write:\nP(1 + i) = P\n \n1 + i(m)\nm\n!m\n∴1 + i =\n \n1 + i(m)\nm\n!m\n395\nChapter 9.\nFinance, growth and decay\n\nWorked example 15: Nominal and effective interest rates\nQUESTION\nInterest on a credit card is quoted as 23% p.a. compounded monthly. What is the\neffective annual interest rate? Give your answer correct to two decimal places.\nSOLUTION\nStep 1: Write down the known variables\nInterest is being added monthly, therefore:\nm = 12\ni(12) = 0,23\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i\n1 + i =\n\u0012\n1 + 0,23\n12\n\u001312\n∴i = 1 −\n\u0012\n1 + 0,23\n12\n\u001312\n= 25,59%\nStep 3: Write the final answer\nThe effective interest rate is 25,59% per annum.\nWorked example 16: Nominal and effective interest rates\nQUESTION\nDetermine the nominal interest rate compounded quarterly if the effective interest rate\nis 9% per annum (correct to two decimal places).\nSOLUTION\nStep 1: Write down the known variables\n396\n9.4.\nNominal and effective interest rates\n\nInterest is being added quarterly, therefore:\nm = 4\ni = 0,09\n1 + i =\n \n1 + i(m)\nm\n!m\nStep 2: Substitute values and solve for i(m)\n1 + 0,09 =\n \n1 + i(4)\n4\n!4\n4p\n1,09 = 1 + i(4)\n4\n4p\n1,09 −1 = i(4)\n4\n4\n\u0010\n4p\n1,09 −1\n\u0011\n= i(4)\n∴i(4) = 8,71%\nStep 3: Write the final answer\nThe nominal interest rate is 8,71% p.a. compounded quarterly.\nExercise 9 – 6: Nominal and effect interest rates\n1. Determine the effective annual interest rate if the nominal interest rate is:\na) 12% p.a. compounded quarterly.\nb) 14,5% p.a. compounded weekly.\nc) 20% p.a. compounded daily.\n2. Consider the following:\n• 16,8% p.a. compounded annually.\n• 16,4% p.a. compounded monthly.\n• 16,5% p.a. compounded quarterly.\na) Determine the effective annual interest rate of each of the nominal rates\nlisted above.\nb) Which is the best interest rate for an investment?\nc) Which is the best interest rate for a loan?\n397\nChapter 9.\nFinance, growth and decay\n\n3. Calculate the effective annual interest rate equivalent to a nominal interest rate\nof 8,75% p.a. compounded monthly.\n4. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every\nfour months on her investment of R 85 000.\nCalculate the effective rate per\nannum.\n5. Determine which of the following would be the better agreement for paying back\na student loan:\na) 9,1% p.a. compounded quarterly.\nb) 9% p.a. compounded monthly.\nc) 9,3% p.a. compounded half-yearly.\n6. Miranda invests R 8000 for 5 years for her son’s study fund. Determine how\nmuch money she will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 6% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 238V\n1b. 238W\n1c. 238X\n2. 238Y\n3. 238Z\n4. 2392\n5. 2393\n6. 2394\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n9.5\nSummary\nEMBJN\nSee presentation: 2395 at www.everythingmaths.co.za\n• Simple interest: A = P(1 + in)\n• Compound interest: A = P(1 + i)n\n• Simple depreciation: A = P(1 −in)\n• Compound depreciation: A = P(1 −i)n\n• Nominal and effective annual interest rates: 1 + i =\n\u0010\n1 + i(m)\nm\n\u0011m\n398\n9.5.\nSummary\n\nExercise 9 – 7: End of chapter exercises\n1. Thabang buys a Mercedes worth R 385 000 in 2007. What will the value of the\nMercedes be at the end of 2013 if:\na) the car depreciates at 6% p.a. straight-line depreciation.\nb) the car depreciates at 6% p.a. reducing-balance depreciation.\n2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R 8900.\nThe interest rate is quoted as 11% per annum based on simple interest. Calculate\nthe required monthly payment for this contract.\n3. A computer is purchased for R 16 000. It depreciates at 15% per annum.\na) Determine the book value of the computer after 3 years if depreciation is\ncalculated according to the straight-line method.\nb) Find the rate according to the reducing-balance method that would yield,\nafter 3 years, the same book value as calculated in the previous question.\n4. Maggie invests R 12 500 for 5 years at 12% per annum compounded monthly\nfor the first 2 years and 14% per annum compounded semi-annually for the next\n3 years. How much will Maggie receive in total after 5 years?\n5. Tintin invests R 120 000. He is quoted a nominal interest rate of 7,2% per an-\nnum compounded monthly.\na) Calculate the effective rate per annum (correct to two decimal places).\nb) Use the effective rate to calculate the value of Tintin’s investment if he\ninvested the money for 3 years.\nc) Suppose Tintin invests his money for a total period of 4 years, but after 18\nmonths makes a withdrawal of R 20 000, how much will he receive at the\nend of the 4 years?\n6. Ntombi opens accounts at a number of clothing stores and spends freely. She\ngets herself into terrible debt and she cannot pay off her accounts. She owes\nFashion World R 5000 and the shop agrees to let her pay the bill at a nominal\ninterest rate of 24% compounded monthly.\na) How much money will she owe Fashion World after two years?\nb) What is the effective rate of interest that Fashion World is charging her?\n7. John invests R 30 000 in the bank for a period of 18 months. Calculate how\nmuch money he will have at the end of the period and the effective annual\ninterest rate if the nominal interest of 8% is compounded:\nCalculation\nAccumulated\namount\nEffective annual\ninterest rate\nyearly\nhalf-yearly\nquarterly\nmonthly\ndaily\n399\nChapter 9.\nFinance, growth and decay\n\n8. Convert an effective annual interest rate of 11,6% p.a. to a nominal interest rate\ncompounded:\na) half-yearly\nb) quarterly\nc) monthly\n9. Joseph must sell his plot on the West Coast and he needs to get R 300 000 on the\nsale of the land. If the estate agent charges him 7% commission on the selling\nprice, what must the buyer pay for the plot?\n10. Mrs. Brown retired and received a lump sum of R 200 000. She deposited the\nmoney in a fixed deposit savings account for 6 years. At the end of the 6 years\nthe value of the investment was R 265 000. If the interest on her investment was\ncompounded monthly, determine:\na) the nominal interest rate per annum\nb) the effective annual interest rate\n11. R 145 000 is invested in an account which offers interest at 9% p.a.\ncom-\npounded half-yearly for the first 2 years. Then the interest rate changes to 4%\np.a. compounded quarterly. Four years after the initial investment, R 20 000 is\nwithdrawn. 6 years after the initial investment, a deposit of R 15 000 is made.\nDetermine the balance of the account at the end of 8 years.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 2396\n2. 2397\n3. 2398\n4. 2399\n5. 239B\n6. 239C\n7. 239D\n8. 239F\n9. 239G\n10. 239H\n11. 239J\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n400\n9.5.\nSummary\n\nCHAPTER\n10\nProbability\n10.1\nRevision\n402\n10.2\nDependent and independent events\n411\n10.3\nMore Venn diagrams\n419\n10.4\nTree diagrams\n426\n10.5\nContingency tables\n431\n10.6\nSummary\n435\n\n10\nProbability\n10.1\nRevision\nEMBJP\nTerminology\nEMBJQ\nOutcome: a single observation of an uncertain or random process (called an experi-\nment). For example, when you accidentally drop a book, it might fall on its cover, on\nits back or on its side. Each of these options is a possible outcome.\nSample space of an experiment: the set of all possible outcomes of the experiment. For\nexample, the sample space when you roll a single 6-sided die is the set {1; 2; 3; 4; 5; 6}.\nFor a given experiment, there is exactly one sample space. The sample space is de-\nnoted by the letter S.\nEvent: a set of outcomes of an experiment. For example, during radioactive decay of\n1 gramme of uranium-234, one possible event is that the number of alpha-particles\nemitted during 1 microsecond is between 225 and 235.\nProbability of an event: a real number between 0 and 1 that describes how likely it\nis that the event will occur. A probability of 0 means the outcome of the experiment\nwill never be in the event set. A probability of 1 means the outcome of the experiment\nwill always be in the event set. When all possible outcomes of an experiment have\nequal chance of occurring, the probability of an event is the number of outcomes in\nthe event set as a fraction of the number of outcomes in the sample space.\nRelative frequency of an event: the number of times that the event occurs during\nexperimental trials, divided by the total number of trials conducted. For example, if\nwe flip a coin 10 times and it landed on heads 3 times, then the relative frequency of\nthe heads event is 3\n10 = 0,3.\nUnion of events: the set of all outcomes that occur in at least one of the events. For\n2 events called A and B, we write the union as “A or B”. Another way of writing the\nunion is using set notation: A ∪B.\nIntersection of events: the set of all outcomes that occur in all of the events. For 2\nevents called A and B, we write the intersection as “A and B”. Another way of writing\nthe intersection is using set notation: A ∩B.\nMutually exclusive events: events with no outcomes in common, that is (A and B) =\n∅. Mutually exclusive events can never occur simultaneously. For example the event\nthat a number is even and the event that the same number is odd are mutually exclu-\nsive, since a number can never be both even and odd.\nComplementary events: two mutually exclusive events that together contain all the\noutcomes in the sample space. For an event called A, we write the complement as\n“not A”. Another way of writing the complement is as A′.\nSee video: 239K at www.everythingmaths.co.za\n402\n10.1.\nRevision\n\nIdentities\nEMBJR\nThe addition rule (also called the sum rule) for any 2 events, A and B is\nP(A or B) = P(A) + P(B) −P(A and B)\nThis rule relates the probabilities of 2 events with the probabilities of their union and\nintersection.\nThe addition rule for 2 mutually exclusive events is\nP(A or B) = P(A) + P(B)\nThis rule is a special case of the previous rule. Because the events are mutually exclu-\nsive, P(A and B) = 0.\nThe complementary rule is\nP(not A) = 1 −P(A)\nThis rule is a special case of the previous rule. Since A and (not A) are mutually\nexclusive, P(A or (not A)) = 1.\nSee video: 239M at www.everythingmaths.co.za\nWorked example 1: Events\nQUESTION\nYou take all the hearts from a deck of cards. You then select a random card from the set\nof hearts. What is the sample space? What is the probability of each of the following\nevents?\n1. The card is the ace of hearts.\n2. The card has a prime number on it.\n3. The card has a letter of the alphabet on it.\nSOLUTION\nStep 1: Write down the sample space\nSince we are considering only one suit from the deck of cards (the hearts), we need to\nwrite down only the letters and numbers on the cards. Therefore the sample space is\nS = {A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K}\nStep 2: Write down the event sets\n• ace of hearts: {A}\n• prime number: {2; 3; 5; 7}\n• letter of alphabet: {A; J; Q; K}\n403\nChapter 10.\nProbability\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are 13 elements in the\nsample space. So the probability of each event is\n• ace of hearts:\n1\n13\n• prime number:\n4\n13\n• letter of alphabet:\n4\n13\nWorked example 2: Events\nQUESTION\nYou roll two 6-sided dice. Let E be the event that the total number of dots on the dice\nis 10. Let F be the event that at least one die is a 3.\n1. Write down the event sets for E and F.\n2. Determine the probabilities for E and F.\n3. Are E and F mutually exclusive? Why or why not?\nSOLUTION\nStep 1: Write down the sample space\nThe sample space of a single 6-sided die is just {1; 2; 3; 4; 5; 6}. To get the sample\nspace of two 6-sided dice, we have to take every possible pair of numbers from 1 to 6.\nS =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(1; 1)\n(1; 2)\n(1; 3)\n(1; 4)\n(1; 5)\n(1; 6)\n(2; 1)\n(2; 2)\n(2; 3)\n(2; 4)\n(2; 5)\n(2; 6)\n(3; 1)\n(3; 2)\n(3; 3)\n(3; 4)\n(3; 5)\n(3; 6)\n(4; 1)\n(4; 2)\n(4; 3)\n(4; 4)\n(4; 5)\n(4; 6)\n(5; 1)\n(5; 2)\n(5; 3)\n(5; 4)\n(5; 5)\n(5; 6)\n(6; 1)\n(6; 2)\n(6; 3)\n(6; 4)\n(6; 5)\n(6; 6)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nStep 2: Write down the events\nFor E the dice have to add to 10.\nE = {(4; 6); (5; 5); (6; 4)}\nFor F at least one die has to be 3.\nF = {(1; 3); (3; 1); (2; 3); (3; 2); (3; 3); (4; 3); (3; 4); (5; 3); (3; 5); (6; 3); (3; 6)}\n404\n10.1.\nRevision\n\nStep 3: Compute the probabilities\nThe probability of an event is defined as the number of elements in the event set\ndivided by the number of elements in the sample space. There are\n• 6 × 6 = 36 outcomes in the sample space, S;\n• 3 outcomes in event E; and\n• 11 outcomes in event F.\nTherefore\nP(E) = 3\n36 = 1\n12\nand\nP(F) = 11\n36\nStep 4: Are they mutually exclusive\nTo test whether two events are mutually exclusive, we have to test whether their in-\ntersection is empty. Since E has no outcomes that contain a 3 on one of the dice,\nthe intersection of E and F is empty: (E and F) = ∅. This means that the events are\nmutually exclusive.\nSee video: 239N at www.everythingmaths.co.za\nExercise 10 – 1: Revision\n1. A bag contains r red balls, b blue balls and y yellow balls. What is the probability\nthat a ball drawn from the bag at random is yellow?\n2. A packet has yellow and pink sweets. The probability of taking out a pink sweet\nis 7\n12. What is the probability of taking out a yellow sweet?\n3. You flip a coin 4 times. What is the probability that you get 2 heads and 2 tails?\nWrite down the sample space and the event set to determine the probability of\nthis event.\n4. In a class of 37 children, 15 children walk to school, 20 children have pets at\nhome and 12 children who have a pet at home also walk to school. How many\nchildren walk to school and do not have a pet at home?\n5. You roll two 6-sided dice and are interested in the following two events:\n• A: the sum of the dice equals 8\n• B: at least one of the dice shows a 1\nShow that these events are mutually exclusive.\n405\nChapter 10.\nProbability\n\n6. You ask a friend to think of a number from 1 to 100. You then ask her the\nfollowing questions:\n• Is the number even?\n• Is the number divisible by 7?\nHow many possible numbers are less than 80 if she answered “yes” to both\nquestions?\n7. In a group of 42 pupils, all but 3 had a packet of chips or a Fanta or both. If 23\nhad a packet of chips and 7 of these also had a Fanta, what is the probability that\none pupil chosen at random has:\na) both chips and Fanta\nb) only Fanta\n8. Tamara has 18 loose socks in a drawer. Eight of these are orange and two are\npink. Calculate the probability that the first sock taken out at random is:\na) orange\nb) not orange\nc) pink\nd) not pink\ne) orange or pink\nf) neither orange nor pink\n9. A box contains coloured blocks. The number of blocks of each colour is given\nin the following table.\nColour\nPurple\nOrange\nWhite\nPink\nNumber of blocks\n24\n32\n41\n19\nA block is selected randomly. What is the probability that the block will be:\na) purple\nb) purple or white\nc) pink and orange\nd) not orange?\n10. The surface of a soccer ball is made up of 32 faces. 12 faces are regular pen-\ntagons, each with a surface area of about 37 cm2. The other 20 faces are regular\nhexagons, each with a surface area of about 56 cm2.\nYou roll the soccer ball. What is the probability that it stops with a pentagon\ntouching the ground?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 239P\n2. 239Q\n3. 239R\n4. 239S\n5. 239T\n6. 239V\n7. 239W\n8. 239X\n9. 239Y\n10. 239Z\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n406\n10.1.\nRevision\n\nVenn diagrams\nEMBJS\nA Venn diagram is used to show how events are related to one another.\nA Venn\ndiagram can be very helpful when doing calculations with probabilities. In a Venn\ndiagram each event is represented by a shape, often a circle or a rectangle. The region\ninside the shape represents the outcomes included in the event and the region outside\nthe shape represents the outcomes that are not in the event.\nS\nA\nB\nA and B\nA Venn diagram representing a sample space, S, as a square; and two events, A and\nB, as circles. The intersection of the two circles contains outcomes that are in both A\nand B.\nVenn diagrams can be used in slightly different ways and it is important to notice the\ndifferences between them. The following 3 examples show how a Venn diagram is\nused to represent\n• the outcomes included in each event;\n• the number of outcomes in each event; and\n• the probability of each event.\nWorked example 3: Venn diagrams with outcomes\nQUESTION\nChoose a number between 1 and 20. Draw a Venn diagram to answer the following\nquestions.\n1. What is the probability that the number is a multiple of 3?\n2. What is the probability that the number is a multiple of 5?\n3. What is the probability that the number is a multiple of 3 or 5?\n4. What is the probability that the number is a multiple of 3 and 5?\nSOLUTION\nStep 1: Draw a Venn diagram\nThe Venn diagram should show the sample space of all numbers from 1 to 20. It should\nalso show an event set that contains all the multiples of 3, let A = {3; 6; 9; 12; 15; 18},\n407\nChapter 10.\nProbability\n\nand another event set that contains all the multiples of 5, let B = {5; 10; 15; 20}. Note\nthat there is one shared outcome between these two events, namely 15.\n3\n18\n15\n12\n9\n6\n10\n20\n5\n1\n2\n4\n7\n8\n11\n13\n14\n16\n17\n19\nStep 2: Compute probabilities\nThe probability of an event is the number of outcomes in the event set divided by the\nnumber of outcomes in the sample space. There are 20 outcomes in the sample space.\n1. Since there are 6 outcomes in the multiples of 3 event set, the probability of a\nmultiple of 3 is P(A) = 6\n20 = 3\n10.\n2. Since there are 4 outcomes in the multiples of 5 event set, the probability of a\nmultiple of 5 is P(B) = 4\n20 = 1\n5.\n3. The event that the number is a multiple of 3 or 5 is the union of the above two\nevent sets. There are 9 elements in the union of the event sets, so the probability\nis 9\n20.\n4. The event that the number is a multiple of 3 and 5 is the intersection of the\ntwo event sets. There is 1 element in the intersection of the event sets, so the\nprobability is 1\n20.\nWorked example 4: Venn diagrams with counts\nQUESTION\nIn a group of 50 learners, 35 take Mathematics and 30 take History, while 12 take\nneither of the two subjects. Draw a Venn diagram representing this information. If a\nlearner is chosen at random from this group, what is the probability that he takes both\nMathematics and History?\nSOLUTION\nStep 1: Draw outline of Venn diagram\nThere are 2 events in this question, namely\n• M: that a learner takes Mathematics; and\n• H: that a learner takes History.\n408\n10.1.\nRevision\n\nWe need to do some calculations before drawing the full Venn diagram, but with the\ninformation above we can already draw the outline.\nS\nM\nH\nStep 2: Write down sizes of the event sets, their union and intersection\nWe are told that 12 learners take neither of the two subjects. Graphically we can\nrepresent this as:\nS\nM\nH\n12\nSince there are 50 elements in the sample space, we can see from this figure that there\nare 50 −12 = 38 elements in (M or H). So far we know\n• n(M) = 35\n• n(H) = 30\n• n(M or H) = 38\nFrom the addition rule,\nn(M or H) = n(M) + n(H) −n(M and H)\n∴n(M and H) = 35 + 30 −38\n= 27\nStep 3: Draw the final Venn diagram\nS\nM\nH\n12\n27\n8\n3\n409\nChapter 10.\nProbability\n\nWorked example 5: Venn diagrams with probabilities\nQUESTION\nDraw a Venn diagram to represent the same information as in the previous example,\nexcept showing the probabilities of the different events, rather than the counts.\nIf a learner is chosen at random from this group, what is the probability that she takes\nboth Mathematics and History?\nSOLUTION\nStep 1: Use counts to compute probabilities\nSince there are 50 elements (learners) in the sample space, we can compute the prob-\nability of any event by dividing the size of the event set by 50. This gives the following\nprobabilities:\n• P(M) = 35\n50 = 7\n10\n• P(H) = 30\n50 = 3\n5\n• P(M or H) = 38\n50 = 19\n25\n• P(M and H) = 27\n50\nStep 2: Draw the Venn diagram\nNext we replace each count from the Venn diagram in the previous example with a\nprobability.\nS\nM\nH\n6\n25\n27\n50\n4\n25\n3\n50\nStep 3: Find the answer\nThe probability that a random learner will take both Mathematics and History is\nP(M and H) = 27\n50.\nSee video: 23B2 at www.everythingmaths.co.za\n410\n10.1.\nRevision\n\nExercise 10 – 2: Venn diagram revision\n1. Given the following information:\n• P(A) = 0,3\n• P(B and A) = 0,2\n• P(B) = 0,7\nFirst draw a Venn diagram to represent this information. Then compute the value\nof P(B and (not A)).\n2. You are given the following information:\n• P(A) = 0,5\n• P(A and B) = 0,2\n• P(not B) = 0,6\nDraw a Venn diagram to represent this information and determine P(A or B).\n3. A study was undertaken to see how many people in Port Elizabeth owned either\na Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60%\nowned a Volkswagen. What percentage of people owned neither car?\n4. Let S denote the set of whole numbers from 1 to 15, X denote the set of even\nnumbers from 1 to 15 and Y denote the set of prime numbers from 1 to 15.\nDraw a Venn diagram depicting S, X and Y .\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B3\n2. 23B4\n3. 23B5\n4. 23B6\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.2\nDependent and independent events\nEMBJT\nSometimes the presence or absence of one event tells us something about other events.\nWe call events dependent if knowing whether one of them happened tells us some-\nthing about whether the others happened. Independent events give us no information\nabout one another; the probability of one event occurring does not affect the probabil-\nity of the other events occurring.\nDEFINITION: Independent events\nTwo events, A and B are independent if and only if\nP(A and B) = P(A) × P(B)\nAt first it might not be clear why we should call events that satisfy the equation above\nindependent. We will explore this further using a number of examples.\n411\nChapter 10.\nProbability\n\nInvestigation: Independence\nRoll a single 6-sided die and consider the following two events:\n• E: you get an even number\n• T: you get a number that is divisible by three\nNow answer the following questions:\n• What is the probability of E?\n• What is the probability of getting an even number if you are told that the number\nwas also divisible by three?\n• Does knowing that the number was divisible by three change the probability that\nthe number was even?\nAre the events E and T dependent or independent according to the definition (hint:\ncompute the probabilities in the definition of independence)?\nSee video: 23B7 at www.everythingmaths.co.za\nSo, why do we call it independence when P(A and B) = P(A) × P(B)? For two\nevents, A and B, independence means that knowing the outcome of B does not affect\nthe probability of A.\nConsider the following Venn diagram.\nS\nA\nB\nA and B\nThe probability of A is the ratio between the number of outcomes in A and the number\nof outcomes in the sample space, S.\nP(A) = n(A)\nn(S)\n412\n10.2.\nDependent and independent events\n\nNow, let’s say that we know that event B happened. How does this affect the proba-\nbility of A? Here is how the Venn diagram changes:\nS\nA\nB\nA and B\nA lot of the possible outcomes (all of the outcomes outside B) are now out of the pic-\nture, because we know that they did not happen. Now the probability of A happening,\ngiven that we know that B happened, is the ratio between the size of the region where\nA is present (A and B) and the size of all possible events (B).\nP(A if we know B) = n(A and B)\nn(B)\nIf P(A) = P(A if we know B) we call them independent, because knowing B does\nnot change the probability of A.\nWith some algebra, we can prove that this statement of independence is the same\nas the definition of independence that we saw at the beginning of this section. For\nindependent events\nP(A and B) = P(A) × P(B)\nThis is equivalent to\nP(A) = P(A and B) ÷ P(B)\n= n(A and B)\nn(S)\n÷ n(B)\nn(S)\n= n(A and B)\nn(B)\n= P(A if we know B)\nThat is why we call events independent!\n(For enrichment only):\nThe ratio\nP(A and B)\nP(B)\nis called a conditional probability and written using the notation P(A | B). This\nnotation is read as “the probability of A given B.”\nIf (and only if) A and B are independent: P(A | B) = P(A) and P(B | A) = P(B).\nTry to prove this using the definition of independence.\n413\nChapter 10.\nProbability\n\nWorked example 6: Independent and dependent events\nQUESTION\nA bag contains 5 red and 5 blue balls. We remove a random ball from the bag, record\nits colour and put it back into the bag. We then remove another random ball from the\nbag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Probability of a red ball first\nSince there are a total of 10 balls, of which 5 are red, the probability of getting a red\nball is\nP(first ball red) = 5\n10 = 1\n2\nStep 2: Probability of a blue ball second\nThe problem states that the first ball is placed back into the bag before we take the\nsecond ball. This means that when we draw the second ball, there are again a total of\n10 balls in the bag, of which 5 are blue. Therefore the probability of drawing a blue\nball is\nP(second ball blue) = 5\n10 = 1\n2\nStep 3: Probability of red first and blue second\nWhen drawing two balls from the bag, there are 4 possibilities. We can get\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nWe want to know the probability of the second outcome, where we have to get a red\nball first. Since there are 5 red balls and 10 balls in total, there are\n5\n10 ways to get a\nred ball first. Now we put the first ball back, so there are again 5 red balls and 5 blue\nballs in the bag. Therefore there are\n5\n10 ways to get a blue ball second if the first ball\nwas red. This means that there are\n5\n10 × 5\n10 = 25\n100\n414\n10.2.\nDependent and independent events\n\nways to get a red ball first and a blue ball second. So, the probability of getting a red\nball first and a blue ball second is 1\n4.\nStep 4: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 1\n4\nSince 1\n4 = 1\n2 × 1\n2, the events are independent.\nSee video: 23B8 at www.everythingmaths.co.za\nWorked example 7: Independent and dependent events\nQUESTION\nIn the previous example, we picked a random ball and put it back into the bag before\ncontinuing. This is called sampling with replacement. In this example, we will follow\nthe same process, except that we will not put the first ball back into the bag. This is\ncalled sampling without replacement.\nSo, from a bag with 5 red and 5 blue balls, we remove a random ball and record its\ncolour. Then, without putting back the first ball, we remove another random ball from\nthe bag and record its colour.\n1. What is the probability that the first ball is red?\n2. What is the probability that the second ball is blue?\n3. What is the probability that the first ball is red and the second ball is blue?\n4. Are the first ball being red and the second ball being blue independent events?\nSOLUTION\nStep 1: Count the number of outcomes\nWe will look directly at the number of possible ways in which we can get the 4 possible\noutcomes when removing 2 balls. In the previous example, we saw that the 4 possible\noutcomes are\n415\nChapter 10.\nProbability\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball;\n• a blue ball and then a red ball;\n• a blue ball and then another blue ball.\nFor the first outcome, we have to get a red ball first. Since there are 5 red balls and\n10 balls in total, there are\n5\n10 ways to get a red ball first. After we have taken out a red\nball, there are now 4 red balls and 5 blue balls left. Therefore there are 4\n9 ways to get\na red ball second if the first ball was also red. This means that there are\n5\n10 × 4\n9 = 20\n90\nways to get a red ball first and a red ball second. The probability of the first outcome\nis 2\n9.\nFor the second outcome, we have to get a red ball first. As in the first outcome, there\nare\n5\n10 ways to get a red ball first; and there are now 4 red balls and 5 blue balls left.\nTherefore there are 5\n9 ways to get a blue ball second if the first ball was red. This means\nthat there are\n5\n10 × 5\n9 = 25\n90\nways to get a red ball first and a blue ball second. The probability of the second\noutcome is 5\n18.\nWe can compute the probabilities of the third and fourth outcomes in the same way as\nthe first two, but there is an easier way. Notice that there are only 2 types of ball and\nthat there are exactly equal numbers of them at the start. This means that the problem\nis completely symmetric in red and blue. We can use this symmetry to compute the\nprobabilities of the other two outcomes.\nIn the third outcome, the first ball is blue and the second ball is red. Because of\nsymmetry this outcome must have the same probability as the second outcome (when\nthe first ball is red and the second ball is blue). Therefore the probability of the third\noutcome is 5\n18.\nIn the fourth outcome, the first and second balls are both blue. From symmetry, this\noutcome must have the same probability as the first outcome (when both balls are red).\nTherefore the probability of the fourth outcome is 2\n9.\nTo summarise, these are the possible outcomes and their probabilities:\n• first ball red and second ball red: 2\n9;\n• first ball red and second ball blue:\n5\n18;\n• first ball blue and second ball red:\n5\n18;\n• first ball blue and second ball blue: 2\n9.\nStep 2: Probability of a red ball first\nTo determine the probability of getting a red ball on the first draw, we look at all of the\noutcomes that contain a red ball first. These are\n416\n10.2.\nDependent and independent events\n\n• a red ball and then another red ball;\n• a red ball and then a blue ball.\nThe probability of the first outcome is 2\n9 and the probability of the second outcome is\n5\n18. By adding these two probabilities, we see that the probability of getting a red ball\nfirst is\nP(first ball red) = 2\n9 + 5\n18 = 1\n2\nThis is the same as in the previous exercise, which should not be too surprising since\nthe probability of the first ball being red is not affected by whether or not we put it\nback into the bag before drawing the second ball.\nStep 3: Probability of a blue ball second\nTo determine the probability of getting a blue ball on the second draw, we look at all\nof the outcomes that contain a blue ball second. These are\n• a red ball and then a blue ball;\n• a blue ball and then another blue ball.\nThe probability of the first outcome is 5\n18 and the probability of the second outcome is\n2\n9. By adding these two probabilities, we see that the probability of getting a blue ball\nsecond is\nP(second ball blue) = 5\n18 + 2\n9 = 1\n2\nThis is also the same as in the previous exercise! You might find it surprising that the\nprobability of the second ball is not affected by whether or not we replace the first ball.\nThe reason why this probability is still 1\n2 is that we are computing the probability that\nthe second ball is blue without knowing the colour of the first ball. Because there are\nonly two equal possibilities for the second ball (red and blue) and because we don’t\nknow whether the first ball is red or blue, there is an equal chance that the second ball\nwill be one colour or the other.\nStep 4: Probability of red first and blue second\nWe have already calculated the probability that the first ball is red and the second ball\nis blue. It is 5\n18.\nStep 5: Dependent or independent?\nAccording to the definition, events are independent if and only if\nP(A and B) = P(A) × P(B)\nIn this problem:\n• P(first ball red) = 1\n2\n• P(second ball blue) = 1\n2\n• P(first ball red and second ball blue) = 5\n18\nSince 5\n18 ̸= 1\n2 × 1\n2, the events are dependent.\n417\nChapter 10.\nProbability\n\nWARNING!\nJust because two events are mutually exclusive does not necessarily mean that they\nare independent. To test whether events are mutually exclusive, always check that\nP(A and B) = 0. To test whether events are independent, always check that P(A and B) =\nP(A) × P(B). See the exercises below for examples of events that are mutually ex-\nclusive and independent in different combinations.\nExercise 10 – 3: Dependent and independent events\n1. Use the following Venn diagram to determine whether events X and Y are\na) mutually exclusive or not mutually exclusive;\nb) dependent or independent.\nS\nX\nY\n11\n7\n3\n14\n2. Of the 30 learners in a class 17 have black hair, 11 have brown hair and 2 have\nred hair. A learner is selected from the class at random.\na) What is the probability that the learner has black hair?\nb) What is the probability that the learner has brown hair?\nc) Are these two events mutually exclusive?\nd) Are these two events independent?\n3. P(M) = 0,45; P(N) = 0,3 and P(M or N) = 0,615. Are the events M and N\nmutually exclusive, independent or neither mutually exclusive nor independent?\n4. (For enrichment)\nProve that if event A and event B are mutually exclusive with P(A) ̸= 0 and\nP(B) ̸= 0, then A and B are always dependent.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23B9\n2. 23BB\n3. 23BC\n4. 23BD\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n418\n10.2.\nDependent and independent events\n\n10.3\nMore Venn diagrams\nEMBJV\nIn the rest of this chapter we will look at tools and techniques for working with proba-\nbility problems.\nWhen working with more complex problems, we can have three or more events that\nintersect in various ways.\nTo solve these problems, we usually want to count the\nnumber (or percentage) of outcomes in an event, or a combination of events. Venn\ndiagrams are a useful tool for recording and visualising the counts.\nInvestigation: Venn diagram for 3 events\nThe diagram below shows a general Venn diagram for 3 events.\nS\nA\nB\nC\nWrite down the sets corresponding to each of the three coloured regions and also\nto the shaded region. Remember that the intersections between circles represent the\nintersections between the different events.\nWhat is the event for\n• the red region;\n• the green region;\n• the blue region; and\n• the shaded region?\n419\nChapter 10.\nProbability\n\nWorked example 8: Venn diagram for 3 events\nQUESTION\nDraw a Venn diagram that shows the following sample space and events:\n• S: all the integers from 1 to 30\n• P: prime numbers\n• M: multiples of 3\n• F: factors of 30\nSOLUTION\nStep 1: Write down the sample space and event sets\nThe sample space contains all the positive integers up to 30.\nS = {1; 2; 3; . . . ; 30}\nThe prime numbers between 1 and 30 are\nP = {2; 3; 5; 7; 11; 13; 17; 19; 23; 29}\nThe multiples of 3 between 1 and 30 are\nM = {3; 6; 9; 12; 15; 18; 21; 24; 27; 30}\nThe factors of 30 are\nF = {1; 2; 3; 5; 6; 10; 15; 30}\nStep 2: Draw the outline of the Venn diagram\nThere are 3 events, namely P, M and F, and the sample space, S. Put this information\non a Venn diagram:\nS\nP\nM\nF\n420\n10.3.\nMore Venn diagrams\n\nStep 3: Place the outcomes in the appropriate event sets\nS\nP\nM\nF\n3\n2\n5\n6\n15\n30\n1\n10\n7\n11\n13\n17\n19\n23\n29\n9\n12\n21\n24\n18\n27\n4\n8\n14\n16\n20\n22\n25\n26\n28\nWorked example 9: Venn diagram for 3 events\nQUESTION\nAt Dawnview High there are 400 Grade 11 learners. 270 do Computer Science, 300\ndo English and 50 do Business studies. All those doing Computer Science do English,\n20 take Computer Science and Business studies and 35 take English and Business\nstudies. Using a Venn diagram, calculate the probability that a pupil drawn at random\nwill take:\n1. English, but not Business studies or Computer Science\n2. English but not Business studies\n3. English or Business studies but not Computer Science\n4. English or Business studies\nSOLUTION\nStep 1: Draw the outline of the Venn diagram\nWe need to be careful with this problem. In the question statement we are told that all\nthe learners who do Computer Science also do English. This means that the circle for\nComputer Science on the Venn diagram needs to be inside the circle for English.\n421\nChapter 10.\nProbability\n\nS\nE\nC\nB\nStep 2: Fill in the counts on the Venn diagram\nS\nE\nC\nB\n20\n250\n15\n15\n15\n85\nStep 3: Compute probabilities\nTo find the number of learners taking English, but not Business studies or Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 15 and there are a total of 400 learners in the grade. There-\nfore the probability that a learner will take English but not Business studies or Computer\nScience is\n15\n400 = 3\n80.\nTo find the number of learners taking English but not Business studies, we need to look\nat this region of the Venn diagram:\n422\n10.3.\nMore Venn diagrams\n\nThe count in this region is 265. Therefore the probability that a learner will take\nEnglish but not Business studies is 265\n400 = 53\n80.\nTo find the number of learners taking English or Business studies but not Computer\nScience, we need to look at this region of the Venn diagram:\nThe count in this region is 45. Therefore the probability that a learner will take English\nor Business studies but not Computer Science is\n45\n400 = 9\n80.\nTo find the number of learners taking English or Business studies, we need to look at\nthis region of the Venn diagram:\nThe count in this region is 315. Therefore the probability that a learner will take\nEnglish or Business studies is 315\n400 = 63\n80.\n423\nChapter 10.\nProbability\n\nThere are some words that tell you which part of the Venn diagram should be filled in.\nThe following table summarises the most important ones:\nWords\nSymbols\nVenn diagram\n“all”\nA and B and C / A ∩B ∩C\n“none”\n“at least one”\nA or B or C / A ∪B ∪C\n“both A and B”\nA and B / A ∩B\n“A or B”\nA or B / A ∪B\nExercise 10 – 4: Venn diagrams\n1. Use the Venn diagram below to answer the following questions. Also given:\nn(S) = 120.\nS\nF\n8\n10\nG\n24\n15\nH\n14\n7\n2\na) Compute P(F).\nb) Compute P(G or H).\nc) Compute P(F and G).\nd) Are F and G dependent or independent?\n424\n10.3.\nMore Venn diagrams\n\n2. The Venn diagram below shows the probabilities of 3 events. Complete the Venn\ndiagram using the additional information provided.\nS\nZ\n1\n25\nY\n17\n100\nX\n17\n100\n3\n20\n• P(Z and (not Y )) =\n31\n100\n• P(Y and X) =\n23\n100\n• P(Y ) =\n39\n100\nAfter completing the Venn diagram, compute the following:\nP (Z and not (X or Y ))\n3. There are 79 Grade 10 learners at school. All of these take some combination of\nMaths, Geography and History. The number who take Geography is 41; those\nwho take History is 36; and 30 take Maths. The number who take Maths and\nHistory is 16; the number who take Geography and History is 6, and there are 8\nwho take Maths only and 16 who take History only.\na) Draw a Venn diagram to illustrate all this information.\nb) How many learners take Maths and Geography but not History?\nc) How many learners take Geography only?\nd) How many learners take all three subjects?\n4. Draw a Venn diagram with 3 mutually exclusive events. Use the diagram to\nshow that for 3 mutually exclusive events, A, B and C, the following is true:\nP(A or B or C) = P(A) + P(B) + P(C)\nThis is the addition rule for 3 mutually exclusive events.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BF\n2. 23BG\n3. 23BH\n4. 23BJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n425\nChapter 10.\nProbability\n\n10.4\nTree diagrams\nEMBJW\nTree diagrams are useful for organising and visualising the different possible outcomes\nof a sequence of events. For each possible outcome of the first event, we draw a line\nwhere we write down the probability of that outcome and the state of the world if that\noutcome happened. Then, for each possible outcome of the second event we do the\nsame thing.\nBelow is an example of a simple tree diagram, showing the possible outcomes of\nrolling a 6-sided die.\n1\n1\n6\n2\n1\n6\n3\n1\n6\n4\n1\n6\n5\n1\n6\n6\n1\n6\noutcomes\nprobabilities\nNote that each outcome (the numbers 1 to 6) is shown at the end of a line; and that\nthe probability of each outcome (all 1\n6 in this case) is shown shown on a line. The\nprobabilities have to add up to 1 in order to cover all of the possible outcomes. In the\nexamples below, we will see how to draw tree diagrams with multiple events and how\nto compute probabilities using the diagrams.\nEarlier in this chapter you learned about dependent and independent events. Tree\ndiagrams are very helpful for analysing dependent events. A tree diagram allows you\nto show how each possible outcome of one event affects the probabilities of the other\nevents.\nTree diagrams are not so useful for independent events since we can just multiply the\nprobabilities of separate events to get the probability of the combined event. Remem-\nber that for independent events:\nP(A and B) = P(A) × P(B)\nSo if you already know that events are independent, it is usually easier to solve a\nproblem without using tree diagrams. But if you are uncertain about whether events\nare independent or if you know that they are not, you should use a tree diagram.\nWorked example 10: Drawing a tree diagram\nQUESTION\nIf it rains on a given day, the probability that it rains the next day is 1\n3. If it does not rain\non a given day, the probability that it rains the next day is 1\n6. The probability that it will\nrain tomorrow is 1\n5. What is the probability that it will rain the day after tomorrow?\nDraw a tree diagram of all the possibilities to determine the answer.\nSOLUTION\nStep 1: Draw the first level of the tree diagram\nBefore we can determine what happens on the day after tomorrow, we first have to\ndetermine what might happen tomorrow. We are told that there is a 1\n5 probability that\n426\n10.4.\nTree diagrams\n\nit will rain tomorrow. Here is how to represent this information using a tree diagram:\n1\n5\nrain\n4\n5\nno rain\ntoday:\ntomorrow:\nStep 2: Draw the second level of the tree diagram\nWe are also told that if it does rain on one day, there is a 1\n3 probability that it will also\nrain on the following day. On the other hand, if it does not rain on one day, there is\nonly a 1\n6 probability that it will also rain on the following day. Using this information\nwe complete the tree diagram:\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nStep 3: Compute the probability\nWe are asked what the probability is that it will rain the day after tomorrow. On the\ntree diagram above we can see that there are 2 situations where it rains on the day\nafter tomorrow. They are marked in red below.\n1\n5\nrain\n4\n5\nno rain\n1\n3\nrain\n2\n3\nno rain\n1\n6\nrain\n5\n6\nno rain\ntoday:\ntomorrow:\nday after tomorrow:\nTo get the probability for the first situation (that it rains tomorrow and the day after\ntomorrow) we have to multiply the probabilies along the first red line.\nP(rain tomorrow and rain day after tomorrow)\n=1\n5 × 1\n3\n= 1\n15\n427\nChapter 10.\nProbability\n\nTo get the probability for the second situation (that it does not rain tomorrow, but it\ndoes rain the day after tomorrow) we have to multiply the probabilies along the second\nred line.\nP(not rain tomorrow and rain day after tomorrow)\n=4\n5 × 1\n6\n= 2\n15\nTherefore the total probability that it will rain the day after tomorrow is the sum of the\nprobabilities along the two red paths, namely\n1\n15 + 2\n15 = 1\n5\nWorked example 11: Drawing a tree diagram\nQUESTION\nYou play the following game. You flip a coin. If it comes up tails, you get 2 points\nand your turn ends. If it comes up heads, you get only 1 point, but you can flip the\ncoin again. If you flip the coin multiple times in one turn, you add up the points. You\ncan flip the coin at most 3 times in one turn. What is the probability that you will get\nexactly 3 points in one turn? Draw a tree diagram to visualise the different possibilities.\nSOLUTION\nStep 1: Write down the events and their symbols\nEach coin toss has on of two possible outcomes, namely heads (H) and tails (T). Each\noutcome has a probability of 1\n2. We are asked to count the number of points, so we\nwill also indicate how many points we have for each outcome.\nStep 2: Draw the first level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\nThis tree diagram shows the possible outcomes after 1 flip of the coin. Remember that\nwe can have up to 3 flips, so the diagram is not complete yet. If the coin comes up\nheads, we flip the coin again. If the coin comes up tails, we stop.\n428\n10.4.\nTree diagrams\n\nStep 3: Draw the second and third level of the tree diagram\n1\n2\nH\n1 pt\n1\n2\nT\n2 pts\n1\n2\nH\n2 pts\n1\n2\nT\n3 pts\n1\n2\nH\n3 pts\n1\n2\nT\n4 pts\nIn this tree diagram you can see that we add up the points we get with each coin flip.\nAfter three coin flips, the game is over.\nStep 4: Find the relevant outcomes and compute the probability\nWe are interested in getting exactly 3 points during the game. To find these outcomes\nwe look only at the tips of the tree. We end with exactly 3 points when the coin flips\nare\n• (H; T) with probability 1\n2 × 1\n2 = 1\n4;\n• (H; H; H) with probability 1\n2 × 1\n2 × 1\n2 = 1\n8.\nNotice that we compute the probability of an outcome by multiplying all the probabil-\nities along the path from the start of the tree to the tip where the outcome is. We add\nthe above two probabilites to obtain the final probability of getting exactly 3 points as\n1\n4 + 1\n8 = 3\n8.\nWorked example 12: Drawing a tree diagram\nQUESTION\nA person takes part in a medical trial that tests the effect of a medicine on a disease.\nHalf the people are given medicine and the other half are given a sugar pill, which has\nno effect on the disease. The medicine has a 60% chance of curing someone. But,\npeople who do not get the medicine still have a 10% chance of getting well. There are\n50 people in the trial and they all have the disease. Talwar takes part in the trial, but\nwe do not know whether he got the medicine or the sugar pill. Draw a tree diagram\nof all the possible cases. What is the probability that Talwar gets cured?\nSOLUTION\nStep 1: Summarise the information in the problem\nThere are two uncertain events in this problem. Each person either receives medicine\n(probability 1\n2) or a sugar pill (probability 1\n2). Each person also gets cured (probability\n429\nChapter 10.\nProbability\n\n3\n5 with medicine and\n1\n10 without) or stays ill (probability 2\n5 with medicine and\n9\n10\nwithout).\nStep 2: Draw the tree diagram\n1\n2\nmedicine\n1\n2\nsugar pill\n3\n5\ncured\n2\n5\nnot cured\n1\n10\ncured\n9\n10\nnot cured\nIn the first level of the tree diagram we show that Talwar either gets the medicine or\nthe sugar pill. The second level of the tree diagram shows whether Talwar is cured or\nnot, depending on which one of the pills he got.\nStep 3: Compute the required probability\nWe multiply the probabilites along each path in the tree diagram that leads to Talwer\nbeing cured:\n1\n2 × 3\n5 = 3\n10\n1\n2 × 1\n10 = 1\n20\nWe then add these probabilites to get the final answer. The probability that Talwar is\ncured is 7\n20.\nExercise 10 – 5: Tree diagrams\n1. You roll a die twice and add up the dots to get a score. Draw a tree diagram to\nrepresent this experiment. What is the probability that your score is a multiple\nof 5?\n2. What is the probability of throwing at least one five in four rolls of a regular\n6-sided die? Hint: do not show all possible outcomes of each roll of the die. We\nare interested in whether the outcome is 5 or not 5 only.\n3. You flip one coin 4 times.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\n430\n10.4.\nTree diagrams\n\n4. You flip 4 different coins at the same time.\na) What is the probability of getting exactly 3 heads?\nb) What is the probability of getting at least 3 heads?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BK\n2. 23BM\n3. 23BN\n4. 23BP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n10.5\nContingency tables\nEMBJX\nA contingency table is another tool for keeping a record of the counts or percentages\nin a probability problem. Contingency tables are especially helpful for figuring out\nwhether events are dependent or independent.\nWe will be studying two-way contingency tables, where we count the number of out-\ncomes for 2 events and their complements, making 4 events in total. A two-way contin-\ngency table always shows the counts for the 4 possible combinations of events, as well\nas the totals for each event and its complement. We can use a contingency table to\ncompute the probabilities of various events by computing the ratios between counts,\nand to determine whether the events are dependent or independent. The example\nbelow shows a two-way contingency table, representing the outcome of a medical\nstudy.\nWorked example 13: Contingency tables\nQUESTION\nA medical trial into the effectiveness of a new medication was carried out. 120 females\nand 90 males took part in the trial. Out of those people, 50 females and 30 males\nresponded positively to the medication. Given below is a contingency table with the\ngiven information filled in.\nFemale\nMale\nTotals\nPositive\n50\n30\nNegative\nTotals\n120\n90\n1. What is the probability that the medicine gives a positive result for females?\n2. What is the probability that the medicine gives a negative result for males?\n3. Was the medication’s success independent of gender? Explain.\n431\nChapter 10.\nProbability\n\nSOLUTION\nStep 1: Complete the contingency table\nThe best place to start is always to complete the contingency table. Because the each\ncolumn has to sum up to its total, we can work out the number of females and males\nwho responded negatively to the medication. Then we can add each row to get the\ntotals on the right hand side of the table.\nFemale\nMale\nTotals\nPositive\n50\n30\n80\nNegative\n70\n60\n130\nTotals\n120\n90\n210\nStep 2: Compute the required probabilities\nThe way the first question is phrased, we need to determine the probability that a\nperson responds positively if she is female. This means that we do not include males\nin this calculation. So, the probability that the medicine gives a positive result for\nfemales is the ratio between the number of females who got a positive response and\nthe total number of females.\nP(positive if female) = n(positive and female)\nn(female)\n= 50\n120\n= 5\n12\nSimilarly, the probability that the medicine gives a negative result for males is:\nP(negative if male) = n(negative and male)\nn(male)\n= 60\n90\n= 2\n3\nStep 3: Independence\nWe need to determine whether the effect of the medicine and the gender of a par-\nticipant are dependent or independent. According to the definition, two events are\nindependent if and only if\nP(A and B) = P(A) × P(B)\nWe will look at the events that a participant is female and that the participant re-\nsponded positively to the trial.\nP(female) =\nn(female)\nn(total trials)\n= 120\n210\n= 4\n7\n432\n10.5.\nContingency tables\n\nP(positive) =\nn(positive)\nn(total trials)\n= 80\n210\n= 8\n21\nP(female and positive) = n(female and positive)\nn(total trials)\n= 50\n210\n= 5\n21\nFrom these probabilities we can see that\nP(female and positive) ̸= P(female) × P(positive)\nand therefore the gender of a participant and the outcome of a trial are dependent\nevents.\nWorked example 14: Contingency tables\nQUESTION\nUse the contingency table below to answer the following questions.\nGrade 11\nGrade 12\nTotals\nHas cellphone\n59\n50\n109\nNo cellphone\n6\n3\n9\nTotals\n65\n53\n118\n1. What is the probability that a learner from Grade 11 has a cellphone?\n2. What is the probability that a learner who does not have a cellphone is from\nGrade 11.\n3. Are the grade of a learner and whether he has a cellphone or not independent\nevents? Explain your answer.\nSOLUTION\n1. There are 65 learners in Grade 11 and 59 of them have a cellphone. Therefore\nthe probability that a learner from Grade 11 has a cellphone is 59\n65.\n2. There are 9 learners who do not have a cellphone and 6 of them are in Grade\n11. Therefore the probability that a learner who does not have a cellphone is\nfrom from Grade 11 is 6\n9 = 2\n3.\n433\nChapter 10.\nProbability\n\n3. To test for independence, we will consider whether a learner is in Grade 11 and\nwhether a learner has a cellphone. The probability that a learner is in Grade 11\nis\n65\n118. The probability that a learner has a cellphone is 109\n118. The probability that\na learner is in Grade 11 and has a cellphone is\n59\n118 = 1\n2. Since 1\n2 ̸=\n65\n118 × 109\n118\nthe grade of a learner and whether he has a cellphone are dependent.\nExercise 10 – 6: Contingency tables\n1. Use the contingency table below to answer the following questions.\nBrown eyes\nNot brown eyes\nTotals\nBlack hair\n50\n30\n80\nRed hair\n70\n80\n150\nTotals\n120\n110\n230\na) What is the probability that someone with black hair has brown eyes?\nb) What is the probability that someone has black hair?\nc) What is the probability that someone has brown eyes?\nd) Are having black hair and having brown eyes dependent or independent\nevents?\n2. Given the following contingency table, identify the events and determine\nwhether they are dependent or independent.\nLocation A\nLocation B\nTotals\nBuses left late\n15\n40\n55\nBuses left on time\n25\n20\n45\nTotals\n40\n60\n100\n3. You are given the following information.\n• Events A and B are independent.\n• P(not A) = 0,3.\n• P(B) = 0,4.\nComplete the contingency table below.\nA\nnot A\nTotals\nB\nnot B\nTotals\n50\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BQ\n2. 23BR\n3. 23BS\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n434\n10.5.\nContingency tables\n\n10.6\nSummary\nEMBJY\nSee presentation: 23BT at www.everythingmaths.co.za\n• Terminology:\n– Outcome: a single observation of an experiment.\n– Sample space of an experiment: the set of all possible outcomes of the\nexperiment.\n– Event: a set of outcomes of an experiment.\n– Probability of an event: a real number between 0 and 1 that describes how\nlikely it is that the event will occur.\n– Relative frequency of an event: the number of times that the event occurs\nduring experimental trials, divided by the total number of trials conducted.\n– Union of events: the set of all outcomes that occur in at least one of the\nevents, written as “A or B”.\n– Intersection of events: the set of all outcomes that occur in all of the events,\nwritten as “A and B”.\n– Mutually exclusive events: events with no outcomes in common, that is\n(A and B) = ∅.\n– Complementary events: two mutually exclusive events that together con-\ntain all the outcomes in the sample space. We write the complement as\n“not A”.\n– Independent events: two events where knowing the outcome of one event\ndoes not affect the probability of the other event. Events are independent if\nand only if P(A and B) = P(A) × P(B).\n• Identities:\n– The addition rule: P(A or B) = P(A) + P(B) −P(A and B)\n– The addition rule for 2 mutually exclusive events: P(A or B) = P(A) +\nP(B)\n– The complementary rule: P(not A) = 1 −P(A)\n• A Venn diagram is a visual tool used to show how events overlap. Each region\nin a Venn diagram represents an event and could contain either the outcomes in\nthe event, the number of outcomes in the event or the probability of the event.\n• A tree diagram is a visual tool that helps with computing probabilities for depen-\ndent events. The outcomes of each event are shown along with the probability\nof each outcome. For each event that depends on a previous event, we go one\nlevel deeper into the tree. To compute the probability of some combination of\noutcomes, we\n– find all the paths that contain the outcome of interest;\n– multiply the probabilities along each path;\n– add the probabilities between different paths.\n• A 2-way contingency table is a tool for organising data, especially when we want\nto determine whether two events, each with only two outcomes, are dependent\nor independent. The counts for each possible combination of outcomes are\nentered into the table, along with the totals of each row and column.\n435\nChapter 10.\nProbability\n\nExercise 10 – 7: End of chapter exercises\n1. Jane invested in the stock market. The probability that she will not lose all her\nmoney is 0,32. What is the probability that she will lose all her money? Explain.\n2. If D and F are mutually exclusive events, with P(not D)\n=\n0,3 and\nP(D or F) = 0,94, find P(F).\n3. A car sales person has pink, lime-green and purple models of car A and purple,\norange and multicolour models of car B. One dark night a thief steals a car.\na) What is the experiment and sample space?\nb) What is the probability of stealing either a model of A or a model of B?\nc) What is the probability of stealing both a model of A and a model of B?\n4. The probability of event X is 0,43 and the probability of event Y is 0,24. The\nprobability of both occurring together is 0,10. What is the probability that X or\nY will occur?\n5. P(H) = 0,62; P(J) = 0,39 and P(H and J) = 0,31. Calculate:\na) P(H′)\nb) P(H or J)\nc) P(H′ or J′)\nd) P(H′ or J)\ne) P(H′ and J′)\n6. The last ten letters of the alphabet are placed in a hat and people are asked to\npick one of them. Event D is picking a vowel, event E is picking a consonant\nand event F is picking one of the last four letters. Draw a Venn diagram showing\nthe outcomes in the sample space and the different events. Then calculate the\nfollowing probabilities:\na) P(not F)\nb) P(F or D)\nc) P(neither E nor F)\nd) P(D and E)\ne) P(E and F)\nf) P(E and D′)\n7. Thobeka compares three neighbourhoods (we’ll call them A, B and C) to see\nwhere the best place is to live. She interviews 80 people and asks them whether\nthey like each of the neighbourhoods, or not.\n• 40 people like neighbourhood A.\n• 35 people like neighbourhood B.\n• 40 people like neighbourhood C.\n• 21 people like both neighbourhoods A and C.\n• 18 people like both neighbourhoods B and C.\n• 68 people like at least one neighbourhood.\n• 7 people like all three neighbourhoods.\n436\n10.6.\nSummary\n\na) Use this information to draw a Venn diagram.\nb) How many people like none of the neighbourhoods?\nc) How many people like neighbourhoods A and B, but not C?\nd) What is the probability that a randomly chosen person from the survey likes\nat least one of the neighbourhoods?\n8. Let G and H be two events in a sample space.\nSuppose that P(G) = 0,4;\nP(H) = h; and P(G or H) = 0,7.\na) For what value of h are G and H mutually exclusive?\nb) For what value of h are G and H independent?\n9. The following tree diagram represents points scored by two teams in a soccer\ngame. At each level in the tree, the points are shown as (points for Team 1;\npoints for Team 2).\n0,75\n(3; 0)\n0,25\n(2; 1)\n0,5\n(2; 1)\n0,5\n(1; 2)\n0,65\n(2; 0)\n0,35\n(1; 1)\n0,4\n(1; 1)\n0,6\n(0; 2)\n0,52\n(1; 0)\n0,48\n(0; 1)\n(0; 0)\nUse this diagram to determine the probability that:\na) Team 1 will win\nb) The game will be a draw\nc) The game will end with an even number of total points\n10. A bag contains 10 orange balls and 7 black balls. You draw 3 balls from the bag\nwithout replacement. What is the probability that you will end up with exactly\n2 orange balls? Represent this experiment using a tree diagram.\n11. Complete the following contingency table and determine whether the events are\ndependent or independent.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\nDid not like living there\n140\n340\nTotals\n230\n500\n12. Summarise the following information about a medical trial with 2 types of multi-\nvitamin in a contingency table and determine whether the events are dependent\nor independent.\n• 960 people took part in the medical trial.\n• 540 people used multivitamin A for a month and 400 of those people\nshowed an improvement in their health.\n437\nChapter 10.\nProbability\n\n• 300 people showed an improvement in health when using multivitamin B\nfor a month.\nIf the events are independent, it means that the two multivitamins have the same\neffect on people. If the events are dependent, it means that one multivitamin is\nbetter than the other. Which multivitamin is better than the other, or are the both\nequally effective?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23BV\n2. 23BW\n3. 23BX\n4. 23BY\n5. 23BZ\n6. 23C2\n7. 23C3\n8. 23C4\n9. 23C5\n10. 23C6\n11. 23C7\n12. 23C8\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n438\n10.6.\nSummary\n\nCHAPTER\n11\nStatistics\n11.1\nRevision\n440\n11.2\nHistograms\n444\n11.3\nOgives\n451\n11.4\nVariance and standard deviation\n455\n11.5\nSymmetric and skewed data\n461\n11.6\nIdentification of outliers\n464\n11.7\nSummary\n467\n\n11\nStatistics\n11.1\nRevision\nEMBJZ\nMeasures of central tendency\nEMBK2\nThe mean and median of a data set both give an indication where the centre of the\ndata distribution is located. The mean, or average, is calculated as\nx =\nPn\ni=1 xi\nn\nwhere the xi are the data and n is the number of data. We read x as “x bar”.\nThe median is the middle value of an ordered data set. To find the median, we first\nsort the data and then pick out the value in the middle of the sorted list. If the middle\nis in between two values, the median is the average of those two values.\nSee video: 23C9 at www.everythingmaths.co.za\nWorked example 1: Computing measures of central tendency\nQUESTION\nCompute the mean and median of the following data set:\n72,5 ; 92,6 ; 15,6 ; 53,0 ; 86,4 ; 89,9 ; 90,9 ; 21,7 ; 46,0 ; 4,1 ; 51,7 ; 2,2\nSOLUTION\nStep 1: Compute the mean\nUsing the formula for the mean, we first compute the sum of the values and then divide\nby the number of values.\nx = 626,6\n12\n≈52,22\nStep 2: Compute the median\nTo find the median, we first have to sort the data:\n2,2 ; 4,1 ; 15,6 ; 21,7 ; 46,0 ; 51,7 ; 53,0 ; 72,5 ; 86,4 ; 89,9 ; 90,9 ; 92,6\nSince there are an even number of values, the median will lie between two values.\nIn this case, the two values in the middle are 51,7 and 53,0. Therefore the median is\n52,35.\n440\n11.1.\nRevision\n\nMeasures of dispersion\nEMBK3\nMeasures of dispersion tell us how spread out a data set is. If a measure of dispersion\nis small, the data are clustered in a small region. If a measure of dispersion is large,\nthe data are spread out over a large region.\nThe range is the difference between the maximum and minimum values in the data\nset.\nThe inter-quartile range is the difference between the first and third quartiles of the\ndata set. The quartiles are computed in a similar way to the median. The median is\nhalfway into the ordered data set and is sometimes also called the second quartile.\nThe first quartile is one quarter of the way into the ordered data set; whereas the third\nquartile is three quarters of the way into the ordered data set.\nSee video: 23CB at www.everythingmaths.co.za\nWorked example 2: Range and inter-quartile range\nQUESTION\nDetermine the range and the inter-quartile range of the following data set.\n14 ; 17 ; 45 ; 20 ; 19 ; 36 ; 7 ; 30 ; 8\nSOLUTION\nStep 1: Sort the values in the data set\nTo determine the range we need to find the minimum and maximum values in the\ndata set. To determine the inter-quartile range we need to compute the first and third\nquartiles of the data set. For both of these requirements, it is easier to order the data\nset first.\nThe sorted data set is\n7 ; 8 ; 14 ; 17 ; 19 ; 20 ; 30 ; 36 ; 45\nStep 2: Find the minimum, maximum and range\nThe minimum value is the first value in the ordered data set, namely 7. The maximum\nis the last value in the ordered data set, namely 45. The range is the difference between\nthe minimum and maximum: 45 −7 = 38.\nStep 3: Find the quartiles and inter-quartile range\nThe diagram below shows how we find the quartiles one quarter, one half and three\nquarters of the way into the ordered list of values.\n441\nChapter 11.\nStatistics\n\n7\n8\n14\n17\n19\n20\n30\n36\n45\n0\n1\n4\n1\n2\n3\n4\n1\nFrom this diagram we can see that the first quartile is at a value of 14, the second\nquartile (median) is at a value of 19 and the third quartile is at a value of 30.\nThe inter-quartile range is the difference between the first and third quartiles. The\nfirst quartile is 14 and the third quartile is 30. Therefore the inter-quartile range is\n30 −14 = 16.\nFive number summary\nEMBK4\nThe five number summary combines a measure of central tendency, namely the me-\ndian, with measures of dispersion, namely the range and the inter-quartile range. This\ngives a good overview of the overall data distribution. More precisely, the five number\nsummary is written in the following order:\n• minimum;\n• first quartile;\n• median;\n• third quartile;\n• maximum.\nThe five number summary is often presented visually using a box and whisker diagram.\nA box and whisker diagram is shown below, with the positions of the five relevant\nnumbers labelled. Note that this diagram is drawn vertically, but that it may also be\ndrawn horizontally.\nmaximum\nupper quartile\nmedian\nlower quartile\nminimum\ninter-quartile range\ndata range\nSee video: 23CC at www.everythingmaths.co.za\n442\n11.1.\nRevision\n\nWorked example 3: Five number summary\nQUESTION\nDraw a box and whisker diagram for the following data set:\n1,25 ; 1,5 ; 2,5 ; 2,5 ; 3,1 ; 3,2 ; 4,1 ; 4,25 ; 4,75 ; 4,8 ; 4,95 ; 5,1\nSOLUTION\nStep 1: Determine the minimum and maximum\nSince the data set is already ordered, we can read off the minimum as the first value\n(1,25) and the maximum as the last value (5,1).\nStep 2: Determine the quartiles\nThere are 12 values in the data set.\n1,25 1,5\n2,5\n2,5\n3,1\n3,2\n4,1 4,25 4,75 4,8 4,95 5,1\n0\n1\n4\n1\n2\n3\n4\n1\nUsing the figure above we can see that the median is between the sixth and seventh\nvalues, making it.\n3,2 + 4,1\n2\n= 3,65\nThe first quartile lies between the third and fourth values, making it\nQ1 = 2,5 + 2,5\n2\n= 2,5\nThe third quartile lies between the ninth and tenth values, making it\nQ3 = 4,75 + 4,8\n2\n= 4,775\nStep 3: Draw the box and whisker diagram\nWe now have the five number summary as (1,25; 2,5; 3,65; 4,775; 5,1). The box and\nwhisker diagram representing the five number summary is given below.\n1,25\n2,5\n3,65\n4,775 5,1\n443\nChapter 11.\nStatistics\n\nExercise 11 – 1: Revision\n1. For each of the following data sets, compute the mean and all the quartiles.\nRound your answers to one decimal place.\na) −3,4 ; −3,1 ; −6,1 ; −1,5 ; −7,8 ; −3,4 ; −2,7 ; −6,2\nb) −6 ; −99 ; 90 ; 81 ; 13 ; −85 ; −60 ; 65 ; −49\nc) 7 ; 45 ; 11 ; 3 ; 9 ; 35 ; 31 ; 7 ; 16 ; 40 ; 12 ; 6\n2. Use the following box and whisker diagram to determine the range and inter-\nquartile range of the data.\n−5,52\n−2,41−1,53\n0,10\n4,08\n3. Draw the box and whisker diagram for the following data.\n0,2 ; −0,2 ; −2,7 ; 2,9 ; −0,2 ; −4,2 ; −1,8 ; 0,4 ; −1,7 ; −2,5 ; 2,7 ; 0,8 ; −0,5\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23CD\n1b. 23CF\n1c. 23CG\n2. 23CH\n3. 23CJ\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.2\nHistograms\nEMBK5\nA histogram is a graphical representation of how many times different, mutually exclu-\nsive events are observed in an experiment. To interpret a histogram, we find the events\non the x-axis and the counts on the y-axis. Each event has a rectangle that shows what\nits count (or frequency) is.\nSee video: 23CK at www.everythingmaths.co.za\nWorked example 4: Reading histograms\nQUESTION\nUse the following histogram to determine the events that were recorded and the rela-\ntive frequency of each event. Summarise your answer in a table.\n444\n11.2.\nHistograms\n\n0\n2\n4\n6\n8\n10\nnot yet\nin school\nin primary\nschool\nin high\nschool\nSOLUTION\nStep 1: Determine the events\nThe events are shown on the x-axis. In this example we have “not yet in school”, “in\nprimary school” and “in high school”.\nStep 2: Read off the count for each event\nThe counts are shown on the y-axis and the height of each rectangle shows the fre-\nquency for each event.\n• not yet in school: 2\n• in primary school: 5\n• in high school: 9\nStep 3: Calculate relative frequency\nThe relative frequency of an event in an experiment is the number of times that the\nevent occurred divided by the total number of times that the experiment was com-\npleted. In this example we add up the frequencies for all the events to get a total\nfrequency of 16. Therefore the relative frequencies are:\n• not yet in school:\n2\n16 = 1\n8\n• in primary school:\n5\n16\n• in high school:\n9\n16\nStep 4: Summarise\nEvent\nCount\nRelative frequency\nnot yet in school\n2\n1\n8\nin primary school\n5\n5\n16\nin high school\n9\n9\n16\n445\nChapter 11.\nStatistics\n\nTo draw a histogram of a data set containing numbers, the numbers first have to be\ngrouped.\nEach group is defined by an interval.\nWe then count how many times\nnumbers from each group appear in the data set and draw a histogram using the counts.\nWorked example 5: Draw a histogram\nQUESTION\nThe following data represent the heights of 16 adults in centimetres.\n162 ; 168 ; 177 ; 147 ; 189 ; 171 ; 173 ; 168\n178 ; 184 ; 165 ; 173 ; 179 ; 166 ; 168 ; 165\nDivide the data into 5 equal length intervals between 140 cm and 190 cm and draw a\nhistogram.\nSOLUTION\nStep 1: Determine intervals\nTo have 5 intervals of the same length between 140 and 190, we need and interval\nlength of 10. Therefore the intervals are (140; 150]; (150; 160]; (160; 170]; (170; 180];\nand (180; 190].\nStep 2: Count data\nThe following table summarises the number of data values in each of the intervals.\nInterval\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\n(180; 190]\nCount\n1\n0\n7\n6\n2\nStep 3: Draw the histogram\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\n446\n11.2.\nHistograms\n\nFrequency polygons\nEMBK6\nA frequency polygon is sometimes used to represent the same information as in a his-\ntogram. A frequency polygon is drawn by using line segments to connect the middle of\nthe top of each bar in the histogram. This means that the frequency polygon connects\nthe coordinates at the centre of each interval and the count in each interval.\nWorked example 6: Drawing a frequency polygon\nQUESTION\nUse the histogram from the previous example to draw a frequency polygon of the same\ndata.\nSOLUTION\nStep 1: Draw the histogram\nWe already know that the histogram looks like this:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n140\n150\n160\n170\n180\n190\nStep 2: Connect the tops of the rectangles\nWhen we draw line segments between the tops of the rectangles in the histogram, we\nget the following picture:\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\n447\nChapter 11.\nStatistics\n\nStep 3: Draw final frequency polygon\nFinally, we remove the histogram to show only the frequency polygon.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n140\n150\n160\n170\n180\n190\nFrequency polygons are particularly useful for comparing two data sets. Comparing\ntwo histograms would be more difficult since we would have to draw the rectangles of\nthe two data sets on top of each other. Because frequency polygons are just lines, they\ndo not pose the same problem.\nWorked example 7: Drawing frequency polygons\nQUESTION\nHere is another data set of heights, this time of Grade 11 learners.\n132 ; 132 ; 156 ; 147 ; 162 ; 168 ; 152 ; 174\n141 ; 136 ; 161 ; 148 ; 140 ; 174 ; 174 ; 162\nDraw the frequency polygon for this data set using the same interval length as in the\nprevious example. Then compare the two frequency polygons on one graph to see the\ndifferences between the distributions.\nSOLUTION\nStep 1: Frequency table\nWe first create the table of counts for the new data set.\nInterval\n(130; 140]\n(140; 150]\n(150; 160]\n(160; 170]\n(170; 180]\nCount\n4\n3\n2\n4\n3\n448\n11.2.\nHistograms\n\nStep 2: Draw histogram and frequency polygon\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\nStep 3: Compare frequency polygons\nWe draw the two frequency polygons on the same axes. The red line indicates the\ndistribution over heights for adults and the blue line, for Grade 11 learners.\nheight (cm)\ncount\n0\n2\n4\n6\n8\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n130\n140\n150\n160\n170\n180\n190\nFrom this plot we can easily see that the heights for Grade 11 learners are distributed\nmore towards the left (shorter) than adults. The learner heights also seem to be more\nevenly distributed between 130 and 180 cm, whereas the adult heights are mostly\nbetween 160 and 180 cm.\n449\nChapter 11.\nStatistics\n\nExercise 11 – 2: Histograms\n1. Use the histogram below to answer the following questions.\nThe histogram\nshows the number of people born around the world each year. The ticks on\nthe x-axis are located at the start of each year.\npeople (millions)\nyear\n79\n80\n81\n82\n83\n84\n85\n86\n87\n1994 1995 1996 1997 1998 1999 2000 2001\na) How many people were born between the beginning of 1994 and the be-\nginning of 1996?\nb) Is the number people in the world population increasing or decreasing?\n(Ignore the rate at which people are dying for this question.)\nc) How many more people were born in 1994 than in 1997?\n2. In a traffic survey, a random sample of 50 motorists were asked the distance (d)\nthey drove to work daily. The results of the survey are shown in the table below.\nDraw a histogram to represent the data.\nd\n0 < d ≤10\n10 < d ≤20\n20 < d ≤30\n30 < d ≤40\n40 < d ≤50\nf\n9\n19\n15\n5\n4\n3. Below is data for the prevalence of HIV in South Africa. HIV prevalence refers to\nthe percentage of people between the ages of 15 and 49 who are infected with\nHIV.\nyear\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nprevalence (%)\n17,7\n18,0\n18,1\n18,1\n18,1\n18,0\n17,9\n17,9\nDraw a frequency polygon of this data set.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CM\n2. 23CN\n3. 23CP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n450\n11.2.\nHistograms\n\n11.3\nOgives\nEMBK7\nCumulative histograms, also known as ogives, are graphs that can be used to deter-\nmine how many data values lie above or below a particular value in a data set. The\ncumulative frequency is calculated from a frequency table, by adding each frequency\nto the total of the frequencies of all data values before it in the data set. The last value\nfor the cumulative frequency will always be equal to the total number of data values,\nsince all frequencies will already have been added to the previous total.\nAn ogive is drawn by\n• plotting the beginning of the first interval at a y-value of zero;\n• plotting the end of every interval at the y-value equal to the cumulative count for\nthat interval; and\n• connecting the points on the plot with straight lines.\nIn this way, the end of the final interval will always be at the total number of data since\nwe will have added up across all intervals.\nWorked example 8: Cumulative frequencies and ogives\nQUESTION\nDetermine the cumulative frequencies of the following grouped data and complete the\ntable below. Use the table to draw an ogive of the data.\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n20 < n ≤30\n7\n30 < n ≤40\n12\n40 < n ≤50\n10\n50 < n ≤60\n6\nSOLUTION\nStep 1: Compute cumulative frequencies\nTo determine the cumulative frequency, we add up the frequencies going down the\ntable. The first cumulative frequency is just the same as the frequency, because we are\nadding it to zero. The final cumulative frequency is always equal to the sum of all the\nfrequencies. This gives the following table:\nInterval\nFrequency\nCumulative frequency\n10 < n ≤20\n5\n5\n20 < n ≤30\n7\n12\n30 < n ≤40\n12\n24\n40 < n ≤50\n10\n34\n50 < n ≤60\n6\n40\n451\nChapter 11.\nStatistics\n\nStep 2: Plot the ogive\nThe first coordinate in the plot always starts at a y-value of 0 because we always start\nfrom a count of zero. So, the first coordinate is at (10; 0) — at the beginning of the\nfirst interval. The second coordinate is at the end of the first interval (which is also the\nbeginning of the second interval) and at the first cumulative count, so (20; 5). The third\ncoordinate is at the end of the second interval and at the second cumulative count,\nnamely (30; 12), and so on.\nComputing all the coordinates and connecting them with straight lines gives the fol-\nlowing ogive.\nn\n0\n10\n20\n30\n40\n10\n20\n30\n40\n50\n60\n•\n•\n•\n•\n•\n•\nOgives do look similar to frequency polygons, which we saw earlier. The most impor-\ntant difference between them is that an ogive is a plot of cumulative values, whereas\na frequency polygon is a plot of the values themselves. So, to get from a frequency\npolygon to an ogive, we would add up the counts as we move from left to right in the\ngraph.\nOgives are useful for determining the median, percentiles and five number summary\nof data. Remember that the median is simply the value in the middle when we order\nthe data. A quartile is simply a quarter of the way from the beginning or the end of an\nordered data set. With an ogive we already know how many data values are above or\nbelow a certain point, so it is easy to find the middle or a quarter of the data set.\nWorked example 9: Ogives and the five number summary\nQUESTION\nUse the following ogive to compute the five number summary of the data. Remember\nthat the five number summary consists of the minimum, all the quartiles (including the\nmedian) and the maximum.\n452\n11.3.\nOgives\n\ncount\nvalue\n0\n10\n20\n30\n40\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\nSOLUTION\nStep 1: Find the minimum and maximum\nThe minimum value in the data set is 1 since this is where the ogive starts on the\nhorizontal axis. The maximum value in the data set is 10 since this is where the ogive\nstops on the horizontal axis.\nStep 2: Find the quartiles\nThe quartiles are the values that are 1\n4, 1\n2 and 3\n4 of the way into the ordered data set.\nHere the counts go up to 40, so we can find the quartiles by looking at the values\ncorresponding to counts of 10, 20 and 30. On the ogive a count of\n• 10 corresponds to a value of 3 (first quartile);\n• 20 corresponds to a value of 7 (second quartile); and\n• 30 corresponds to a value of 8 (third quartile).\nStep 3: Write down the five number summary\nThe five number summary is (1; 3; 7; 8; 10). The box-and-whisker plot of this data set\nis given below.\n1\n3\n7\n8\n10\n453\nChapter 11.\nStatistics\n\nExercise 11 – 3: Ogives\n1. Use the ogive to answer the questions below. Note that marks are given as a\npercentage.\nnumber of students\nmark\n0\n10\n20\n30\n40\n50\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n•\n•\n•\n•\n•\n•\n•\n•\n•\na) How many students got between 50% and 70%?\nb) How many students got at least 70%?\nc) Compute the average mark for this class, rounded to the nearest integer.\n2. Draw the histogram corresponding to this ogive.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n−25\n−15\n−5\n5\n15\n25\n•\n•\n•\n•\n•\n•\n3. The following data set lists the ages of 24 people.\n2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53; 38\n4; 28; 5; 73; 79; 17; 15; 5; 34; 37; 45; 56\nUse the data to answer the following questions.\na) Using an interval width of 8 construct a cumulative frequency plot.\nb) How many are below 30?\nc) How many are below 60?\nd) Giving an explanation state below what value the bottom 50% of the ages\nfall.\ne) Below what value do the bottom 40% fall?\nf) Construct a frequency polygon.\n454\n11.3.\nOgives\n\n4. The weights of bags of sand in grams is given below (rounded to the nearest\ntenth):\n50,1; 40,4; 48,5; 29,4; 50,2; 55,3; 58,1; 35,3; 54,2; 43,5\n60,1; 43,9; 45,3; 49,2; 36,6; 31,5; 63,1; 49,3; 43,4; 54,1\na) Decide on an interval width and state what you observe about your choice.\nb) Give your lowest interval.\nc) Give your highest interval.\nd) Construct a cumulative frequency graph and a frequency polygon.\ne) Below what value do 53% of the cases fall?\nf) Below what value of 60% of the cases fall?\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CQ\n2. 23CR\n3. 23CS\n4. 23CT\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n11.4\nVariance and standard deviation\nEMBK8\nMeasures of central tendency (mean, median and mode) provide information on the\ndata values at the centre of the data set. Measures of dispersion (quartiles, percentiles,\nranges) provide information on the spread of the data around the centre. In this section\nwe will look at two more measures of dispersion called the variance and the standard\ndeviation.\nSee video: 23CV at www.everythingmaths.co.za\nVariance\nEMBK9\nDEFINITION: Variance\nLet a population consist of n elements, {x1; x2; . . . ; xn}. Write the mean of the data as\nx.\nThe variance of the data is the average squared distance between the mean and each\ndata value.\nσ2 =\nPn\ni=1 (xi −x)2\nn\nNOTE:\nThe variance is written as σ2. It might seem strange that it is written in squared form,\nbut you will see why soon when we discuss the standard deviation.\n455\nChapter 11.\nStatistics\n\nThe variance has the following properties.\n• It is never negative since every term in the variance sum is squared and therefore\neither positive or zero.\n• It has squared units. For example, the variance of a set of heights measured in\ncentimetres will be given in centimeters squared. Since the population variance\nis squared, it is not directly comparable with the mean or the data themselves. In\nthe next section we will describe a different measure of dispersion, the standard\ndeviation, which has the same units as the data.\nWorked example 10: Variance\nQUESTION\nYou flip a coin 100 times and it lands on heads 44 times. You then use the same\ncoin and do another 100 flips. This time in lands on heads 49 times. You repeat this\nexperiment a total of 10 times and get the following results for the number of heads.\n{44; 49; 52; 62; 53; 48; 54; 49; 46; 51}\nCompute the mean and variance of this data set.\nSOLUTION\nStep 1: Compute the mean\nThe formula for the mean is\nx =\nPn\ni=1 xi\nn\nIn this case, we sum the data and divide by 10 to get x = 50,8.\nStep 2: Compute the variance\nThe formula for the variance is\nσ2 =\nPn\ni=1 (xi −x)2\nn\nWe first subtract the mean from each datum and then square the result.\nxi\n44\n49\n52\n62\n53\n48\n54\n49\n46\n51\nxi −x\n−6,8\n−1,8\n1,2\n11,2\n2,2\n−2,8\n3,2\n−1,8\n−4,8\n0,2\n(xi −x)2\n46,24\n3,24\n1,44\n125,44 4,84\n7,84\n10,24\n3,24\n23,04\n0,04\nThe variance is the sum of the last row in this table divided by 10, so σ2 = 22,56.\n456\n11.4.\nVariance and standard deviation\n\nStandard deviation\nEMBKB\nSince the variance is a squared quantity, it cannot be directly compared to the data val-\nues or the mean value of a data set. It is therefore more useful to have a quantity which\nis the square root of the variance. This quantity is known as the standard deviation.\nDEFINITION: Standard deviation\nLet a population consist of n elements, {x1; x2; . . . ; xn}, with a mean of x. The stan-\ndard deviation of the data is\nσ =\nsPn\ni=1 (xi −x)2\nn\nIn statistics, the standard deviation is a very common measure of dispersion. Standard\ndeviation measures how spread out the values in a data set are around the mean. More\nprecisely, it is a measure of the average distance between the values of the data in the\nset and the mean. If the data values are all similar, then the standard deviation will be\nlow (closer to zero). If the data values are highly variable, then the standard variation\nis high (further from zero).\nThe standard deviation is always a positive number and is always measured in the\nsame units as the original data. For example, if the data are distance measurements in\nkilogrammes, the standard deviation will also be measured in kilogrammes.\nThe mean and the standard deviation of a set of data are usually reported together. In\na certain sense, the standard deviation is a natural measure of dispersion if the centre\nof the data is taken as the mean.\nInvestigation: Tabulating results\nIt is often useful to set your data out in a table so that you can apply the for-\nmulae easily.\nComplete the table below to calculate the standard deviation of\n{57; 53; 58; 65; 48; 50; 66; 51}.\n• Firstly, remember to calculate the mean, x.\n• Complete the following table.\nindex: i\ndatum: xi\ndeviation: xi −x\ndeviation\nsquared: (xi −x)2\n1\n57\n2\n53\n3\n58\n4\n65\n5\n48\n6\n50\n7\n66\n8\n51\nP xi = . . .\nP(xi −x) = . . .\nP(xi −x)2 = . . .\n• The sum of the deviations is always zero. Why is this? Find out.\n• Calculate the variance using the completed table.\n• Then calculate the standard deviation.\n457\nChapter 11.\nStatistics\n\nWorked example 11: Variance and standard deviation\nQUESTION\nWhat is the variance and standard deviation of the possibilities associated with rolling\na fair die?\nSOLUTION\nStep 1: Determine all the possible outcomes\nWhen rolling a fair die, the sample space consists of 6 outcomes. The data set is\ntherefore x = {1; 2; 3; 4; 5; 6} and n = 6.\nStep 2: Calculate the mean\nThe mean is:\nx = 1\n6 (1 + 2 + 3 + 4 + 5 + 6)\n= 3,5\nStep 3: Calculate the variance\nThe variance is:\nσ2 =\nP (x −x)2\nn\n= 1\n6 (6,25 + 2,25 + 0,25 + 0,25 + 2,25 + 6,25)\n= 2,917\nStep 4: Calculate the standard deviation\nThe standard deviation is:\nσ =\np\n2,917\n= 1,708\nSee video: 23CW at www.everythingmaths.co.za\n458\n11.4.\nVariance and standard deviation\n\nInterpretation and application\nEMBKC\nA large standard deviation indicates that the data values are far from the mean and a\nsmall standard deviation indicates that they are clustered closely around the mean.\nFor example, consider the following three data sets:\n{65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\n{85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\n{43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nEach of these data sets has the same mean, namely 67. However, they have different\nstandard deviations, namely 8,97, 17,75 and 21,23. The following figures show plots\nof the data sets with the mean and standard deviation indicated on each. You can see\nhow the standard deviation is larger when the data are more spread out.\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 8,97\ndata:\n{xi} = {65; 75; 73; 50; 60; 64; 69; 62; 67; 85}\nmean:\nx = 67\nstandard deviation:\nσ ≈8,97\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 17,75\ndata:\n{xi} = {85; 79; 57; 39; 45; 71; 67; 87; 91; 49}\nmean:\nx = 67\nstandard deviation:\nσ ≈17,75\n30\n40\n50\n60\n70\n80\n90\n100\n110\nx = 67\nσ = 21,23\ndata:\n{xi} = {43; 51; 53; 110; 50; 48; 87; 69; 68; 91}\nmean:\nx = 67\nstandard deviation:\nσ ≈21,23\nThe standard deviation may also be thought of as a measure of uncertainty. In the phys-\nical sciences, for example, the reported standard deviation of a group of repeated mea-\nsurements represents the precision of those measurements. When deciding whether\n459\nChapter 11.\nStatistics\n\nmeasurements agree with a theoretical prediction, the standard deviation of those mea-\nsurements is very important: if the mean of the measurements is too far away from the\nprediction (with the distance measured in standard deviations), then we consider the\nmeasurements as contradicting the prediction. This makes sense since they fall outside\nthe range of values that could reasonably be expected to occur if the prediction were\ncorrect.\nExercise 11 – 4: Variance and standard deviation\n1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban.\nThe data, in rands per litre, are given below.\nCape Town\n3,96\n3,76\n4,00\n3,91\n3,69\n3,72\nDurban\n3,97\n3,81\n3,52\n4,08\n3,88\n3,68\na) Find the mean price in each city and then state which city has the lowest\nmean.\nb) Find the standard deviation of each city’s prices.\nc) Which city has the more consistently priced petrol? Give reasons for your\nanswer.\n2. Compute the mean and variance of the following set of values.\n150 ; 300 ; 250 ; 270 ; 130 ; 80 ; 700 ; 500 ; 200 ; 220 ; 110 ; 320 ; 420 ; 140\n3. Compute the mean and variance of the following set of values.\n−6,9 ; −17,3 ; 18,1 ; 1,5 ; 8,1 ; 9,6 ; −13,1 ; −14,0 ; 10,5 ; −14,8 ; −6,5 ; 1,4\n4. The times for 8 athletes who ran a 100 m sprint on the same track are shown\nbelow. All times are in seconds.\n10,2 ; 10,8 ; 10,9 ; 10,3 ; 10,2 ; 10,4 ; 10,1 ; 10,4\na) Calculate the mean time.\nb) Calculate the standard deviation for the data.\nc) How many of the athletes’ times are more than one standard deviation away\nfrom the mean?\n5. The following data set has a mean of 14,7 and a variance of 10,01.\n18 ; 11 ; 12 ; a ; 16 ; 11 ; 19 ; 14 ; b ; 13\nCompute the values of a and b.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23CX\n2. 23CY\n3. 23CZ\n4. 23D2\n5. 23D3\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n460\n11.4.\nVariance and standard deviation\n\n11.5\nSymmetric and skewed data\nEMBKD\nWe are now going to classify data sets into 3 categories that describe the shape of the\ndata distribution: symmetric, left skewed, right skewed. We can use this classification\nfor any data set, but here we will look only at distributions with one peak. Most of\nthe data distributions that you have seen so far have only one peak, so the plots in this\nsection should look familiar.\nDistributions with one peak are called unimodal distributions.\nUnimodal literally\nmeans having one mode. (Remember that a mode is a maximum in the distribution.)\nSymmetric distributions\nEMBKF\nA symmetric distribution is one where the left and right hand sides of the distribution\nare roughly equally balanced around the mean. The histogram below shows a typical\nsymmetric distribution.\nmean ≈median\nbalanced left and right tails\nFor symmetric distributions, the mean is approximately equal to the median. The tails\nof the distribution are the parts to the left and to the right, away from the mean. The\ntail is the part where the counts in the histogram become smaller. For a symmetric\ndistribution, the left and right tails are equally balanced, meaning that they have about\nthe same length.\nThe figure below shows the box and whisker diagram for a typical symmetric data set.\nmedian halfway\nbetween\nfirst and third quartiles\nAnother property of a symmetric distribution is that its median (second quartile) lies\nin the middle of its first and third quartiles. Note that the whiskers of the plot (the\nminimum and maximum) do not have to be equally far away from the median. In the\nnext section on outliers, you will see that the minimum and maximum values do not\nnecessarily match the rest of the data distribution well.\n461\nChapter 11.\nStatistics\n\nSkewed\nEMBKG\nA distribution that is skewed right (also known as positively skewed) is shown below.\nmean\nmedian\nmean > median\nlong right tail\nshort left tail\nNow the picture is not symmetric around the mean anymore.\nFor a right skewed\ndistribution, the mean is typically greater than the median. Also notice that the tail of\nthe distribution on the right hand (positive) side is longer than on the left hand side.\nmedian closer to first quartile\nFrom the box and whisker diagram we can also see that the median is closer to the first\nquartile than the third quartile. The fact that the right hand side tail of the distribution\nis longer than the left can also be seen.\nA distribution that is skewed left has exactly the opposite characteristics of one that is\nskewed right:\n• the mean is typically less than the median;\n• the tail of the distribution is longer on the left hand side than on the right hand\nside; and\n• the median is closer to the third quartile than to the first quartile.\nThe table below summarises the different categories visually.\nSymmetric\nSkewed right (positive)\nSkewed left (negative)\n462\n11.5.\nSymmetric and skewed data\n\nExercise 11 – 5: Symmetric and skewed data\n1. Is the following data set symmetric, skewed right or skewed left? Motivate your\nanswer.\n27 ; 28 ; 30 ; 32 ; 34 ; 38 ; 41 ; 42 ; 43 ; 44 ; 46 ; 53 ; 56 ; 62\n2. State whether each of the following data sets are symmetric, skewed right or\nskewed left.\na) A data set with this histogram:\nb) A data set with this box and whisker plot:\nc) A data set with this frequency polygon:\n• • • • • • • • •\n•\n•\n•\n• • • •\nd) The following data set:\n11,2 ; 5 ; 9,4 ; 14,9 ; 4,4 ; 18,8 ; −0,4 ; 10,5 ; 8,3 ; 17,8\n3. Two data sets have the same range and interquartile range, but one is skewed\nright and the other is skewed left. Sketch the box and whisker plot for each of\nthese data sets. Then, invent data (6 points in each data set) that matches the\ndescriptions of the two data sets.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23D4\n2a. 23D5\n2b. 23D6\n2c. 23D7\n2d. 23D8\n3. 23D9\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n463\nChapter 11.\nStatistics\n\n11.6\nIdentification of outliers\nEMBKH\nAn outlier in a data set is a value that is far away from the rest of the values in the\ndata set. In a box and whisker diagram, outliers are usually close to the whiskers of\nthe diagram. This is because the centre of the diagram represents the data between\nthe first and third quartiles, which is where 50% of the data lie, while the whiskers\nrepresent the extremes — the minimum and maximum — of the data.\nWorked example 12: Identifying outliers\nQUESTION\nFind the outliers in the following data set by drawing a box and whisker diagram and\nlocating the data values on the diagram.\n0,5 ; 1 ; 1,1 ; 1,4 ; 2,4 ; 2,8 ; 3,5 ; 5,1 ; 5,2 ; 6 ; 6,5 ; 9,5\nSOLUTION\nStep 1: Determine the five number summary\nThe minimum of the data set is 0,5. The maximum of the data set is 9,5. Since there\nare 12 values in the data set, the median lies between the sixth and seventh values,\nmaking it equal to 2,8+3,5\n2\n= 3,15. The first quartile lies between the third and fourth\nvalues, making it equal to 1,1+1,4\n2\n= 1,25. The third quartile lies between the ninth\nand tenth values, making it equal to 5,2+6\n2\n= 5,6.\nStep 2: Draw the box and whisker diagram\n0,5 1,25\n3,15\n5,6\n9,5\n• •• •\n• •\n•\n••\n• •\n•\nIn the figure above, each value in the data set is shown with a black dot.\nStep 3: Find the outliers\nFrom the diagram we can see that most of the values are between 1 and 6. The only\nvalue that is very far away from this range is the maximum at 9,5. Therefore 9,5 is the\nonly outlier in the data set.\nYou should also be able to identify outliers in plots of two variables. A scatter plot\nis a graph that shows the relationship between two random variables. We call these\ndata bivariate (literally meaning two variables) and we plot the data for two different\nvariables on one set of axes. The following example shows what a typical scatter plot\nlooks like. For Grade 11 you do not need to learn how to draw these 2-dimensional\n464\n11.6.\nIdentification of outliers\n\nscatter plots, but you should be able to identify outliers on them. As before, an outlier\nis a value that is far removed from the main distribution of data.\nWorked example 13: Scatter plot\nQUESTION\nWe have a data set that relates the heights and weights of a number of people. The\nheight is the first variable and its value is plotted along the horizontal axis. The weight\nis the second variable and its value is plotted along the vertical axis. The data values\nare shown on the plot below. Identify any outliers on the scatter plot.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\nSOLUTION\nWe inspect the plot visually and notice that there are two points that lie far away from\nthe main data distribution. These two points are circled in the plot below.\n130\n140\n150\n160\n170\n180\nheight (cm)\n30\n40\n50\n60\n70\n80\n90\nweight (kg)\n465\nChapter 11.\nStatistics\n\nExercise 11 – 6: Outliers\n1. For each of the following data sets, draw a box and whisker diagram and deter-\nmine whether there are any outliers in the data.\na) 30 ; 21,4 ; 39,4 ; 33,4 ; 21,1 ; 29,3 ; 32,8 ; 31,6 ; 36 ;\n27,9 ; 27,3 ; 29,4 ; 29,1 ; 38,6 ; 33,8 ; 29,1 ; 37,1\nb) 198 ; 166 ; 175 ; 147 ; 125 ; 194 ; 119 ; 170 ; 142 ; 148\nc) 7,1 ; 9,6 ; 6,3 ; −5,9 ; 0,7 ; −0,1 ; 4,4 ; −11,7 ; 10 ; 2,3 ; −3,7 ; 5,8 ; −1,4\n; 1,7 ; −0,7\n2. A class’s results for a test were recorded along with the amount of time spent\nstudying for it. The results are given below. Identify any outliers in the data.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1a. 23DB\n1b. 23DC\n1c. 23DD\n2. 23DF\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n466\n11.6.\nIdentification of outliers\n\n11.7\nSummary\nEMBKJ\nSee presentation: 23DG at www.everythingmaths.co.za\n• Histograms visualise how many times different events occurred. Each rectangle\nin a histogram represents one event and the height of the rectangle is relative to\nthe number of times that the event occurred.\n• Frequency polygons represent the same information as histograms, but using\nlines and points rather than rectangles. A frequency polygon connects the mid-\ndle of the top edge of each rectangle in a histogram.\n• Ogives (also known as cumulative histograms) show the total number of times\nthat a value or anything less than that value appears in the data set. To draw an\nogive you need to add up all the counts in a histogram from left to right.\n– The first count in an ogive is always zero.\n– The last count in an ogive is always the sum of all the counts in the data\nset.\n• The variance and standard deviation are measures of dispersion.\n– The standard deviation is the square root of the variance.\n– Variance: σ2 = 1\nn\nPn\ni=1(xi −x)2\n– Standard deviation: σ =\nq\n1\nn\nPn\ni=1(xi −x)2\n– The standard deviation is measured in the same units as the mean and the\ndata, but the variance is not. The variance is measured in the square of the\ndata units.\n• In a symmetric distribution\n– the mean is approximately equal to the median; and\n– the tails of the distribution are balanced.\n• In a right (positively) skewed distribution\n– the mean is greater than the median;\n– the tail on the right hand side is longer than the tail on the left hand side;\nand\n– the median is closer to the first quartile than the third quartile.\n• In a left (negatively) skewed distribution\n– the mean is less than the median;\n– the tail on the left hand side is longer than the tail on the right hand side;\nand\n– the median is closer to the third quartile than the first quartile.\n• An outlier is a value that is far away from the rest of the data.\n467\nChapter 11.\nStatistics\n\nExercise 11 – 7: End of chapter exercises\n1. Draw a histogram, frequency polygon and ogive of the following data set. To\ncount the data, use intervals with a width of 1, starting from 0.\n0,4 ; 3,1 ; 1,1 ; 2,8 ; 1,5 ; 1,3 ; 2,8 ; 3,1 ; 1,8 ; 1,3 ;\n2,6 ; 3,7 ; 3,3 ; 5,7 ; 3,7 ; 7,4 ; 4,6 ; 2,4 ; 3,5 ; 5,3\n2. Draw a box and whisker diagram of the following data set and explain whether\nit is symmetric, skewed right or skewed left.\n−4,1 ; −1,1 ; −1 ; −1,2 ; −1,5 ; −3,2 ; −4 ; −1,9 ; −4 ;\n−0,8 ; −3,3 ; −4,5 ; −2,5 ; −4,4 ; −4,6 ; −4,4 ; −3,3\n3. Eight children’s sweet consumption and sleeping habits were recorded. The data\nare given in the following table and scatter plot.\nNumber of sweets\nper week\n15\n12\n5\n3\n18\n23\n11\n4\nAverage sleeping\ntime (hours per day)\n4\n4,5\n8\n8,5\n3\n2\n5\n8\n5\n10\n15\n20\n25\nnumber of sweets\n1\n2\n3\n4\n5\n6\n7\n8\n9\nsleeping time (hours per day)\na) What is the mean and standard deviation of the number of sweets eaten per\nday?\nb) What is the mean and standard deviation of the number of hours slept per\nday?\nc) Make a list of all the outliers in the data set.\n4. The monthly incomes of eight teachers are as follows:\nR 10 050;\nR 14 300;\nR 9800;\nR 15 000;\nR 12 140;\nR 13 800;\nR 11 990;\nR 12 900.\na) What is the mean and standard deviation of their incomes?\nb) How many of the salaries are less than one standard deviation away from\nthe mean?\nc) If each teacher gets a bonus of R 500 added to their pay what is the new\nmean and standard deviation?\nd) If each teacher gets a bonus of 10% on their salary what is the new mean\nand standard deviation?\ne) Determine for both of the above, how many salaries are less than one stan-\ndard deviation away from the mean.\n468\n11.7.\nSummary\n\nf) Using the above information work out which bonus is more beneficial fi-\nnancially for the teachers.\n5. The weights of a random sample of boys in Grade 11 were recorded. The cumu-\nlative frequency graph (ogive) below represents the recorded weights.\n0\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100 110 120\n0\n10\n20\n30\n40\n50\n60\nWeight (in kilogrammes)\nCumulative frequency\nCumulative frequency curve showing weight of boys\na) How many of the boys weighed between 90 and 100 kilogrammes?\nb) Estimate the median weight of the boys.\nc) If there were 250 boys in Grade 11, estimate how many of them would\nweigh less than 80 kilogrammes?\n6. Three sets of 12 learners each had their test scores recorded. The test was out of\n50. Use the given data to answer the following questions.\nSet A\nSet B\nSet C\n25\n32\n43\n47\n34\n47\n15\n35\n16\n17\n32\n43\n16\n25\n38\n26\n16\n44\n24\n38\n42\n27\n47\n50\n22\n43\n50\n24\n29\n44\n12\n18\n43\n31\n25\n42\na) For each of the sets calculate the mean and the five number summary.\nb) Make box and whisker plots of the three data sets on the same set of axes.\nc) State, with reasons, whether each of the three data sets are symmetric or\nskewed (either right or left).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DH\n2. 23DJ\n3. 23DK\n4. 23DM\n5. 23DN\n6. 23DP\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n469\nChapter 11.\nStatistics\n\n\nCHAPTER\n12\nLinear programming\n12.1\nIntroduction\n472\n\n12\nLinear programming\n12.1\nIntroduction\nEMBKK\nIn everyday life people are interested in knowing the most efficient way of carrying out\na task or achieving a goal. For example, a farmer wants to know how many hectares to\nplant during a season in order to maximise the yield (produce), a stock broker wants to\nknow how much to invest in stocks in order to maximise profit, an entrepreneur wants\nto know how many people to employ to minimise expenditure. These are optimisation\nproblems; we want to to determine either the maximum or the minimum in a specific\nsituation.\nTo describe this mathematically, we assign variables to represent the different factors\nthat influence the situation. Optimisation means finding the combination of variables\nthat gives the best result.\nSee video: 23DQ at www.everythingmaths.co.za\nWorked example 1: Mountees and Roadees\nQUESTION\nInvestigate the following situation and use your knowledge of mathematics to solve the\nproblem:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make the maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nStep 2: Organise the information given\nWrite down a summary of the information given in the problem so that we consider\n472\n12.1.\nIntroduction\n\nall the different components in the situation.\nmaximum number for M\n= 5\nmaximum number for R\n= 3\nnumber of technicians needed for M = 1\nnumber of technicians needed for R = 2\ntotal number of technicians\n= 8\nprofit per M\n= 800\nprofit per R\n= 2400\nStep 3: Draw up a table\nUse the summary to draw up a table of all the possible combinations of the number of\nMountees and Roadees that can be manufactured per day:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n(4; 3)\n5\n(5; 0)\n(5; 1)\n(5; 2)\n(5; 3)\nNote that there are 24 possible combinations.\nStep 4: Consider the limitation of the number of technicians\nIt takes 1 technician to assemble a Mountee and 2 technicians to assemble a Roadee.\nThere are a total of 8 technicians in the assembly department, therefore we can write\nthat 1(M) + 2(R) ≤8.\nWith this limitation, we are able to eliminate some of the combinations in the table\nwhere M + 2R > 8:\nM\nR\n0\n1\n2\n3\n0\n(0; 0)\n(0; 1)\n(0; 2)\n(0; 3)\n1\n(1; 0)\n(1; 1)\n(1; 2)\n(1; 3)\n2\n(2; 0)\n(2; 1)\n(2; 2)\n(2; 3)\n3\n(3; 0)\n(3; 1)\n(3; 2)\n\b\b\b\n(3; 3)\n4\n(4; 0)\n(4; 1)\n(4; 2)\n\b\b\b\n(4; 3)\n5\n(5; 0)\n(5; 1)\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nThese combinations have been excluded as possible answers. For example, (5; 3) gives\n5 + 2(3) = 11 technicians.\n473\nChapter 12.\nLinear programming\n\nStep 5: Consider the profit on the bicycles\nWe can express the profit (P) per day as: P = 800(M)+2400(R). Notice that a higher\nprofit is made on a Roadee.\nBy substituting the different combinations for M and R, we can find the values that\ngive the maximum profit:\nFor (5; 0)\nP = 800(5) + 2400(0)\n= R 4000\nFor (3; 1)\nP = 800(3) + 2400(1)\n= R 4800\nM\nR\n0\n1\n2\n3\n0\n(0; 0) ⇒R 0\n(0; 1) ⇒R 2400\n(0; 2) ⇒R 4800\n(0; 3) ⇒R 7200\n1\n(1; 0) ⇒R 800\n(1; 1) ⇒R 3200\n(1; 2) ⇒R 5600\n(1; 3) ⇒R 8000\n2\n(2; 0) ⇒R 1600\n(2; 1) ⇒R 4000\n(2; 2) ⇒R 6400\n(2; 3) ⇒R 8800\n3\n(3; 0) ⇒R 2400\n(3; 1) ⇒R 4800\n(3; 2) ⇒R 7200\n\b\b\b\n(3; 3)\n4\n(4; 0) ⇒R 3200\n(4; 1) ⇒R 5600\n(4; 2) ⇒R 8000\n\b\b\b\n(4; 3)\n5\n(5; 0) ⇒R 4000\n(5; 1) ⇒R 6400\n\b\b\b\n(5; 2)\n\b\b\b\n(5; 3)\nStep 6: Write the final answer\nTherefore the maximum profit of R 8800 is obtained if 2 Mountees and 3 Roadees are\nmanufactured per day.\nExercise 12 – 1: Optimisation\n1. Furniture store opening special:\nAs part of their opening special, a furniture store has promised to give away at\nleast 40 prizes with a total value of at least R 4000. They intend to give away\nkettles and toasters. They decide there will be at least 10 units of each prize. A\nkettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the\ncompany. Calculate how much this combination of kettles and toasters will cost.\nUse a suitable strategy to organise the information and solve the problem.\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DR\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n474\n12.1.\nIntroduction\n\nOptimisation using graphs\nA more efficient way to solve optimisation problems is using graphs.\nWe write the limitations in the situation, called constraints, as inequalities. Some con-\nstraints can be modelled by an equation, which needs to be maximised or minimized.\nWe sketch the inequalities and indicate the region above or below the line that is to be\nconsidered in determining the solution. This method of solving optimisation problems\nis called linear programming.\nSee video: 23DS at www.everythingmaths.co.za\nWorked example 2: Optimisation using graphs\nQUESTION\nConsider again the example of Mr. Hunter who manufactures Mountees and Roadees:\nMr. Hunter manufactures bicycles. He produces two different models of bicycles;\nstrong mountain bikes called Mountees and fast road bikes called Roadees. He cannot\nproduce more than 5 Mountees on a day and he can manufacture a maximum of 3\nRoadees a day. He needs 1 technician to assemble a Mountee and 2 technicians to\nassemble a Roadee. The company has 8 technicians in the assembly department. The\nprofit on a Mountee is R 800 and R 2400 on a Roadee. The demand is such that he\ncan sell all the bikes he manufactures.\nDetermine the number of each model of bicycle that must be manufactured in order\nto make a maximum profit.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nMountees produced be M and let the number of Roadees produced be R.\nNotice that the values of M and R are limited to positive integers; Mr. Hunter cannot\nsell negative numbers of bikes nor can he sell a fraction of a bike.\nStep 2: Organise the information\nWe can write these constraints as inequalities:\nnumber of Mountees: 0 ≤M ≤5\nnumber of Roadees: 0 ≤R ≤3\ntotal number of technicians: M + 2R ≤8\nWe also know that P = 800M + 2400R. This is called the objective function, some-\ntimes also referred to as the search line, because the objective (goal) is to determine\nthe maximum value of P.\n475\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nWe represent the number of Mountees manufactured daily on the horizontal axis and\nthe number of Roadees manufactured daily on the vertical axis. Since M and R are\npositive integers, we only use the first quadrant of the Cartesian plane. Note that the\ngraph only includes the integer values of M between 0 and 5 and R between 0 and 3.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nFor the number of technicians in the assembly department M + 2R ≤8. If we make\nR (represented on the y-axis) the subject of the inequality we get R ≤−1\n2M + 4.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR = −1\n2M + 4\nThe arrows indicate the region in which the solution will lie, where R ≤−1\n2M + 4.\nThis area is called the feasible region.\nMountees (M)\nRoadees (R)\n0\n1\n2\n3\n4\n0\n1\n2\n3\n4\n5\n6\n7\n8\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nR ≤−1\n2M + 4\nA\n476\n12.1.\nIntroduction\n\nWe substitute the possible combinations into the profit equation P = 800M + 2400R,\nand find the combination that gives the maximum profit.\nAt A(2; 3) :\nP = 800(2) + 2400(3)\n= R 8800\nStep 4: Write the final answer\nTherefore the maximum profit is obtained if 2 Mountees and 3 Roadees are manufac-\ntured per day.\nSee video: 23DT at www.everythingmaths.co.za\nWorked example 3: Optimisation using graphs\nQUESTION\nSolve the “furniture store opening special” problem using graphs:\nAs part of their opening special, a furniture store has promised to give away at least 40\nprizes. They intend to give away kettles and toasters. They decide there will be at least\n10 units of each prize. A kettle costs the company R 120 and a toaster costs R 100.\nDetermine how many of each prize will represent the cheapest option for the company.\nCalculate how much this combination of kettles and toasters will cost.\nSOLUTION\nStep 1: Assign variables\nIn this situation there are two variables that we need to consider: let the number of\nkettles be k and the number of toasters be t, with k, t ∈Z.\nStep 2: Organise the information\nWe can write the given information as inequalities:\nnumber of kettles: k ≥10\nnumber of toasters: t ≥10\ntotal number of prizes: k + t ≥40\nWe make t the subject of the inequality:\nt ≥−k + 40\n477\nChapter 12.\nLinear programming\n\nStep 3: Solve using graphs\nRepresent the constraints on a set of axes:\nKettles (k)\nToasters (t)\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nt ≥−k + 40\nt ≥10\nk ≥10\nWe shade the feasible region as shown in the diagram. Remember that in this situation\nonly the points with integer coordinates inside or on the border of the feasible region\nare possible solutions. The combination giving the minimum cost will lie towards or\non the lower border of the feasible region, which gives us many points to consider. To\nfind the optimum value of C, we use the graph of the objective function\nC = 120k + 100t\nTo draw the line, we make t the subject of the formula\nt = −6\n5k + C\n100\nWe see that the gradient of the objective function is −6\n5, but we do not know the exact\nvalue of the t-intercept ( C\n100). To find the minimum value of C, we need to determine\nthe position of the objective function where it first touches the feasible region and also\ngives the lowest t-intercept.\nKettles (k)\nToasters (t)\nA\n0\n10\n20\n30\n40\n0\n10\n20\n30\n40\nb\nB\nb\nWe indicate the gradient of the objective function on the graph (the green search\nline). Keeping the gradient the same, we “slide” the objective function towards the\nlower border of the feasible region and find that it touches the feasible region at point\n478\n12.1.\nIntroduction\n\nA(10; 30). This optimum position of the objective function is indicated on the graph\nby the dotted line passing through point A.\nWe substitute the coordinates of A into the cost equation C = 120k + 100t:\nAt A(10; 30) :\nC = 120(10) + 100(30)\n= R 4200\nThe minimum cost can also be determined graphically by reading off the coordinates\nof the t-intercept of the objective function in the optimum position:\ntint = 42\n∴C\n100 = 42\n∴C = R 4200\nStep 4: Write the final answer\nTherefore the minimum cost to the company is R 4200 with 10 kettles and 30 toast-\ners.\nExercise 12 – 2: Optimisation\n1. You are given a test consisting of two sections. The first section is on algebra and\nthe second section is on geometry. You are not allowed to answer more than 10\nquestions from any section, but you have to answer at least 4 algebra questions.\nThe time allowed is not more than 30 minutes. An algebra problem will take 2\nminutes and a geometry problem will take 3 minutes to solve.\nLet x be the number of algebra questions and y be the number of geometry\nquestions.\na) Formulate the equations and inequalities that satisfy the above constraints.\nb) The algebra questions carry 5 marks each and the geometry questions carry\n10 marks each. If T is the total marks, write down an expression for T.\n2. A local clinic wants to produce a guide to healthy living. The clinic intends to\nproduce the guide in two formats: a short video and a printed book. The clinic\nneeds to decide how many of each format to produce for sale. Estimates show\nthat no more than 10 000 copies of both items together will be sold. At least\n4000 copies of the video and at least 2000 copies of the book could be sold,\nalthough sales of the book are not expected to exceed 4000 copies. Let x be the\nnumber of videos sold, and y the number of printed books sold.\na) Write down the constraint inequalities that can be deduced from the given\ninformation.\nb) Represent these inequalities graphically and indicate the feasible region\nclearly.\n479\nChapter 12.\nLinear programming\n\nc) The clinic is seeking to maximise the income, I, earned from the sales of\nthe two products. Each video will sell for R 50 and each book for R 30.\nWrite down the objective function for the income.\nd) What maximum income will be generated by the two guides?\n3. A certain motorcycle manufacturer produces two basic models, the Super X and\nthe Super Y. These motorcycles are sold to dealers at a profit of R 20 000 per\nSuper X and R 10 000 per Super Y. A Super X requires 150 hours for assembly,\n50 hours for painting and finishing and 10 hours for checking and testing. The\nSuper Y requires 60 hours for assembly, 40 hours for painting and finishing and\n20 hours for checking and testing. The total number of hours available per month\nis: 30 000 in the assembly department, 13 000 in the painting and finishing\ndepartment and 5000 in the checking and testing department.\nThe above information is summarised by the following table:\nDepartment\nHours for\nSuper X\nHours for\nSuper Y\nHours available\nper month\nAssembly\n150\n60\n30 000\nPainting and\nfinishing\n50\n40\n13 000\nChecking and testing\n10\n20\n5000\nLet x be the number of Super X and y be the number of Super Y models manu-\nfactured per month.\na) Write down the set of constraint inequalities.\nb) Use graph paper to represent the set of constraint inequalities.\nc) Shade the feasible region on the graph paper.\nd) Write down the profit generated in terms of x and y.\ne) How many motorcycles of each model must be produced in order to max-\nimise the monthly profit?\nf) What is the maximum monthly profit?\n4. A group of students plan to sell x hamburgers and y chicken burgers at a rugby\nmatch. They have meat for at most 300 hamburgers and at most 400 chicken\nburgers. Each burger of both types is sold in a packet. There are 500 packets\navailable. The demand is likely to be such that the number of chicken burgers\nsold is at least half the number of hamburgers sold.\na) Write the constraint inequalities and draw a graph of the feasible region.\nb) A profit of R 3 is made on each hamburger sold and R 2 on each chicken\nburger sold. Write the equation which represents the total profit P in terms\nof x and y.\nc) The objective is to maximise profit. How many of each type of burger\nshould be sold?\n5. Fashion-Cards is a small company that makes two types of cards, type X and type\nY. With the available labour and material, the company can make at most 150\ncards of type X and at most 120 cards of type Y per week. Altogether they cannot\nmake more than 200 cards per week.\n480\n12.1.\nIntroduction\n\nThere is an order for at least 40 type X cards and 10 type Y cards per week.\nFashion-Cards makes a profit of R 5 for each type X card sold and R 10 for each\ntype Y card.\nLet the number of type X cards manufactured per week be x and the number of\ntype Y cards manufactured per week be y.\na) One of the constraint inequalities which represents the restrictions above is\n0 ≤x ≤150. Write the other constraint inequalities.\nb) Represent the constraints graphically and shade the feasible region.\nc) Write the equation that represents the profit P (the objective function), in\nterms of x and y.\nd) Calculate the maximum weekly profit.\n6. To meet the requirements of a specialised diet a meal is prepared by mixing\ntwo types of cereal, Vuka and Molo. The mixture must contain x packets of\nVuka cereal and y packets of Molo cereal. The meal requires at least 15 g of\nprotein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains\n4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains\n3 g of protein and 24 g of carbohydrates. There are at most 5 packets of cereal\navailable. The feasible region is shaded on the attached graph paper.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\nNumber of packets of Vuka\nNumber of packets of Molo\na) Write down the constraint inequalities.\nb) If Vuka cereal costs R 6 per packet and Molo cereal also costs R 6 per\npacket, use the graph to determine how many packets of each cereal must\nbe used so that the total cost for the mixture is a minimum.\nc) Use the graph to determine how many packets of each cereal must be used\nso that the total cost for the mixture is a maximum (give all possibilities).\nThink you got it? Get this answer and more practice on our Intelligent Practice Service\n1. 23DV\n2. 23DW\n3. 23DX\n4. 23DY\n5. 23DZ\n6. 23F2\nwww.everythingmaths.co.za\nm.everythingmaths.co.za\n481\nChapter 12.\nLinear programming\n\n\nSolutions to exercises\n1\nExponents and surds\nExercise 1 – 1: The number system\n1. R; Q′\n2. R; Q\n3. R; Q\n4. R; Q\n5. R; Q; Z; N0\n6. R′Q′\n7. R; Q\n8. R; Q′\n9. R′\n10. R; Q′\n11. R; Q\n12. R; Q; Z\n13. R; Q\n14. R; Q′\n15. R; Q\n16. R; Q; Z\nExercise 1 – 2: Laws of exponents\n1. 43a+3\n2. 72\n3. 9p10\n4. k2x−2\n5. 52z−2 + 5z\n6. 1\n7. x10\n8.\nb2\na2\n9.\n1\nm+n\n10. 2pts\n11.\n1\na\n12. k\n13. 2a+1\n14. h4\n15.\na4b6\nc6d2\n16. 4\n17.\nm2n2\n2\n18. 400\n19.\n1\ny7\n20. 8\n21. 26a+2\n22. 2pt\n23. 81q2sy8a+2\nExercise 1 – 3: Rational exponents and surds\n1.\na) 7\nb)\n1\n6\nc)\n1\n3√\n36\nd) −4\n3\ne) 8x3\n2.\na) s\n1\n6\nb) 16m4\nc)\n3\n2 m2\nd) 8\n3. x\n31\n16\n483\nChapter 12.\nLinear programming\n\nExercise 1 – 4: Simplification of surds\n1.\na) 4\nb) ab4c2\nc) 2\nd) xy4\n2.\na)\nab\nb−a\nb) −\n\u0010\na\n1\n2 + b\n1\n2\n\u0011\nExercise 1 – 5: Rationalising the denominator\n1. 2\n√\n5\n2.\n√\n6\n2\n3.\n√\n6\n4.\n3\n√\n5 + 3\n4\n5.\nx√y\ny\n6.\n√\n6 +\n√\n14\n2\n7.\n3p −4√p\np\n8.\n√\nt −2\n9.\n1−√m\n1−m\n10.\n√\nab\nExercise 1 – 6: Solving surd equations\n1. x = 4\n2. p = 3\n3. y = 1\n4. t = 3\n5. z = 9 or z = 1\n4\n6. x = 8 or x = −27\n7. n = −1\n4\n8. d = 3 or d = −5\n9. y = 1 or y = 81\n10. f = 5\nExercise 1 – 7: Applications of exponentials\n1. 9,7%\n2. 4 254 691\n3. 7\n4. 26 893\n484\n12.1.\nIntroduction\n\nExercise 1 – 8: End of chapter exercises\n1.\na)\n1\n4\nb) 4 1\n4\n2.\na) x4\nb) s\nc) m\n25\n3\nd) m\n8\n3\ne) −m\n8\n3\nf) 81y\n16\n3\n3.\na)\n3b\n45\n2\n(a12c\n5\n2\nb) 3a3b2\nc) a24b12\nd) x\n7\n2\ne) x\n4\n3 b\n5\n3\n4.\n1\nx\n1\n16\n5. x −2\n6.\n10√x + 10\nx −1\n7.\n3√x + 2x√x\n2x\n8.\na) 6\n√\n2\nb) 7\n√\n5\nc) 2\nd)\n1\n4\n√\n2\ne) 2\nf)\n16\n√\n15\n5\n9.\na) 6 + 4\n√\n2\nb) 6 + 5\n√\n2\nc) 4+2\n√\n2+2\n√\n3+2\n√\n6\n10.\na) 55\nb) 1\n11. 15\n√\n2x3\n12.\na) 1 + 2\n√\n5\n5\nb)\n2y + y√y −4√y −8\ny −4\nc) 2√x + 2\n√\n10\n13.\n3\n2\n15. 3\n16. −\n√\n288\n17.\na) 4\nb) −1\n3\nc) 3\nd) No solution\ne) x = 1\n8 or x = −8\n18.\nb) x = 1\n2\nEquations and inequalities\nExercise 2 – 1: Solution by factorisation\n1. t = 0 or t = −2\n2. y = −1\n3. s = ±5\n4. y = 3 or y = 2\n5. y = 4 or y = −9\n6. p = −2\n7. y = −3 or y = −8\n8. y = 6 or y = 7\n9. x = −7 or x = −2\n10. y = 4k or y = k\n11. y = 9 or y = −9\n12. y = ±\n√\n5\n13. h = ±6\n14. y = ±\n√\n14\n15. p = −2\n16. y = ±6\n√\n2\n17. f = 5\n2 or f = −3\n18. x = 1\n4\n19. y = 1\n7\n20. x ∈R, x ̸= ±3\n21. y = −13 or y = −1\n22. t = 3\n2 or t = −2\n23. m = −6\n24. t = 0 or t = 3\nExercise 2 – 2: Solution by completing the square\n1.\na) x = −5 −3\n√\n3 or x = −5 + 3\n√\n3\nb) x = −1 or x = −3\nc) p = −4 ±\n√\n21\nd) x = −3 ±\n√\n7\ne) No real solution\nf) t = −8 ± 3\n√\n6\ng) x = −1 ±\nq\n5\n3\nh) z = −4 ±\n√\n22\ni) z = 11\n2 or z = 0\nj) z = 5 or z = −1\n2. k = −3 ± √9 −a\n3. y = −q±√\nq2−4pr\n2p\n485\nChapter 12.\nLinear programming\n\nExercise 2 – 3: Solution by the quadratic formula\n1. t = 1 or t = −4\n3\n2. x = 5+\n√\n37\n2\nor t = 5−\n√\n37\n2\n3. No real solution\n4. p = 1\n2 or p = −1\n5. No real solution\n6. t = −3+\n√\n69\n10\nor t = −3−\n√\n69\n10\n7. t = 2 ±\n√\n2\n8. k = 7+\n√\n373\n18\nor k = 7−\n√\n373\n18\n9. f = 1\n2 or f = −2\n10. No real solution\nExercise 2 – 4:\n1. x = −1, x = −4, x = −2 and x = −3\n2. x = 1, x = 4 and x = −2\n3. x = −7, x = 4, x = −1 and x = −2\n4. x = −4, x = 3, x = −3 and x = 2\n5. x = 8±\n√\n40\n4\n6. x = −5, x = 3, x = −1 +\n√\n10 and\nx = −1 −\n√\n10\nExercise 2 – 5: Finding the equation\n1. x2 −x −6 = 0\n2. x2 −16 = 0\n3. 2x2 −5x −3 = 0\n4. k = 3 and x = 3\n4\n5. p = 5 and x = −1\nExercise 2 – 6: Mixed exercises\n1. y = 1\n8 or y = −8\n3\n2. x = 3\n2 or x = −7\n2\n3. t = 2\n3 or t = 2\n4. y = 1 or y = −1\n5. m = 1 or m = 4\n6. y = ± 5\n7\n7. w = 3\n2 or w = 4\n8. y = 6\n5 or y = 1\n4\n9. n = 8\n3 or n = −9\n8\n10. y = −8\n3 or y = 3\n2\n11. x = −1\n2 or x = 3\n12. y = −5\n2 or y = −5\n9\n13. y = 4\n5 or y = 1\n5\n14. g = −1\n4 or g = 1\n15. y = 2 or y = −5\n9\n16. p = 3\n7 or p = −1\n5\n17. y = −2\n9 or y = −1\n18. y = 9\n2 or y = 9\n7\n486\n12.1.\nIntroduction\n\nExercise 2 – 7: From past papers\n1.\na) Real, unequal and rational\nb) Real and equal\nc) Real, unequal and irrational\nd) Real, unequal and rational\ne) Real, unequal and irrational\nf) Non-real\ng) Real, unequal and rational\nh) Real, unequal and irrational\ni) Non-real\nj) Real and equal\n2.\nb) real and unequal\nc) k = −6 ± 2\n√\n6\n4.\na) k = 6\nb) k = 1\n3\n5.\na) k = 4 or k = 1\nb) k = 0 or k = 5\n6.\na) all real values of a, b and p\nb) a = b and p = 0\nExercise 2 – 8: Solving quadratic inequalities\n1.\na) −3 < x < 4\nb) x < −4\n3 or when x > 1\nc) no real solutions\nd) −1 < t < 3\ne) All real values of s.\nf) All real values of x.\ng) x ≤−1\n4 or x ≥0\ni) x < 3 or x > 6 with x ̸= 3\nj) −2 ≤x ≤2 and x > 7 with x ̸= 7\nk) x > 0 with x ̸= 0\n2.\na) x < −3 or x > 3\nb) −\n√\n5 ≤x ≤\n√\n5\nc) no solution\nd) All real values of x\nExercise 2 – 9: Solving simultaneous equations\n1.\na) (0; 5) and (2; 3)\nb) x = 3 ±\n√\n2 and y = 2 ±\n√\n2\nc) (−1; 0) and ( 1\n4 ; 5\n8 )\nd) b = 2 ±\n√\n88\n6\nand a = 11 ±\n√\n88\n6\ne) (−3; −20) and (2; 0)\nf) x = 6 ±\n√\n264\n2\nand y = 70 ±\n√\n264\n2\n2.\na) (−3; 8) and (2; 3)\nb) (−4; 14) and (3; 7)\nc) (3; 4) and (4; 3)\nExercise 2 – 10:\n1. b = 2 m, l = 4 m\n2. 187\n3. t = 10,5 s\n4. t = 5d; 105 minutes; 1,4 km\n5. 24 A; 70 W; 12 A\n487\nChapter 12.\nLinear programming\n\nExercise 2 – 11: End of chapter exercises\n1. x = 1,62 or x = −0,62\n2. x = ±4 or x = −1\n3. y = 0 or y = ±1\n4. x = ±2\n5.\na) x = 7 or x = 2\nb) x = 2,3 or x = −1,3\nc) x = 1,65 or x = −3,65\nd) x = 0 or x = −3\n6. x =\n√\n16+p2−2\n2\n7. a = 3; b = 10 and c = −8\n8. p = ±16\n9. x2 + 2x −15\n10. Undefined:b = −2 Zero:b = 2 or b = 3\n13. a ≥4\n14. x = −3\n2 or x = 1\n15.\na) x < 3 or x ≥7:\nb) x < 1 or x > 5:\nc) 3 < x < 7:\nd) x < −1 or x > 3\ne) 0,5 < x < 2,5\nf) x ≤−3 or 0 < x ≤5\n2\ng) x < 2\n3\nh) −1 ≤x < 0 or x ≥3\ni) −4 ≤x ≤1\nj) 2 1\n2 ≤x < 3\n16.\na) x = ±\n√\n3 and y = ±2\n√\n3\nb) a = −3 and b = −1 or a = 12 and b = 4\nc) x = −5 and y = 0 or x = 2 and y = 14\nd) p = 5\n3 and q = 2\n9 or p = −1 and q = −2\n3\ne) b = 3±\n√\n5\n2\nand a = 7±3\n√\n5\n2\nf) b = −10±\n√\n140\n4\nand a = −12±\n√\n140\n2\ng) x = 3,4 and y = 5,4 or x = 3 and y = 5\nh) b = −1,4 and a = 23,6 or\nb = 3 and a = 6\n17.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\nx\n0\ny\nb\nb\nb)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n1\n2\n3\n4\n5\n6\n−1\n−2\nx\n0\ny\nb\nb\n18. 35 m\n20.\na) y = −5\n4 or y = −9\nb) x = −9\n4 or x = 1\nc) p = −8\n7 or p = −4\n3\nd) y = −1\n4 or y = 1\n2\ne) y = −2\n9 or y = −1\nf) y = 7\n3 or y = −1\n2\ng) y = 9\n4 or y = −9\n4\nh) y = 8\n3 or y = −6\ni) y = 9\n5 or y = −7\nj) x = ±4\nk) y = ±7\n21. k = 76 and 4\n9\n22. x = 3 or x = −2 and y = 1±√−7\n2\n23. x = 4 or x = −1\n24. y = 3\n2 , y = 1\n2 and p = 9\n2 , p = 7\n2\n25.\n69\n4\n26. 7\n27.\n2±\n√\n12\n2\n28. t = 1\n2 , t = 1 or t = 3±\n√\n33\n4\n488\n12.1.\nIntroduction\n\n3\nNumber patterns\nExercise 3 – 1: Linear sequences\n1. −19; −35; −51\n2.\na) −19\nb) T2 = 15; T4 = 33\n3.\na) Tn = 10 + 3n; T10 = 40; T15 = 55;\nT30 = 100\nb) Tn = 12 + 6n; T10 = 72; T15 = 102;\nT30 = 192\nc) Tn = −5 −5n; T10 = −55; T15 = −80;\nT30 = −155\n4. T9 = 36\n5.\na) 44; 66; 121\nExercise 3 – 2: Quadratic sequences\n1.\na) 10\nb) 2\nc) 2\nd) −2\ne) 2\nf) −4\ng) 4\nh) −2\ni) 6a\nj) 6\nk) 2t\n2.\na) T4 = 53\nb) T2 = 30\nc) T1 = 17\nd) T2 = −3\ne) T4 = 63\nf) T1 = 2\n3.\na) 3; 9; 17; 27\nb) −6; −9; −14; −21\nc) 1; 8; 21; 40\nd) 0; −5; −14; −27\nExercise 3 – 3: Quadratic sequences\n1.\na) 1\nb) 2\nc) 4\nd) 8\ne) −2\n2. 12; 30; 58; 96; 144\n3. T9 = 379\n4. n = 4\n5.\na) T5 = 84; T6 = 111\nb) Tn = 2n2 + 5n + 9\n489\nChapter 12.\nLinear programming\n\nExercise 3 – 4: End of chapter exercises\n1. −4; 9; 16; 25; 36\n2.\na) Quadratic sequence\nb) Quadratic sequence\nc) Quadratic sequence\nd) Quadratic sequence\ne) Quadratic sequence\nf) Quadratic sequence\ng) Linear sequence\nh) Linear sequence\ni) Quadratic sequence\nj) Quadratic sequence\nk) Quadratic sequence\nl) Linear sequence\nm) Quadratic sequence\n3. x = 31\n4. n = 11\n5. T11 = 363\n6. n = 9\n7. T5 = 114\n8. n = 8\n9.\na) T5 = 19;\nTn = 4n −1;\nT10 = 39\nb) T5 = −3;\nTn = 22 −5n;\nT10 = −28\nc) T5 = 2 1\n2 ; Tn = 1\n2 n;\nT10 = 5\nd) T5 = a + 4b;\nTn = a −b + bn;\nT10 = a + 9b\ne) T5 = −7;\nTn = 3 −2n;\nT10 = −17\n10.\na) Tn = n2 + 3;\nT100 = 10 003\nb) Tn = 6n −4;\nT100 = 596\nc) Tn = 2n2 + 5;\nT100 = 20 005\nd) Tn = 3n2 + 2;\nT100 = 30 002\n11.\na) 2; 5; 8; 11; 14\nb) Constant difference,\nd = 3\nc) Yes\n12.\na) Tn = 4n −19\nb) n = 48\n13.\na) Incorrect\nb) Correct\n14.\nc) Linear\n15.\nb) Linear\nd) Quadratic\ne) Tn = 1\n2 n2 + 3\n2 n + 1\nf) T21 = 253\ng) 31 cm\n16.\na) −1\nb) 7\n17.\nb) 2\nc) Tn = n2 −n\nd) 210\ne) 25\n18. 4; 14; 34; 64; 104; 154\n4\nAnalytical geometry\nExercise 4 – 1: Revision\n1.\na) 2\n√\n26units\nb) 7 units\nc) x + 1units\n2. p = 6 or p = 2\n3.\na) −1\n2\nb) 3\n5. 2\n6.\na) (1; 2)\nb)\n\u0000 −1\n2 ; −1\n2\n\u0001\n7. B(4; 2)\n8.\na) y = −4x + 3 and\ny = −4x + 19\nc) AD =\n√\n17units and\nBC =\n√\n17units\nd) y = 4\n3 x −7\n3\ne) Parallelogram (one\nopposite side equal\nand parallel)\n9. N(0; 3)\n10.\na) PQ =\n√\n20 and\nSR =\n√\n20\nb) M( 3\n2 ; 1)\nd) PS: y = −2\n5 x −1\n5\nand SR: y = 1\n2 x −2\ne) No\nf) Parallelogram\nExercise 4 – 2: The two-point form of the straight line equation\n1. y = 2\n3 x + 5\n2. y = −3x + 1\n4\n3. y = x + 3\n4. y = 2x −1\n5. y = −5\n6. y = 3\n4 x + 3\n7. y = −x + (s + t)\n8. y = 5x + 2\n9. y = q\npx −q\n490\n12.1.\nIntroduction\n\nExercise 4 – 3: Gradient–point form of a straight line equation\n1. y = 2\n3 x + 4\n2. y = −x −2\n3. y = −1\n3 x\n4. y = 11\n5. y = −2x + 7\n6. x = −3\n2\n7. y = −4\n5 x + 1\n8. x = 4\n9. y = 3ax + b\nExercise 4 – 4: The gradient–intercept form of a straight line equation\n1. y = 2x + 3\n2. y = 4x −4\n3. y = −x −1\n4. y = −3\n7 x\n5. y = 1\n2 x −1\n5\n6. y = 2x −2\n7. y = −3\n2\n8. y = 3x + 4\n9. y = −5x\nExercise 4 – 5: Angle of inclination\n1.\na) 1,7\nb) −1\nc) 0\nd) 1,4\ne) Undefined\nf) 1\ng) −0,8\nh) 0\ni) 3,7\n2.\na) 36,8◦\nb) 26,6◦\nc) 45◦\nd) Horizontal line\ne) 18,4◦\nf) Vertical line\ng) 71,6◦\nh) 30◦\nExercise 4 – 6: Inclination of a straight line\n1.\na) 38,7◦\nb) 135◦\nc) 80◦\nd) 80◦\ne) 102,5◦\nf) 45◦\ng) 56,3◦\nh) 63,4◦\ni) 161,6◦\nj) Gradient undefined\n2. 85,2◦\n3. 90◦\n4. 81,8◦\nExercise 4 – 7: Parallel lines\n1.\na) Parallel\nb) Parallel\nc) Parallel\nd) Not parallel\ne) Parallel\nf) Parallel\n2. y = −2x −3\n3. y = 3x\n4. y = 3\n2 x + 1\n5. y = −7\n10 x −1\n491\nChapter 12.\nLinear programming\n\nExercise 4 – 8: Perpendicular lines\n1.\na) Perpendicular\nb) Not perpendicular\nc) Perpendicular\nd) Perpendicular\ne) Perpendicular\nf) Not perpendicular\ng) Not perpendicular\n2. y = 1\n2 x −3\n3. y = −5x + 3\n4. y = −x + 2\n5. x = −2\nExercise 4 – 9: End of chapter exercises\n1.\na) y = 1\n2 x + 7\n2\nb) y = −x + 4\nc) y = 1\n2 x + 4\nd) y = 2x + 4\ne) y = 3x\n2.\na) θ = 63,4◦\nb) θ = 18,4◦\nc) θ = 36,9◦\nd) θ = 146,3◦\ne) θ = 161,6◦\n3.\na) y = −2x + 7\nb)\n\u0000 7\n2 ; 0\n\u0001\nc) θ = 116,6◦\nd) m = 1\n2\ne) Q ˆPR = 90◦\nf) y = −2x\ng)\n\u0000 1\n2 ; −3\n2\n\u0001\nh) y = −2x −1\n2\n4.\na)\n1\n2\n3\n4\n−1\n−2\n1\n2\n3\n−1\n−2\n−3\nb\nb\nb\nb\ny\nx\nA(−3; 5)\nB(−7; −4)\nC(2; 0)\nD(x; y)\nb) D (6; 9)\n5.\na) (−1; −2)\nb) (8; 3)\nc) x = −1\nd) MN = 5 units\ne) M ˆ\nNP = 21,8◦\nf) y = 5\n2 x + 11\n2\n6.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nA(−2; 3)\nB(2; 4)\nC(3; 0)\ny\nx\n0\nb\nb\nb\nc) y = 1\n4 x + 7\n2\nd) D(−1; −1)\ne) E\n\u0000 5\n2 ; 2\n\u0001\n7.\na) y = 3\n2 x + 2\nb) T ˆSV = 49,6◦\n8.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n4\n5\n−1\n−2\nb\nb\nb\nF(−1; 3)\nH(4; 4)\nG(2; 1)\ny\nx\n0\nc) y = −5x + 11\nd) Yes\ne) y = 3\n2 x + 9\n2\n9.\na)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n−5\n−6\n1\n2\n3\n4\n5\n6\n−1\n−2\nb\nb\nb\nb\nb\nA(−1; 5)\nB(5; −3)\nC(0; −6)\nx\ny\nM\nN\n492\n12.1.\nIntroduction", "chapter_id": "4.6" }, { "title": "Functions", "content": "5\nFunctions\nExercise 5 – 1: Revision\n1.\na)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nc)\n1\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nd)\n1\n2\n3\n4\n5\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nExercise 5 – 2: Domain and range\n1. {x : x ∈R} ; {y : y ≥−1, y ∈R}\n2. {x : x ∈R} ; {y : y ≤4, y ∈R}\n3. {x : x ∈R} ; {y : y ≥0, y ∈R}\n4. {x : x ∈R} ; {y : y ≤0, y ∈R}\n5. {x : x ∈R} ; {y : y ≤2, y ∈R}\nExercise 5 – 3: Intercepts\n1. (0; 15) and (−5; 0); (−3; 0)\n2. (0; 16) and (4; 0)\n3. (0; −3) and (1; 0); (3; 0)\n4. (0; 35) and (−7\n2 ; 0); (−5\n2 ; 0)\n5. (0; 37) and no x-intercepts\n6. (0; −4) and\n(−0,85; 0); (−2,35; 0)\nExercise 5 – 4: Turning points\n1. (3; −1)\n2. (2; 1)\n3. (−2; −1)\n4. (−1\n2 ; 1\n2 )\n5. (1; 21)\n6. (−1; −6)\nExercise 5 – 5: Axis of symmetry\n1.\na) Axis of symmetry:\nx = 5\n4\nb) Axis of symmetry:\nx = 2\nc) Axis of symmetry:\nx = 2\n2. y = ax2 + q\n493\nChapter 12.\nLinear programming\n\nExercise 5 – 6: Sketching parabolas\n1.\na)\n1\n2\n3\n4\n5\n6\n7\n8\n9\n−1\n1\n2\n3\n4\n5\n6\n−1\ny\nx\n0\nIntercepts: (−1; 0), (5; 0), (0; 5)\nTurning point: (2; 9)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≤9, y ∈R}\nb)\n1\n2\n3\n4\n5\n−1\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−1; 0), (0; 2)\nTurning point: (−1; 0)\nAxes of symmetry: x = −1\nDomain: {x : x ∈R}\nRange: {y : y ≥0, y ∈R}\nc)\n1\n2\n3\n4\n−1\n−2\n−3\n−4\n1\n2\n3\n−1\n−2\n−3\ny\nx\n0\nIntercepts: (−0,87; 0), (1,54; 0), (0; −4)\nTurning point: (0,33; −4,33)\nAxes of symmetry: x = −0,33\nDomain: {x : x ∈R}\nRange: {y : y ≥4,33, y ∈R}\nd)\n1\n2\n3\n4\n5\n6\n−1\n1\n2\n3\n4\n−1\ny\nx\n0\nIntercepts: (0; 13) Turning point: (2; 1)\nAxes of symmetry: x = 2\nDomain: {x : x ∈R}\nRange: {y : y ≥1, y ∈R}\n3.\na)\ny\nx\n0\nb)\ny\nx\n0\nc)\ny\nx\n0\nd)\ny\nx\n0\ne)\ny\nx\n0\nf)\ny\nx\n0\n4.\na) yshifted = 2x2 + 16x + 32\nb) yshifted = −x2 −2x\nc) yshifted = 3x2 −16x + 22\n494\n12.1.\nIntroduction\n\nExercise 5 – 7: Finding the equation\n1. y = −3(x + 1)2 + 6 or y = −3x2 −6x + 3\n2. y = 1\n2 x2 −5\n2 x\n3. y = 2\n3 (x + 2)2\n4. y = −x2 + 3x + 4\nExercise 5 – 8:\n1.\na) 11\n2.\na)\n1\n2\n3\n4\n1\n2\n−1\n−2\nf(x)\nx\n0\nA(1; 3)\nb\nb) 6\nc) y = 6x −3\n3.\na)\n1\n2\n−1\n−2\n−3\n−4\n1\n2\n−1\n−2\ng(x)\nx\n0\nb) 1\nc) 4\nd) 0\nExercise 5 – 10: Domain and range\n1. {x : x ∈R, x ̸= 0} ; {y : y ∈R, y ̸= 1}\n2. {x : x ∈R, x ̸= 8} ; {y : y ∈R, y ̸= 4}\n3. {x : x ∈R, x ̸= −1} ; {y : y ∈R, y ̸= −3}\n4. {x : x ∈R, x ̸= 5} ; {y : y ∈R, y ̸= 3}\n5. {x : x ∈R, x ̸= −2} ; {y : y ∈R, y ̸= 2}\nExercise 5 – 11: Intercepts\n1. (0; −1 3\n4 ) and\n\u0000−3 1\n2 ; 0\n\u0001\n2.\n\u0000 5\n2 ; 0\n\u0001\n3. (0; 1) and\n\u0000 1\n3 ; 0\n\u0001\n4.\n\u00000; 3\n2\n\u0001\nand\n\u0000 1\n3 ; 0\n\u0001\n5. (0; 2) and (8; 0)\nExercise 5 – 12: Asymptotes\n1. y = −2 and x = −4\n2. y = 0 and x = 0\n3. y = 1 and x = 2\n4. y = −8 and x = 0\n5. y = 0 and x = 2\n495\nChapter 12.\nLinear programming\n\nExercise 5 – 13: Axes of symmetry\n1.\na) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (0; 1); y1 = x + 1 and\ny2 = −x + 1\nb) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (−1; 0); y1 = x + 1 and\ny2 = −x −1\nc) For f(x): (0; 0); y1 = x and y2 = −x\nFor g(x): (1; −1); y1 = x −2 and\ny2 = −x\n2. k(x) =\n5\nx+1 + 2\nExercise 5 – 14: Sketching graphs\n1.\na) Asymptotes: x = 0; y = 2\nIntercepts:\n\u0000−1\n2 ; 0\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= 0}\nRange: {y : y ∈R, y ̸= 2}\nb) Asymptotes: x = −4; y = −2\nIntercepts:\n\u0000−3 1\n2 ; 0\n\u0001\nand\n\u00000; −1 3\n4\n\u0001\nAxes of symmetry: y = x + 2 and\ny = −x −6\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\nc) Asymptotes: x = −1; y = 3\nIntercepts:\n\u0000−2\n3 ; 0\n\u0001\nand (0; 2)\nAxes of symmetry: y = x + 4 and\ny = −x + 2\nDomain: {x : x ∈R, x ̸= −1}\nRange: {y : y ∈R, y ̸= 3}\nd) Asymptotes: x = −2 1\n2 ; y = −2\nIntercepts: (0; 0)\nAxes of symmetry: y = x −4 1\n2 and\ny = −x + 1\n2\nDomain: {x : x ∈R, x ̸= −4}\nRange: {y : y ∈R, y ̸= −2}\ne) Asymptotes: x = 8; y = 4\nIntercepts: (6; 0) and (0; 3)\nAxes of symmetry: y = x −4 and\ny = −x + 12\nDomain: {x : x ∈R, x ̸= 8}\nRange: {y : y ∈R, y ̸= 4}\n2. y =\n1\nx+2 −1\n3. y = −4\nx + 2\n4.\na)\nb) Average gradient = 1\nc) Average gradient = 12\nExercise 5 – 16: Domain and range\n1. {x : x ∈R} ; {y : y > 0, y ∈R}\n2. {x : x ∈R} ; {y : y < 1, y ∈R}\n3. {x : x ∈R} ; {y : y > −3, y ∈R}\n4. {x : x ∈R} ; {y : y > n, y ∈R}\n5. {x : x ∈R} ; {y : y > 2, y ∈R}\nExercise 5 – 17: Intercepts\n1. (0; −6) and (2; 0)\n2. (0; −17 1\n3 ) and (3; 0)\n3. (0; −20) and (−1; 0)\n4. (0; 15\n16 ) and (−2; 0)\n496\n12.1.\nIntroduction\n\nExercise 5 – 18: Asymptote\n1. y = 0\n2. y = 1\n3. y = −2\n3\n4. y = −2\n5. y = −2\nExercise 5 – 19: Mixed exercises\n1.\nb)\ni. y = 3\nx + 3\nii. y =\n3\nx−3\niii. y = −3\nx\niv. y = 3\nx −1\n4\nv. y = 3\nx + 4\nvi. y =\n3\nx+2 −1\n2.\na) M(−2; 2)\nb) g(x) = −4\nx\nc) f(x) = 2(x + 1)2\nd) −2 < x < 0\ne) Range: {y : y ∈R, y ≥0}\n3.\na) For k(x) :\nIntercepts:\n(−2; 0), (1; 0) and (0; −4)\nTurning point:\n\u0000−1\n2 ; −4 1\n2\n\u0001\nAsymptote:\nnone\nFor h(x) :\nIntercepts:\n(1,41; 0)\nTurning point:\nnone\nAsymptote:\ny = 0\n6.\na) f(x) = −3\n4 (x −2)2 + 3 ;\nAxes of symmetry: x = 2 ;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≤3}\nb) g(x) = 1\n4 x2 −2;\nAxes of symmetry: x = 0;\nDomain: {x : x ∈R};\nRange: {y : y ∈R, y ≥−2}; h(x) = 2\nx ;\nAxes of symmetry: y = x\nDomain: {x : x ∈R, x < 0};\nRange: {y : y ∈R, y < 0};\nc) k(x) =\n\u0000 1\n2\n\u0001x + 1\n2 ;\nDomain: {x : x ∈R};\nRange:\n\b\ny : y ∈R, y > 1\n2\n\t\n7.\nb) p = 9\nc) Average gradient = −2 8\n9\nd) y =\n\u0000 1\n3\n\u0001x+2 −2\n8.\na) f(x) = 2x −3\n2 and g(x) = −1\n4 x −1\n2\nb) h(x) = −\n3\nx+2 + 1\n9.\na) AO = 2 units OB = 5 units\nOC = 10 units DE = 12,25 units\nb) DE = 12 1\n4\nc) h(x) = −2x + 10\nd) {x : x ∈R, x < −2 and x > 5}\ne) {x : x ∈R, 0 ≤x ≤5}\nf) 5,25 units\n497\nChapter 12.\nLinear programming\n\nExercise 5 – 20: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\nPeriod: = 360◦\nAmplitude: = 1\nDomain: = [0◦; 360◦]\nRange: = [−1; 1]\nx-intercepts: = (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: = (0◦; 0)\nMax. turning point: = (90◦; 1)\nMin. turning point: = (270◦; −1)\n2.\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny2 = −2 sin θ\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 2\nDomain: [0◦; 360◦]\nRange: [−2; 2]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nMax. turning point: (270◦; 2)\nMin. turning point: (90◦; −2)\n3.\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny3 = sin θ + 1\ny1 = sin θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [0; 2]\nx-intercepts: (270◦; 0)\ny-intercepts: (0◦; 1)\nMax. turning point: (90◦; 2)\nMin. turning point: (270◦; 0)\n4.\n1\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny4 = 1\n2 sin θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (90◦; 1\n2 )\nMin. turning point: (270◦; −3\n2 )\nExercise 5 – 21: Sine functions of the form y = sin kθ\n2.\na) k = 2\nb) k = −3\n4\n498\n12.1.\nIntroduction\n\nExercise 5 – 23: The sine function\n1.\na)\n1\n2\n−1\n−2\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 2 sin( θ\n2 )\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = 1\n2 sin(θ −45◦)\nc)\n1\n2\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(θ + 90◦) + 1\nd)\n1\n−1\n60◦\n120◦\n180◦\n−60◦\n−120◦\n−180◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(−3θ\n2 )\ne)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = sin θ\ny2 = sin(30◦−θ)\n2. a = 2; p = 90◦∴y = 2 sin(θ + 90◦) and\ny = 2 cos θ\nExercise 5 – 24: Revision\n1.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\n2.\n1\n2\n3\n−1\n−2\n−3\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny2 = −3 cos θ\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [−3; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; −3)\n3.\n1\n2\n3\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny3 = cos θ + 2\nPeriod: 360◦\nAmplitude: 1\nDomain: [0◦; 360◦]\nRange: [1; 3]\nx-intercepts: none\ny-intercepts: (0◦; 3)\nMax. turning point: (0◦; 3); (360◦; 3)\nMin. turning point: (180◦; 1)\n4.\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny1 = cos θ\ny4 = 1\n2 cos θ −1\nPeriod: 360◦\nAmplitude: 1\n2\nDomain: [0◦; 360◦]\nRange: [−1\n2 ; −3\n2 ]\nx-intercepts: none\ny-intercepts: (0◦; −1\n2 )\nMax. turning point: (0◦; −1\n2 ); (360◦; −1\n2 )\nMin. turning point: (180◦; −3\n2 )\nExercise 5 – 25: Cosine functions of the form y = cos kθ\n2.\na) k = 3\n2\nb) k = 2\n3\n499\nChapter 12.\nLinear programming\n\nExercise 5 – 27: The cosine function\n1.\na)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦\n−30◦\n−60◦\n−90◦\n−120◦\n−150◦\n−180◦\nθ\n0◦\ny\ny = cos θ\ny = cos(θ + 15◦)\nb)\n1\n−1\n30◦\n60◦\n90◦\n−30◦\n−60◦\n−90◦\nθ\n0◦\ny\ny = cos θ\nf(θ) = 1\n3 cos(θ −60◦)\nc)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ny = −2 cos θ\nd)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(30◦−θ)\ne)\n1\n2\n−1\n−2\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nθ\n0◦\ny\ny = cos θ\ng(θ) = 1 + cos(θ −90◦)\nf)\n1\n−1\n60◦\n120◦\n180◦\n240◦\n300◦\n360◦\n−60◦\n−120◦\n−180◦\n−240◦\n−300◦\n−360◦\nθ\n0◦\ny\ny1 = cos θ\ny = cos(2θ + 60◦)\n2.\na) a = −1\nb) p = −180◦\nc) cos(θ −180◦) = −cos θ\nExercise 5 – 28: Revision\n1.\n1\n2\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1\n2 )\nAsymptotes: 90◦; 270◦\n2.\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0); (180◦; 0); (360◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: 90◦; 270◦\n3.\n1\n2\n3\n4\n5\n−1\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (116,6◦; 0); (296,6◦; 0)\ny-intercepts: (0◦; 2)\nAsymptotes: 90◦; 270◦\n4.\n1\n2\n−1\n−2\n−3\n−4\n90◦\n180◦\n270◦\n360◦\ny\n0◦\nθ\nPeriod: 180◦\nDomain: [0◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (26,6◦; 0); (206,6◦; 0)\ny-intercepts: (0◦; −1)\nAsymptotes: 90◦; 270◦\n500\n12.1.\nIntroduction\n\nExercise 5 – 29: Tangent functions of the form y = tan kθ\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦] Range: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 240◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −120◦; 120◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 90◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (−180◦; 0); (−90◦; 0); (0◦; 0);\n(90◦; 0); (180◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; −45◦; 45◦; 135◦\n4.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\nf(θ)\nPeriod: 270◦\nDomain: [−180◦; 180◦]\nRange: [−∞; ∞]\nx-intercepts: (0◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −135◦; 135◦\n501\nChapter 12.\nLinear programming\n\nExercise 5 – 30: Tangent functions of the form y = tan(θ + p)\n1.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−225◦; 0); (−45◦; 0); (135◦; 0);\n(315◦; 0)\ny-intercepts: (0◦; 0)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n2.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−330◦; 0); (−150◦; 0); (30◦; 0);\n(210◦; 0)\ny-intercepts: (0◦; −0,58)\nAsymptotes: −315◦; −135◦; 45◦; 225◦\n3.\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\n−45◦\n−90◦\n−135◦\n−180◦\n−225◦\n−270◦\n−315◦\n−360◦\nθ\nf(θ)\nPeriod: 180◦\nDomain: [−360◦; 360◦]\nRange: [−∞; ∞]\nx-intercepts: (−240◦; 0); (−60◦; 0); (120◦; 0);\n(300◦; 0)\ny-intercepts: (0◦; 1.73)\nAsymptotes: −330◦; −150◦; 30◦; 210◦\nExercise 5 – 31: The tangent function\n1.\na)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n−45◦\n−90◦\nθ\ny\nb)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\nθ\ny\nc)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\nd)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n−45◦\n−90◦\n−135◦\n−180◦\nθ\ny\n2. a = −1; k = 1\n2\n502\n12.1.\nIntroduction\n\nExercise 5 – 32: Mixed exercises\n1.\na) f(θ) = 3\n2 sin 2θ and g(θ) = −3\n2 tan θ\nb) f(θ) = −2 sin θ and\ng(θ) = 2 cos(θ + 360◦\nc) y = 3 tan θ\n2\nd) y = y = 2 cos θ + 2\n2.\na)\n1\n2\n3\n−1\n−2\n−3\n90◦\n180◦\n270◦\n360◦\nθ\n0\ng\nf\ny\nb\nb\nb\nb\n(90◦; 2)\n(270◦; −2)\nb) 360◦\nc) 1\nd) At θ = 180◦\n3.\na) a = 2, b = −1 and c = 240◦\nb) 180◦\nc) θ = 60◦; 300◦\nd) y = −tan(θ −45◦)\n4.\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny1\ny2\ny\nb\nb\nExercise 5 – 33: End of chapter exercises\n2. a = −2; k = −1\n4.\na) y = x + 22 + 2\nb) y = x −12 + 5\n5. (−1; 0)\n6.\ny\nx\n0\n4\n−4\n4\n−4\ny =\n2\nx−3 −1\n7. y =\n1\n(x−1) + 2\n9.\na) a = −1\nb) f(−15) = 0,99997\nc) x = −1\nd) h(x) = −2(x−2) + 1\n10.\na) a = 256\nb) f(x) = 256\n\u0000 3\n4\n\u0001x\nc) f(13) = 6,08\n11.\na)\n1\n−1\n90◦\n180◦\n−90◦\n−180◦\nθ\n0\ny\nb)\n1\n−1\n90◦\n180◦\nθ\n0\ny\nd)\n1\n−1\n90◦\n180◦\n270◦\n360◦\n−90◦\n−180◦\n−270◦\n−360◦\nθ\n0\ny\ne)\n1\n−1\n90◦\n180◦\n270◦\n360◦\nθ\n0\ny\nf)\n1\n2\n3\n4\n5\n6\n−1\n−2\n−3\n−4\n−5\n−6\n45◦\n90◦\n135◦\n180◦\n225◦\n270◦\n315◦\n360◦\nθ\nf(θ)\n503\nChapter 12.\nLinear programming\n\n6\nTrigonometry\nExercise 6 – 1: Revision\n1.\na) True\nb) True\nc) False\nd) True\n2.\na) 50,2◦\nb) 40,5◦\nc) 26,6◦\nd) 109,8◦\ne) No solution\nf) 17,7◦\ng) 69,4◦\n3.\na) 17,3 cm\nb) 10 cm\nc) 64,8◦\n4.\na) 10 cm\nb) 5,2 cm and 19,3 cm\nc) 50 cm2\n5.\na) 2\nb) 0\nc) −1 1\n2\nd) 1\ne) 1\n6.\na) 60◦\nb)\n1\n2\nc) 1\n7. No\nExercise 6 – 2: Trigonometric identities\n1.\na) cos α\nb) tan2 θ\nc) cos2 θ\nd) 0\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1.\na)\n√\n3\n3\nb)\n1\n8\nc) 1\n2.\na)\n1−cos2 θ\ncos θ\nb) −1\n3.\na) 2t\nb) −1\nt\nExercise 6 – 4: Using reduction formula\n1.\na) −tan θ\nb) 1\nc) 1\n2. −cos β\n3.\na)\n1\n√\n3\nb) 2\nc) 2\nd) −3\n2\ne) −4\n√\n3\n5\n5.\na) −t\nb) 1 −t2\nc) ±\nt\n√\n1−t2\nExercise 6 – 5: Co-functions\n1.\na) cos θ\nb)\n3\n2\n2.\na) p\nb)\np\n1 −p2\nc) −\np\n√\n1−p2\nd) p\n504\n12.1.\nIntroduction\n\nExercise 6 – 6: Reduction formulae\n1.\na) sin2 θ\nb) cos2 θ\nc)\ni. 1\nii. tan2 θ\n2.\na) sin 17◦\nb) cos 33◦\nc) tan 68◦\nd) −cos 33◦\n3.\na)\n√\n3\nb)\n√\n3\n2\nc)\n1\n4\nd) 1\nExercise 6 – 7: Solving trigonometric equations\n1.\na) α = 60◦; 300◦\nb) α = 220,5◦; 319,5◦\nc) α = 79,2◦; 259,2◦\nd) α = 200,1◦; 339,9◦\ne) α = 36,9◦; 143,1◦\nf) α = 109,7◦; 289,7◦\n2.\na) θ = −323,1◦; −216,9◦; 36,9◦; 143,1◦\nb) θ = −221,4◦; −138,6◦; 138,6◦; 221,4◦\nc) θ = −278,5◦; −98,5◦; 81,5◦; 261,5◦\nd) θ = −90◦; 270◦\ne) θ = −293,6◦; −66,4◦; 66,4◦; 293,6◦\nExercise 6 – 8: General solution\n1.\na) θ = −128,36◦; −101,64◦; 51,64◦\nb) θ = −80,45◦; −9,54◦; 99,55◦; 170,46◦\nc) θ = −53,27◦; 126,73◦\nd) α = 0◦\ne) θ = −180◦; 0◦; 180◦\nf) θ = −180◦; 180◦\ng) θ = 84◦\nh) θ = −120◦; 120◦\ni) θ = −60◦; −30◦; 120◦; 150◦\n2.\na) θ = −20◦+ n . 360◦\nb) α = 30◦+ n . 120◦\nc) β = 10,25◦+ n . 45◦or\nβ = 55,25◦+ n . 45◦\nd) α = 70◦+ n . 360◦or\nα = 340◦+ n . 360◦\ne) θ = 140◦+ n . 240◦or\nθ = 220◦+ n . 240◦\nf) β = 15◦+ n . 180◦\nExercise 6 – 9: Solving trigonometric equations\n1.\na) θ = 45◦+ k . 180◦or\nθ = 135◦+ k . 180◦\nb) α = 50◦+ k . 360◦or\nα = 110◦+ k . 360◦\nc) θ = 60◦+ k . 720◦or\nθ = 660◦+ k . 720◦\nd) β = 146,6◦+ k . 180◦\ne) θ = 110,27◦+ k . 360◦or\nθ = 249,73◦+ k . 360◦\nf) α = 210◦+ k . 360◦or\nα = 330◦+ k . 360◦\ng) β = 23,3◦+ k . 120◦\nh) θ = 122◦+ k . 180◦\ni) α = 21◦+ k . 180◦or\nα = 39,5◦+ k . 90◦\nj) β = 22,5◦+ k . 90◦\n2. θ = 0◦, 180◦, 210◦, 330◦or 360◦\n3.\na) θ = 120◦+ k . 360◦or\nθ = 240◦+ k . 360◦\nb) θ = 0◦+ k . 180◦or\nθ = 146,3◦+ k . 180◦\nc) α = 36,9◦+ k . 360◦or\nα = 143,1◦+ k . 360◦or\nα = 216,9◦+ k . 360◦or\nα = 323,1◦+ k . 360◦\nd) β = 15◦+ k . 120◦or\nβ = 75◦+ k . 120◦\ne) α = 48,4◦+ k . 180◦\nf) θ = 63,4◦+ k . 180◦or\nθ = 116,6◦+ k . 180◦\ng) θ = 54,8◦+ k . 180◦or\nθ = 95,25◦+ k . 180◦\n4. β = −70,5◦or β = 109,5◦\n505\nChapter 12.\nLinear programming\n\nExercise 6 – 10: The area rule\n1.\na)\nP\nQ\nR\n30◦\n10\n7\nArea △PQR = 17,5 square units\nb)\nP\nQ\nR\n110◦\n9\n8\nArea △PQR = 33,8 square units\n2. Area △XY Z = 645,6 square units\n3. Area = 106,5 square units\n4.\nˆC = 72,2◦or ˆC = 107,8◦\nExercise 6 – 11: Sine rule\n1.\na)\nˆP = 92◦, q = 6,6, p = 7,4\nb)\nˆL = 87◦, l = 1,3, k = 0,89\nc)\nˆB = 76,8◦, b = 94,3, c = 91,3\nd)\nˆY = 84◦, y = 60, z = 38,8\n2.\nˆB = 32◦, AB = 23, BC = 39\n3. ST = 78,1 km\n4. m = 26,2\n5. BC = 3,2\nExercise 6 – 12: The cosine rule\n1.\na) a = 8,5, ˆC = 83,9◦, ˆB = 26,1◦\nb)\nˆR = 120◦, ˆS = 32,2◦, ˆT = 27,8◦\nc)\nˆ\nM = 27,7◦, ˆL = 40,5◦, ˆ\nK = 111,8◦\nd) h = 19,1, ˆJ = 18,2◦, ˆ\nK = 31,8◦\ne)\nˆD = 34◦, ˆE = 44,4◦, ˆF = 101,6◦\n2.\na) x = 4,4 km\nb) y = 63,5 cm\n3.\na)\nˆ\nK = 117,3◦\nb)\nˆQ = 78,5◦\nExercise 6 – 13: Area, sine and cosine rule\n1.\na) 7,78 km\nb) 6 km\n2. XZ = 1,73 km, XY = 0,87 km\n3.\na) 1053 km\nb) 4,42◦\n4. DC = x sin a sin(b+c)\nsin(a+c) sin b\n5.\nb) 438,5 km\n6. 9,38 m2\n7. DC = x sin α\nsin β\n506\n12.1.\nIntroduction\n\nExercise 6 – 14: End of chapter exercises\n1. sin2 A\n2. 1 1\n4\n3. cos α\n4. 3\n7.\na) −1\nb) θ = 135◦or θ = 315◦\n8.\na)\nb\nx\ny\n0\nθ\n(−12; −5)\nb) −5\n13 and 12\n13\nc) θ = 202,62◦\n9.\na) a = 1 and b = −\n√\n3\nb) −\n√\n3\n2\n10.\na) x = 50,9◦or x = 309,1◦\nb) x = 127,3◦or x = 307,3◦\nc) x = 26,6◦; 153,4◦206,6◦or 333,4◦\n11.\na) x = 55◦+ k . 360◦or\nx = 175◦+ k . 360◦\nb) x = 180◦+ k . 360◦\n12.\na) x = 28,6◦+ k . 180◦or\nx = 61,4◦+ k . 180◦\nb)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nb\nb\nb\nb\nθ\n0◦\ny\ny = sin 2α\nc) 28,6◦; 61,4◦; 208,6◦; 241,4◦\n13.\na) A ˆGN = α −β\nb)\nˆ\nA = 90◦−α\nd) H = 5 m\n14.\na) AC = 9,43 m\nb) AD = 6,2 m\nc) Area = 49,25 m2\nd) Area = 49,23 m2\n7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "29" }, { "title": "Quadratic functions", "content": "", "chapter_id": "5.1" }, { "title": "Average gradient", "content": "", "chapter_id": "5.2" }, { "title": "Hyperbolic functions", "content": "", "chapter_id": "5.3" }, { "title": "Exponential functions", "content": "", "chapter_id": "5.4" }, { "title": "The sine function", "content": "", "chapter_id": "5.5" }, { "title": "The cosine function", "content": "", "chapter_id": "5.6" }, { "title": "The tangent function", "content": "", "chapter_id": "5.7" }, { "title": "Summary", "content": "", "chapter_id": "5.8" }, { "title": "Trigonometry", "content": "6\nTrigonometry\nExercise 6 – 1: Revision\n1.\na) True\nb) True\nc) False\nd) True\n2.\na) 50,2◦\nb) 40,5◦\nc) 26,6◦\nd) 109,8◦\ne) No solution\nf) 17,7◦\ng) 69,4◦\n3.\na) 17,3 cm\nb) 10 cm\nc) 64,8◦\n4.\na) 10 cm\nb) 5,2 cm and 19,3 cm\nc) 50 cm2\n5.\na) 2\nb) 0\nc) −1 1\n2\nd) 1\ne) 1\n6.\na) 60◦\nb)\n1\n2\nc) 1\n7. No\nExercise 6 – 2: Trigonometric identities\n1.\na) cos α\nb) tan2 θ\nc) cos2 θ\nd) 0\nExercise 6 – 3: Reduction formulae for function values of 180◦± θ\n1.\na)\n√\n3\n3\nb)\n1\n8\nc) 1\n2.\na)\n1−cos2 θ\ncos θ\nb) −1\n3.\na) 2t\nb) −1\nt\nExercise 6 – 4: Using reduction formula\n1.\na) −tan θ\nb) 1\nc) 1\n2. −cos β\n3.\na)\n1\n√\n3\nb) 2\nc) 2\nd) −3\n2\ne) −4\n√\n3\n5\n5.\na) −t\nb) 1 −t2\nc) ±\nt\n√\n1−t2\nExercise 6 – 5: Co-functions\n1.\na) cos θ\nb)\n3\n2\n2.\na) p\nb)\np\n1 −p2\nc) −\np\n√\n1−p2\nd) p\n504\n12.1.\nIntroduction\n\nExercise 6 – 6: Reduction formulae\n1.\na) sin2 θ\nb) cos2 θ\nc)\ni. 1\nii. tan2 θ\n2.\na) sin 17◦\nb) cos 33◦\nc) tan 68◦\nd) −cos 33◦\n3.\na)\n√\n3\nb)\n√\n3\n2\nc)\n1\n4\nd) 1\nExercise 6 – 7: Solving trigonometric equations\n1.\na) α = 60◦; 300◦\nb) α = 220,5◦; 319,5◦\nc) α = 79,2◦; 259,2◦\nd) α = 200,1◦; 339,9◦\ne) α = 36,9◦; 143,1◦\nf) α = 109,7◦; 289,7◦\n2.\na) θ = −323,1◦; −216,9◦; 36,9◦; 143,1◦\nb) θ = −221,4◦; −138,6◦; 138,6◦; 221,4◦\nc) θ = −278,5◦; −98,5◦; 81,5◦; 261,5◦\nd) θ = −90◦; 270◦\ne) θ = −293,6◦; −66,4◦; 66,4◦; 293,6◦\nExercise 6 – 8: General solution\n1.\na) θ = −128,36◦; −101,64◦; 51,64◦\nb) θ = −80,45◦; −9,54◦; 99,55◦; 170,46◦\nc) θ = −53,27◦; 126,73◦\nd) α = 0◦\ne) θ = −180◦; 0◦; 180◦\nf) θ = −180◦; 180◦\ng) θ = 84◦\nh) θ = −120◦; 120◦\ni) θ = −60◦; −30◦; 120◦; 150◦\n2.\na) θ = −20◦+ n . 360◦\nb) α = 30◦+ n . 120◦\nc) β = 10,25◦+ n . 45◦or\nβ = 55,25◦+ n . 45◦\nd) α = 70◦+ n . 360◦or\nα = 340◦+ n . 360◦\ne) θ = 140◦+ n . 240◦or\nθ = 220◦+ n . 240◦\nf) β = 15◦+ n . 180◦\nExercise 6 – 9: Solving trigonometric equations\n1.\na) θ = 45◦+ k . 180◦or\nθ = 135◦+ k . 180◦\nb) α = 50◦+ k . 360◦or\nα = 110◦+ k . 360◦\nc) θ = 60◦+ k . 720◦or\nθ = 660◦+ k . 720◦\nd) β = 146,6◦+ k . 180◦\ne) θ = 110,27◦+ k . 360◦or\nθ = 249,73◦+ k . 360◦\nf) α = 210◦+ k . 360◦or\nα = 330◦+ k . 360◦\ng) β = 23,3◦+ k . 120◦\nh) θ = 122◦+ k . 180◦\ni) α = 21◦+ k . 180◦or\nα = 39,5◦+ k . 90◦\nj) β = 22,5◦+ k . 90◦\n2. θ = 0◦, 180◦, 210◦, 330◦or 360◦\n3.\na) θ = 120◦+ k . 360◦or\nθ = 240◦+ k . 360◦\nb) θ = 0◦+ k . 180◦or\nθ = 146,3◦+ k . 180◦\nc) α = 36,9◦+ k . 360◦or\nα = 143,1◦+ k . 360◦or\nα = 216,9◦+ k . 360◦or\nα = 323,1◦+ k . 360◦\nd) β = 15◦+ k . 120◦or\nβ = 75◦+ k . 120◦\ne) α = 48,4◦+ k . 180◦\nf) θ = 63,4◦+ k . 180◦or\nθ = 116,6◦+ k . 180◦\ng) θ = 54,8◦+ k . 180◦or\nθ = 95,25◦+ k . 180◦\n4. β = −70,5◦or β = 109,5◦\n505\nChapter 12.\nLinear programming\n\nExercise 6 – 10: The area rule\n1.\na)\nP\nQ\nR\n30◦\n10\n7\nArea △PQR = 17,5 square units\nb)\nP\nQ\nR\n110◦\n9\n8\nArea △PQR = 33,8 square units\n2. Area △XY Z = 645,6 square units\n3. Area = 106,5 square units\n4.\nˆC = 72,2◦or ˆC = 107,8◦\nExercise 6 – 11: Sine rule\n1.\na)\nˆP = 92◦, q = 6,6, p = 7,4\nb)\nˆL = 87◦, l = 1,3, k = 0,89\nc)\nˆB = 76,8◦, b = 94,3, c = 91,3\nd)\nˆY = 84◦, y = 60, z = 38,8\n2.\nˆB = 32◦, AB = 23, BC = 39\n3. ST = 78,1 km\n4. m = 26,2\n5. BC = 3,2\nExercise 6 – 12: The cosine rule\n1.\na) a = 8,5, ˆC = 83,9◦, ˆB = 26,1◦\nb)\nˆR = 120◦, ˆS = 32,2◦, ˆT = 27,8◦\nc)\nˆ\nM = 27,7◦, ˆL = 40,5◦, ˆ\nK = 111,8◦\nd) h = 19,1, ˆJ = 18,2◦, ˆ\nK = 31,8◦\ne)\nˆD = 34◦, ˆE = 44,4◦, ˆF = 101,6◦\n2.\na) x = 4,4 km\nb) y = 63,5 cm\n3.\na)\nˆ\nK = 117,3◦\nb)\nˆQ = 78,5◦\nExercise 6 – 13: Area, sine and cosine rule\n1.\na) 7,78 km\nb) 6 km\n2. XZ = 1,73 km, XY = 0,87 km\n3.\na) 1053 km\nb) 4,42◦\n4. DC = x sin a sin(b+c)\nsin(a+c) sin b\n5.\nb) 438,5 km\n6. 9,38 m2\n7. DC = x sin α\nsin β\n506\n12.1.\nIntroduction\n\nExercise 6 – 14: End of chapter exercises\n1. sin2 A\n2. 1 1\n4\n3. cos α\n4. 3\n7.\na) −1\nb) θ = 135◦or θ = 315◦\n8.\na)\nb\nx\ny\n0\nθ\n(−12; −5)\nb) −5\n13 and 12\n13\nc) θ = 202,62◦\n9.\na) a = 1 and b = −\n√\n3\nb) −\n√\n3\n2\n10.\na) x = 50,9◦or x = 309,1◦\nb) x = 127,3◦or x = 307,3◦\nc) x = 26,6◦; 153,4◦206,6◦or 333,4◦\n11.\na) x = 55◦+ k . 360◦or\nx = 175◦+ k . 360◦\nb) x = 180◦+ k . 360◦\n12.\na) x = 28,6◦+ k . 180◦or\nx = 61,4◦+ k . 180◦\nb)\n1\n−1\n30◦\n60◦\n90◦\n120◦150◦180◦210◦240◦270◦300◦330◦360◦\nb\nb\nb\nb\nθ\n0◦\ny\ny = sin 2α\nc) 28,6◦; 61,4◦; 208,6◦; 241,4◦\n13.\na) A ˆGN = α −β\nb)\nˆ\nA = 90◦−α\nd) H = 5 m\n14.\na) AC = 9,43 m\nb) AD = 6,2 m\nc) Area = 49,25 m2\nd) Area = 49,23 m2\n7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "38" }, { "title": "Revision", "content": "", "chapter_id": "6.1" }, { "title": "Trigonometric identities", "content": "", "chapter_id": "6.2" }, { "title": "Reduction formula", "content": "", "chapter_id": "6.3" }, { "title": "Trigonometric equations", "content": "", "chapter_id": "6.4" }, { "title": "Area, sine, and cosine rules", "content": "", "chapter_id": "6.5" }, { "title": "Summary", "content": "", "chapter_id": "6.6" }, { "title": "Measurement", "content": "7\nMeasurement\nExercise 7 – 1: Area of a polygon\n1.\nb) 240 cm\nc) 0,6 m2\ne) Wood: 233,2 cm and paper: 0,6 m2\n2.\na) 25π units2\nb) 20π units2\n3.\na) 1,2 m2\nb) Perimeter: 414,8 cm; Area 11 700 cm2\nc) 108 × 108cm2\nExercise 7 – 2: Calculating surface area\n1. 273 cm2\n2. Yes\nExercise 7 – 3: Calculating volume\n1.\na) 67,5 m2\nb) 3,39 ℓ\n2.\nb) 13,86 cm\nc) 554,24 m3\n507\nChapter 12.\nLinear programming\n\nExercise 7 – 4: Finding surface area and volume\n1.\na) 120 cm2\nb) 124 cm3\nc) 40\nd)\ni. 120 mm\nii. 165 mm\niii. 589 mm\nExercise 7 – 5: The effects of k\n1.\na) Is halved\nb) Approx. 50 times bigger\n2.\na) 0,5W 3\nb) 0,93 × W\nExercise 7 – 6: End of chapter exercises\n2. a and d\n3.\na) Triangular prism\nb) Triangular pyramid\nc) Rhombic prism\n4.\na)\ni. 856 cm2\nii. Rectangular\nprism\niii. 960 cm3\nb) 600 cm2\n5.\n√\n5x2\n6.\na) 72 000 cm3\nb) H = 54 cm and\nh = 60,2 cm\nc) 12 732 cm2\n7. No\n8.\na) 10 cm × 10 cm ×\n10 cm\nb) 12,6 cm\n9.\na) Volume triples\nb) Surface area ×9\nc) Volume ×27\n8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "45" }, { "title": "Area of a polygon", "content": "", "chapter_id": "7.1" }, { "title": "Right prisms and cylinders", "content": "", "chapter_id": "7.2" }, { "title": "Right pyramids, right cones and spheres", "content": "", "chapter_id": "7.3" }, { "title": "Multiplying a dimension by a constant factor", "content": "", "chapter_id": "7.4" }, { "title": "Summary", "content": "", "chapter_id": "7.5" }, { "title": "Euclidean geometry", "content": "8\nEuclidean geometry\nExercise 8 – 1: Perpendicular line from center bisects chord\n1. x =\n√\n41\n2. x =\n√\n84\n3. x = 10 units\n4. TU = 2,66 units\n5. x = 3,6 units\nExercise 8 – 2: Angle at the centre of circle is twice angle at circumference\n1. b = 90◦\n2. c = 22,5◦\n3. d = 200◦\n4. e = 55◦\n5. f = 120◦\n508\n12.1.\nIntroduction\n\nExercise 8 – 3: Subtended angles in the same segment\n1.\na) a = 21◦\nb) c = 24◦d = 78◦\nc) d = 28◦\n2.\na) e = 85◦\n3. f = 35◦\nExercise 8 – 4: Cyclic quadrilaterals\n1.\na) a = 93◦, b = 74◦\nb) a = 114◦\nc) a = 29◦\nExercise 8 – 5: Tangents to a circle\n1. d = 9,4 cm\n2. e = 2,5 cm\n3. f = 3 cm\nExercise 8 – 6: Tangent-chord theorem\n1.\na) a = 33◦, b = 33◦\nb) c = 72◦, d = 54◦\nc) f = 38◦, g = 47◦\nd) l = 48◦\ne) i = 40◦, j = 101◦,\nk = 40◦\nf) m = 56◦, n = 34◦,\no = 56◦\ng) p = 38◦, q = 52◦,\nr = 90◦\n2. a = 26◦, b = 64◦, c = 128◦\nExercise 8 – 7: End of chapter exercises\n1.\na)\nˆ\nA = x\nb) C ˆOD = 2x\nc)\nˆD = 90◦−x\n2.\na)\nˆD1 = 78◦\nb)\nˆ\nM1 = 39◦\nc)\nˆF2 = 51◦\nd)\nˆG = 58◦\ne)\nˆE1 = 32◦\n3.\na)\nˆD2 = x\nb) O ˆ\nAB = x\nc) O ˆBA = x\nd) A ˆOB = 180◦−2x\ne)\nˆC = 90◦−x\n4.\na) OM = 3 cm\nb) AM = 8 cm\nc) AB = 4\n√\n5cm\n5. x = 35◦\n6.\na) R ˆQS, Q ˆSO\nb) P ˆOS = 2x\n7.\na) O ˆDC = 35◦\nb) C ˆOD = 110◦\nc) C ˆBD = 55◦\nd) B ˆ\nAD = 90◦\ne) A ˆDB = 45◦\n8. x = 4y\n10. OQ = 17 mm\n11.\na) Q ˆRP, Q ˆSP, R ˆST\nb) S ˆRT = 80◦, S ˆTR =\n30◦, P ˆQS = 30◦\nd) P ˆ\nMQ = 110◦\n13.\na) 90◦−x\n2\nb)\nx\n2\nc) 90◦−x\n2\n14.\nc) 90◦−2x\nd) AO = 13 cm\n509\nChapter 12.\nLinear programming\n\n9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "51" }, { "title": "Revision", "content": "", "chapter_id": "8.1" }, { "title": "Circle geometry", "content": "", "chapter_id": "8.2" }, { "title": "Summary", "content": "", "chapter_id": "8.3" }, { "title": "Finance, growth and decay", "content": "9\nFinance, growth and decay\nExercise 9 – 1: Revision\n1. R 13 630\n2. R 10 246,59\n3.\na) R 2536\nb) R 2468,27\n4. 9,38%\n5. 4,56%\nExercise 9 – 2: Simple decay\n1. R 112 000\n2. R 941,18\n3. 11,66%\n4. 7 years\nExercise 9 – 3: Compound depreciation\n1. R 23 766,73\n2. 2229 cormorants\n3. R 131 072\n4. 132 221\n5. 7,62 kg\n6. R 85 997,13\nExercise 9 – 4: Finding i\n1. 14,9%\n2. 16,4%\n3. 12,0%\n4. 9,4%\nExercise 9 – 5: Timelines\n1. R 38 588,25\n2. R 35 308,00\n3. R 2600\n4. R 19 950,62\n5. R 1 149 283,50\n6. R 7359,83\nExercise 9 – 6: Nominal and effect interest rates\n1.\na) 12,6%\nb) 15,5%\nc) 22,1%\n2.\na) 16,8%; 17,7%;\n17,5%\nb) 17,7%\nc) 16,8%\n3. 9,1%\n4. 9,4%\n5.\na) 9,42% is the better\nrate.\nb) 9,38%\nc) 9,52%\n510\n12.1.\nIntroduction\n\nExercise 9 – 7: End of chapter exercises\n1.\na) R 246 400\nb) R 265 599,87\n2. R 229,92\n3.\na) R 8800\nb) 18,1%\n4. R 238 191,17\n5.\na) 7,44%\nb) R 148 826,15\nc) R 135 968,69\n6.\na) R 8042,19\nb) 26,82%\n8.\na) 11,3%\nb) 11,1%\nc) 11,0%\n9. R 322 580,65\n10.\na) 4,7%\nb) 4,8%\n11. R 212 347,69\n10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "55" }, { "title": "Revision", "content": "", "chapter_id": "9.1" }, { "title": "Simple and compound depreciation", "content": "", "chapter_id": "9.2" }, { "title": "Timelines", "content": "", "chapter_id": "9.3" }, { "title": "Nominal and effective interest rates", "content": "", "chapter_id": "9.4" }, { "title": "Summary", "content": "", "chapter_id": "9.5" }, { "title": "Probability", "content": "10\nProbability\nExercise 10 – 1: Revision\n1.\ny\nr+b+y\n2.\n5\n12\n3.\n3\n8\n4. 3\n6. 5\n7.\na)\n1\n6\nb)\n8\n21\n8.\na)\n4\n9\nb)\n5\n9\nc)\n1\n9\nd)\n8\n9\ne)\n5\n9\nf)\n4\n9\n9.\na)\n24\n116 ≈0,21\nb)\n65\n116 ≈0,56\nc) 0\nd)\n21\n29 ≈0,72\n10. 0,28\nExercise 10 – 2: Venn diagram revision\n1. 0,5\n2. 0,7\n3. 0,18\nExercise 10 – 3: Dependent and independent events\n1.\na) not mutually\nexclusive\nb) dependent\n2.\na)\n17\n30\nb)\n11\n30\nc) Yes\nd) No\n3. independent and not\nmutually exclusive\n511\nChapter 12.\nLinear programming\n\nExercise 10 – 4: Venn diagrams\n1.\na)\n9\n40\nb)\n3\n5\nc)\n1\n10\nd) dependent\n2. P (Z and not (X or Y )) =\n4\n25\n3.\na)\n2\nM\n8\n14\nH\n16\n4\nG\n29\n6\nb) 6\nc) 29\nd) 2\n4.\nA\nB\nC\nExercise 10 – 5: Tree diagrams\n1.\n7\n36\n2.\n671\n1296\n3.\na)\n1\n4\nb)\n5\n16\n4.\na)\n1\n4\nb)\n5\n16\nExercise 10 – 6: Contingency tables\n1.\na)\n5\n8\nb)\n8\n23\nc)\n12\n23\nd) dependent\n2. The events are whether a bus leaves from\nLocation A or not and whether a bus left late or\nnot.\nThe events are dependent.\n3.\nA\nnot A\nTotals\nB\n14\n6\n20\nnot B\n21\n9\n30\nTotals\n35\n15\n50\n512\n12.1.\nIntroduction\n\nExercise 10 – 7: End of chapter exercises\n1. 0,68\n2. 0,24\n3.\na) The experiment is the outcome of\nselecting a particular model and colour\ncar from the sample space of available\ncars.\nThe sample space is {pink model A;\nlime-green model A; purple model A;\npurple model B; orange model B;\nmulticolour model B}.\nb) 1\nc) 0\n4. 0,57\n5.\na) 0,38\nb) 0,7\nc) 0,69\nd) 0,69\ne) 0,3\n6.\na)\n3\n5\nb)\n1\n2\nc)\n1\n10\nd) 0\ne)\n2\n5\nf)\n9\n10\n7.\na)\nS\nA\n11\n8\nB\n9\n11\nC\n8\n14\n7\n12\nb) 12\nc) 8\nd)\n17\n20\n8.\na) h = 0,3\nb) h = 0,5\n9.\na) 0,434\nb) 0,182\nc) 0,47\n10.\n8\n15\n{O; O; O}\n7\n15\n{O; O; B}\n9\n15\n{O; B; O}\n6\n15\n{O; B; B}\n9\n15\n{B; O; O}\n6\n15\n{B; O; B}\n10\n15\n{B; B; O}\n5\n15\n{B; B; B}\n9\n16\n{O; O}\n7\n16\n{O; B}\n10\n16\n{B; O}\n6\n16\n{B; B}\n10\n17\n{O}\n7\n17\n{B}\nP(two orange balls) =\n63\n136\n11.\nDurban\nBloemfontein\nTotals\nLiked living there\n130\n30\n160\nDid not like living there\n140\n200\n340\nTotals\n270\n230\n500\nThe events are dependent.\n12.\nMultivitamin A\nMultivitamin B\nTotals\nImprovement in health\n400\n300\n700\nNo improvement in health\n140\n120\n260\nTotals\n540\n420\n960\nMultivitamin A is more effective.\n11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "10" }, { "title": "Revision", "content": "", "chapter_id": "10.1" }, { "title": "Dependent and independent events", "content": "", "chapter_id": "10.2" }, { "title": "More Venn diagrams", "content": "", "chapter_id": "10.3" }, { "title": "Tree diagrams", "content": "", "chapter_id": "10.4" }, { "title": "Contingency tables", "content": "", "chapter_id": "10.5" }, { "title": "Summary", "content": "", "chapter_id": "10.6" }, { "title": "Statistics", "content": "11\nStatistics\nExercise 11 – 1: Revision\n1.\na) mean = −4,3; first quartile = −6,2;\nsecond quartile = −3,4; third quartile =\n−2,9.\nb) mean = −5,6; first quartile = −60;\nsecond quartile = −6; third quartile =\n65.\nc) mean = 18,5; first quartile = 7; second\nquartile = 11,5; third quartile = 33.\n2. range = 9,6; inter-quartile range = 2,51.\n3.\n−4,2\n−1,8\n−0,2\n0,4\n2,9\n513\nChapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.\nIntroduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "11" }, { "title": "Revision", "content": "", "chapter_id": "11.1" }, { "title": "Histograms", "content": "", "chapter_id": "11.2" }, { "title": "Ogives", "content": "", "chapter_id": "11.3" }, { "title": "Variance and standard deviation", "content": "", "chapter_id": "11.4" }, { "title": "Symmetric and skewed data", "content": "", "chapter_id": "11.5" }, { "title": "Identification of outliers", "content": "", "chapter_id": "11.6" }, { "title": "Summary", "content": "", "chapter_id": "11.7" }, { "title": "Linear programming", "content": "Chapter 12.\nLinear programming\n\nExercise 11 – 2: Histograms\n1.\na) 170 million\nb) increasing\nc) 3 million\n2.\nd\ncount\n0\n4\n8\n12\n16\n20\n10\n20\n30\n40\n50\n3.\nyear\nprevalence\n17,5%\n17,6%\n17,7%\n17,8%\n17,9%\n18,0%\n18,1%\n18,2%\n•\n•\n•\n•\n•\n•\n•\n•\n2002\n2003\n2004\n2005\n2006\n2007\n2008\n2009\nExercise 11 – 3: Ogives\n1.\na) 20\nb) 15\nc) 60%\n2.\n0\n10\n20\n30\n40\n−25\n−15\n−5\n5\n15\n25\n3.\na)\nage (years)\n0\n5\n10\n15\n20\n25\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\nb) 11 people\nc) 19 people\nd) 34\ne) 25,5\nf)\nage (years)\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n8\n16\n24\n32\n40\n48\n56\n64\n72\n80\n4.\ne) 49,25\nf) 49,7\nExercise 11 – 4: Variance and standard deviation\n1.\na) Cape Town: 3,84. Durban: 3,82.\nb) Cape Town: 0,121. Durban: 0,184.\nc) Cape Town\n2. Mean = 270,7. Variance = 27 435,2.\n3. Mean = −1,95. Variance = 127,5.\n4.\na) 10,4\nb) 0,27\nc) 3\n5. 13 and 20\nExercise 11 – 5: Symmetric and skewed data\n1. skewed left\n2.\na) skewed right\nb) skewed right\nc) skewed left\nd) symmetric\n514\n12.1.", "chapter_id": "12" }, { "title": "Introduction", "content": "Introduction\n\nExercise 11 – 6: Outliers\n1.\na)\n21,1\n33,8\n39,4\n29,1\n30,0\nThere are two outliers on the left.\nb)\n119\n144,5\n157\n172,5\n198\nThere are no outliers.\nc)\n−11,7\n−1,05\n1,7\n6,05\n10\nThere is one outlier on the left.\n2.\n60\n90\n120\n150\n180\nstudy time (minutes)\n40\n50\n60\n70\n80\n90\n100\nmark (percentage)\nExercise 11 – 7: End of chapter exercises\n1.\n0\n1\n2\n3\n4\n5\n6\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n1\n2\n3\n4\n5\n6\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n0\n5\n10\n15\n20\n•\n•\n•\n•\n•\n•\n•\n•\n•\n0\n1\n2\n3\n4\n5\n6\n7\n8\n2.\n−4,6\n−4,1\n−3,3\n−1,5\n−0,8\nSkewed right\n3.\na) Mean = 11 3\n8 . Standard deviation = 6,69.\nb) Mean = 5 3\n8 . Standard deviation = 2,33.\nc) There are no outliers.\n4.\na) Mean = R 12 497,50. Standard deviation\n= R 1768,55.\nb) 4\nc) Mean = R 12 997,50. Standard deviation\n= R 1768,55.\nd) Mean = R 13 747,25. Standard deviation\n= R 1945,41.\ne) 4\nf) 10%\n5.\na) 14\nb) approximately 88 kg.\nc) 75\n6.\na) A. Mean = 23,83. Five number summary\n= [ 12 ; 16,5 ; 24 ; 26,5 ; 47 ].\nB. Mean = 31,17. Five number summary\n= [ 16 ; 25 ; 32 ; 36,5 ; 47 ].\nC. Mean = 41,83. Five number summary\n= [ 16 ; 42 ; 43 ; 45,5 ; 50 ].\nb)\n10\n15\n20\n25\n30\n35\n40\n45\n50\nSet A\nSet B\nSet C\nc) Set A: skewed left. Set B: slightly skewed\nleft. Set C: skewed right.\n515\nChapter 12.\nLinear programming\n\n12\nLinear programming\nExercise 12 – 2: Optimisation\n1.\na) x ≤10; y ≤10; x ≥4; 2x + 3y ≤30\nb) T = 5x + 10y\n2.\na) x + y ≤10 000; x ≥4000; y ≥2000;\ny ≤4000\nc) I = 50x + 30y\nd) R 460 000\n3.\na) 150x + 60y ≥30 000;\n50x + 40y ≥13 000;\n(10)x + (20)y ≥5000\nd) E = (20 000)x + (10 000)y\ne) 140 Super X and 150 Super Y\nf) R 4 300 000\n4.\na) x ≤300; y ≥0.5x; x + y ≤500\nb) P = (3)x + (2)y\nc) 300 hamburgers and 200 chicken burgers\n5.\na) x ≤150; y ≤120; x + y ≤200; x ≥\n40; y ≥10\nb)\n50\n100\n150\n200\n250\n50\n100\n150\n200\n250\nx\ny\nA\nB\nC\nD\nE\nc) P = (5)x + (10)y\nd) 80 of card X and 120 of card Y\n6.\na) 4x + 3y ≥15; 16x + 24y ≥72;\nx + y ≤5\nb) 3 packets of Vuka and 1 packets of Molo\nc) 0; 5 or 5; 0\n516\n12.1.\nIntroduction", "chapter_id": "12.1" }, { "title": "Solutions to exercises", "content": "", "chapter_id": "78" } ] }