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3A Sadler Solutions

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Solutions to Unit 3A Specialist Mathematics by A.J. Sadler Prepared by: Glen Prideaux 2009
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Solutions toUnit 3A Specialist Mathematicsby A.J. SadlerPrepared by:Glen Prideaux2009iiPrefaceThe answers in the Sadler text book sometimes are not enough. For those times when your really need to seea fully worked solution, look here.It is essential that you use this sparingly!You should not look here until you have given your best eort to a problem. Understand the problem here,then go away and do it on your own.ContentsChapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Exercise 1A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Exercise 1B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Miscellaneous Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Exercise 2A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Exercise 2B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Exercise 2C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Exercise 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Exercise 2E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Miscellaneous Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Exercise 3A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Exercise 3B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Exercise 3C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31Exercise 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Miscellaneous Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Exercise 4A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Exercise 4B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Exercise 4C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48Miscellaneous Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Exercise 5A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Exercise 5B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Exercise 5C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Miscellaneous Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Exercise 6A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Exercise 6B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Exercise 6C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Exercise 6D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Exercise 6E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Miscellaneous Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76Exercise 7A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76Exercise 7B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76Miscellaneous Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84Exercise 8A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84Exercise 8B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84Exercise 8C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86Exercise 8D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91Miscellaneous Exercise 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96Exercise 9A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96Miscellaneous Exercise 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97iiiivCHAPTER 1Chapter 1Exercise 1A1.-2 -1 0 1 2 3 4 5 6 7 8 9 102 2x = 3 or x = 72.-2 -1 0 1 2 3 4 5 6 7 8 9 101 1x = 5 or x = 73.-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 22 2x = 6 or x = 24.-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 35 5x = 8 or x = 25.-2 -1 0 1 2 3 4 5 6 7 8 9 103 3x = 46.-3 -2 -1 0 1 2 3 4 5 6 7 8 95 5x = 37. x + 3 = 7x = 4or x + 3 = 7x = 108. x 3 = 5x = 8or x 3 = 5x = 29. No solution (absolute value can never be nega-tive).10. x 2 = 11x = 13or x 2 = 11x = 911. 2x + 3 = 72x = 4x = 2or 2x + 3 = 72x = 10x = 512. 5x 8 = 75x = 15x = 3or 5x 8 = 75x = 1x = 1513. Find the appropriate intersection and read thex-coordinate.(a) Intersections at (3,4) and (7,4) so x = 3 orx = 7.(b) Intersections at (-2,4) and (6,4) so x = 2or x = 6.(c) Intersections at (4,2) and (8,6) so x = 4 orx = 8.14.xy-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8-2-112345678910y = 5y = [x[y = 3 0.5xy = [2x + 3[(a) Intersections at (-4,5) and (1,5) so x = 4or x = 1.(b) Intersections at (-6,6) and (2,2) so x = 6or x = 2.(c) Intersections at (-4,5) and (0,3) so x = 4or x = 0.(d) Intersections at (-3,3) and (-1,1) so x = 3or x = 1.15.-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 21 1x = 7 or x = 516. No solution (absolute value can never be nega-tive).17.-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 22.5 2.5x = 5.518.-12 -10 -8 -6 -4 -2 0 2 4 68 8x = 319.0 1 2 3 4 5 6 7 8 9 10 11 122 2x = 820.-10 -8 -6 -4 -2 0 2 4 6 8 10 12 148 8x = 221.-6 -5 -4 -3 -2 -1 01.5 1.5x = 3.51Exercise 1A22.-1 0 1 2 3 4 5 6 7 8 9 10 115 5x = 523. x + 5 = 2x 14x = 19[19 + 5[ = [2 19 14[[24[ = [24[ orx + 5 = (2x 14)x + 5 = 2x + 143x = 9x = 3[3 + 5[ = [2 3 14[[8[ = [ 8[ 24. 3x 1 = x + 92x = 10x = 5[3 5 1[ = [5 + 9[[14[ = [14[ or(3x 1) = x + 93x + 1 = x + 94x = 8x = 2[3 2 1[ = [ 2 + 9[[ 7[ = [7[ 25. 4x 3 = 3x 11x = 8[4 8 3[ = [3 8 11[[ 35[ = [ 35[ or4x 3 = (3x 11)4x 3 = 3x + 117x = 14x = 2[4 2 3[ = [3 2 11[[5[ = [ 5[ 26. 5x 11 = 5 3x8x = 16x = 2[5 2 11[ = [5 3 2[[ 1[ = [ 1[ or(5x 11) = 5 3x5x + 11 = 5 3x6 = 2xx = 3[5 3 11[ = [5 3 3[[4[ = [ 4[ 27. x 2 = 2x 6x = 4x = 4[4 2[ = 2 4 6[2[ = 2 or(x 2) = 2x 6x + 2 = 2x 63x = 8x = 83[83 2[ = 2 83 6[23[ , = 23The second solution is not valid. The only so-lution is x = 4.28. x 3 = 2xx = 3[ 3 3[ = 2 3[ 6[ , = 6or(x 3) = 2xx + 3 = 2x3x = 3x = 1[1 3[ = 2 1[ 2[ = 2 The rst solution is not valid. The only solutionis x = 1.29. x 2 = 0.5x + 10.5x = 3x = 6[6[ 2 = 0.5 6 + 14 = 4 or2Exercise 1Ax 2 = 0.5x + 11.5x = 3x = 2[ 2[ 2 = 0.5 2 + 10 = 0 .30. x + 2 = 3x + 64x = 4x = 1[1 + 2[ = 3 1 + 6[3[ = 3 or(x + 2) = 3x + 6x 2 = 3x + 62x = 8x = 4[4 + 2[ = 3 4 + 6[6[ , = 3 6The second solution is invalid. The only solutionis x = 1.31. x 1:x + 5 +x 1 = 72x + 4 = 72x = 3x = 1.5 5 x 1:x + 5 (x 1) = 7x + 5 x + 1 = 76 ,= 7 = no solnx 5:(x + 5) (x 1) = 7x 5 x + 1 = 72x 4 = 72x = 11x = 5.5 32. x 4:x + 3 +x 4 = 22x 1 = 22x = 3x = 1.5= no soln (out of domain)3 x 4:x + 3 (x 4) = 2x + 3 x + 4 = 27 ,= 2 = no solnx 3:(x + 3) (x 4) = 2x 3 x + 4 = 22x + 1 = 22x = 1x = 0.5= no soln (out of domain)The equation has no solution.33. x 3:x + 5 +x 3 = 82x + 2 = 82x = 6x = 3 5 x 3:x + 5 (x 3) = 8x + 5 x + 3 = 88 = 8= all of 5 x 3 is a solution.x 5:(x + 5) (x 3) = 8x 5 x + 3 = 82x 2 = 82x = 10x = 5Solution is 5 x 3.34. x 8:x 8 = (2 x) 6x 8 = 2 +x 68 = 8= all of x 8 is a solution.2 x 8:(x 8) = (2 x) 6x + 8 = 2 +x 62x = 16x = 8x 2:(x 8) = 2 x 6x + 8 = x 48 ,= 4 = no solnSolution is x 8.3Exercise 1BExercise 1B1.-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 62 22 < x < 22.-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 65 55 x 53.-10 -8 -6 -4 -2 0 2 4 6 8 107 7x < 7 or x > 74.-5 -4 -3 -2 -1 0 1 2 3 4 5 6 73 3x < 1 or x > 55.-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 44 47 x 16.xy-5 -4 -3 -2 -1 1 2 3 4 5-1123456789 y = [5x 3[y = 7Algebraically:For 5x 3 0: For 5x 3 0:5x 3 < 75x < 10x < 5(5x 3) < 75x 3 > 75x > 4x > 4545 < x < 27.xy-4 -3 -2 -1 1 2 3 4 5 6-1123456789 y = [2x 3[y = 5Algebraically:For 2x 3 0: For 2x 3 0:2x 3 > 52x > 8x > 4(2x 3) > 52x 3 < 52x < 2x < 1x < 1 or x > 48.xy-6 -4 -2 2 4 6 8 10-1135791113 y = [5 2x[y = 11Algebraically:For 5 2x 0: For 5 2x 0:5 2x 112x 6x 3(5 2x) 115 + 2x 112x 16x 83 x 89. Centred on 0, no more than 3 units from centre:[x[ 310. Centred on 0, less than 4 units from centre:[x[ < 411. Centred on 0, at least 1 unit from centre: [x[ 112. Centred on 0, more than 2 units from centre:[x[ > 213. Centred on 0, no more than 4 units from centre:[x[ 414. Centred on 0, at least 3 units from centre: [x[ 34Exercise 1B15. Distance from 3 is greater than distance from 7.Distance is equal at x = 5 so possible values arex R : x > 5.16. Distance from 1 is less than or equal to distancefrom 8. Distance is equal at x = 4.5 so possiblevalues are x R : x 4.5.17. Distance from 2 is less than distance from 2.Distance is equal at x = 0 so possible values arex R : x < 0.18. Distance from 5 is greater than or equal to dis-tance from 1. Distance is equal at x = 2 sopossible values are x R : x 2.19. Distance from 13 is greater than distance from5. (Note [5 x[ = [x 5[.) Distance is equal atx = 9 so possible values are x R : x < 9.20. Distance from 12 is greater than or equal todistance from 2. Distance is equal at x = 5 sopossible values are x R : x 5.21. Centred on 2, no more than 3 units from centre:[x 2[ 322. Centred on 3, less than 1 unit from centre:[x 3[ < 123. Centred on 2, at least 2 units from centre:[x 2[ 224. Centred on 1, more than 2 units from centre:[x 1[ > 225. Centred on 1, no more than 4 units from centre:[x 1[ 426. Centred on 1, at least 4 units from centre:[x 1[ 427.-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 85 5x 5 or x 528. For 2x > 0: For 2x < 0:2x < 8x < 42x < 82x > 8x > 44 < x < 429. [x[ > 3 is true for all x (since the absolute valueis always positive).30. Distance from 3 is greater than or equal to dis-tance from 5. Distance is equal at 1 sox 1.31.xy-5 -4 -3 -2 -1 1 2 3-1123456y = [x + 1[y = [2x + 5[Algebraically:First solve [x + 1[ = [2x + 5[x + 1 = 2x + 5x = 4or (x + 1) = 2x + 5x 1 = 2x + 56 = 3xx = 2Now consider the three intervals delimited bythese two solutions. x < 4Try a value, say -5:Is it true that [ 5 + 1[ [2(5) + 5[ ?Yes (4 5). 4 < x < 2Try a value, say -3:Is it true that [ 3 + 1[ [2(3) + 5[ ?No (2 1). x > 2Try a value, say 0:Is it true that [0 + 1[ [2(0) + 5[ ?Yes (1 5).Solution set isx R : x 4 x R : x 232. No solution (absolute value can not be negative.)33. First solve [5x + 1[ = [3x + 9[5x + 1 = 3x + 92x = 8x = 4or (5x + 1) = 3x + 95x 1 = 3x + 910 = 8xx = 1.25Now consider the three intervals delimited bythese two solutions. x < 1.25Try a value, say -2:Is it true that [5(2) + 1[ > [3(2) + 9[ ?Yes (9 > 3).5Exercise 1B 1.25 < x < 4Try a value, say 0:Is it true that [5(0) + 1[ > [3(0) + 9[ ?No (1 9). x > 4Try a value, say 5:Is it true that [5(5) + 1[ > [3(5) + 9[ ?Yes (26 > 24).Solution set isx R : x < 1.25 x R : x > 434. First solve [2x + 5[ = [3x 1[2x + 5 = 3x 1x = 6or (2x + 5) = 3x 12x 5 = 3x 14 = 5xx = 0.8Now consider the three intervals delimited bythese two solutions. x < 0.8Try a value, say -2:Is it true that [2(2) + 5[ [3(2) 1[ ?No (1 7). 0.8 < x < 6Try a value, say 0:Is it true that [2(0) + 5[ [3(0) 1[ ?Yes (5 1). x > 6Try a value, say 7:Is it true that [2(7) + 5[ [3(7) 1[ ?No (19 20).Solution set isx R : 0.8 x 6Actually we only need to test one of the threeintervals. At each of the two initial solutions wehave lines crossing so if the LHSRHSon the other side, and vice versa. Well use thisin the next questions.35. First solve [6x + 1[ = [2x + 5[6x + 1 = 2x + 54x = 4x = 1or (6x + 1) = 2x + 56x 1 = 2x + 58x = 6x = 0.75Now test one of the three intervals delimited bythese two solutions. x < 0.75Try a value, say -1:Is it true that [6(1) + 1[ [2(1) + 5[ ?No (5 3).Solution set isx R : 0.75 x 136. First solve [3x + 7[ = [2x 4[3x + 7 = 2x 4x = 11or (3x + 7) = 2x 43x 7 = 2x 45x = 3x = 0.6Now test one of the three intervals delimited bythese two solutions. x < 11Try a value, say -12:Is it true that [3(12) +7[ > [2(12) 4[ ?Yes (29 > 28).Solution set isx R : x < 11 x R : x > 0.637. This is true for all x R since the absolute valueis never negative, and hence always greater than-5.38. First solve [x 1[ = [2x + 7[x 1 = 2x + 7x = 8or (x 1) = 2x + 7x + 1 = 2x + 73x = 6x = 2Now test one of the three intervals delimited bythese two solutions. x < 8Try a value, say -10:Is it true that [ 10 1[ [2(10) + 7[ ?Yes (11 13).Solution set isx R : x < 8 x R : x > 239. Distance from 11 is greater than or equal to dis-tance from 5. 3 is equidistant, so x 340. First solve [3x + 7[ = [7 2x[3x + 7 = 7 2x5x = 0x = 0or (3x + 7) = 7 2x3x 7 = 7 2xx = 14x = 14Now test one of the three intervals delimited bythese two solutions. x < 14Try a value, say -20:Is it true that [3(20) +7[ > [7 2(20)[ ?Yes (53 > 47).6Exercise 1BSolution set isx R : x < 14 x R : x > 041. No solution (LHS=RHS x R)42. True for all x (LHS=RHS x R)43. We can rewrite this as 3[x + 1[ [x + 1[ whichcan only be true at x + 1 = 0, i.e. x = 1.44. We can rewrite this as 2[x 3[ < 5[x 3[ whichsimplies to 2 < 5 for all x ,= 3, so the solutionset isx R : x ,= 345. First solve x = [2x 6[x = 2x 6x = 6or x = (2x 6)x = 2x + 63x = 6x = 2Now test one of the three intervals delimited bythese two solutions. x < 2Try a value, say 0:Is it true that 0 > [2(0) 6[ ?No (0 6).Solution set isx R : 2 < x < 646. First solve [x 3[ = 2xx 3 = 2xx = 3or (x 3) = 2xx + 3 = 2x3x = 3x = 1The rst of these is not really a solution, becauseit was found based on the premise of x 3 beingpositive which is not true for x = 3. As a re-sult we really have only one solution. (Graph iton your calculator if youre not sure of this.)Now test one of the two intervals delimited bythis solution. x < 1Try a value, say 0:Is it true that [0 3[ 2(0) ?No (3 0).Solution set isx R : x 147. First solve 2x 2 = [x[2x 2 = xx = 2or 2x 2 = x3x = 2x = 23The second of these is not really a solution, be-cause it was found based on the premise of xbeing negative which is not true for x = 23. As aresult we really have only one solution.Now test one of the two intervals delimited bythis solution. x < 2Try a value, say 0:Is it true that 2(0) 2 < [0[) ?Yes (2 < 0).Solution set isx R : x < 248. First solve [x[ + 1 = 2x. If you sketch the graphof LHS and RHS it should be clear that this willhave one solution with positive x:x + 1 = 2xx = 1The LHS is clearly greater than the RHS for neg-ative x so we can conclude that the solution setisx R : x 149. Apart from having a > instead of this problemcan be rearranged to be identical to the previousone, so it will have a corresponding solution set:x R : x < 150. First solve [x + 4[ = x + 2x + 4 = x + 2No Solutionor (x + 4) = x + 2x 4 = x + 22x = 6x = 3The second of these is not really a solution, be-cause it was found based on the premise of x +4being negative which is not true for x = 3. Asa result we have no solution. Graphically, thegraphs of the LHS and RHS never intersect, sothe inequality is either always true or never true.Test a value to determine which:Try a value, say 0:Is it true that [(0) + 4[ > 0 + 2 ?Yes (4 > 2).Solution set is R.51. *must be > because we are including all valuesof x greater than some distance from the centralpoint.7Miscellaneous Exercise 1At the value x = 3 we must have[2x + 5[ = a[2 3 + 5[ = aa = 11Then at x = b(2b + 5) = 112b 5 = 112b = 16b = 852. Since 3 is a member of the solution set, result-ing in the LHS being zero, the smallest possibleabsolute value, the inequality must be either .[2(3) + 5[ [(3) + 1[1 2and we see that * is [ax[b[x[ > a[x[Because [x[ is positive we can divide both sides by[x[ without being concerned about the inequalitychanging direction. This is, of course, only validfor x ,= 0.b > awhich is true for all x so we can conclude thatthe original inequality is true for all x ,= 0.9CHAPTER 2Chapter 2Exercise 2A1. (a) 035 (read directly from the diagram)(b) 35 + 45 = 080(c) 35 + 45 + 30 = 110(d) 180 35 = 145(e) 180 + 20 = 200(f) 360 60 = 300(g) Back bearings:35 + 180 = 215(h) 80 + 180 = 260(i) 110 + 180 = 290(j) 145 + 180 = 325(k) 200 180 = 020(l) 300 180 = 1202. No working required. Refer to the answers inSadler.3. tan28 = h22.4h = 22.4 tan28= 11.9m4. tan = 24.1 = tan1 24.1= 262.0m4.1m5. tan24 = h22.5h = 22.5 tan24= 10.0m24h22.5m6. After one and a half hours, the rst ship has trav-elled 6km and the second 7.5km.d2= 62+ 7.52d = 9.6km 10kmNorth110d6km7.5km7.351840mA CBdBdAdABtan18 = 40dAdA = 40tan18tan35 = 40dBdB = 40tan35dAB = 40tan18 40tan35= 66m8.AB12.2kmCN302N239N212dCAB = 302 239= 63CBA = 212 (302 180)= 90cos 63 = 12.2dd = 12.2cos 63= 26.9km10Exercise 2A9. 28 1742mA CBdA dCdACtan28 = 42dAdA = 42tan28tan17 = 42dCdC = 42tan17dAC = 42tan28 + 42tan17= 216m10.15 4036mA CBdAdBdABtan15 = 36dAdA = 36tan15tan40 = 36dBdB = 36tan40dAB = 36tan15 36tan40= 91m11.203040mA BE CDDistance between towers:tan30 = 40ABAB = 40tan30= 403Additional height of second tower:tan20 = DE403DE = 403 tan20= 25.22mTotal height of second tower:DB = 25.22 + 40 65.2m12.193920513935mhABCFirst determine the angles in the triangle madeby the tree, the hillslope and the suns ray.ACB = 39 20= 19CAB = 90 39= 51Now use the sine rule:hsin19 = 35sin51h = 35 sin19sin51= 14.7m13.5.3kmN33565A BFdABF = 335 270= 65tan65 = d5.3d = 5.3 tan65= 11.4km11Exercise 2A14.500m4030Ship1 Ship2Planed1d2 dtan30 = 500d1d1 = 500tan30= 866mtan40 = 500d2d2 = 500tan40= 596md = d1 d2= 270m15.C DAB10 60m40mAC =_602+ 402= 2013tanDAC = 6040DAC = 56.3BAC = 180 56.3= 123.7ABC = 180 123.7 10= 46.3ABsin10 = 2013sin46.3AB = 2013 sin10sin46.3= 17.3m16. sin17 = 540dd = 540sin17= 1847cmA B540cm17dRounded up to the next metre this is 19m.17. tan = 1.619.6 = 4.7DCE = 40 4.7= 35.3tan35.3 = DE19.6DE = 19.6 tan35.3= 13.9mh = 13.9 + 1.6= 15.5mA B 19.6m1.6m40 hC DE18. Let the height of the agpole be h and the dis-tance from the base be d. Let be the angle ofelevation of the point 34 of the way up the ag-pole.tan40 = hdtan = 0.75hd= 0.75hd= 0.75 tan40 = tan1(0.75 tan40)= 3219. x y1 y2ytan = xy1y1 = xtantan = xy2y2 = xtany = y1 +y2= xtan + xtan= x_ 1tan + 1tan_= x_ tantan tan + tantan tan_= x_tan + tantan tan_

12Exercise 2A20. x y y2y1tan = xy1y1 = xtantan = xy2y2 = xtany = y1 y2= xtan xtan= x_ 1tan 1tan_= x_ tantan tan tantan tan_= x_tan tantan tan_

21.xzysin = xzz = xsintan = yzy = z tan=_ xsin_tan= xtansin

22.xzycos = zxz = xcos cos = yzy = z cos = (xcos ) cos = xcos cos

23. (a)x zysin = xzz = xsinsin = yzy = z sin=_ xsin_sin= xsinsin(b)xzytan = xzz = xtancos = yzy = z cos =_ xtan_cos = xcos tan13Exercise 2BExercise 2B1. (a) AM = 12AC= 12_AB2+ BC2= 12_5.22+ 5.22= 3.68cm(b) tanEAM = EMAMEAM = tan1 EMAM= tan1 6.33.68= 59.7(c) DEM = AEM= 180 90 EAM= 90 59.7= 30.3(d) Let F be the midpoint of AB. The angle be-tween the face EAB and the base ABCD isEFM.FM = 12AB= 2.6cmtanEFM = EMFMEFM = tan1 EMFM= tan1 6.32.6= 67.62. (a) tanGDC = GCDCGDC = tan1 GCDC= tan1 3562= 29.4(b) tanGBC = GCBCGBC = tan1 GCBC= tan1 3538= 42.6(c) tanGAC = GCACGAC = tan1 GCAC= tan1 35622+ 382= 25.7(d) AG =_AC2+ GC2=_622+ 382+ 352= 80.7mm(e) The angle between the plane FADG and thebase ABCD is equal to GDC = 29.4.(f) The angle between skew lines DB and HEis equal to ADB.tanADB = ABADADB = tan1 ABAD= tan1 6238= 58.53. The key to this problem and others like it isa clear diagram that captures the informationgiven.A BCDEF100mm30mm 120mm(a) BF2= BE2+ EF2= (AB2+ AE2) + EF2= 1002+ 302+ 1202BF = 25300= 159mm(b) sinFBD = DFBF= 30159FBD = sin1 30159= 10.94. (a) tan50 = 186ACAC = 186tan50= 156cm(b) tan24 = 186ABAB = 186tan24= 418cmBC =_AB2+ AC2=_1562+ 4182= 446cm14Exercise 2B(c) tanACB = ABACACB = tan1 418156= 69.55. (a) DCA = DBA (SAS) soDCA = DBA andDBA = 50(b) DBA = DAABAB = DAtanDBA= 7.4tan50= 6.2cm(c) There are a couple of ways this could bedone. Since we now know all three sides oftriangle ABC we could use the cosine ruleto nd ACB. Alternatively, since we havean isosceles triangle, we can divide it in halfto create a right triangle, like this:ACB4.6cm2.3cm6.2cmcos ACB = 2.36.2ACB = cos1 2.36.2= 686.A BCDE FG H I6cm(a) BI =_BF2+ FI2=_BF2+ FG2+ GI2=_62+ 62+ 32= 9(b) cos IBF = BFFI= 69= 23IBF = cos1 23= 48(c) The angle between IAB and the baseABCD is the same as the angle betweenrectangle ABGH and the base ABCD (sincethe triangle and the rectangle are coplanar).This is the same as GBC: 45.7.A BCDE FG80mm50mm30mm(a) tanEBC = CEBCEBC = tan1 3050= 31(b) tanEGC = CEGCEBC = tan1 30402+ 502= 25(c) tanEAC = CEACEBC = tan1 30802+ 502= 188.P QRST UV W2.3cm3020(a) tan30 = QUPQPQ = 2.3tan30= 3.98cmtan20 = VRQRQR = 2.3tan20= 6.32cmVolume = 2.3 3.98 6.32= 57.9cm315Exercise 2B(b) PV =_PQ2+ QR2+ RV2=_3.982+ 6.322+ 2.32= 7.8cm(c) tanUSW = UWSWUSW = tan13.982+ 6.3222.3= 739. The angle between plane VAB and plane ABCis equal to the angle between lines that are bothperpendicular to AB. Consider point D the mid-point of AB such that VD and VC are both per-pendicular to AB.40mm30mmABVCDsin45 = DCACDC = 40 sin45= 28.28mmtanVDC = VCDCVDC = tan1 3028.28= 4710.2030PQR20mPQ = 20tan20= 54.9mPR = 20tan30= 34.6mQR =_PQ2+ PR2= 65m11. 1710atdTownAireld750ma = 750tan17= 2453mt = 750tan30= 4253md =_a2+t2= 4910m 5km12. 2010120mabda = 120tan20= 330mb = 120tan30= 681md =_b2a2= 595mSpeed = 59510= 59.5m/min= 59.5 60m/hr= 3572m/hr= 3.6km/hr16Exercise 2C13.2340m70mABCDh(a) h2 = 40 tan23h = 33.96mABD = tan1 h40= 40(b) ACD = tan1 h70= 2614. BD = hAB = hsin28cos 35 = ABACAC = ABcos 35=hsin 28cos 35= hsin28cos 35sin = hAC= h1 sin28cos 35h= sin28cos 35 = sin1(sin28cos 35)= 23Exercise 2C1. (a) c2= a2+b22ab cos C10.22= x2+ 6.922 x 6.9 cos 50x = 4.29 or x = 13.16Reject the negative solution and round to1d.p.: x = 13.2cm.(b) sinAa = sinCcsinA = a sinCcA = sin1 a sinCc= sin1 6.9 sin5010.2= 31.2or A = 180 31.2= 148.8Reject the obtuse solution since it results inan internal angle sum greater than 180.B = 180 AC= 180 50 31.2= 98.8 xsin98.8 = 10.2sin50x = 10.2 sin98.8sin50= 13.2cm2. sinx11.2 = sin5012.1x = sin1 11.2 sin5012.1= 45No need to consider the obtuse solution since theopposite side is not the longest in the triangle (xmust be less than 50).3. x2= 6.82+ 14.322 6.8 14.3 cos 20x =_6.82+ 14.322 6.8 14.3 cos 20= 8.2cm17Exercise 2C4. 19.72= 9.82+ 14.322 9.8 14.3 cos xcos x = 9.82+ 14.3219.722 9.8 14.3x = cos1 9.82+ 14.3219.722 9.8 14.3= 1085. xsin(180 105 25) = 11.8sin105x = 11.8 sin50sin105= 9.4cm6. sinx7.2 = sin404.8x = sin1 7.2 sin404.8= 75 or x = 180 75= 1057. c2= a2+b22ab cos C11.82= x2+ 8.722 x 8.7 cos 80x = 9.6(rejecting the negative solution)8. The smallest angle is opposite the shortest side,so272= 333+ 5522 33 55 cos = cos1 332+ 5522722 33 55= 219. a2= b2+c22bc cos A9.12= 7.32+x22 7.3xcos 72x = 8.1 ABC727.3cm9.1cm x(rejecting the negative solution)AB=8.1cm10.ABC4312.4cm14.3cma2= b2+c22bc cos Aa =_12.42+ 14.322 12.4 14.3 cos 43a = 9.9cmsinC12.4 = sin43aC = sin1 12.4 sin439.9= 58(Cannot be obtuse because c is not the longestside.)B = 180 43 58= 7911.GHI5519.4cm 18.2cmsinI19.4 = sin5518.2I = sin1 19.4 sin5518.2= 61 or 119H = 180 55 61 or 180 55 119= 64 =6hsinH = gsinGh = g sinHsinG= 18.2 sin64sin55 or 18.2 sin6sin55= 20.0cm =2.3cm12.North30100LAB15.2km12.1kmALB = 100 30= 70AB =_15.22+ 12.122 15.2 12.1 cos 70= 15.9km13.North70North150130LPQ7.3kmPQL = 150 130= 20PLQ = 130 70= 60LPQ = 180 20 60= 10018Exercise 2CLQsinLPQ = LPsinPQLLQ = LPsinLPQsinPQL= 7.3 sin100sin20= 21.0km14.BTHh25m3052HBT = 90 30= 60BTH = 180 60 52= 68hsin52 = 25sin68h = 25 sin52sin68= 21m15. N30182012BCAT40mtan20 = 40ABAB = 40tan20= 109.9mtan12 = 40ACAC = 40tan12= 188.2mBAC = 30 + 18= 48BC2= AB2+ AC22AB ACcos BACBC =_109.92+ 188.222 109.9 188.2 cos 48= 141m16.A BCD20 3540mADB = 35 20= 15BDsin20 = 40sin15BD = 40 sin20sin15= 52.9msinDBC = DCBDDC = BDsinDBC= 52.9 sin35= 30m17. There are a couple of ways you could approachthis problem. You could use the cosine rule todetermine an angle, then use the formula Area=12ab sinC. Alternatively you could use Heronsformula:A =_s(s a)(s b)(s c)where s = a+b+c2 and determine the area withoutresort to trigonometry at all. Ill use trigonom-etry for the rst block, and Herons formula forthe second.First blockIll start by nding the largest angle: = cos1 252+ 4825322 25 48= 87.1Area = 12ab sin= 12 25 48 sin87.1= 599.2m2Second block:s = 33 + 38 + 452= 58Area =_58(58 33)(58 38)(58 45)= 614.0m2The second block is larger by 15m2.19Exercise 2C18.A BCT30171237mtan17 = 37ABAB = 37tan17= 121.0mtan12 = 37ACAC = 37tan12= 174.0mBC2= AB2+ AC22 AB ACcos 30BC =_121.02+ 174.022 121.0 174.0 cos 30= 92.0m19.A BCD7.2cm6.1cm8.2cm100(a) BCD = 180 100= 80(b) BD2= 6.12+ 7.222 6.1 7.2 cos 100= 104.3BD = 10.2cmsinADB7.2 = sin10010.2ADB = sin1 7.2 sin10010.2= 44.0sinCDB8.2 = sin8010.2CDB = sin1 8.2 sin10010.2= 52.3ADC = 44.0 + 52.3= 96(c) BD2= BC2+ CD22 BC CDcos 80104.3 = 8.22+ CD22 8.2 CDcos 80CD = 7.7cmP = 7.2 + 8.2 + 7.7 + 6.1= 29.2cm(d) AABD = 12 6.1 7.2 sin100= 21.6cm2ACBD = 12 8.2 7.7 sin80= 31.1cm2AABCD = 21.6 + 31.1= 52.7cm220.A BCD10cm14cm 12cm9cm xcm(a) x2= 102+ 1222 10 12 cos = 100 + 144 240 cos = 244 240 cos (b) x2= 142+ 922 14 9 cos = 196 + 81 252 cos = 277 252 cos (c) = 180 cos = cos(180 )= cos 244 240 cos = 277 252 cos = 277 + 252 cos 240 cos = 33 + 252 cos 492 cos = 33cos = 33492 = 9420Exercise 2DExercise 2DQuestions 115 are single step problems. Noworked solutions necessary.Note: My exact values are given with rational de-nominators. I write22 rather than 12. Your answersmay appear dierent without being wrong.16. 120 makes an angle of 60 with the x-axisand is in quadrant II (where sine is positive) sosin120 = sin60 =32 .17. 135 makes an angle of 45 with the x-axis andis in quadrant II (where cosine is negative) socos 135 = cos 45 = 22 .18. 150 makes an angle of 30 with the x-axis andis in quadrant II (where cosine is negative) socos 150 = cos 30 = 32 .19. 120 makes an angle of 60 with the x-axis andis in quadrant II (where cosine is negative) socos 120 = cos 60 = 12.20. 180 makes an angle of 0 with the x-axis and ison the negative x-axis (where cosine is negative)so cos 180 = cos 0 = 1.21. 135 makes an angle of 45 with the x-axis andis in quadrant II (where tangent is negative) sotan135 = tan45 = 1.22. 120 makes an angle of 60 with the x-axis andis in quadrant II (where tangent is negative) sotan120 = tan60 = 3.23. 150 makes an angle of 30 with the x-axis andis in quadrant II (where tangent is negative) sotan150 = tan30 =33 .24. 180 lies on the negative x-axis (where tangentis zero) so tan180 = 0.25. 180 lies on the negative x-axis (where sine iszero) so sin180 = 0.26. 150 makes an angle of 30 with the x-axisand is in quadrant II (where sine is positive) sosin150 = sin30 = 12.27. 135 makes an angle of 45 with the x-axisand is in quadrant II (where sine is positive) sosin135 = sin45 =22 .28. 20 = 4 5 = 45 = 2529. 45 = 9 5 = 95 = 3530. 32 = 16 2 = 162 = 4231. 72 = 36 2 = 362 = 6232. 50 = 25 2 = 252 = 5233. 200 = 100 2 = 1002 = 10234. 2 3 = 2 3 = 635. 5 3 = 5 3 = 1536. 5 5 = (5)2= 537. 15 3 = 15 3 = 9 5 = 95 = 3538. 8 6 = 8 6 = 16 3 = 163 = 4339. 32 42 = 1222 = 12(2)2= 12 2 = 2440. (52)(38) = 1528 = 152 8 = 1516 =15 4 = 6041. (63)(12) = 63 12 = 636 = 6 6 = 3642. (35)(72) = 215 2 = 211043. (52) (8) = 52 4 2 = 52 (22) =5 2 = 2.544. (53)2= 52(3)2= 25 3 = 7545. (32)2= 32(2)2= 9 2 = 1846. 12 = 12 22 =2247. 13 = 13 33 =3348. 15 = 15 55 =5549. 32 = 32 22 = 32250. 27 = 27 77 = 27751. 63 = 63 33 = 633 = 2352. 13+5 = 13+5 3535 = 3595 = 35453. 132 = 132 3+23+2 = 3+292 = 3+2754. 13+2 = 13+2 3232 = 3292 = 32755. 23+2 = 23+2 3232 = 2(32)32 = 23 2256. 332 = 332 3+23+2 = 3(3+2)32 = 33 +3257. 65+2 = 65+2 5252 = 6(52)52 =6(52)3 = 25 2221Exercise 2D58. sin60 = 9x32 = 9x3x = 18x = 183= 183 33= 1833= 6359. x2+ 32= 72x2+ 9 = 49x2= 40x = 40= 4 10= 4 10= 21060. Label the vertical in the diagram as y, thensin45 = y1022 = y10y = 52sin60 = xy32 = x52x = 521 32= 5322= 56261. Use the cosine rule:x2= 42+ (23)22 4 23 cos 150= 16 + 22(3)2163 (cos 30)= 16 + 4 3 163 _32_= 16 + 12 + 163 32= 28 + 8 3= 52x = 52= 4 13= 21362. Label the diagonal in the diagram as y, thenysin60 = 10sin45y = 10 sin60sin45= 10 32 12= 531 21= 532= 56tan30 = xyx = y tan30= 56 13= 5323= 5263.xyab3060 45cos 30 = xa32 = xa3a = 2xa = 2x3sin60 = ab32 = ab3b = 2a= 2 2x3= 4x3b = 4x3 13= 4x322Exercise 2Eysin = bsin45y = b sin sin45= b sin 12= b sin 21= 2b sin= 2 4x3 sin= 42xsin3

64.422 60y45 wxw2= 42+ (22)2= 16 + 4 2= 24w = 24= 4 6= 26xsin = wsin60x = wsinsin60= 26 sin32= 26 sin1 23= 432 sin3= 42 sinysin45 = xsiny = xsin45sin= x 12sin= 42 sin 12sin= 4 sinsin

Exercise 2E1. 43 19 = 24d = 24360 2 6350= 2660km2. 32 21 = 11d = 11360 2 6350= 1219km3. 39 (32) = 71d = 71360 2 6350= 7869km4. 51.5 5 = 46.5d = 46.5360 2 6350= 5154km5. 41 4 = 37d = 37360 2 6350= 4101km6. 134 114 = 20d = 20360 2 6350 cos 25= 2009km23Exercise 2E7. 119 77 = 42d = 42360 2 6350 cos 39= 3617km8. 105 75 = 30d = 30360 2 6350 cos 40= 2547km9. 122 117 = 5d = 5360 2 6350 cos 34= 459km10. 175 (73) = 248Longitude dierence is greater than 180 so it isshorter to go the other way and cross the dateline.360 248 = 112d = 112360 2 6350 cos 40= 9509km11. 360 = 5552 6350 = 5552 6350 360= 5latitude = 29 + 5= 34SAugusta: 34S, 115E12. 360 = 33002 6350 cos 34 = 33002 6350 cos 34 360= 36longitude = 115 + 36= 151ESydney: 34S, 151E13. 360 = 78702 6350 = 78702 6350 360= 71latitude = 71 36= 35SAdelaide: 35S, 138E14. 360 = 96002 6350 cos 35 = 96002 6350 cos 35 360= 106longitude = 135 + 106= 241E= 360 241= 119WBakerseld: 35N, 119W15. 360 = 8202 6350 cos 35 = 8202 6350 cos 35 360= 9longitude = 135 9= 126W360 = 20002 6350 = 20002 6350 360= 18latitude = 35 + 18= 53SNew position: 53S, 126WIf the ship rst heads south, the new latitude re-mains 53S.360 = 8202 6350 cos 53 = 8202 6350 cos 53 360= 12longitude = 135 12= 123WNew position: 53S, 123W16. First nd the length of the chord LS from LosAngeles to Shimoneski through the earth usingthe angle subtended at the middle of the latitudecircle:r = 6350 cos 34= 5264km = 360 (131 + 118)= 111sin 2 = 0.5LSr0.5LS = 5264 sin55.5LS = 2 5264 sin55.5= 8677kmNow consider the angle that same chord sub-tends at the centre of the earth (i.e. the centre ofthe great circle passing through the two points).Lets call this angle .sin 2 = 0.5LSR= 0.5 867763502 = 43 = 86Now use this angle to determine the arc lengthalong this great circle:d = 86360 2 6350= 9553km24Miscellaneous Exercise 2Miscellaneous Exercise 21. See the answer in Sadler.2. (a) tan20 = 15ACAC = 15tan20= 41.2m(b) tan30 = 15ABAB = 15tan30= 26.0m(c) BC2= AC2+ AB2BC =_41.22+ 26.02= 48.7m(d) tanABC = ACABABC = tan1 41.226.0 = 58bearing = 270 + 58= 3283.North40100LAB6.2km10.8kmAB =_6.22+ 10.822 6.2 10.8 cos 60= 9.4kmsinLBA6.2 = sin609.4LBA = sin1 6.2 sin609.4= 35bearing = (100 + 180) + 35= 3154. Let l be the length of the ladder.cos 75 = all = acos 75cos =5a4l= 5a4 1l= 5a4 cos 75a= 5 cos 754 = cos1 5 cos 754= 715. 3 65 + 26 = 3 65 + 26 5 265 26= (3 6)(5 26)(5 + 26)(5 26)= 15 66 56 + 1225 24= 27 1161= 27 1166. (a) Read the question as distance from 3 is lessthan distance from 11. The midpoint be-tween 11 and 3 is 4, so the solution isx > 4.(b) Read the question as distance from 0 is lessthan distance from 6. The midpoint be-tween 0 and 6 is 3, so the solution is x < 3.(c) First solve the equation [3x 17[ = [x 3[3x 17 = x 3 or 3x 17 = (x 3)2x = 14 3x 17 = x + 3x = 7 4x = 20x = 5Now test a value for x, say x = 6, to deter-mine whether the inequality holds at thatpoint.Is it true that [3(6) 17[ [(6) 3[1 3 : no.Conclude that the solution lies outside theinterval 57:x R : x 5 x R : x 7(d) This is the complementary case to the pre-vious question, so it will have the comple-mentary solution:x R : 5 < x < 77.xyy = [x a[ay = [2x a[a225Miscellaneous Exercise 2From the graph it appears that [2xa[ [xa[is true for 0 x 2a3 . (You should conrm thatthese are the interval endpoints by substitution.)8. (a) AH =_AG2+ GH2=42+_12 62_2= 5m(b) EH =_AE2AH2=_8252= 39m 6.2m(c) cos EAH = AHAE= 58EAH = 51(d) tanEGH = EHGH= 6.23EGH = 64(e) tan = EHGB= 6.24 = 579. 360 = 4402 6350 cos 37 = 4402 6350 cos 37 360= 5longitude = 126 + 5= 131E360 = 3302 6350 = 3302 6350 360= 3latitude = 37 3= 34SNew position: 34S, 131W10. For the triangle to have an obtuse angle, thelongest side must be longer than the hypotenuseif it were right-angled, i.e. c2> a2+ b2. Thisyields two possibilities.If x is the longest side, thenx2> 52+ 92x >106Since it must also satisfy the triangle inequality xmust be less than the sum of the other two sides.The solution in this case is 106 < x < 14.If x is not the longest side, then92> 52+x2x 238Miscellaneous Exercise 3(c) Graphically:xy-1 1 2 3 4 5 6 7 8 9 101112-112345678910y = [x 10[y = 2x + 1x 3Algebraically:First solve [x 10[ = 2x + 1x 10 = 2x + 1x = 11or (x 10) = 2x + 1x + 10 = 2x + 13x = 9x = 3However, x = 11 is not actually a solu-tion, as you can see by substituting into theequation, so we are left with two intervals(either side of x = 3).Now test one of these intervals delimited bythese two solutions. Try a value, say x = 0:Is it true that [(0) 10[ 2(0) + 1 ?No (10 1).Solution set isx R : x 32.2.4km4.4kmdN60N190 = 369 190 (180 60)= 50d =_2.42+ 4.422 2.4 4.4 cos 50= 3.4kmIts tempting to nd angle using the sine rule,but because its opposite the longest side of thetriangle, it could be either acute or obtuse: itsthe ambiguous case. Finding instead is un-ambiguous. can not be obtuse because it isopposite a shorter side.sin2.4 = sin503.4 = sin1 2.4 sin503.4= 33bearing = 190 + (180 33)= 3273.xy-10 -8 -6 -4 -2 2 4 6 8 102468101214161820y = [x 5[y = [x + 5[y = [x 5[ +[x + 5[[x 5[ +[x + 5[ 14 for x R : 7 x 74.c(10units)N160d (12 units)c+d20d (12 units)cd 2d (24 units)c+2dIn each case below, let be the angle formedbetween c and the resultant.(a) [c +d[ =_102+ 1222 10 12 cos 110= 18.1 unitssin12 = sin11018.1 = sin1 12 sin11018.1= 39direction = 160 39= 12139Miscellaneous Exercise 3(b) [c d[ =_102+ 1222 10 12 cos 70= 12.7 unitssin12 = sin7012.7 = sin1 12 sin7012.7= 62direction = 160 + 62= 222(c) [c + 2d[ =_102+ 2422 10 24 cos 110= 29.0 unitssin24 = sin11029.0 = sin1 24 sin11029.0= 51direction = 160 51= 1095. First, rearrange the equation to[x a[ +[x + 3[ = 5and read this as distance from a plus distancefrom 3 is equal to 5. If the distance between a and 3 is greaterthan 5 then the equation has no solution. If the distance between a and 3 is equalto 5 then every point between a and 3 isa solution. If the distance between a and 3 is less than5 then there will be two solutions, one lyingabove the interval between 3 and a andone lying below it.(a) For exactly two solutions,[a + 3[ < 55 < a + 3 < 58 < a < 2(b) For more than two solutions,[a + 3[ = 5a + 3 = 5 or a + 3 = 5a = 2 a = 86. Let l be the length of the ladder.8075a20cmlcos 80 = ala = l cos 80cos 75 = a + 20la + 20 = l cos 75a = l cos(75) 20l cos 80 = l cos(75) 20l cos(75) l cos 80 = 20l(cos(75) cos 80) = 20l = 20cos(75) cos 80= 235cma = l cos 80= 41cm7. (a) h = k = 0(b) ha +b = kbha = kb b= (k 1)bh = 0 k 1 = 0k = 1(c) (h 3)a = (k + 1)bh 3 = 0 k + 1 = 0h = 3 k = 1(d) ha + 2a = kb 3aha + 5a = kb(h + 5)a = kbh + 5 = 0 k = 0h = 5(e) 3ha +ka +hb 2kb = a + 5b3ha +ka a = 5b hb + 2kb(3h +k 1)a = (5 h + 2k)b3h +k 1 = 0 5 h + 2k = 03h +k = 1 h 2k = 5h = 1k = 240Miscellaneous Exercise 3(Note: the nal step in the solution above isdone by solving the simultaneous equations3h + k = 1 and h 2k = 5. You shouldbe familiar with doing this by eliminationor substitution. (Either would be suitablehere.) You should also know how to do iton the ClassPad:In the Main application, select the simulta-neous equations icon in the 2D tab. Enterthe two equations to the left of the verticalbar, and the two variables to the right:(f) h(a +b) +k(a b) = 3a + 5b(h +k)a + (h k)b = 3a + 5b(h +k 3)a = (h k 5)bh +k 3 = 0 h k 5 = 0h +k = 3 h k = 5solving by elimination:2h = 8h = 44 +k = 3k = 18.2820ABC D 65mtreeLet the height of the tree be h. Let A be thepoint at the base of the tree and B the point atthe apex.tan28 = hACAC = htan28tan20 = hADAD = htan20ACD is right-angled at C, soAD2= AC2+ CD2h2tan220 = h2tan228 + 652h2_ 1tan220 1tan228_= 652Solving this and discarding the negative root:h = 32.5mAC = htan28= 61.0m41CHAPTER 4Chapter 4Exercise 4A1.7N6N10NNorth2614.3N2.4m/s8m/s7m/s3m/sNorth7413.2m/s3.12units10units15units20unitsNorth14210.5units4.5N12N6N8N10NNorth4215.7N5. a = 3i + 2j;b = 3i + 1j = 3i +j;c = 2i + 2j;d = 1i + 3j = i + 3j;e = 0i + 2j = 2j;f = 1i + 2j = i + 2j;g = 1i 2j = i 2j;h = 4i + 0j = 4i;k = 2i 4j;l = 4i 1j = 4i j;m = 4i 1j = 4i j;n = 9i + 2j;6. [a[ = 32+ 22= 13;[b[ = 32+ 12= 10;[c[ = 22+ 22= 22;[d[ = 12+ 32= 10;[e[ = 2;[f [ = 12+ 22= 5;[g[ = 12+ 22= 5;[h[ = 4;[k[ = 22+ 42= 25;[l[ = 42+ 12= 17;[m[ = 42+ 12= 17;[n[ = 92+ 22= 85;7. [(7i + 24j)[ = 72+ 242= 25Newtons8. (a) (5 cos(30)i + 5 sin(30)j)units (4.3i + 2.5j)units(b) (7 cos(60)i + 7 sin(60)j)units (3.5i + 6.1j)units(c) (10 cos(25)i + 10 sin(25)j)units (9.1i + 4.2j)units(d) (7 sin(50)i + 7 cos(50)j)N (5.4i + 4.5j)N(e) (5 8 cos(60)i + 8 sin(60)j)m/s (4.0i + 6.9j)m/s(f) (10 cos(20)i 10 sin(20)j)N (9.4i 3.4j)N(g) (4 cos(50)i + 4 sin(50)j)units (2.6i + 3.1j)units(h) (8 cos(24)i 8 sin(24)j)units (7.3i 3.3j)units(i) (6 sin(50)i 6 cos(50)j)units (4.6i 3.9j)units(j) (10 cos(50)i + 10 sin(50)j)m/s (6.4i + 7.7j)m/s(k) (8 cos(25)i 8 sin(25)j)N (7.3i 3.4j)N(l) (5 cos(35)i + 5 sin(35)j)m/s (4.1i + 2.9j)m/s9. (a) [a[ =_32+ 42= 5 = tan1 43 53.142Exercise 4A(b) [b[ =_52+ 22= 29 = tan1 25 21.8(c) [c[ =_22+ 32= 13 = 180 tan1 32 123.7(d) [d[ =_42+ 32= 5 = tan1 43 53.1(e) [e[ =_52+ 42= 41 = tan1 45 38.7(f) [f [ =_42+ 42= 42 = tan1 44 = 45.010.N160350km/h20Northerly component =350 cos 160 (or 350 cos 20)= 328.9km/hEasterly component =350 sin160 (or 350 sin20)= 119.7km/h11.5units8unitsN329.4unitsMagnitude= 52+ 82=89 9.4unitsDirection= 360 tan1 58 360 32.0 = 328T12. (a) a+b = 2i +3j +i +4j = (2+1)i +(3+4)j =3i + 7j(b) a b = (2 1)i + (3 4)j = i j(c) b a = (1 2)i + (4 3)j = i +j(d) 2a = 2(2i) + 2(3j) = 4i + 6j(e) 3b = 3(i) + 3(4j) = 3i + 12j(f) 2a+3b = (22+31)i +(23+34)j =7i + 18j(g) 2a 3b = (4 3)i + (6 12)j = i 6j(h) 2a+3b = (4+3)i +(6+12)j = i +6j(i) [a[ = 22+ 32= 13 3.61(j) [b[ = 12+ 42= 17 4.12(k) [a[ +[b[ = 13 +17 7.73(l) [a+b[ = [3i +7j[ = 32+ 72= 58 7.6213. (a) 2c +d = (2 + 2)i + (2 + 1)j = 4i j(b) c d = (1 2)i + (1 1)j = i 2j(c) d c = i + 2j(d) 5c = 5i 5j(e) 5c +d = (5 + 2)i + (5 + 1)j = 7i 4j(f) 5c + 2d = (5 + 4)i + (5 + 2)j = 9i 3j(g) 2c + 5d = (2 + 10)i + (2 + 5)j = 12i + 3j(h) 2c d = (2 2)i + (2 1)j = 3j(i) [d 2c[ = [(2 2)i + (1 2)j[ = [3j[ = 3(j) [c[+[d[ = 12+ 12+22+ 12= 2+5 3.65(k) [c +d[ = [(1 + 2)i + (1 + 1)j[ = [3i[ = 3(l) [cd[ = [(12)i+(11)j[ = [ 1i2j[ =5 2.2414. (a) a +b = 5 + 2, 4 +3) = 7, 1)(b) a +b = 5 2, 4 3) = 3, 7)(c) 2a = 2 5, 4) = 10, 8)(d) 3a +b = 3 5 + 2, 3 4 +3) = 17, 9)(e) 2b a = 4 5, 6 4) = 1, 10)(f) [a[ = 52+ 42= 41 6.40(g) [a +b[ = 72+ 12= 50 = 52 7.07(h) [a[ +[b[ = 41 +22+ 32= 41 +13 10.0115. (a)_ 34_+_ 10_=_ 24_(b)_ 34__ 10_=_ 44_(c)_ 10__ 34_=_ 44_(d) 2_ 34_+_ 10_=_ 68_+_ 10_=_ 58_(e)_ 34_+ 2_ 10_=_ 34_+_ 20_=_ 14_(f)_ 34_2_ 10_=_ 34__ 20_=_ 54_(g)_ 34_2_ 10_=_ 54_=_52+ 42= 41 6.40(h)2_ 10__ 34_=_ 54_=_52+ 42= 41 6.4016. (a)_ 27_= 22+ 72= 53(b)_ 23_= 22+ 32= 13(c)2_ 27_= 42+ 142= 212 = 25343Exercise 4B(d)_ 27_+_ 23_=_ 010_= 10(e)_ 27__ 23_=_ 44_=_42+ 42= 32= 4217.204000N20lift= 4000 cos 20 = 3759Ndrag= 4000 sin20 = 1368N18. (12 cos 50i + 12 sin50j) + 10i (17.7i + 9.2j)N19. (12 cos 50i +12 sin50j) +10i (2.3i +9.2j)N20._ 8 sin408 cos 40_+_ 5 cos 305 sin30_+_ 100_=_ 8 sin40 + 5 cos 30 + 108 cos 40 + 5 sin30_ 9.2i + 8.6jN21._ 0 + 10 cos 30 8 sin206 + 10 sin30 8 cos 20_= 5.9i + 3.5jm/s22. 0i + 5j+ 10 cos 30i + 10 sin30j+ 4i + 0j+ 7 cos 60i 7 sin60j (16.2i + 3.9j)N23. 10 sin40i + 10 cos 40j+ 10 cos 30i + 10 sin30j+ 10 cos 10i 10 sin10j+ 10 sin10i 10 cos 10j (10.3i + 1.1j)N24. F1 +F2 +F3 = (2 + 4 + 2)i + (3 + 3 4)j= (8i + 2j)N[F1 +F2 +F3[ = [8i + 2j[=_82+ 22= 217N25. (a +b) + (a b) = (3i +j) + (i 7j)2a = 4i 6ja = 2i 3j(a +b) (a b) = (3i +j) (i 7j)2b = 2i + 8jb = i + 4j26. 2(2c +d) 2(c +d) = 2(i + 6j) (2i 10j)2c = 4i + 22jc = 2i + 11j(2c +d) 2(c +d) = (i + 6j) (2i 10j)d = 3i + 16jd = 3i 16jExercise 4B1. (a) a = 4i + 3j(b) 2a = 8i + 6j(c) a|a| = 4i+3j5 = 0.8i + 0.6j(d) 2 a|a| = 2(0.8i + 0.6j) = 1.6i + 1.2j (a) b = 4i 3j(b) 2b = 8i 6j(c) b|b| = 4i3j5 = 0.8i 0.6j(d) 2 b|b| = 2(0.8i 0.6j) = 1.6i 1.2j (a) c = 2i + 2j(b) 2c = 4i + 4j(c) c|c| = 2i+2j22 = 12i + 12j(d) 2 c|c| = 2( 12i + 12j) = 2i +2j (a) d = 3i 2j(b) 2d = 6i 4j(c) d|d| = 3i2j13 = 313i 213j(d) 2 d|d| = 2( 313i 213j) = 613i 413j2. (a) b|b| = 2i+j5 = 25i + 15j(b) [a[ b|b| = 5( 25i + 15j) = 25i +5j(c) [c[ a|a| = 133i+4j5 = 3135 i + 4135 j(d) a +b +c = 2i + 3j[a +b +c[ = 13[a[ = 5[a[ a+b+c|a+b+c| = 52i+3j13 = 1013i + 1513j3. (a) a and d are parallel since a = 2d.44Exercise 4B(b) a +b +c +d +e= (2 +4 +1 +1 +4)i +(4 +2 8 2 2)j= 12i 14j(c) [a +b +c +d +e[= 122+ 142= 340= 285(d)1214North13918.4bearing=90 + tan1 1412 1394. a is of magnitude 5 units and w is negative.[a[ = 5[wi + 3j[ = 5_w2+ 32= 5w2+ 9 = 25w2= 16w = 4 b is parallel to ab = kai +xj = k(wi + 3j)i +xj = k(4i + 3j)i +xj = 4ki + 3kj(1 + 4k)i = (3k x)j1 + 4k = 0k = 143k x = 0x = 34 c is a unit vector[c[ = 1[0.5i +yj[ = 1_0.52+y2= 10.25 +y2= 1y2= 34y = 32 the resultant of a and d has a magnitude of13 units[a +d[ = 13[(w 1)i + (3 z)j[ = 13[(4 1)i + (3 z)j[ = 13_52+ (3 z)2= 1325 + (9 6z +z2) = 169z26z + 9 + 25 169 = 0z26z 135 = 0(z 15)(z + 9) = 0z = 15or z = 9w = 4; x = 34; y = 32 ; z = 15 or 9.5. p is a unit vector and a is positive[0.6i aj = 10.62+a2= 12a = 0.8 q is in the same direction as p and ve timesthe magnitude.q = 5pbi +ci = 5(0.6i 0.8j)= 3i 4jb = 3c = 4 r + 2q = 11i 20j(di +ej) + 2(3i 4j) = 11i 20j(d + 6)i + (e 8)j = 11i 20jd + 6 = 11d = 5e 8 = 20e = 12 s is in the same direction as r but equal inmagnitude to qs = [q[ r[r[fi +gj = 5 5i 12j52+ 122= 2513i 6013jf = 2513g = 6013a = 0.8, b = 3, c = 4, d = 5, e = 12, f = 2513and g = 601345Exercise 4B6. R = a +b +c +dR = 7 cos 30i +7 sin30j+ 0i +6j+ 10 cos 45i +10 sin45j+ 4 cos 145i +4 sin145jR = 9.9i + 18.9j[R[ =_9.92+ 18.92= 21.3 e = 9.9i 18.9j7. P = [(6i + 5j)[ = 62+ 52 7.8 = tan1 65 508. Horizontal components:P sin = 8 sin50Vertical components:P cos = 8 cos 50 + 5Dividing gives:P sinP cos = 8 sin508 cos 50 + 5tan = 8 sin508 cos 50 + 5 = tan1 8 sin508 cos 50 + 5 31Substituting:P sin = 8 sin50P = 8 sin50sin31 11.99. Horizontal components:P sin = 12 10 sin40Vertical components:P cos = 10 cos 40Dividing gives:P sinP cos = 12 10 sin4010 cos 40tan = 12 10 sin4010 cos 40 = tan1 12 10 sin4010 cos 40 36Substituting:P cos = 10 cos 40P = 10 cos 40cos 36 9.510. T1 sin30 +T2 sin30 = 0 T1 = T2T1 cos 30 +T2 cos 30 = 100T1 cos 30 = 50T1 = 50cos 30= 1003 T1 = T2 = 1003N11. T1 sin30 +T2 sin30 = 0 T1 = T2T1 cos 60 +T2 cos 60 = 100T1 cos 60 = 50T1 = 50cos 60= 100 T1 = T2 = 100N12. First the horizontal components:T1 sin30 +T2 sin60 = 0T1 sin30 = T2 sin6012T1 =32 T2T1 = 3T2Now the vertical components:T1 cos 30 +T2 cos 60 = 10032 T1 + 12T2 = 1003T1 +T2 = 200Substituting:3(3T2) +T2 = 2003T2 +T2 = 2004T2 = 200T2 = 50NT1 = 503N13. Speed of A is 212+ 172= 730m/s.Speed of B is 262+ 22= 680m/s.Particle A is moving fastest.14. Speed=52+ 22= 29m/s.In one minute it will move 6029 323.1m.15. (a) When there is no wind blowing, the piloties due North with velocity vector 75jm/sfor 300000 75 = 4000 seconds=1hr 6min40sec.46Exercise 4B(b)ABt(21i + 10j)t(ai +bj)300kmWe must add the helicopters own velocityto the wind velocity to produce a resultantheaded due North.Easterly (i) components:21 +a = 0a = 21Now nd the northerly (j) component togive the correct speed:_a2+b2= 75212+b2= 752b = _752212= 72We know were heading north, so we dis-regard the negative solution and concludeb = 72.To calculate time, we use the total northerlycomponent (i.e. wind plus plane):10t +bt = 30000082t = 300000t = 30000082 3659s 61minThe velocity vector is (21i + 72j)m/s andthe trip will take about one hour and oneminute.16.ABt(21i + 10j)t(ai +bj)300kmEasterly components must total zero, so as in theprevious question a = 21.Calculation of the northerly component is thesame as in the previous question, but this timewe are heading southwards, so we reject the pos-itive solution and conclude b = 72.We now have a total southwards speed of 72 10 = 62m/s so the time ist = 30000062 4839s 81minThe velocity vector is (21i 72j)m/s and thetrip will take about 81 minutes.17. No working is required for this question. Referto the answers in Sadler.18.60 10Naibja = 10 cos 60 = 5N, b = 10 sin60 = 53N.The weight is (5i 53j)N.19. (a) x(2i + 3j) +y(i j) = 3i + 2j(2x +y)i + (3x y)j = 3i + 2j2x +y = 33x y = 2solving simultaneously:x = 1y = 13i + 2j = a +b(b) (2x +y)i + (3x y)j = 5i + 5j2x +y = 53x y = 5x = 2y = 15i + 5j = 2a +b(c) (2x +y)i + (3x y)j = i + 9j2x +y = 13x y = 9x = 2y = 3i + 9j = 2a 3b(d) (2x +y)i + (3x y)j = 4i + 7j2x +y = 43x y = 7x = 115y = 254i + 7j = 115 a 25b47Exercise 4C(e) (2x +y)i + (3x y)j = 3i j2x +y = 33x y = 1x = 25y = 1153i j = 25a + 115 b(f) (2x +y)i + (3x y)j = 3i + 7j2x +y = 33x y = 7x = 2y = 13i + 7j = 2a b20.ABwind (13i9j)tplane (400t)250i+750jTo y directly from A to B, the resultant of theplanes velocity and the wind must be in the samedirection as AB. That is,(ai +bj) + (13i 9j) = (a 13)i + (b 9)jin the same direction as250i + 750jLet represent this direction (an angle measuredfrom the positive i direction) thentan = 750250 = 3andtan = b 9a 13henceb 9a 13 = 3b 9 = 3(a 13)= 3a + 39b = 48 3aNow consider speeda2+b2= 4002and substitute for b:a2+ (48 3a)2= 160 000a2+ 2304 288a + 9a2= 160 00010a2288a 157696 = 0a = 112 or a = 140.8b = 48 3(112) b = 48 3(140.8)= 384 = 374.4It should be clear that the rst of these so-lutions takes us in the correct direction to gofrom A to B. The pilot should set a vector of(112i + 384j)km/h for the trip from A to B.For the return trip the same calculations apply(since tan( + 180) = tan) so we will get thesame solutions for a and b, but here we will rejectthe rst and accept the second.The pilot should set a vector of (140.8i 374.4j)km/h for the return trip from B to A.Exercise 4C1. (a) OA = 2i + 5j(b) OB = 3i + 6j(c) OC = 0i 5j(d) OD = 3i + 8j2. (a) AB = AO +OB= (3i +j) + (2i j)= i 2j(b) BA = AB= i + 2j3. (a) AB = OA +OB= (i + 4j) + (2i 3j)= 3i 7j(b) BC = OB +OC= (2i 3j) + (i + 5j)= i + 8j48Exercise 4C(c) CA = (i + 5j) + (i + 4j)= 2i j4. (a) AB = (i + 2j) + (4i 2j)= 3i 4j(b) BC = (4i 2j) + (i + 11j)= 5i + 13j(c) CD = (i + 11j) + (6i 13j)= 7i 24j(d) [CD[AB[AB[= 253i 4j5= 15i 20j5. (a) [OA[ = [3i + 7j[=_32+ 72= 58(b) [OB[ = [ 2i +j[=_22+ 12= 5(c) [AB[ = [ (3i + 7j) + (2i +j)[= [ 5i 6j[=_52+ 62= 616. (a) [AB[ = [3i 4j[ = 5(b) [BA[ = [ 3i + 4j[ = 5(c) [AC[ = [i + 4j[ = 17(d) [BC[ = [ 2i + 8j[ = 68 = 2177. (a) OA = 12+ 62= 37(b) OB = 52+ 32= 34(c) BA =_(5 1)2+ (3 6)2= 45 = 358. (a) AB = (1 2)i + (2 3)j= i + 5j(b) BC = (9 1)i + (21 2)j= 8i + 19j(c) CD = (6 9)i + (2 21)j= 3i 23j(d) [AC[ = [(9 2)i + (21 3)j[= 72+ 242= 25(e) OA +AB = OB= i + 2j(f) OA + 2AC= (2i 3j) + 2 ((9 2)i + (21 3)j)= 16i + 45j9.OBA (3i + 4j)AB(7i j)OB = OA +AB= (3i + 4j) + (7i j)= 10i + 3j10.OA (i + 7j)BCAB(2i+3j)AC(4i3j) BC(a) OB = OA +AB= (i + 7j) + (2i + 3j)= i + 10j(b) OC = OA +AC= (i + 7j) + (4i 3j)= 3i + 4j(c) BC = OB +OC= (i + 10j) + (3i + 4j)= 2i 6j11.OA (i + 9j)BC (7i j)DBC(4i6j)DC(3i+2j)BDAD(a) OB = OC +BC= (7i j) (4i 6j)= 3i + 5j(b) OD = OC +DC= (7i j) (3i + 2j)= 4i 3j(c) BD = BC DC= (4i 6j) (3i + 2j)= i 8j49Miscellaneous Exercise 4(d) [AD[ = [ OA +OD[= [ (i + 9j) + (4i 3j)[= [5i 12j[= 1312. (a) (2i + 9j) + (2i 5j) = (4i + 4j)m(b) (2i + 9j) + 2(2i 5j) = (6i j)m(c) (2i + 9j) + 10(2i 5j) = (22i 41j)m(d) [(2i + 9j) + 5(2i 5j)[ = [12i 16j[ = 20m13. (a) (5i 6j) + 2(i + 6j) = (7i + 6j)m(b) (5i 6j) + 3(i + 6j) = (8i + 12j)m(c) (5i 6j) + 7(i + 6j) = (12i + 36j)m(d) [(5i 6j) + 5(i + 6j)[ = [10i + 24j[ = 26m(e) [(5i 6j) +t(i + 6j)[ = 50[(5 +t)i + (6 + 6t)j[ = 50_(5 +t)2+ (6 + 6t)2= 50(5 +t)2+ (6 + 6t)2= 250025 + 10t +t2+ 36 72t + 36t2= 250037t262t 2439 = 0t = 9or t = 27137The particle is 50m from the origin after 9seconds.14. If A, B and C are collinear, vectors AB, AC andBC will all be parallel; showing that any pairof these are scalar multiples of each other willdemonstrate collinearity.AB = (3i j) + (i + 15j) = 4i + 16jAC = (3i j) + (9i 25j) = 6i 24jAC = 32AB = collinear.15. DE = (9i 7j) + (11i + 8j) = 20i + 15jDF = (9i 7j) + (25i 19j) = 16i 12jDE = 54DF = collinear.16.OA (2i + 5j)B (12i + 10j)ABOA + 45AB = (2i + 5j) + 45 ((2i + 5j) + (12i + 10j))= (2i + 5j) + 45(10i + 5j)= (2i + 5j) + (8i + 4j)= 10i + 9j17.OA (2i + 2j)B (10i 1j)ABOA + 13AB = (2i + 2j) + 13 ((2i + 2j) + (10i j))= (2i + 2j) + 13(12i 3j)= (2i + 2j) + (4i j)= 2i +j18.OA (1i + 8j)B (19i + 2j)ABOA + 15AB = (i + 8j) + 15 ((i + 8j) + (19i + 2j))= (i + 8j) + 15(18i 6j)= (i + 8j) + (3.6i 1.2j)= 4.6i + 6.8jMiscellaneous Exercise 41.A BCDE FG10cm4cm3cm7cm(a) tanEBC = 34EBC = tan1 34 36.9(b) To nd EGC we must rst nd the lengthGC.Consider BCG.GB = 10 AG = 3cm.Using Pythagoras GC = 42+ 32= 5cm.Now in CGE,tanEGC = 35EGC = tan1 35 31.0(c) To nd EAC we must rst nd the lengthAC.50Miscellaneous Exercise 4Consider BCA.Using Pythagoras AC = 42+ 102=229cm.Now in CAE,tanEAC = 3229EAC = tan1 3229 15.62.10000Nij2070ab(a) a = 10000 cos 70 3400b = 10000 sin70 9400Weight=(3400i 9400j)N.(b) The resistance force the brakes must applyis equal and opposite the i component of theweight, that is 3 400 N.3. c = a +b2i +j = (2i + 3j) +(3i 4j)2i +j = 2i + 3j + 3i 4j(2 2 3)i = (1 + 3 4)jSince i and uj are not parallel, LHS and RHSmust evaluate to the zero vector:2 + 3 = 2 x3 4 = 1 y17 = 4 (3x2y) = 4173 1617 = 1 (subst. into y)3 = 3317 = 11174.3560A BCO 30mhdFirst consider BCO:tan60 = hd3 = hdd = h3 0.577hNow consider ACO:tan35 = hd + 30d + 30 = htan35d = htan35 30 1.428h 30combining these two results . . . 1.428h 30 = 0.577h(1.428 0.577)h = 300.851h = 30h = 300.851 35m5. OABCAC = 23CBOC OA = 23(OB OC)(4i 3j) (ai 15j) = 23 ((10i +bj) (4i 3j))(4 a)i + 12j = 23 (6i + (b + 3)j)= 4i + 2(b + 3)3 ji components:4 a = 4a = 0j components:12 = 2(b + 3)318 = b + 3b = 1551Miscellaneous Exercise 46. (a) The ball speed is [7i + 24j[ = 72+ 242=25m/s.The time the ball takes to reach the bound-ary is t = 6025 = 2.4s.(b)OA(2,3)(7i + 24j)m/sPLet P be the point of closest approach.tan = 32 = 56.3tan = 247 = 73.7POA = = 17.4OA =_22+ 32= 13sin17.4 = APOAAP = OAsin17.4= 13 sin17.4= 1.08m52CHAPTER 5Chapter 5Exercise 5A1. minor arc AB = 50360 2 12.4 10.8cm2. major arc AB = 235360 2 14.7 60.3cm3. minor arc AB = 360290360 2 6.7 8.2cm4. major arc AB = 360120360 2 8 = 23 16 =323 cm5. minor arc AB = 150360 2 10 = 253 cm6. major arc AB = 280360 2 6 = 283 cm7. minor sector= 60360 122= 24cm28. minor sector= 110360 62= 11cm29. major sector= 360120360 82= 1283 cm210. minor sector= 360205360 15.42 321cm211. First, from arc length l to angle l = 3602r = 360l2rThen from angle to sector area aa = 360r2=360l2r360r2= 360lr2720r= lr2So for question 11minor sector= 12.317.62 108cm212. major sector= 40102 = 200cm213. minor segment = minor sector triangle= 100360 152152sin1002 86cm214. = 360l2r= 288minor segment = 360r2 r2sin2= 288360 102 102sin_288_2 80.0 50.0 30cm215. = 90 102 360= 324minor segment = 90 102sin 3242 41cm216. minor segment = 60360 122 12122sin60= 24 72 32= 24 363= 12(2 33)cm217. minor segment = 135360 62 1262sin135= 272 18 22= 272 92= 9_32 2_cm218. = 360 210= 150minor segment = 150360 102 12102sin(150)= 1253 50 sin150= 1253 25= 253 (5 3) cm219. OAB15cm 112(a) minor arc=112360 2 15 = 283 29.3cm(b) major arc=360112360 2 15 = 623 64.9cm53Exercise 5A20.OABr 75 24cm75360 2r = 24r = 242 36075= 2885 18.3cm21. OAB15cm 50a = 50360 152= 1254 98.2cm222. OAB18cm 140OAB = 20= OAB = 180 2 20= 140a = 140360 182 12182sin140 292cm223.378cm2OAB12cm378 = 360 360 122360 = 378 360144 300.8 5924. OAB12cm 10cm = cos1 122+ 1221022 12 12 49.2a = 49.2360 122 12122sin49.2 7.3cm225. In half an hour the minute hand sweeps out 180or half a circle. Its tip travelsd = 12 2 12= 12cmIn half an hour the hour hand sweeps out 124 ofa full circle. Its tip travelsd = 124 2 8= 23 cm26. The ship is travelling through 3 degrees of lati-tude. 3 = 3 60 = 180

. The ship travels 180nautical miles.One nautical mile in kilometres isd =160360 2 6350 1.85km54Exercise 5B27. The circumference of the base of the cone is equalto the arc length of the sector:2r = 240360 2 10r = 203The slant height of the cone is the radius of thesector:h2+r2= 102h =102_203_2=_100 4009= 10_1 49= 10_59= 1053Exercise 5B1. = 31 = 3 rads2. = 32 = 1.5 rads3. = 51 = 54. = 52 = 2.55. = 41 = 46. = 82 = 47. 5 = 5 180 = 368. 18 = 18 180 = 109. 30 = 30 180 = 610. 80 = 80 180 = 4911. 144 = 144 180 = 4512. 40 = 40 180 = 2913. 145 = 145 180 = 293614. 108 = 108 180 = 3515. 165 = 165 180 = 111216. 9 = 9 180 = 2017. 65 = 65 180 = 133618. 110 = 110 180 = 111819. 130 = 130 180 = 131820. 126 = 126 180 = 71021. 99 = 99 180 = 112022. 155 = 155 180 = 313623. 6 rads = 6 180 = 3024. 12 rads = 12 180 = 1525. 518 rads = 518 180 = 5026. 310 rads = 310 180 = 5427. 25 rads = 25 180 = 7228. 89 rads = 89 180 = 16029. rads = 18030. 3536 rads = 3536 180 = 17531. 2 rads = 2 180 = 9032. 38 rads = 38 180 = 67.533. 3 rads = 3 180 = 6034. 5 rads = 5 180 = 3635. 1736 rads = 1736 180 = 8536. 34 rads = 34 180 = 13537. 1160 rads = 1160 180 = 3338. 718 rads = 718 180 = 7039. 32 = 32 180 0.5640. 63 = 63 180 1.1041. 115 = 115 180 2.0142. 170 = 170 180 2.9743. 16 = 16 180 0.2844. 84 = 84 180 1.4745. 104 = 104 180 1.8246. 26 = 26 180 0.4555Exercise 5B47. 76 = 76 180 1.3348. 51 = 51 180 0.8949. 152 = 152 180 2.6550. 158 = 158 180 2.7651. 1.5R= 1.5 180 8652. 2.3R= 2.3 180 13253. 1.4R= 1.4 180 8054. 0.6R= 0.6 180 3455. 0.2R= 0.2 180 1156. 0.32R= 0.32 180 1857. 1.21R= 1.21 180 6958. 3.1R= 3.1 180 178For exact value problems, you should work towardsknowing these in radians as well as degrees, so youdont have to rst convert to degrees. This will comewith time and eort. You should deliberately memo-rise the following table: sin cos tan0 0 1 0612321341212 13321232 1 0 undened59. sin 4 = 1260. 56 makes an angle of 6 with the x-axis and is inquadrant 2, so sin 56 = 1261. 34 makes an angle of 4 with the x-axis and is inquadrant 2, so cos 34 = 1262. sin 2 = 163. 23 makes an angle of 3 with the x-axis and is inquadrant 2, so sin 23 =3264. 34 makes an angle of 4 with the x-axis and is inquadrant 2, so sin 34 = 1265. cos 4 = 1266. 23 makes an angle of 3 with the x-axis and is inquadrant 2, so tan 23 = 367. cos 2 = 068. tan 2 is undened69. 23 makes an angle of 3 with the x-axis and is inquadrant 2, so cos 23 = 1270. 56 makes an angle of 6 with the x-axis and is inquadrant 2, so tan 56 = 1371. 56 makes an angle of 6 with the x-axis and is inquadrant 2, so cos 56 = 3272. makes an angle of 0 with the x-axis and is inquadrant 2, so tan = 073. cos 3 = 1274. makes an angle of 0 with the x-axis and is inquadrant 2, so sin = 0Questions 7590 are single-step calculator exercises, sotheres no point in reproducing the solutions here. Re-fer to the answers in Sadler.91. (a) 3 revolutions/second= 3 2 = 6 radi-ans/second.(b) 15 revolutions/minute= 1560 2 = 2 radi-ans/second.(c) 90 degrees/second= 4 radians/second.92. (a) 2 radians/minute = 1 revolution/minute(b) 34 radians/second = 34 60 = 45radians/minute = 452 = 22.5 revolu-tions/minute(c) 3 radians/second = 3 60 = 20 radi-ans/minute = 202 = 10 revolutions/minute93. sin1 = 6xx = 6sin1 7.194. tan1.2 = 8xx = 8tan1.2 3.195. Let h be the perpendicular height in cm.sin0.6 = h20h = 20 sin0.6 11.3x =_h2+ 62 12.896. xsin1.1 = 14sin1.8x = 14 sin1.1sin1.8= 12.897. = 0.64 2.50x =_72+ 1022 7 10 cos 2.50 16.256Exercise 5C98. 7.22= 5.02+ 6.122 5.0 6.1 cos xx = cos1_5.02+ 6.127.222 5.0 6.1_ 1.499. (a) 14 2 = 2(b) 23 2 = 43(c) 56 2 = 53(d) 5560 2 = 116100. (a) 50 grads= 0.50 2 = 4(b) 75 grads= 0.75 2 = 38(c) 10 grads= 0.10 2 = 20(d) 130 grads= 0.10 2 = 1320101. (a) C BD EAO0.51Let d be the diameter of the pipe.DODA = tan0.5DA = DOtan0.5= d2 tan0.5= d1.09= 0.915dThe scale along AB must be set up so that1cm units are 0.915cm apart, starting with0 at A.(b) If BAC = 2 then ODA is isosceles soOD=AD or AD = 0.5d. This would besimpler to construct as 1cm units along ABwould be exactly 0.5cm apart.Exercise 5C1. l = r = 5 0.8 = 4cm2. l = r = 10 2.5 = 25cm3. l = r = 7.8 (2 4.5) 13.9cm4. a = 12r2= 12 421= 8cm25. a = 12r2= 12 622.5= 45cm26. a = 12r2= 12 102(2 4)= 114.2cm27. a = 12r2( sin)= 12 592(1 sin1)= 275.9cm28. a = 12r2( sin)= 12 52((2 3.5) sin(2 3.5))= 30.4cm29. a = 12r2( sin)= 12 7.52(2.2 sin2.2)= 39.1cm210. OAB15cm 1.2 ll = r = 15 1.2 = 18cm57Exercise 5C11. OAB15cm 0.8(a) a = 12r2 = 12 1520.8 = 90cm2(b) a = 15290 616.9cm212. OAB8cm 1 l(a) l = r = 8 1 = 8cm(b) a = 12r2( sin)= 12 82(1 sin1) 5.1cm213.15cm2OAB5cm l(a) 12r2 = 15252 = 15 = 65l = r= 5 65= 6cm(b) a = 12r2( sin)= 12 52(65 sin65) 3.35cm214. OAB8cm 20cm = l/r= 20/8= 2.5a = 12r2= 12 822.5= 80cm215. OAB6cm12r2 = 912 62 = 918 = 9 = 12a = 12r2( sin)= 12 62_12 sin 12_ 0.37cm216. a = segment BOC segment AOD= 12R2 12r2= 2(R2r2)= 1.52 (12262)= 81cm217. a = 2(R2r2)= 1.52 (9252)= 42cm258Exercise 5C18.ABO1 O28cm6cm10cmConsider O1AO2.AO1O2 = cos1 82+ 102622 8 10 0.644AO1B = 2AO1O2 1.287segment AO1B = 12 82(1.287 sin1.287) 10.46SimilarlyAO2O1 = cos1 62+ 102822 6 10 0.927AO2B = 2AO2O1 1.855segment AO2B = 12 62(1.855 sin1.855) 16.10total area = 10.46 + 16.10= 26.57cm219. areasegment BOC = 12 820.8= 25.6areaAOD = 12 52sin0.8 8.97areaABCD = 25.6 8.97= 16.60cm220. The AC and BC are perpendicular to AO andBO respectively, since a tangent is perpendicularto a radius to the same point. This makes calcu-lating the area of the halves of the quadrilateralsimple.areaAOC = 12 6 8= 24areaAOBC = 48AOC = tan1 86 0.927AOB = 2AOC 1.855areasector AOB = 12 621.855= 33.38area = 48 33.38= 14.62cm221. Area ABCD is equal to the area of the segmentcreated by chord AD minus the area of the seg-ment created by chord BC.a = 12 52(2 sin2) 12 52(1 sin1)= 12 52(2 sin2 (1 sin1))= 252 (1 sin2 + sin1)= 11.65cm222. (a) l = r = 75 0.8 = 60cm each way, or120cm total.(b) BC=2 75 sin0.4 58.4cmArc BC exceeds chord BC by 1.6cm.23.ABO1 O250mm40mm70mmConsider O1AO2.AO1O2 = cos1 502+ 7024022 50 70 0.594AO1B = 2AO1O2 1.188segment AO1B = 12 502(1.188 sin1.188) 325.9SimilarlyAO2O1 = cos1 402+ 7025022 40 70 0.775AO2B = 2AO2O1 1.550segment AO2B = 12 402(1.550 sin1.550) 440.5total area = 325.9 + 440.5 770mm259Exercise 5C24.ABO1 O210cm7cm15cmAO1O2 = cos1 102+ 152722 10 15 0.403AO1B = 2AO1O2 0.805arc AO1B = 10 0.805 8.05AO2O1 = cos1 72+ 1521022 7 15 0.594AO2B = 2AO2O1 1.188arc AO2B = 7 1.188 8.32perimeter = 8.05 + 8.32 16.4cm25.ABO P10cm10+12=22cmcos 2 = 1010 + 12 = 2 cos1 1022 2.20percentage = 2.202 100% 35%26.10m6mfencecos 2 = 610 = 2 cos1 610 1.85Tond the major segment, use 2 :a = 12 102(2 sin(2 )) 269m227.12m 5mfencefenceIt may be simplest to deal with this as the area ofthe circle minus the area of two identical minorsegments.a = 122 452.39m2cos 2 = 512 = 2 cos1 512 2.28aseg = 12r2( sin)= 1222 (2.28 sin2.28) 109.76m2a = a 2aseg 452.39 2 109.76 233m260Exercise 5C28.16+4+6=26cm16cm6cm 10cmABCDThis is similar to the Situation at the beginningof the chapter.Straight segments AB and CDAB =_262102= 24cmAngle cos 2 = 1026 = 2 cos1 1026 2.35Major arc ACAC = r(2 )= 16(2 2.35) 62.90Minor arc BDBD = r= 6 2.35 14.11Total length = 2 24 + 62.90 + 14.11 125cm29. 5+14+20=39cm20+8+2=30cm5cm20cm2cm15cm18cmABCDEF GHStraight segments AB and EFAB =_392152= 36cmStraight segments CD and GHCD =_302182= 24cmAngle cos 2 = 1539 = 2 cos1 1539 2.35Minor arc AEAE = r= 5 2.35 11.76Angle cos 2 = 1830 = 2 cos1 1830 1.85Minor arc DHDH = r= 2 1.85 3.71Minor arcs BC and FG = 12(2 ) 1.04BG = r= 20 1.04 20.77Total length = 2(AB + BC + CD) + AE + DH= 2(36 + 20.77 + 24) + 11.76 + 3.71 177cm30. perimeter: 2r +r = 14r = 14 2r = 14r 2area: 12r2 = 10subst. for : 12r2_14r 2_= 107r r2= 10r27r + 10 = 0(r 5)(r 2) = 0r = 5or r = 2for r = 5, = 145 2= 0.8 (an acute angle)for r = 2, = 142 2= 5 (a reex angle)61Exercise 5C(a) Radius = 5 cm.(b) Radius = 2 cm.31.a

= 12 10 10 sin 3= 5032= 253asector = 12 52

3= 256ashaded = a

3asector= 253 3 256= 253 252= 25_3 2_cm232.OAB5cm10cma

= 12 10 5= 25 = tan1 105 1.11asector = 12 521.11 13.84ashaded = a

asector= 25 13.84 11.2cm233. r is the slant height of the cone, and is suchthat the arc length of the sector equals the cir-cumference of the base of the cone.r =_282+ 82= 453 29.1cmr = 2 8 = 16453 1.73RADIANS34. First, nd the area of the major segment, thennd the capacity:cos 2 = 1060 = 2 cos1 16 2.81a = 12 602(2 sin(2 )) 6849cm2V = al= 6849 120 821915cm3 822L35. (a) aI = 12bh= 12 15 40= 300cm2aII = aI = 300cm2aIII = aI +aII = 600cm2for segment IV:r =_402+ 152= 573 42.72 = 2 tan1 1540 0.718aIV = 12r2 aIII= 12 1825 0.718 600 55cm2(b) l = 40 2 +30 2+ 573 2 +573 0.718256cm62Exercise 5C36.O1O2C DA B5cm15cm10cm5cmECD =_15252= 102aO1O2DC = 12CD(r1 +r2)= 12 102(10 + 5)= 752 106.07cos CO1E = 515CO1E 1.23aCO1E = 12 1021.23 61.55sinDO2E 4 = 515DO2E 0.34 + 4 1.91aDO2E = 12 521.91 23.88ashaded = aO1O2DC aCO1E adO2E 106.07 61.55 23.88 20.64cm237.A BC4cm 3cm2cmA = cos1 62+ 72522 6 7 0.775B = cos1 52+ 72622 5 7 0.997C = cos1 62+ 52722 6 5 1.369a

= 12ab sinC= 12 5 6 sin1.369 14.70asector A = 12 420.775 6.20asector B = 12 320.997 4.49asector C = 12 221.369 2.74ashaded = a

asector A asector Basector C= 14.70 6.20 4.49 2.74 1.27ashadeda

= 1.2714.70= 8.6%38.OQPRShxuv1The lowest position of the piston is r above thetop of the wheel. Therefore it is 3r above thebottom of the wheel. Thus the xed length ofthe drive rod PR= 3r.It should be clear that h = 2r since at the low po-sition the arm goes straight down from the pistonto the bottom of the wheel, and at the high posi-tion the arm goes straight down from the pistonto the top of the wheel.If arc length PQ is equal to r, the angle it sub-tends is POQ=1RADIAN.The height of point P relative to point O isu = r cos 1.Similarly the horizontal position of point P isv = r sin1.The length from point S to R can be determined63Miscellaneous Exercise 5by Pythagoras Theorem:SR =_(3r)2(r sin1)2=_9r2r2sin21= r_9 sin21= 2.88rx = SR 2r u= 2.88r 2r r cos 1= r(2.88 2 cos 1)= 0.33rxh = 0.33r2r 17%Miscellaneous Exercise 51. No worked solution necessary.2. No worked solution necessary.3. (a) x < 2.5: 2x 5 < 0; 4 x > 0(2x 5) = 4 x2x + 5 = 4 xx = 1(which is in the part of the domain we areconsidering, so we do not reject it.)2.5 x < 4: 2x 5 0; 4 x > 02x 5 = 4 x3x = 9x = 3x 4: 2x 5 0; 4 x 02x 5 = (4 x)x = 1We dont really need to do the third part(which yields a solution outside the part ofthe domain were considering), as there canonly be two solutions. (You can see that ifyou graph the left and right hand sides ofthe equation and see where they intersect.)Solution: x = 1 or x = 3.(b) x < 2.5: 2x 5 < 0; 4 x > 0(2x 5) < 4 x2x + 5 < 4 xx > 1(which is in the part of the domain we areconsidering, so we have 1 < x < 2.5.)2.5 x < 4: 2x 5 0; 4 x > 02x 5 < 4 x3x < 9x < 3(which is in the part of the domain we areconsidering, so we have 2.5 x < 4.)Combining the two parts of solution we get:Solution: 1 < x < 3.(c) This inequality must be true where the pre-vious inequality is false, i.e.:Solution: x 1 or x 3.There is another way we could look at this. Weknow the left and right hand sides are equal at 1and 3, so the inequality must be true either be-tween 1 and 3 or less than 1/greater than 3. Wecan decide which by testing a suitable value: sub-stitute (for example) 2 for x and decide whetherthe inequality holds true. Here [2 2 5[ = 1and [4 2[ = 2 so LHS x3 > xx < 3So the inequality is satised for all x < 0.For x 0, [x[ = x3 2x > x3 > 3xx < 1Solution: x < 1(c) For x < 3, [x + 3[ = (x + 3)(x + 3) < x + 1x 3 < x + 14 < 2x2 < xx > 2No solution for x < 3. For x 3,[x + 3[ = x + 3x + 3 < x + 13 < 1No solution for x 3.No value of x satises the inequality.(d) For x > 1 this problem is essentially thesame as the previous one which has no so-lution, so we only need to consider x 1and the problem becomes[x + 3[ (x + 1)For x < 3, [x + 3[ = (x + 3)(x + 3) (x + 1)x 3 x 13 1The inequality is satised for all x < 3.For 3 x 1, [x + 3[ = x + 3x + 3 (x + 1)x + 3 x 12x 4x 2Solution: x 24. (a) d = r = 25 1.8 = 2.26m(b) d = r = 25 1 = 1.26m81Miscellaneous Exercise 75.Lookout1Lookout2FireNorth120North50 North20North6010kmd1d230d1sin80 = 10sin30d1 = 10 sin80sin30= 19.7kmd2sin70 = 10sin30d2 = 10 sin70sin30= 18.8km6. a = kb2i + 2j = k(xi + 5j)x = 5[b[ = [c[[5i + 5j[ = [7i +yj[_52+ 52=_72+y250 = 49 +y2y = 17. (a) BC = b a + 3b = 2b a(b) BD = 13BC = 23b 13a(c) OD = a +b +BD= a +b + 23b 13a= 23a + 53bab3bhbOA BCDEOE = OA +AB +BD= a +b +hb= a + (h + 1)bOE = OD +OE= (k + 1)OD= (k + 1)(23a + 53b) a + (h + 1)b = (k + 1)(23a + 53b)a 2(k + 1)3 a = 5(k + 1)3 b (h + 1)b_1 2k3 23_a =_5k3 + 53 h 1_bLHS: 13 2k3 = 0k = 12RHS: 5k3 + 23 h = 056 + 23 h = 096 h = 0h = 328. (a) logax + logay = logaxy so p = xy(b) p = xy(c) 3 logax logay = logaplogax3logay = logaplogax3y = logapp = x3y(d) 2 +.5 log10y = log10plog10 102+.5 log10y = log10plog10 100 + log10y0.5= log10plog10 100 + log10y = log10plog10(100y) = log10pp = 100y82Miscellaneous Exercise 79. Let point P be the position of the object after 2seconds.PrB = 2(3i 2j)= 6i 4jPrA = PrB + BrA= 6i 4j + 8i + 3j= 14i jrP = PrA + rA= 14i j +2i + 7j= 12i + 6j[rP[ = [12i + 6j[= 6[2i +j[= 6_22+ 12= 65 m from the origin.10.AB40twind300tairspeed600km45NorthFlying from A to B:(300t)2= (40t)2+ 60022 40t 600 cos 4590000t2= 1600t2+ 360000 48000t 22225t2= 4t2+ 900 120t 220 = 221t2+ 602t 900t = 1.835hourst = 110minutesFlying from B to A:(300t)2= (40t)2+ 60022 40t 600 cos 13590000t2= 1600t2+ 360000 + 48000t 22225t2= 4t2+ 900 + 120t 220 = 221t2602t 900t = 2.219hourst = 133minutes83CHAPTER 8Chapter 8Exercise 8A1. Cubic, y = x3(by observation)2. Quadratic, y = x21 (by observation)3. Note the common product. Relationship is recip-rocal xy = 24 so y = 24x4. There is a common ratio of 12 so the relationshipis exponential. y = 128(0.5)x5. There is a common rst dierence of -5, so therelationship is linear. y = 5x + 136. First dierences are -5, -3, -1, 1, 3,5, 7, 9. Com-mon second dierence of 2, so the relationship isquadratic: y = x2+ 2x 37. Looks like it might be a common ratio. Checkinggives a common ratio of 1.5; the relationship isexponential. y = 1280(1.5)x8. Common rst dierence of 2.5 = linear.y = 2.5x + 1.59. First dierences are -12,-10,-8,-6,-4,-2,0,2. Com-mon second dierence of 2 so its a quadratic witha = 1. When x = 0, y = 1 so c = 1. a + b = 4so b = 5 giving y = x25x + 110. Looks like it could be a cubic. Construct a tableof dierences.x -4 -3 -2 -1 0 1 2 3 4y -69 -32 -13 -6 -5 -4 3 22 5937 19 7 1 1 7 19 37-18 -12 -6 0 6 12 186 6 6 6 6 6The common third dierence conrms that thisis a cubic. From here you could use your calcu-lator to nd the cubic of best t, or proceed asfollows.For the general cubic y = ax3+ bx2+ cx + d.Constructing a dierence table for this gives:x 0 1 2 3y d a +b +c +d 8a + 4b + 2c +d 27a + 9b + 3c +da +b +c 7a + 3b +c 19a + 5b +c6a + 2b 12a + 2b6aCompare this with the corresponding part of thedierence table above and the following resultsappear:6a = 6 a = 16a + 2b = 6 b = 0a +b +c = 1 c = 0d = 5giving y = x35.You might have spotted this by observation with-out going through all this eort, but this is ageneric approach that will work for any cubic.11. This is the absolute value function y = [x[ and isnone of linear, quadratic, cubic, exponential norreciprocal.12. This is is none of linear, quadratic, cubic, expo-nential nor reciprocal. (It is actually a log func-tion, y = log2x)13. Its a linear pattern, increasing by 6 each time.The nthterm is 6n 114. First dierences are 4, 6, 8, 10,12,14. Its a qua-dratic pattern where the nthterm is n(n + 1) orn2+n.15. Its an exponential pattern with each term 3times the previous term. The nthterm is 2 3n.16. Its an exponential pattern with each term twicethe previous term. The nthterm is 32 2n.Exercise 8B1. (a) One-to-one . . . function(b) One-to-many . . . not a function(c) Many-to-one . . . function(d) Many-to-many . . . not a function(e) Many-to-one . . . function(f) Many-to-many . . . not a function2. (a) x maps to unique y . . . function(b) x maps to unique y . . . function(c) x maps to either zero, one or two values ofy . . . not a function(d) x maps to either zero, one or two values ofy . . . not a function(e) x maps to unique y . . . function(f) x maps to either one, two or three values ofy . . . not a function84Exercise 8B3. (a) Range=1 2 + 3, 2 2 + 3, 3 2 + 3,4 2 + 3=5, 7, 9, 11(b) Range=(1 +3) 2, (2 +3) 2, (3 +3) 2,(4 + 3) 2=8, 10, 12, 14(c) Range=1 1, 2 2, 3 3, 4 4=1(d) The function machine maps x x2. Therange is the set of non-negative real num-bers.4. (a) f(4) = 5 4 2 = 18(b) f(1) = 5 1 2 = 7(c) f(3) = 5 3 2 = 13(d) f(1.2) = 5 1.2 2 = 4(e) f(3) + f(2) = (5 3 2) + (5 2 2) = 21(f) f(5) = 5 5 2 = 23(g) f(5) = 5 5 2 = 27(h) f(a) = 5 a 2 = 5a 2(i) f(2a) = 5 2a 2 = 10a 2(j) f(a2) = 5 a22 = 5a22(k) 3f(2) = 3(5 2 2) = 3 8 = 24(l) f(a +b) = 5 (a +b) 2 = 5a + 5b 2(m) f(p) = 335p 2 = 335p = 35p = 7(n) f(q) = 125q 2 = 125q = 10q = 25. (a) f(4) = 4(4) 7 = 9(b) f(0) = 4(0) 7 = 7(c) g(3) = 3212 = 3(d) g(3) = (3)212 = 3(e) h(5) = (5)23(5) + 3 = 43(f) h(5) = 523(5) + 3 = 13(g) h(2) = (2)23(2) + 3 = 13(h) 3f(a) = 3(4a 7) = 12a 21(i) f(3a) = 4(3a) 7 = 12a 7(j) 3g(a) = 3(a212) = 3a236(k) g(3a) = (3a)212 = 9a212(l) g(p) = 24p212 = 24p2= 36p = 6(m) g(q) = h(q)q212 = q23q + 312 = 3q + 315 = 3qq = 5(n) h(r) = f(r) + 28r23r + 3 = (4r 7) + 28r23r + 3 = 4r 7 + 28r23r + 3 = 4r + 21r27r + 3 = 21r27r 18 = 0(r 9)(r + 2) = 0r = 9or r = 26. Add 5 to domain to get range.y R : 5 y 87. Subtract 3 from domain to get range.y R : 3 y 08. Multiply domain by 3 to get range.y R : 6 y 159. Multiply domain by 4 to get range.y R : 20 y 4010. Multiply domain by 2 and subtract 1 to getrange.y R : 1 y 911. Here the minimum for the domain maps to themaximum for the range and vice versa, so therange is y R : (1 5) y (1 0) = y R : 4 y 112. Here the minimum value of the function is zero(when x = 0) and the maximum is 32. The rangeis y R : 0 y 913. Here the minimum value of the function is zero(when x = 1) and the maximum is (3 + 1)2.The range is y R : 0 y 1614. The minimum value of x2+ 1 is 1 (when x = 0.The maximum for the given domain is 32+ 1.The range is y R : 1 y 1015. Here the minimum for the domain maps to themaximum for the range and vice versa, so therange is y R : 14 y 116. This function has a minimum value of 1 but ithas no maximum. y R : y 117. Absolute value has a minimum of 0. y R :0 y 318. Absolute value has a minimum of 0. y R :y 019. [x[ has a minimum of 0 so [x[ +2 has a minimumof 2. y R : y 220. x2has a minimum of 0 so x21 has a minimumof -1. y R : y 121. x2has a minimum of 0 so x2+4 has a minimumof 4. y R : y 485Exercise 8C22. This function has no minimum or maximum, butit cannot have a value of zero. y R : y ,= 023. Its dicult to visualise this function withoutgraphing it. Use your Classpad to graph it. Itshould be clear that the output of the functioncan be any real number except 1. y R : y ,= 124. one-to-one25. one-to-one (since every positive number has itsown unique square)26. many-to-one (since a positive and negative num-ber can map to the same square)27. many-to-one (as for the previous question)28. one-to-one (any number can be the square rootof at most one other number)29. The natural domain of the function is x R :x 0 (since the function is not dened for neg-ative x), and it is a one-to-one function (like theprevious question).30. Both x and y can take any real value.x R, y R31. x can take any real value, but y can not be neg-ative.x R, y R : y 032. Neither x nor y can be negative.x R : x 0, y R : y 033. x must not be less than 3 (so that x 3 is notnegative) and y can not be negative.x R : x 3, y R : y 034. x must not be less than 3 (so that x + 3 is notnegative) and y can not be negative.x R : x 3, y R : y 035. x must not be less than 3 (so that x 3 is notnegative) and y can not be less than 5.x R : x 3, y R : y 536. Neither x nor y can be zero.x R : x ,= 0, y R : y ,= 037. Neither x 1 nor y can be zero.x R : x ,= 1, y R : y ,= 038. Neither x 3 nor y can be zero.x R : x ,= 3, y R : y ,= 039. x 3 must be non-negative and non-zero and ysimilarly can be neither zero nor a negative num-ber.x R : x > 3, y R : y > 0Exercise 8C1. (a) x f(x) g f(x)0 1 11 2 12 3 33 4 54 5 7Range is -1, 1, 3, 5, 7.(b) x g(x) f g(x)0 3 21 1 02 1 23 3 44 5 6Range is -2, 0, 2, 4, 6.(c) x g(x) g g(x)0 3 91 1 52 1 13 3 34 5 7Range is -9, -5, -1, 3, 7.2. (a) x f(x) g f(x)1 4 92 5 163 6 25Range is 9, 16, 25.(b) x h(x) g h(x) f g h(x)1 1 0 32 8 49 523 27 676 679Range is 3, 52, 679.(c) x f(x) g f(x) hg f(x)1 4 9 7292 5 16 40963 6 25 15 625Range is 729, 4096, 15 625.3. (a) Domain: x R; Range: y R(b) Domain: x R; Range: y R(c) f(x) + g(x) = (x + 5) + (x 5) = 2xDomain: x R; Range: y R(d) f(x) g(x) = (x + 5) (x 5) = 10Domain: x R; Range: 1086Exercise 8C(e) f(x) g(x) = (x + 5)(x 5)= x225Domain: x R; Range: y R : y 25(f) f(x)g(x) = x + 5x 5= (x 5) + 10x 5= 1 + 10x 5Domain: x R : x ,= 5; Range: y R :y ,= 14. (a) 23x+2 = 2f(x) = g f(x)(b) 3x + 2 =_f(x) = hf(x)(c) 6x + 2 = 3 2x + 2 = 3g(x) + 2 = f g(x)(d) 3x + 2 = 3h(x) + 2 = f h(x)(e) 2x = 2h(x) = g h(x)(f)_2x =_g(x) = hg(x)(g) 9x+8 = (9x+6)+2 = 3(3x+2)+2 = f f(x)(h) x0.25=_x = hh(x)(i) 27x + 26 = 3(9x + 8) + 2 = f f f(x)5. (a) f f(x) = 2(f(x)) 3= 2(2x 3) 3= 4x 6 3= 4x 9(b) g g(x) = 4(g(x)) + 1= 4(4x + 1) + 1= 16x + 4 + 1= 16x + 5(c) h h(x) = (h(x))2+ 1= (x2+ 1)2+ 1= x4+ 2x2+ 1 + 1= x4+ 2x2+ 2(d) f g(x) = 2(g(x)) 3= 2(4x + 1) 3= 8x + 2 3= 8x 1(e) g f(x) = 4(f(x)) + 1= 4(2x 3) + 1= 8x 12 + 1= 8x 11(f) f h(x) = 2(h(x)) 3= 2(x2+ 1) 3= 2x2+ 2 3= 2x21(g) h f(x) = (f(x))2+ 1= (2x 3)2+ 1= 4x212x + 9 + 1= 4x212x + 10(h) g h(x) = 4(h(x)) + 1= 4(x2+ 1) + 1= 4x2+ 4 + 1= 4x2+ 5(i) h g(x) = (g(x))2+ 1= (4x + 1)2+ 1= 16x2+ 8x + 1 + 1= 16x2+ 8x + 26. (a) f f(x) = 2(f(x)) + 5= 2(2x + 5) + 5= 4x + 10 + 5= 4x + 15(b) g g(x) = 3(g(x)) + 1= 3(3x + 1) + 1= 9x + 3 + 1= 9x + 4(c) h h(x) = 1 + 2h(x)= 1 + 21 + 2x= 1 + 2x+2x= 1 + 2xx + 2(d) f g(x) = 2(g(x)) + 5= 2(3x + 1) + 5= 6x + 2 + 5= 6x + 7(e) g f(x) = 3(f(x)) + 1= 3(2x + 5) + 1= 6x + 15 + 1= 6x + 16(f) f h(x) = 2(h(x)) + 5= 2(1 + 2x) + 5= 2 + 4x + 5= 4x + 7(g) h f(x) = 1 + 2f(x)= 1 + 22x + 5(h) g h(x) = 3(h(x)) + 1= 3(1 + 2x) + 1= 3 + 6x + 1= 6x + 487Exercise 8C(i) h g(x) = 1 + 2g(x)= 1 + 23x + 17. g [f(x)] = x 4x 4 0x 48. g [f(x)] = 4 x4 x 0x 49. g [f(x)] = 4 x24 x2 0x2 42 x 210. g [f(x)] =_4 [x[4 [x[ 0[x[ 44 x 411. g [f(x)] =_(x + 3) 5(x + 3) 5 0x 2 0x 212. g [f(x)] =_(x 6) + 3(x 6) + 3 0x 3 0x 313. (a) f(3) = (3)2+ 3 = 12(b) f(3) = (3)2+ 3 = 12(c) g(2) = 12(d) fg(1) = f(11)= f(1)= (1)2+ 3= 4(e) gf(1) = g_(1)2+ 3_= g(4)= 14(f) R f(x) = x2+ 3 y R : y 3(g) x R : x ,= 0 g(x) = 1x y R : y ,= 0(h) It may be useful to think of these com-pound functions as sequential mappings,something likex f(x) u g(u) ywhere x is the domain of the compoundfunction and y is its range. The interme-diate set u is the intersection of the rangeof f and the domain of g. Thus to deter-mine the natural domain we work right toleft, then to determine the range we workleft to right.The whole of the range of f(x) lies within thedomain of g(x) (since the only real numberexcluded from the domain of g(x) is 0 andthis is outside the range of f(x)) so there isno additional restriction to the domain andthe domain of gf(x) is the same the domainof f(x): x R.When the domain of g(x) is restricted to therange of f(x) (i.e. x R : x 3) the rangeis y R : 0 < y 13.R gf(x) _y R : 0 < y 13_There are two main ways of graphing thison the ClassPad 330.From the Main appInteractive->DeneFunc name: fVariable/s: xExpression: x2+3OKInteractive->DeneFunc name: gVariable/s: xExpression: 1/xOKg(f(x))Tap the graph icon then highlight g(f(x))and drag and drop it onto the graph.88Exercise 8CFrom the Graph&Tab appy1=x2+3y2=1/xy3=y2(y1(x))Tap the graph icon.Note that to enter y2(y1(x)) you must usey in the abc tab, not the y button on thekeyboard or under the VAR tab.Once you have obtained the graph you canexamine it in all the usual ways.(i) The whole of the range of g(x) lies withinthe domain of f(x) (since f(x) is dened forall real numbers) so there is no additionalrestriction to the domain and the domainof fg(x) is the same the domain of g(x):x R : x ,= 0.When the domain of f(x) is restricted to therange of g(x) (i.e. x R : x ,= 0) the rangeis y R : y > 3.x R : x ,= 0 fg(x) y R : y > 314. (a) f(5) = 25 (5)2= 0(b) f(5) = 25 (5)2= 0(c) g(4) = 4 = 2(d) fg(4) = f(4)= f(2)= 25 (2)2= 21(e) gf(4) = g(25 (4)2)= g(9)= 9= 3(f) R f(x) = 25 x2 y R : y 25(g) x R : x 0 g(x) = x y R : y 0(h) gf(x) :x R : 5 x 5f(x)=25x2u R : 0 u 25g(x)=xy R : 0 y 5Hence the domain is x R : 5 x 5and the range is y R : 0 y 5.(i) fg(x) :x R : x 0g(x)=xu R : u 0f(x)=25x2y R : 0 y 25Hence the domain is x R : x 0 andthe range is y R : 0 y 25.15. (a) g(x) is not dened for x = 3 so we mustexclude x = 1 from the domain. g f(x) :x R : x ,= 1f(x)=x+2u R : u ,= 3g(x)= 1x3y R : y ,= 0Hence the domain is x R : x ,= 1 andthe range is y R : y ,= 0.(b) f g(x) :x R : x ,= 3g(x)= 1x3u R : u ,= 0f(x)=x+2y R : y ,= 2Hence the domain is x R : x ,= 3 andthe range is y R : y ,= 2.16. (a) g f(x) :x R : x 0f(x)=xu R : u 0g(x)=2x1y R : y 1Hence the d


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