Trigonometric ratios 9B 1 a d b 2 a c b

Trigonometric ratios 9B 1 a a Using sin A sin B 8 sin 72 sin 30 8sin 72 sin 30 As 72 > 30, > 8 cm 15.2 cm ( ) ABC 180 68.4 + 83.7 27.9 Using a 9.8 sin 27.9 sin 83.7 9.8sin 27.9 a sin 83.7 4.61 cm ( ) 2 a c Using c 24 sin 22 sin110 24sin 22 c sin110 As 110 > 22, 24 cm > c. 9.57 cm ( ) x 180 57 + 39 84 a Using sin A sin B y 7.5 sin 57 sin84 7.5sin 57 y sin84 6.32 cm ( ) ABC 180 30 + 95 55 a Using sin A sin B a 14.7 sin 30 sin 55 14.7sin 30 a sin 55 8.97 cm ( ) Using Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 1

2 x 25 sin 30 sin112 25sin 30 x sin112 B 180 112 + 30 13.5 cm ( ) 38 Using sin B sin C y 25 sin 38 sin112 25sin 38 y sin112 16.6 cm ( ) e y 8 sin 50 sin 80 8sin 50 y 6.22 cm ( ) sin80 (Note: You coul use the line of symmetry to split the triangle into two right-angle 4 triangles an use cos 50. y c x 180 60 + 35 85 a Using sin B sin A y 8 sin85 sin 35 8sin 85 y sin 35 13.9 cm ( ) C 180 56.4 + 72 51.6 Using x 5.9 sin 56.4 sin 51.6 5.9sin 56.4 x sin 51.6 6.27 cm ( ) Using sin B sin C y 5.9 sin 72 sin 51.6 5.9sin 72 y sin 51.6 7.16 cm ( ) f x 180 50 + 50 80 Using Using x 6 sin 36.8 sin 53.2 Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 2

2 f 3 a 6sin 36.8 x sin 53.2 B 180 36.8 + 53.2 4.49 cm ( ) 90 Using sin B sin C 6 y sin 53.2 sin 90 6sin 90 y sin 53.2 7.49 cm ( ) (Note: The thir angle is 90 so you coul solve the prolem using sine or cosine; the sine rule is not necessary.) 3 c sin C sin A Using c a sin x sin 60 6 8 6sin 60 sin x ( 0.6495 ) 8 6sin 60 x sin 40.5 8 x 40.5 sin C sin A Using c a sin x sin117 6 9 6sin117 sin x 0.5940 9 1 6sin117 x sin 9 x 36.4 36.4 ( ) sin C sin B Using c sin C sin 28 8.7 10.8 8.7sin 28 sin C 0.3781 10.8 8.7sin 28 C sin 10.8 C 22.2 x 180 28 + 22.2 130 x 130 sin A sin B Using a sin x sin 40 10 11 10sin 40 sin x ( 0.5843 ) 11 10sin 40 x sin 35.8 11 x 35.8 4 a Using sin x sin 67.5 5.8 7.2 Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 3

5.8sin 67.5 7.2 1 5.8sin 67.5 x sin 48.09 7.2 x 48.1 4 a sin x ( 0.7442 ) Angle ACB 180 70 110 Using sin x sin110 8 10 8sin110 sin x 0.7517 10 8sin110 x sin 10 48.74 x 48.7 e c Using sin x sin80 4.5 6.2 4.5sin 80 sin x ( 0.7147 ) 6.2 4.5sin 80 x sin 45.63 6.2 x 45.6 sin C sin B Using c sin C sin 55 7.9 10.4 7.9sin 55 sin C ( 0.6222 ) 10.4 7.9sin 55 C sin 38.48 10.4 x 180 55 + C x 86.52 86.5 ( ) Using sin x sin 50 2 3 2 2 sin 50 sin x 0.2553 3 2 sin 50 x sin 14.79 3 x 14.8 f sin B sin C Using sin B sin 60 9.7 12.4 9.7sin 60 sin B 0.6774 12.4 B 42.65 x 180 60 + B 77.35 x 77.4 () Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 4

5 7 a 6 q p a Using sin Q sin P PR 3 sin 45 sin 60 3 sin 45 PR 1.41 cm sin 60 ( The exact answer is 2 cm. ) r p Using sin R sin P R 180 60 + 45 75 ( ) PQ 3 sin 75 sin 60 3 sin 75 PQ 1.93 cm sin 60 Using sin x sin 75 3.9 5.5 3.9sin 75 sin x 5.5 3.9sin 75 x sin 5.5 43.23 x 43.2 So ABC 180 75 + 43.2 61.8 Using sin B sin C y 5.5 sin 61.8 sin 75 5.5sin 61.8 y 5.018 sin 75 y 5.02 sin P sin R Using p r sin P sin 75 12 15 12sin 75 sin P ( 0.7727 ) 15 12sin 75 P sin 50.60 15 Angle QPR 50.6 Angle PQR 180 75 + 50.6 54.4 Using sin A sin 45 8.5 10.8 8.5sin 45 sin A 10.8 8.5sin 45 A sin 33.815 10.8 x 180 45 + A 101.2 x 101 Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 5

7 Using sin B sin C y 10.8 sin x sin 45 10.8sin x y 14.98 sin 45 y 15.0 c 1 5sin102 x sin 54.599 6 x 54.6 In triangle ABC: BAC 180 102 10 + x 13.4 So ADB 180 10 13.4 156.6 a Using in ABD sin D sin A y 6 sin156.6 sin13.4 6sin156.6 y sin13.4 10.28 10.3 x In ABC, cos 20 7 x 7cos 20 6.58 sin D sin A Using in ADC a sin y sin100 x 12.2 xsin100 sin y 12.2 xsin100 y sin 32.07 12.2 y 32.1 e sin C sin A Using in ABC c a sin x sin 24 6.4 7 x 21.8 ( ) a Using in ABD sin A sin D y 6.4 sin 24 sin120 6.4sin 24 y sin120 y 3.01 3.0058 In triangle BDC: C 180 78 102 sin B sin C Using sin x sin102 5 6 5sin102 sin x 6 f (The aove approach fins the two values inepenently. You coul fin y first an then use it to fin x, ut if your answer for y is wrong then x will e wrong as well.) Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 6

7 f 8 sin D sin B Using in BDC sin x sin80 6.2 8.5 6.2sin 80 sin x 8.5 1 6.2sin 80 x sin 45.92 8.5 x 45.9 In triangle ABC: ACB 180 80 + x 54.08 Using sin A sin 54.08 6.2 7.5 6.2sin 54.08 sin A 7.5 6.2sin 54.08 A sin 7.5 42.03 So y 180 42.03 + 134.1 y 3.87 ( ) 8 Using BC 6 sin 35 sin 65 6sin 35 BC 3.80 km sin 65 9 a In triangle ABD: DAB 43 isosceles ( 43 ) So ADB 180 2 94 As the triangle is isosceles you coul work with right-angle triangles, ut using the sine rule a sin D sin A AB 5 sin 94 sin 43 5sin 94 AB 7.31cm ( ) sin 43 BAC 55 20 35 ABC 20 + 60 80 (Alternate angles an angles on a straight line.) ACB 180 80 + 35 65 a Using sin B sin C AC 6 sin80 sin 65 6sin 80 AC 6.52 km sin 65 In triangle ADC: ADC 180 94 86 So CAD 180 72 + 86 22 Using CD 5 sin 22 sin 72 5sin 22 CD 1.97 cm sin 72 10 a In triangle ABD: sin B sin A a sin B sin 66 76 136 76sin 66 So sin B 136 B 30.6978... Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 7

10 a So the angle etween AB an BD is 30.7. Using triangle BCD: sin B sin C sin B sin 98 80 136 80sin 98 So sin B 136 B 35.6273... So the angle etween BC an BD is 35.6. The angle etween the fences AB an BC is 30.7 + 35.6 66.3. 11 In triangle ABD: Angle ADB 180 66 30.7 83.3 Using the cosine rule: 2 2 2 a + 2acos D 2 2 2 136 + 76 2 136 76 cos83.3 2 18 496 + 5776 2411.817 477 2 21860.182 52 So 147.851... So the length of the fence AB is 148 m (). 11 Multiply top an ottom y 2 1: x 4( 2) ( 2 1)( 2+ 1) 4( 2) 2 4 2 12 a Using the left-han triangle, the angles are 40, 128 an 12. (128 180 52 an 180 (128 + 40 ) 12 ) a sin A sin B a 15 sin128 sin12 15sin128 a sin12 a 56.8518... Using the larger right-angle triangle: height sin 40 56.8518 Height 56.8518 sin 40 36.54... The height of the uiling is 36.5 m (). Assume that the angles of elevation have een measure from groun level. Using 4 x x sin y sin 30 4 x sin 30 xsin y 1 1 ( 4 x) x 2 2 Multiply throughout y 2: 4 x x 2 x+ 2x 4 x 1+ 2 4 4 x 1 + 2 Pearson Eucation Lt 2017. Copying permitte for purchasing institution only. This material is not copyright free. 8