Chapter 4 — Triangle trigonometry - Weebly€¦ · Triangle trigonometry MB11 Qld-4 63 Exercise...

T r i a n g l e t r i g o n o m e t r y M B 1 1 Q l d - 4 63

Exercise 4A — Calculating trigonometric ratios 1 a tan 57° = 1.540 b 9 × tan 63° = 17.663

c 8.6tan12°

= 40.460

d tan 33°19′ = 0.657 2 a sin 37° = 0.602 b 9.3 × sin 13° = 2.092

c 14.5sin 72°

= 15.246

d 48sin 67 40′°

= 51.893

3 a cos 45° = 0.707 b 0.25 × cos 9° = 0.247

c 6cos 24°

= 6.568

d 5.9 × cos 2°3′ = 5.896 4 a sin 30° = 0.5 b cos 15° = 0.9659 c tan 45° = 1 d 48 × tan 85° = 548.6 e 128 × cos 60° = 64 f 9.35 × sin 8° = 1.301

g 4.5cos32°

= 5.306

h 0.5tan 20°

= 1.374

i 15sin 72°

= 15.77

5 a sin 24°38′ = 0.42 b tan 57°21′ = 1.56 c cos 84°40′ = 0.09 d 9 × cos55°30′ = 5.10 e 4.9 × sin 35°50′ = 2.87 f 2.39 × tan 8°59′ = 0.38

g 19tan 67 45′°

= 7.77

h 49.6cos 47 25′°

= 73.30

i 0.84sin 75 5′°

= 0.87

6 sin θ = 0.167 θ = sin−1 0.167 = 9.61° ≈ 10° 7 a sin θ = 0.698 θ = sin−1 0.698 ≈ 44° b cos θ = 0.173 θ = cos−1 0.173 ≈ 80° c tan θ = 1.517 θ = tan−1 1.517 ≈ 57° 8 cos θ = 0.058 θ = cos−1 0.058 ≈ 86°40′

9 a tan θ = 0.931 θ = tan−1 0.931 ≈ 42°57′ b cos θ = 0.854 θ = cos−1 0.854 θ ≈ 31°21′ c sin θ = 0.277 θ = sin−1 0.277 ≈ 16°05′

Exercise 4B — Finding an unknown side 1 a

b

c

2 tan θ = oppadj

tan71° = 51x

x = 51 tan 71° = 148.1 mm

3 sinθ = opphyp

sin 23° = 13a

a = 13 sin 23 ° = 5.08 m

4 cos θ = adjhyp

cos 31° = 35d

d = 35 cos 31 ° = 30.0 cm

5 a sin θ = opphyp

sin 68° = 13x

x = 13 sin 68° = 12.1 cm

b tan θ = oppadj

tan 49° = 48y

y = 48 tan 49° = 55.2 m

c cosθ = adjhyp

cos 41° = 12.5

z

z = 12.5 cos 41° = 9.43 km

6 a tan θ = oppadj

tan 21° = 4.8t

t = 4.8tan 21°

= 12.5 m

b sin θ = opphyp

sin 77° = 87p

p = 87sin 77°

= 89.3 mm

c cos θ = adjhyp

cos 36° = 8.2q

q = 8.2cos 36o

= 10.1m

7 a tan θ = oppadj

tan 23° = 2.3a

a = 2.3tan 23°

= 5.42 m

b sin θ = opphyp

sin 39° = 0.85b

b = 0.85sin39°

= 1.35 km

c cos θ = adjhyp

cos 76° = 8.5x

x = 8.5 × cos 76° = 2.06 km

Chapter 4 — Triangle trigonometry

M B 1 1 Q l d - 4 64 T r i a n g l e t r i g o n o m e t r y

d tan 9° = 116m

m = 116 × tan 9° = 18.4 mm

e sin 1116.75

d° =

d = 16.75 sin 11° = 3.20 cm

f cos 13° = 64.75x

x = 64.75cos 13°

= 66.5 m

g cos 83° = 44.3

x

x = 44.3 cos 83° = 5.40 m

h sin 20° = 15.75

g

g = 15.75 sin 20° = 5.39 km

i tan 84°9′ = 2.34m

m = 2.34tan84 9′o

= 0.240 m

j cos 60°32′ =84.6

q

q = 84.6 cos 60°32′ = 41.6 km

k cos 75°19′ = 21.4t

t = 21.4cos75 19′°

= 84.4 m

l sin 29°32′ = 26.8

r

t = 26.8 sin 29° 32′ = 13.2 cm

8 cos θ = adjhyp

cos 69° = 9.2x

x = 9.2cos69°

The answer is D

9 tanφ = oppadj

= 815

The answer is A 10

tan 59° = 3.6x

x = 3.6 tan 59° = 6 m

11

cos 65° = 10x

x = 10 cos 65° = 4.2 m 12 Distance From A to B

Sin 60° = AB23

AB = 23 sin 60° = 20 km 13 a

b tan 24° = 13.5length

length = 13.5tan 24°

= 30.3 cm 14 a

b sin 60° = 1.4x

x = 1.4sin 60°

The brace needs to be 1.6 m long. 15

cos 70° = 3.3x

x = 3.3cos70°

= 9.65 m 16 a

b cos 15° = depth60

depth = 60 cos 15° = 58 m

c sin 15° = 60x

x = 60 sin 15° = 15.5 m The ship had drifted 15.5 m

Exercise 4C — Finding angles

1 a tan θ = 712

θ = tan−1 712⎛ ⎞⎜ ⎟⎝ ⎠

= 30°

b tan φ = 113

φ = tan−1 113

⎛ ⎞⎜ ⎟⎝ ⎠

= 75°

c tan r = 16225

r = tan−1 16225

⎛ ⎞⎜ ⎟⎝ ⎠

= 81°

2 a sin θ = 1324

θ = sin−1 1324

⎛ ⎞⎜ ⎟⎝ ⎠

= 32° 48′

b sin θ = 4.66.5

θ = sin−1 4.66.5

⎛ ⎞⎜ ⎟⎝ ⎠

= 45°3′

c sin α = 5.69.7

α = sin−1 5.69.7

⎛ ⎞⎜ ⎟⎝ ⎠

= 35°16′

3 a cos θ = 915

θ = sin−1 915⎛ ⎞⎜ ⎟⎝ ⎠

= 53°8′

b cosα = 2.64.6

α = cos−1 2.64.6

⎛ ⎞⎜ ⎟⎝ ⎠

= 55° 35′

c cos β = 19.527.8

β = cos−1 19.527.5

⎛ ⎞⎜ ⎟⎝ ⎠

= 45° 27′

4 a cos θ = 711

θ = cos−1

711⎛ ⎞⎜ ⎟⎝ ⎠

≈ 50°

b sin θ = 815

θ = sin−1 815⎛ ⎞⎜ ⎟⎝ ⎠

≈ 32°


c tan θ = 914

θ = tan−1 914⎛ ⎞⎜ ⎟⎝ ⎠

≈ 33°

d tan θ = 3.69.2

θ = tan−1 3.69.2

⎛ ⎞⎜ ⎟⎝ ⎠

≈ 21°

e cosθ = 32196

θ = cos−1 32196⎛ ⎞⎜ ⎟⎝ ⎠

≈ 81°

f sin θ = 14.926.8

θ = sin−1 14.926.5

⎛ ⎞⎜ ⎟⎝ ⎠

≈ 34°

5 a θ = sin−1 19.230

⎛ ⎞⎜ ⎟⎝ ⎠

≈ 39° 48′

b θ = tan−1 6310⎛ ⎞⎜ ⎟⎝ ⎠

≈ 80° 59′

c θ = tan−1 0.62.5

⎛ ⎞⎜ ⎟⎝ ⎠

≈ 13° 30′

d θ = cos−1 3.518.5⎛ ⎞⎜ ⎟⎝ ⎠

≈ 79° 6′

e θ = tan−1 16.38.3

⎛ ⎞⎜ ⎟⎝ ⎠

≈ 63° 1′

f θ = sin−1 6.318.9⎛ ⎞⎜ ⎟⎝ ⎠

≈ 19° 28′

6 θ = sin−1 510⎛ ⎞⎜ ⎟⎝ ⎠

θ = 30° ∠ ABC = 30° The answer is A.

7 sin θ = 32

θ = sin−1 32

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 60° The answer is D. 8

sin θ = 610

θ = sin−1 0.6 ≈ 37°

9 θ = cos−1 1040

⎛ ⎞⎜ ⎟⎝ ⎠

= 75° 31′

10 error = sin−1 1.510

⎛ ⎞⎜ ⎟⎝ ⎠

= 8° 38′ 11

θ = tan−1 3.530

⎛ ⎞⎜ ⎟⎝ ⎠

= 6.65 ≈ 7°

Between the two posts the footballer should kick within an

angle of 2 × 7 = 14° 12

θ = sin−1 20250

⎛ ⎞⎜ ⎟⎝ ⎠

= 4°35′

Exercise 4D — Applications of right-angled triangles 1

tan θ = oppadj

tan 6° = 60x

x tan 6° = 60

x = 60tan 6°

= 570.86 The fire is 571 metres away from

the tower. 2

tan θ = oppadj

tan 55° = 20x

x = 20 tan 55° x = 28.6 h = x + 1.7 = 28.6 + 1.7 = 30.3 The building is 30 metres high.

3

tan θ = oppadj

tan 25° = 57x

x tan 25° = 57

x = 57tan 25°

= 122.2

tan θ = oppadj

tan 15° = 57y

y tan 15° = 57

y = 57tan 15°

= 212.7 Distance travelled = y − x = 212.7 − 122.2 = 90.5 The ship travels 91 metres. 4

tan θ = oppadj

tan 27° = 22x

x = 22tan 27°

= 43.18 The river is 43.18 m wide. 5

a tan θ = oppadj

tan 62° = 42x

x tan 62° = 42

x = 42tan 62°

x = 22.33 Denis is 22.33 m from the

building.


b tan θ = oppadj

tan 68° = yx

tan 68° = 22.33

y

y = 22.33 tan 68° = 55.27 h = y − 42 = 55.27 − 42 = 13.27 The height of the crane is

13.27 metres. 6 4.2 km = 4200 m

tan θ = oppadj

tan θ = 2004200

= 0.0472 θ = tan−1 (0.0472) = 2.72631 = 2°44′ The angle of depression is 2°44′. 7 a

b

tan θ = oppadj

tan 42° = 2500

x

x = 2500 tan 42° x = 2251.01

tan θ = oppadj

tan 55° = 2500

y

y = 2500 tan 55° y = 3570.37 Distance apart = y − x = 3570.37 − 2251.01 = 1319.36 The two survivors are 1319.36 m

apart.

8

tan θ = oppadj

tan 15.7° = 5.8x

x = 5.8 tan 15.7° = 1.6303

tan 15.9° = 5.8y

y = 5.8 tan 15.9° = 1.6522 h = y − x = 1.6522 − 1.6303 = 0.0219 km h = 0.0219 × 1000 m h = 21.9 m h ≈ 22 The height of the lookout tower

is 22 m. 9

tan θ = oppadj

tan 60° = 50x

x tan 60° = 50

x = 50tan 60°

= 503

= 50 33

tan θ = oppadj

tan 45° = 50y

y tan 45° = 50

y = 50tan 45°

= 501

= 50 d = y − x

= 50 − 50 33

The distance from A to B

is 50 3503

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠ m.

10 a N 35° W is (360 − 35)° T = 325° T b S 47° W is (180 + 47)° T = 227° T c N 58° E is 058° T d S 17° E is (180 − 17)° T = 163° T 11 a 246° T is S(246 − 180)° W = S66° W b 107° T is S(180 − 107)° E = S73° E c 321° T is N(360 − 321)° W = N39° W d 074° T is N 74° E 12 a S30° E = (180 − 30)° T = 150° T The answer is C. b 280° T = N(360 − 280)° W = N80° W The answer is D. 13

cos θ = adjhyp

cos 20° = 1800

x

x = 1800 cos 20° = 1691.4 She is 1691 m North of her starting

point. 14

a x2 = 22 + 52

= 4 + 25 = 29 x = 29 = 5.385 The final leg is 5.39 km long.

b tan θ = oppadj

tan θ = 25

tan θ = 0.4 θ = tan−1 (0.4) = 21.801°


θ = 21°48′ The bearing is N 21°48′ W. 15

tan θ = oppadj

= 820

= 0.4 θ = tan−1 (0.4) = 21.8014 = 21°48′ Bearing from starting point is

180 + 21°48′ = 201°48′T. 16

a

sin θ = opphyp

sin 30° = 2x

x = 2 sin 30° = 1.0

cos 30° = 2y

y = 2 cos 30° = 1.732

d2 = 42 + (1.732)2

=16 + 3 =19 d = 19 = 4.36 The competitor is 4.36 km from his

starting point.

b tan θ = oppadj

tan θ = 1.7324

= 0.433 θ = tan−1 (0.433) = 23.41 θ = 23°25′ The true bearing is 180 − 23°25′ = 156°35′T 17

a d 2 = 72 + 102

= 49 + 100 = 149 d = 149 = 12.2 The two hikers are 12.2 km apart.

b tan θ = oppadj

tan θ = 710

θ = 1 7tan10

− ⎛ ⎞⎜ ⎟⎝ ⎠

= tan− 1 0.7 = 34.99° ≈ 35°

∝ = 43° + 35° = 78° Bearing is 270° + 78° = 348°T or 360° − 348° = 12° N 12° W

18

a

cos θ = adjhyp

cos 40° = 3x

x = 30 cos 40° x = 22.98

sin θ = opphyp

sin 20°= 20y

y = 20 sin 20° y = 6.84 Distance south = x + y = 22.98 + 6.84 = 29.82 Ship is 29.82 km south of its starting

point. b

sin θ = opphyp

sin 40° = 30a

a = 30 sin 40° a = 19.284


cos θ = adjhyp

cos 20° = 20b

b = 20 cos 20° b = 18.794 Distance west = a + b = 19.284 + 18.794 = 38.078 ≈ 38.08 Ship is 38.08 km west of its starting

point. c

tan θ = oppadj

tan θ = 29.8238.08

= 0.783 09 θ = tan−1 (0.783 09) = 38.06 θ ≈ 38° The ship is (270 − 38)° True. That is, 232° T from its original

position. 19

a Distance = speed × time d = 40 × 3 = 120

sin θ = opphyp

sin 70° = xd

= 120

x

x = 120 sin 70° = 112.76 The distance between Arley and

Bracknaw is 112.76 km.

b time = distancespeed

Time between

A and B = 112.7645

= 2.506 hours = 2 hours and 30 minutes

Total time = 2 hours and 30 minutes + 3 hours

= 5 hours and 30 minutes

20 a

tan θ = oppadj

tan 46° = 85d

d tan 46° = 85

d = 85tan 46°

= 82.08 From A, the tower is 82.08 m away. b

tan θ = oppadj

tan 32° = 85x

x tan 32° = 85

x = 85tan 32°

= 136.03 From B, the tower is 136.03 m away. c

tan θ = oppadj

tan θ = 82.08136.03

= 0.6034 θ = tan−1 (0.6034) = 31.107 θ = 31°6′ The tower is (270° + 31°6′)

= 301°6′ T from B.

Exercise 4E — Using the sine rule to find side lengths

1 a sin sin sin

a b cA B C

= =

b sin sin sin

x y zX Y Z

= =

c sin sin sin

p q rP Q R

= =

2 a sin

bB

= sin

c

c

16sin50°

= sin 45

x°

x = 16sin 45sin50

°°

= 14.8 cm

b sin

lL

= sin

nN

sin 63

q°

= 1.9sin59°

q = 1.9sin 63sin59

°°

= 1.98 km

c sin

tT

= sin

rR

sin 84

t°

= 89sin52°

t = 89 sin84sin52

°°

= 112 mm 3 a ∠ HIG = 180° −(74° + 74°) = 32°

sin32

x°

= 18.2sin 74°

x = 18.2sin32sin 74

°°

= 10.0 m b ∠ NMP = 180° − (80° + 62°) = 38°

sin38

m°

= 35.3sin80°

m = 35.3sin35sin 50

°°

= 22.1 cm c ∠ BAC = 180° − (85° + 27°) = 68°

sin 68

y°

= 19.4sin 27°

y = 19.4sin 68sin 27

°°

= 39.6 km 4


sin 62

x°

= 9sin54°

x = 9sin 62sin54

°°

= 9.8 cm 5

∠ YXZ = 180° − (42° + 28°) = 110°

sin110

x°

= 19.2sin 42°

x = 19.2sin110sin 42

°°

= 27.0 m 6 ∠ XZY = 180° − (59° + 72°) = 49°

sin 72

y°

= 30sin 49°

y = 30sin 72sin 49

°°

= 37.8 m

7 a

b

Angle at B is 180° − (131° + 34°) =15°

sin34

x°

= 20sin15°

x = 20sin34sin15

°°

= 43.2 Distance NB is 43.2 m

c sin 49° = height43.2

height = 43.2 × sin 49° = 33 The building’s height is 33 m. 8

B = 180° − 64° = 116°

C = 180° − (48° + 116°) = 16°

sin

bB

= sin

cC

sin 116

b°

= 18sin 16°

b = 18 sin 116sin 16

°°

= 58.69

sin θ = opphyp

sin 48° = 58.69

h

h = 58.69 sin 48° = 43.62 The height of the building is

43.62 metres. 9

a C =180° − (43° + 35°) = 102°

sin

aA

= sin

cC

sin 43

x°

= 10sin 102°

x = 10 sin 43sin 102

°°

= 6.97 The distance from the second

bearing to the tree is 6.97 metres. b

cos θ = adjhyp

cos 55° = 6.97

w

w = 6.97 cos 55° = 3.9978 The width of the river is 4 metres.

10

a B = 30° + 53° = 83° C = 180° − (60° + 83°) = 37° Distance = speed × time

c = 8 × 4560

= 6 km

sin

aA

= sin

cC

sin 60

a°

= 6sin 37°

a = 6 sin 60sin 37

°°

= 8.634 The second leg is 8.63 km long.

b Speed = distancetime

= 8.6348060

= 8.634 6080

×

= 6.48 The speed is 6.48 km/h.

c sin

bB

= sin

aA

sin 83

b°

= 6sin 37°

b = 6 sin 83sin 37

°°

= 9.896 He needs to run 9.90 km to get

back to the start. 11

A = 180° − (42° + 63°) = 75° B = 63° − 12° = 51°


C = 180° − (75° + 51°) = 54°

sin

aA

= sin

cC

sin 75

a°

= 23sin 54°

a = 23 sin 75sin 54

°°

= 27.46

sin

bB

= sin

cC

sin 51

b°

= 23sin 54°

b = 23 sin 51sin 54

°°

= 22.09 The fire is 22.09 km from A and

27.46 km from B. 12

B = 85° + 15° = 100° A = 90° − 15° = 75° C = 180° − (100° + 75°) = 5°

sin

bB

= sin

cC

sin 100

b°

= 10sin 5°

b = 10 sin 100sin 5

°°

= 112.99 or b = 113 km The answer is D. 13

A = 90° − 30° = 60° C = 30° + (90° − 10°) = 110°

sin

cC

= sin

aA

sin 110

c°

= 8sin 60°

c = 8 sin 110sin 60

°°

= 8.68 The answer is B. 14

A = 180° − (40° + 25°) = 115° B = 25° C = 40°

sin

bB

= sin

aA

sin 25

x°

= 37sin 115°

x = 37 sin 25sin 115

°°

= 17.25

sin θ = opphyp

sin 50° = 17.25

y

y = 17.25 sin 50° = 13.21 h = 37 − y = 37 − 13.21 = 23.79 Total length of rope required = 2 + x + h = 2 + 17.25 + 23.79 = 43.04 metres

Therefore 45 m is enough rope since only 43 m is required.

Exercise 4F — Using the sine rule to find angle sizes

1 a sin Aa

= sin Bb

sin10046

° = sin32

θ

sinθ = 32sin10046

°

θ = 43°

b sin18.9

φ = sin 6029.5

°

sin φ = 18.9sin 6029.5

°

φ = 34°

c sin79

α = sin117153

°

sin α = 79sin117153

°°

α = 27°

d sin23.6

θ = sin 7523.6

°

sin θ = 23.6sin 7523.6

°

θ = 75°

e sin16.5

β = sin8627.6

°

sin β = 16.5sin8627.6

°

β = 37°

f sin27

θ = sin170156

°

sin θ = 27sin170156

°

θ = 2°

2 sin7

θ = sin3613

°

sin θ = 7sin3613

°

The answer is B. 3 There is not enough information to

solve triangle B. The answer is B. 4

sin12

θ = sin5616

°

sin θ = 12sin5616

θ = 38° 5

sin4.2

θ = sin 275.6

°

sin θ = 4.2sin 275.6

°

θ = 20° 6


sin45

θ = sin 4532

sin θ = 45sin 4532

θ = 84° 7 a

Let ∠ XZY = θ

sin13.7

θ = sin 6014.2

°

sin θ = 13.7sin 6014.2

°

θ = 57° b ∠ YXZ = 180° − (60° + 57°) = 63° 8

sin8

θ = sin 659

°

sin θ = 8sin 659

°

θ = 54° (to the nearest degree.) 9

a sin14

θ = sin9520

°

sin θ = 14sin9520

°

θ = 44°13′ Third angle is 180° − (95° + 44°13′) = 40°47′

sin 40 47

x′°

= 20sin95°

x = 20sin 40 47sin95

′°°

= 13.11 km b

The bearing for the 20 km leg is N 20°47′ W.

Exercise 4G — Using the cosine rule to find side lengths 1 a a2 = b2 + c2 − 2 bc cos A b r2 = p2 + q2 − 2 pq cos R c n2 = l2 + m2 − 2 lm cos N 2

c2 = a2 + b2 − 2ab cos C = 3.42 + 7.82 − 2 × 3.4 ×

7.8 × cos 80° = 11.56 + 60.84 − 9.2103 = 63.1897 c = 63.1897 = 7.9492 c = 7.95 3 a b2 = a2 + c2 − 2ac cos B

x2 = 142 + 122 − 2 × 14 × 12 cos 35°

= 196 + 144 − 275.235 = 64.765 x = 64.765 = 8.05 m b r2 = p2 + q2 − 2pq cos R

= 212 + 132 − 2 × 21 × 13 cos 42°

= 441 + 169 − 405.757 = 204.243 r = 204.243 = 14.3 cm c x2 = y2 + z2 − 2yz cos X

= 122 + 122 − 2 × 12 × 12 cos 60°

= 144 + 144 − 144 = 144 x = 144 = 12 m 4 a x2 = z2 + y2 − 2zy cos X

= 1122 + 1142 − 2 × 112 × 114 cos 110°

= 34273.826 x = 34273.826 = 185.1 cm b b2 = a2 + c2 − 2ac cos B

= 9.72 + 6.12 − 2 × 9.7 × 6.1 cos 130°

= 207.3675 b = 207.3675 = 14.4 m c q2 = p2 + r2 − 2pr cos Q

= 632 + 432 − 2 × 63 × 43 cos 160°

= 10909.2546 q = 10909.2546 = 104.4 mm

5

a2 = b2 + c2 − 2bc cos A = 64.52 + 38.12 − 2 × 64.5

× 38.1 × cos 58°34′ = 4160.25 + 1451.61 − 4914.9

× 0.5215 = 5611.86 − 2563.1504 = 3048.7096 a = 3048.7096 = 55.215 a = 55.22 6

b2 = a2 + c2 − 2ac cos B = 102 + 172 − 2 × 10 × 17

× cos 115° = 100 + 289 − 340 ×

(− 0.4226) = 389 + 143.6902 b2 = 532.6902 b = 532.6902 = 23.080 = 23.08

cos A = 2 2 2

2b c a

bc+ −

= 2 2 223.08 10 17

2 23.08 10+ −

× ×

= 532.6902 100 289461.6

+ −

= 343.6902461.6

= 0.744 563 A = cos−1 (0.744 563) = 41.878 = 41°53′ C = 180° − (41°53′ + 115°) = 23°7′ 7

x2 = 7.92 + 8.62 − 2 × 7.9 × 8.6 cos 48°

= 45.4485 x = 45.4485 = 6.7 km = 7 km

The two walkers are 7 km apart to the nearest metre.

8


x2 = 202 + 552 − 2 × 20 × 55 cos35°

= 1622.8655 x = 1622.8655 = 40.3 m The cricketer must run 40 metres

to field the ball. 9

C = 180° − (30° + 40°) = 110° c2 = a2 + b2 − 2ab cos C

= 12002 + 15002 − 2 × 1200 × 1500 cos 110°

= 1 440 000 + 2 250 000 − 3 600 000 × (−0.342 021 43)

= 3 690 000 + 1 231 272.516 = 4 921 272.516 C = 492127 2.516 = 2218.394 The two rowers are 2218 m apart. 10

C = 117° − 53° = 64° c2 = a2 + b2 − 2ab cos C

= 16.22 + 31.62 − 2 × 16.2 × 31.6 cos 64°

= 1261 − 448.8219 = 812.1781 c = 812.1781 = 28.499 The ships are 28.5 km apart. 11

sin θ = opphyp

sin 47° = 68x

a sin 47° = 68

a = 68sin 47°

a = 92.98

sin θ = opphyp

sin 15° = 68b

b sin 15° = 68

b = 68sin 15°

b = 262.73 47° − 15° = 32° y2 = a2 + b2 − 2ab cos 32° = 92.982 + 262.732 − 2

× 92.98 × 262.73 × cos 32° = 77 672.333 − 41 433.315 = 36 239.018 y = 36 239.018 = 190.37

Speed = distancetime

= 190.37 10001060

÷

≈1.14 Speed of the yacht is 1.14 km/h.

Sin θ = opphyp

Sin 47° = 68x

x = 68sin 47

x = 92.98

sin32

x°

= 92.98sin15°

x = 92.98 sin32sin15

× °°

= 190.368 m

Speed = distancetime

= 0.19037km0.167hour

= 1.14 kmh

12

c2 = a2 + b2 − 2ab cos C = 52 + 62 − 2 × 5 × 6 cos 105° = 76.5291 c = 76.5291 = 8.748 The answer is E. 13

Angle at B = 150°−20° = 130° x2 = a2 + c2 − 2ac cos B = 422 + 582 − 2 × 42 × 58 cos130° = 8259.6612 x = 8259.6612 = 90.88 km The answer is B. 14

x2 = 292 + 292 −2 × 29 × 29 cos 145° = 3059.8137 x = 3059.5137 = 55.316 cm Length of backing is 55 cm.

Exercise 4H — Using the cosine rule to find angles

1 a cos A = 2 2 2

2b c a

bc+ −

b cos Q = 2 2 2

2p r q

pr+ −

c cos P = 2 2 2

2a m p

am+ −

2 a cos A = 2 2 2

2b c a

bc+ −

= 2 2 211 8 132 11 8+ −× ×

= 16176

A = cos−1 16176⎛ ⎞⎜ ⎟⎝ ⎠

= 85°


b cos B = 2 2 2

2a c b

ac+ −

= 2 2 22.8 3.2 4.02 2.8 3.2

+ −× ×

= 2.0817.92

B = cos−1 2.0817.92⎛ ⎞⎜ ⎟⎝ ⎠

= 83°

c cos O = 2 2 2

2n m o

n m+ −× ×

= 2 2 25.4 6.2 4.52 5.4 6.2

+ −× ×

= 47.3566.96

O = cos−1 47.3566.96

⎛ ⎞⎜ ⎟⎝ ⎠

= 45°

3 a cos θ = 2 2 26 5 112 6 8+ −× ×

= 2196−

θ = cos−1 2196−⎛ ⎞

⎜ ⎟⎝ ⎠

= 103°

b cos α = 2 2 24.2 6.1 9.62 4.2 6.1

+ −× ×

= 37.3151.24−

α = cos–1 37.3154.24−⎛ ⎞

⎜ ⎟⎝ ⎠

= 137°

c cos θ = 2 2 29.2 12.9 4.22 9.2 12.9

+ −× ×

= 233.41237.36

θ = cos−1 233.41237.36

⎛ ⎞⎜ ⎟⎝ ⎠

= 10° 4

Largest angle is opposite the longest

side.

cos A = 2 2 2

2b c a

bc+ −

= 2 2 2207 296 356

2 207 296+ −

× ×

= 42 849 87 616 126 736122 544

+ −

= 3729122 544

= 0.030 43

A = cos−1 (0.030 43) = 88.256 = 88°15′ 5

Smallest angle is opposite the shortest side.

cos B = 2 2 2

2a c b

ac+ −

= 2 2 26 8 42 6 8+ −× ×

= 36 64 1696

+ −

= 8496

= 0.875 B = cos−1 (0.875) = 28.955 = 28°57′ 6

cos A = 2 2 2

2b c a

bc+ −

= 2 2 217.3 26.4 23.62 17.3 26.4

+ −× ×

= 439.29913.44

= 0.480 92 A = cos−1 (0.480 92) = 61.2546 A = 61°15′

cos B = 2 2 2

2a c b

ac+ −

= 2 2 223.6 26.4 17.3

2 23.6 26.4+ −

× ×

= 954.631246.08

= 0.7661 B = cos−1 (0.7661) = 39.994 = 39°60′ = 40° C = 180° − (61°15′ + 40°) = 78°45′ 7 Angle opposite the 12-cm side

cos θ = 2 2 214 17 122 14 17

+ −× ×

= 341476

θ = cos−1 341476

⎛ ⎞⎜ ⎟⎝ ⎠

= 44° Angle opposite the 14-cm side

cos α = 2 2 212 17 142 12 17

+ −× ×

= 237408

α = cos−1 237408

⎛ ⎞⎜ ⎟⎝ ⎠

= 54° Third angle = 180° − (44° + 54°) = 82° 8

cos θ = 2 2 28 5 4.6

2 8 5+ −

× ×

= 67.8480

θ = cos−1 67.8480

⎛ ⎞⎜ ⎟⎝ ⎠

= 32° The two roads diverge at 32° 9

a B = 180° − (34° + 68°) = 78° b2 = a2 + c2 − 2ac cos B = 72 + 122 − 2 × 7 × 12 cos 78° = 49 + 144 − 168 × 0.2079 = 193 − 34.9292 = 158.0708 b = 158.0708 = 12.573 She is 12.57 km from her starting

point.

b cos C = 2 2 2

2a b c

ab+ −

= 2 2 27 12.57 122 7 12.57+ −× ×

= 63.0049175.98

= 0.358 023


C = cos−1 (0.358 023) = 69.021 = 69°1′

θ = 69°1′ − 34° θ = 35°1′ The bearing of the starting point

from the finishing point is S 35°1′ E. 10

cos A = 2 2 2

26b c a

c+ −

= 2 2 27 5.2 32 7 5.2+ −× ×

= 67.0472.8

= 0.920 88 A = cos−1 (0.920 88) = 22.945 = 22°57′ ≈ 23° The shot must be made within 23°. 11

cos A = 2 2 2

2b c a

bc+ −

= 2 2 25 110 1002 35 120

3 + −× ×

= 56258400

= 0.669 64 A = cos−1 (0.669 64) = 47.960 = 47°58′

sin θ = opphyp

sin 47°58′ = 120

h

h = 120 sin 47°58′ h = 120 sin (47.96) h = 89.12 The balloon can fly 89.12 m. 12

a B = 70° − 10° = 60° b2 = a2 + c2 − 2ac cos B

= 1502 + 802 − 2 × 150 × 80 × cos 60°

= 28 900 − 12 000 = 16 900 b = 16 900 = 130 The plane is 130 km from its starting

point.

b cos A = 2 2 2

2b c a

bc+ −

= 2 2 2130 180 1502 130 80

+ −× ×

= 80020 800

= 0.038 462 A = cos−1 (0.038 462) = 87.796 = 87°48′ θ = 180° − (87°48′ + 70°) = 22°12′ The plane is on a bearing of

S 22°12′ E from its starting point. 13

C = 35° + 90° + 10° = 135° Distance travelled by plane A

= 120 × 2560

= 50 km

Distance travelled by plane B

= 90 × 2060

= 30 km So a = 50 km, b = 30 km. c2 = a2 + b2 − 2ab cos C = 502 + 302 − 2 × 50 × 30 cos 135° = 3400 − 3000 × (−0.70711) = 3400 + 2121.3203 = 5521.3203 c = 5521.3203 = 74.3 At 10.25 am the planes are 74.3 km

apart. 14

a = 5 + 6 = 11 b = 6 + 8 = 14 c = 5 + 8 = 13 Largest angle is opposite the longest

side.

cos B = 2 2 2

2a c b

ac+ −

= 2 2 211 13 142 11 13

+ −× ×

= 94286

= 0.328 67 B = cos−1 (0.328 67) = 70.8118 B = 70°49′ The largest angle is 70°49′.

Chapter review 1 a sin46° = 0.7193 b tan76°42′ = 4.2303 c 4.9 × cos56° = 2.7400 d 8.9 × sin67°3′ = 8.1955

e 5.69cos75°

= 21.9845

f 2.5tan9 55′°

=14.2998

2 a θ = cos−1(0.5874) = 54° b θ = tan−1(1.23) = 51° c θ = sin−1(0.8) =53° 3 a θ = cos−1(0.199) = 78°31′ b θ = tan−1(0.5) = 26°34′ c θ = sin−1(0.257) = 14°54′


4 a tan 9° = 6q

q = 6tan9°

= 37.9 cm

b sin 78° =3.9x

x = 3.9 sin 78° = 3.8 m

c cos 22° = 12.6m

m = 12.6cos 22°

= 13.6 cm

d cos 22° =12.56

n

n = 12.6 cos 22° = 11.7 cm

e sin 32° = 7.8q

q = 7.8sin32°

= 14.7 cm

f tan 65° =6.8t

t = 6.5 tan 65° = 14.6 m

g tan 26°42′ =2.9g

g = 2.9 tan 26°42′ = 1.5 m

h sin77°18′ = 4.8h

h = 4.8sin 77 18′°

= 4.9 cm

i cos 83°30′ =138

z

z = 138 cos 83°30′ = 15.6 mm

j tan 29°51′ = 4.32j

j = 4.32tan 29 51′°

= 7.5 m

k sin 16°8’ =38.5

k

k = 38.5 sin 16°8′ = 10.7 m

l cos 85°12′ =63m

m = 63 cos 85°12′ = 5.3 km 5

tanθ = oppadj

tan70° =3.1x

x = 8.1 tan 70° = 8.5 m The height of the flagpole is 8.5 m. 6

sin θ = opphyp

sin34° = 4.5x

x = 4.5 sin 34° = 2.5 km The shortest distance is 2.5 km. 7

sin70° = 60x

x = 60sin 70°

= 63.9 m long. The ladder must be 63.9 m long.

8 a sin θ = 1619

θ = sin–1 1619⎛ ⎞⎜ ⎟⎝ ⎠

= 57°

b tan θ = 2.34.6

θ = tan–1 2.34.6

⎛ ⎞⎜ ⎟⎝ ⎠

= 27°

c cos θ = 43116

θ = cos–1 43116⎛ ⎞⎜ ⎟⎝ ⎠

= 68°

9 a tan θ = 4.610.8

θ =tan–1 4.610.8⎛ ⎞⎜ ⎟⎝ ⎠

= 23°4′

b cos θ = 2.96.1

θ = cos–1 2.96.1

⎛ ⎞⎜ ⎟⎝ ⎠

= 61°37′

c sin θ = 11.913.8

θ = sin–1 11.913.8⎛ ⎞⎜ ⎟⎝ ⎠

= 59°35′

10

sin θ = 5080

θ = sin–1 5080

⎛ ⎞⎜ ⎟⎝ ⎠

= 39° 11

sin θ = 2050

θ = sin–1 2050

⎛ ⎞⎜ ⎟⎝ ⎠

= 24° 12

tan θ = oppadj

tan38° =12x

x = 12 tan 38° = 9.38 m The river is 9.38 m wide. 13

a B = 90° + 25° = 115° b2 = a2 + c2 − 2ac cos B = 8.22 + 6.72 − 2 × 8.2 × 6.7 × cos 115° = 112.13 + 46.44 = 158.57 b = 158.57 = 12.59 km


b sin

aA

= sin

bB

6.7sin A

= 12.59sin 115°

6.7 sin 115° = 12.59 sin A

sin A = 6.7 sin 11512.59

°

= 0.4823 A = sin−1 (0.4823) = 28.836 = 28°50′ θ = 65° − 28°50′ = 36°10′ Direction is S36°10′E. 14

tan θ = oppadj

tan 47° = 3500

hx−

1.072 37 = 3500

hx−

h = (3500 − x) × 1.072 37

h = 3753.295 − 1.072 37x [1]

tan 72° = hx

3.077 68 = hx

3.077 68x = h

x = 3.077 68

h [2]

Substitute x into equation [1] h = 3753.295 1.072 37−

3.07768

h⎛ ⎞× ⎜ ⎟⎝ ⎠

= 3753.295 − 0.348 43h 1.348 43h = 3753.295

h = 3753.2951.348 43

= 2783.46 The height is 2783 m.

15 a sin

aA

=sin

bB

sin 20

a°

= 4.6sin 70°

a = 4.6sin 20sin 70

°°

= 1.67cm b Third angle = 180° − (31° + 28°) = 121°

sin

dD

=sin

aA

sin31

d°

= 136sin121°

d = 136 sin31sin 121

°°

= 81.7 mm c Third angle = 180°– (117° + 19°) = 44°

sin

eE

=sin

aA

sin 44

e°

= 4.6sin19°

e = 4.6sin 44sin19

°°

= 9.81 km 16

sin

yY

= sin

xX

sin56

y°

= 9.2sin38°

y = 9.2sin56sin38

°°

= 12.4 cm

17 a sin8

θ = sin 639

°

sin θ = 8 sin 639

°

= 52°

b sin4.1

α = sin1239.7

°

sin α = 4.1sin1239.7

°

α = 21°

c sin7.1

φ = sin91.2

°

sin φ = 7.1 sin91.2

°

φ = 68° 18

C = 180° − (50° + 120°) = 10°

sin

bB

= sin

aA

sin 50

b°

= 25sin 120°

b = 25 sin 50sin 120

°°

b = 22.11 m

sin

cC

= sin

aA

sin 10

c°

= 25sin 120°

c = 25 sin 10sin 120

°°

c = 5.01 m 19 a a2 = 92 + 112 −2 × 9 × 11 cos 50° = 74.728 a = 74.728 = 8.64 m b b2 = 5.72 + 4.62–2 × 5.7 × 4.6

cos 117° = 77.4573 b = 77.4573 = 8.80 m c c2 = 6.22 + 6.92 − 2 × 6.2 × 6.9 cos 125° = 138.726 c = 138.726 = 11.8 cm 20

m2 = 632 + 842 – 2 × 63 × 84 cos 68° = 7060.1638 m = 7060.1638 = 84.0 cm 21

x2 = 5002 + 5002 − 2 × 500 × 500 cos 160°

= 969846.3104 x = 969846.3104 = 984.8 The planes are 985 m apart

22 a cos θ =2 2 26 6 62 6 6+ −× ×

= 3672

θ = cos–1 3672

⎛ ⎞⎜ ⎟⎝ ⎠

= 60°

b cos θ =2 2 24.2 5.3 7.92 4.2 5.3

+ −× ×

= 16.6844.52

−

θ = cos–1 16.6844.52

−⎛ ⎞⎜ ⎟⎝ ⎠

= 112°


c cos θ =2 2 27 9 152 7 9+ −× ×

= 95126−

θ = cos–1 95126−⎛ ⎞

⎜ ⎟⎝ ⎠

= 139° 23

cos θ =2 2 28.3 12.45 7.22 8.3 12.45

+ −× ×

= 172.0525206.67

θ = cos–1 172.0525206.67

⎛ ⎞⎜ ⎟⎝ ⎠

= 34° 24

cos θ =2 2 250 80 442 50 80

+ −× ×

= 69648000

θ = cos–1 69648000

⎛ ⎞⎜ ⎟⎝ ⎠

= 29°

Modelling and problem solving 1 a

cos θ = 2 2 212 12 82 12 12

+ −× ×

= 224288

θ = cos-1 224288

⎛ ⎞⎜ ⎟⎝ ⎠

= 39° b

cos θ=2 2 212 17 82 17 12

+ −× ×

= 369408

θ = cos–1 369408

⎛ ⎞⎜ ⎟⎝ ⎠

= 25° 2 a ABT∠ =180° − 35° (straight angle) = 145°

so, angle ATB = 180° − (20°+145°) = 15°

b sin

aA

=sin

tT

sin 20

BT°

= 30sin15°

BT = 30sin 20sin15

°°

c In Δ BTC,

sin θ = opphyp

sin 35° = 30sin 20sin15

h°

°

h = 30sin 20 sin 35sin15

° °°

d height = 30sin 20 sin 35sin15

° °°

= 22.7 m

3 a i CM = 1202

= 60 mm

ii BD = 2 2200 200 2 200 200 cos160+ − × × × = 394 mm iii ∠MBC = 10° b i CM = 125 mm

BD = 2 22 200 125× − = 312 mm ii BD = 2 × BM = 2 × 200 cos 70 = 136.8 ≈ 137 mm AC = 2 × CM = 2 × 200 sin 70 = 375.88 ≈ 376 mm 4

Tower B is closer.

17sin 40 sin 70

b =

b = 11.6 km The second tower is closer at 11.6 km from the fire. 5

20sin80 sin 60

a =

a = sin8020sin 60

×

= 22.7 km

20sin 40 sin 60

b =

sin 4020sin 60

b = ×

= 14.9 km

Chapter 4 — Triangle trigonometry - Weebly€¦ · Triangle trigonometry MB11 Qld-4 63 Exercise...

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Transcript of Chapter 4 — Triangle trigonometry - Weebly€¦ · Triangle trigonometry MB11 Qld-4 63 Exercise...