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°
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 65
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.
M B 1 1 Q l d - 4 66 T r i a n g l e t r i g o n o m e t r y
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°
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 67
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
M B 1 1 Q l d - 4 68 T r i a n g l e t r i g o n o m e t r y
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
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 69
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°
M B 1 1 Q l d - 4 70 T r i a n g l e t r i g o n o m e t r y
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
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 71
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
M B 1 1 Q l d - 4 72 T r i a n g l e t r i g o n o m e t r y
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°
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 73
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
M B 1 1 Q l d - 4 74 T r i a n g l e t r i g o n o m e t r y
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′
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 75
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
M B 1 1 Q l d - 4 76 T r i a n g l e t r i g o n o m e t r y
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°
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 77
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