New Century Maths 12 Mathematics General 2 HSC course
Worked Solutions
Chapter 5: The sine and cosine rules
SkillCheck
O
H
12

37
1 a sin θ 
A
H
35

37
b cos θ 
O
A
12

35
c tan θ 
2 sin θ 
12
37
 12 
θ  sin 1  
 37 
θ  19
3 0.15  0.15  60
 9
37.15  379
4 Using a scientific calculator:
a 25 sin 1825  7.8981...
 7.90
b
9
 10.612...
cos 32
 10.61
c 42  82  2  4  8 cos 116  108.055...
 108.06
d
17 sin 40
 11.785...
sin 68
 11.79
e
62  7 2  112
 0.4285...
267
  0.43
f
1
 9  5 sin 38  13.8523...
2
 13.85
5 cos (40) 
k
6
k  6 cos  40 
 4.5962...
 4.60
6 a x  360  95 133
 132
b y  180  60  57
(Angles in a revolution add to 360°.)
(Angle sum of a triangle is 180°.)
 63
c z  180  48
 132
7 a cos A 
(Angles on a straight line are supplementary.)
15
17
 15 
A  cos 1  
 17 
 28.072...
 28
b sin B = 0.7329
B  sin 1  0.7329 
 47.130...
 47748.2
 478
8
hm
73 o
80 m
h
80
h  80 tan 73
tan 73 
 261.668...
 261.7 m
Exercise 5-01: Right-angled triangle trigonometry
x
14
x  14 sin  36 
1 a sin (36) 
 8.2289...
 8.23 m
20
t
b sin (51) 
t
20
sin  51 
 25.7351...
 25.74 cm
y
5
y  5 cos  48 
c cos (48) 
 3.3456...
 3.35 m
d tan  206  
p
34
p
34
tan  206 
 92.9093...
 92.91 mm
e tan  62  
d
1.4
d
1.4
tan  62 
 0.7443...
 0.74 m
8
h
f cos (68) 
8
cos  68 
h
 21.3557...
 21.36 cm
w
27
w  27 sin 14 
g sin (14) 
 6.5318...
 6.53 mm
k
2.4
k  2.4 tan  4032 
h tan  4032  
 2.0522...
 2.05 m
i
sin (77.3) 
p
48
p
48
sin  77.3 
 49.2037...
 49.20 m
2 a sin (θ) 
7
12
7
θ  sin 1  
 12 
 35.6853...
 36
b tan (θ) 
6.5
7.1
 6.5 
θ  tan 1 

 7.1 
 42.4738...
 42
c tan (θ) 
22
10
 22 
θ  tan 1  
 10 
 65.5560...
 66
d cos (θ) 
3
5.4
 3 
θ  cos 1 

 5.4 
 56.2510...
 56
e sin (θ) 
11
19
 11 
θ  sin 1  
 19 
 35.3765...
 35
f
tan (θ) 
2
15
 2
θ  tan 1  
 15 
 7.5946...
 8
3 a cos (α) 
6
23
 6 
α  cos 1  
 23 
 74.8783...
 745242
 7453
b cos (α) 
12.5
4.8
 12.5 
α  cos 1 

 14.8 
 32.3714...
 322217
 3222
c tan (α) 
10
9
 10 
α  tan 1  
9
 48.0127...
 48046
 481
4
5 The diagram should have a right angle at T, so C is eliminated.
U = 50°, so B is eliminated.
TV = 10 cm, so A is eliminated.
The correct answer is D.
6
120 m
90 m
x
sin  x  
96
120
 96 
x  sin 1 

 120 
 53.1301...
 53
The correct answer is D.
7 a
5.8 m
72
x
o
x
5.8
x  5.8 sin  72 
sin  72  
 5.5161...
 5.5 m
The ladder reaches 5.5 m up the wall.
b
5.8 m
72
o
y
y
5.8
y  5.8 cos  72 
cos  72  
 1.7922...
 1.8 m
The foot of the ladder is 1.8 m from the base of the wall.
8 tan (θ) 
15.3
3.2
 15.3 
θ  tan 1 

 3.2 
 78.1868...
 7811 13
 7811
x
38
x  38 sin  56 
9 sin (56) 
 31.5034...
 31.5 m
10 sin  2812  
h
46
h  46 sin  2812 
 21.7373...
 21.7 km
11
o
o 8
82
12.5 m
d
d
12.5
d  12.5 tan  82 
tan  82  
 88.9421...
 88.9 m
12
8.5
o
x
6.1 m
x
6.1
x  6.1 tan  8.5 
tan  8.5  
 0.9116... m
h  2.4  x
 2.4  0.9116...
 3.3 m
The correct answer is B.
13
28 o
62 o
h
420 m
420
h
420
h
tan  62 
tan  62  
 223.3179...
 223 m
x
25
x  25 tan  217 
14 tan  217  
 0.9968
d  1.3  x
 1.3  0.9968
 2.2968...
 2.3 m
15
280 m
22 o
y
280
y
280
y
tan  22 
tan  22  
 693.0243...
280 m
30
o
z
280
z
280
z
tan  30 
tan  30  
 484.9742...
x yz
 693.0243...  484.9742...
 208 m
16
63 o
a
63 o
145 m
VT
145
VT  145 tan  63 
tan  63  
49 o
b
49 o
145 m
VW
145
VW  145 tan  49 
tan  49  
d = VT – VW = 145 tan 63° – 145 tan 49°
= 145(tan 63° – tan 49°)
(as required)
17
10 m
73
x
o
tan  73  
10
x
10
tan  73 
x
 3.0573...
16 m
34
o
y
tan  34  
y
16
y
16
tan  34 
 23.7209...
d  x y
 3.0573...  23.7209...
 26.78 m
18
450
x
12
o
x
450
x  450 sin 12 
sin 12  
 95.5602...
1100 m
95.56...
a
95.5602...
1100
 95.5602... 
a  sin 1 

 1100 
 4.879...
 4.9
sin  a  
Exercise 5-02: Bearings
1 a 332° lies between 270° and 360°, so it is in the NW quadrant.
b 245° lies between 180° and 270°, so it is in the SW quadrant.
c 169° lies between 90° and 180°, so it is in the SE quadrant.
d 007° lies between 0° and 90°, so it is in the NE quadrant.
e 218° lies between 180° and 270°, so it is in the SW quadrant.
f 095° lies between 90° and 180°, so it is in the SE quadrant.
2 a 270° – 33° = 237°
b 059°
c 270° + 24° = 294°
d 180° – 50° = 130°
e 180° + 16°46ʹ = 196°46ʹ
f 360° – 42°15ʹ = 317°45ʹ
3 a
b
c
d
e
f
4
a The true bearing of NW is 315°.
b The true bearing of SW is 225°.
c The true bearing of NE is 045°.
5 a The bearing of B from A is 097°.
b The bearing of B from A is 180° – 108° = 072°.
c The bearing of B from A is 180° – 37° = 143°.
6 In each diagram, T is 38 km due north of U.
The angle at T must be 180° – 145° = 35°.
The correct answer is C.
7
7.2 km
24
o
204 - 180
s
w
w
7.2
w  7.2 sin  24 
a sin  24  
 2.928...
Amelia is 2.9 km west of the camp.
s
7.2
s  7.2 cos  24 
b cos  24  
 6.577...
Amelia is 6.6 km south of the camp.
c
Camp
7.2 km
24
24
o
o
s
w
The bearing of the camp from Amelia’s current position is 024°.
8
R
x
4.5 km
x
M
6.4 km
tan  x  
4.5
6.4
 4.5 
x  tan 1 

 6.4 
 35
The bearing of Minto Bay from Rutu Point is 090° + 035° = 125°.
The correct answer is A.
9
18.5 km
52
d km
o
The angle in the triangle is 322° – 270°.
18.5
d
18.5
d
tan  52 
a tan  52  
 14.5
The distance travelled by the ship is 14.5 km.
b Time =
distance
speed
14.4537...
48
 0.3011... h

 0.3011...  60 min
 18 min
10 a The plane flies for 3.5 hours.
Distance travelled = 3.5  700 = 2450 km.
A
60 o
s
2450 km
P
s
sin  60  
2450
s  2450 sin  60 
 2121.762...
The plane is 2122 km south of the airport.
b
A
60 o
2450 km
o
30
P
The bearing of the airport from the plane is 360° – 30° = 330°.
11
C
x
11 km
B
1.1 km
tan  x  
1.1
11
 1.1 
x  tan 1  
 11 
 5.7105...
The true bearing from Camden is 180° + 5.7105…°  186°.
12
L
x
1.5 km
S
3 km
tan  x  
3
1.5
 3 
x  tan 1 

 1.5 
 63.434...
The true bearing from the lookout to her starting point is 180° – 63.434…°  117°.
13 The angle in the triangle at the Harriside vertex is 180° – 38° – 76° = 66°.
The bearing of Landers from Harriside is 90° – 66° = 024°.
14
a
Buoy 1
45 o
b
4.2 km
Buoy 2
a
4.2
a  4.2 cos  45 
cos  45  
 2.9698
b
4.2
b  4.2 sin  45 
sin  45  
 2.9698
Buoy 3
2.8 km
d
45
o
Buoy 2
c
c
2.8
c  2.8 cos  45 
cos  45  
 1.9799
d
2.8
d  2.8 sin  45 
sin  45  
 1.9799
Buoy 1
2.9698-1.9799
=0.9899 km
x
Buoy 3
2.9698+1.9799=4.9497 km
0.9899
tan  x  
4.9497
 0.9899 
x  tan 1 

 4.9497 
 11
The bearing of the start from the finish is 270° + 11° = 281°.
Exercise 5-03: Trigonometry with obtuse angles
1 a sin θ is positive for obtuse θ.
b sin θ is positive for acute θ.
c tan θ is negative for obtuse θ.
d cos θ is negative for obtuse θ.
e cos θ is positive for acute θ.
f tan θ is positive for acute θ.
2 Using a scientific calculator:
a cos 146° = –0.8290
b tan 120° = –1.7321
c sin 163.7° = 0.2807
d tan 175° = –0.0875
e cos 105°23ʹ = –0.2653
f sin 116.2° = 0.8973
g cos 129°57ʹ = –0.6421
h sin 100.29° = 0.9839
i tan 143°38ʹ = –0.7364
3 a tan θ is positive if θ is acute.
b cos θ is negative if θ is obtuse.
c sin θ is positive if θ is acute or obtuse.
d cos θ is positive if θ is acute.
e tan θ is negative if θ is obtuse.
4 a sin A = 0.8756
A  sin 1  0.8756 
 61.116...
 617
b cos A 
15
17
 15 
A  cos 1  
 17 
 28.072...
 284
c tan A = 2.0341
A  tan 1  2.0341
 63.8204...
 6349
d cos A = 0.2925
A  cos 1  0.2925
 72.992...
 730
e tan A 
15
4
 15 
A  tan 1  
4
 75.068...
 754
f sin A 
12
18
 12 
A  sin 1  
 18 
 41.810...
 4149
g tan A = 0.8042
A  tan 1  0.8042 
 38.806...
 3848
h cos A 
3
10
 3
A  cos 1  
 10 
 72.542...
 7233
i
tan A 
2
14
 2
A  tan 1  
 14 
 8.130...
 88
5 a cos θ = –0.2566
θ  cos 1  0.2566 
 104.868...
 105
b tan θ  
6
11
 6
θ  tan 1   
 11 
 28.610...
For obtuse θ:
θ  28.610...  180
 151.389...
 151
c sin θ = 0.7890
θ  sin 1  0.7890 
 52.092...
For obtuse θ:
θ  180  52.092...
 127.907...
 128
d cos θ  
3
5
 3
θ  cos 1   
 5
 126.869...
 127
e sin θ 
1
4
1
θ  sin 1  
4
 14.477...
For obtuse θ:
θ  180  14.477...
 165.522...
 166
f tan θ = –3.8522
θ  tan 1  3.8522 
 75.447...
For obtuse θ:
θ  75.447...  180
 104.552...
 105
g sin θ 
4.7
5
 4.7 
θ  sin 1 

 5 
 70.051...
For obtuse θ:
θ  180  70.051...
 109.948...
 110
h cos θ  
6
15
 6
θ  cos 1   
 15 
 113.578...
 114o
i
tan θ  
7.3
14.1
 7.3 
θ  tan 1  

 14.1 
 27.371...
For obtuse θ:
θ  27.371...  180
 152.628...
 153
j tan θ = –1.4285
θ  tan 1  1.4285 
 55.006...
For obtuse θ:
θ  55.006...  180
 124.993...
 125
k cos θ = –0.5
θ  cos 1  0.5 
 120
l tan θ = –1
θ  tan 1  1
 45
For obtuse θ:
θ  45  180
 135
6 Using a scientific calculator.
a
14 sin 55
 19.055...
sin 143
 19.06
b 52  72  2  5  7 cos 80  61.844...
 61.84
c 42  112  2  4  11 cos 13210  196.073...
 196.07
d
e
f
12 sin 2517
 0.269...
19
 0.27
1
 3  10 sin 151  7.272...
2
 7.27
6 sin 38.5
 4.039...
sin 67.6
 4.04
g
62  52  122
 1.383...
265
 1.38
h
i
j
6.7 sin 118
 0.448...
13.2
 0.45
1
 6  7 sin 8440  20.909...
2
 20.91
132  82  102
 0.639...
2  13  8
 0.64
Exercise 5-04: The sine rule
1 a
x
18

sin  37 sin 108
x
18 sin  37 
sin 108 
 11.39
b
y
3.6

sin  51  sin  22 
y
3.6sin  51 
sin  22 
 7.47
d
5

sin  325 sin  4043
c
5 sin  325 
sin  4043 
d
 4.07
w
10

sin  38  sin  40 
d
w
10 sin  38 
sin  40 
 9.58
z
9.4

sin 111  sin  28 
e
z
9.4 sin 111 
sin  28 
 18.69
b
72

sin  60.1  sin  52.7 
f
b
72 sin  60.1 
sin  52.7 
 78.46
2
D
133
47
o
DEF  180  47  124
 9
o
x
18.1 km
124
o
E
y
a Let EF be y.
F
y
18.1

sin  47  sin 124 
y
18.1 sin  47 
sin 124 
 15.9673...
 16.0
The distance between Everett and Fiskell is 16.0 km.
b Let DE be x.
x
sin  9 

x
18.1
sin 124 
18.1 sin  9 
sin 124 
 3.415...
The distance between Everett and Doomben is 3.4 km.
3
T
46
o
18.3 cm
71
o
M
x
x
18.3

sin  46  sin  71 
x
V
18.3 sin  46 
sin  71 
 13.9 cm
4 The missing angle in the triangle is 180°  55° – 38° = 87°.
p
62

sin  87  sin  38 
p
62 sin  87 
sin  38 
 100.566...
The length of the pathway is 101 m.
5 a
b
A = 180° – 163° = 17°
(angles on a straight line)
 ABF = 180° – 90° – 17°
= 73°
FBC = 270° – 195° = 75°
 ABC = 73° + 75°
= 148°
c Using triangle ABC:
Let AC = x
x
31

sin 148  sin 15 
x
31 sin 148 
sin 15 
 63.4709...
Ian has travelled 63 nm south.
6 a
T
36 o
78
o
66
S
o
N
230 m
ST
230

sin 66 sin 36
230 sin 66
ST 
sin 36
 357.469...
 357 m
b
NT
230

sin 78 sin 36
230 sin 78
sin 36
 382.748...
 383 m
NT 
c
T
357.469 m
78
o
S
66
P
TP
357.489
TP  357.489 sin  78 
sin  78  
 349.677...
 350 m
7 a
b A  128  90
 38
B   90  38   45
 97
o
N
C  180  97  38
 45
AC
8

sin 97 sin 45
c
8 sin 97
sin 45
 11.229....
 11.2 km
AC 
8
4.8 m
x
19
o
140 o
x
4.8

sin 19 sin 140
4.8 sin 19
x
sin 140
 2.431... m
The plank is approximately 2.4 m up the ramp.
The correct answer is A.
9
R
73
37
o
o
53
o
324 o - 270 o
54 o
P
Q
8.5 km
PR
8.5

sin 54 sin 73
8.5 sin 54
sin 73
 7.190...
PR 
The distance between Pleasantville and Rowley is approximately 7.2 km.
10 a
b BTG = 180° – 90° – 38° = 52° (angle sum of ΔTGB)
BTH = 180° – 52° = 128°
c
(angles on a straight line)
BH
800

sin 128 sin 42
BH 
800 sin 128
sin 42
 942.130...
The distance between Boun and the fireworks is 942 m.
d BHT = 180° – 42° – 128° = 10°
BT
800

sin 10 sin 42
800 sin 10
sin 42
 207.610...
BT 
The distance between Boun and the top of the tower is 208 m.
GT
207.610
GT  207.610 sin  38 
e sin  38  
 127.817...
The height of the tower is 128 m.
Exercise 5-05: Using the sine rule to find an unknown angle
1 a
sin θ sin 78

15
22
15 sin 78
sin θ 
22
 0.6669...
θ  sin 1  0.6669...
 41.8297...
 4150
b
sin θ sin 99

3
18
3 sin 99
sin θ 
18
 0.1646...
θ  sin 1  0.1646...
 9.4748...
 928
c
sin θ sin 6014 '

27
24
27 sin 6014 '
sin θ 
24
 0.9765...
θ  sin 1  0.9765... 
 77.5704...
 7734 '
d
sin θ sin 133

92
140
92 sin 133
sin θ 
140
 0.4806...
θ  sin 1  0.4806...
 28.7248...
 2843
e
sin θ sin 64

5.2
8.8
5.2 sin 64
8.8
 0.5311...
sin θ 
θ  sin 1  0.5311...
 32.0801...
 325
f
sin θ sin 3048

12.5
7.1
12.5 sin 3048
sin θ 
7.1
 0.9014...
θ  sin 1  0.9014...
 64.3538...
 6421
g
sin θ sin 28.3

7.4
6.1
7.4 sin 28.3
6.1
 0.5751...
sin θ 
θ  sin 1  0.5751...
 35.1082...
 356
h Find the unnamed angle first. Call this α.
sin α sin 56

42
40
42 sin 56
sin α 
40
 0.8704...
α  sin 1  0.8704...
 60.5155...
θ  180  56  60.5155
 63.4845
 63 29
i Find the unnamed angle first. Call this α.
sin α sin 122

13.5
19
13.5 sin 122
sin α 
19
 0.6025...
α  sin 1  0.6025...
 37.0535...
θ  180  122  37.0535
 20.9465
 2057
2 a
sin  sin 24

17
8
17 sin 24
8
 0.8643...
sin  
  sin 1  0.8643...
 59.8044...
For obtuse ϕ:
  180  59.8046...
 120.1953...
 120
b
sin  sin 61

14.3
13.4
14.3 sin 61
13.4
 0.9333...
sin  
  sin 1  0.9333...
 68.9652...
For obtuse ϕ:
  180  68.9652...
 111.0347...
 111
c
sin  sin 3350

35
26
35 sin 3350
26
 0.7495...
sin  
  sin 1  0.7495...
 48.5479...
For obtuse ϕ:
  180  48.5479...
 131.4520...
 131
3
W
28 mm
40 mm
U
31
a
o
sin U sin 31

40
28
S
40 sin 31
28
 0.7357...
sin U 
U  sin 1  0.7357...
 47.372...
 4722
b For obtuse U:
U  180  47.372...
 132.627...
 13238
4
N
57
o
728 m
M
P
638 m
sin P sin 57


728
638
728 sin 57
sin P 
638
 0.956...
P  sin 1  0.956...
 73.132
 73
N  180  57  73
 50
The correct answer is A.
5
sin θ sin 14

100
27
100 sin 14
sin θ 
27
 0.896...
θ  sin 1  0.896...
 63.638...
 64
6
Q
R
50 o
21 km
31 km
S
sin R sin 50

21
31
21sin 50
sin R 
31
 0.518...
R  sin 1  0.518...
 31.260...
The bearing of Surrey from Roscoe is 270° – R.
Bearing  270  31.260...
 238.739...
 239
7
C
X
13.2 km
x
B
8.5 km
15
o
A
The bearing of C from B is the same as the angle in the triangle at C (alternate angles).
sin x sin 15

8.5
13.2
8.5 sin 15
sin x 
13.2
 0.166...
x  sin 1  0.166...
 9.593...
 10
The bearing of C from B is 010°.
8 Find THD first.
sin H sin 22

112
78
112 sin 22
sin H 
78
 0.537...
H  sin 1  0.537...
 32.540...
 33
TDH  180  22  33
 125
9 James walks from home, H, to A then B.
B
4.9 km
154
o
A
2.8 km
H
Find HBA first.
sin B sin 154

2.8
4.9
2.8 sin 154
sin B 
4.9
 0.250...
B  sin 1  0.250...
 14.506...
 15
BHA  180  154  15
 11
The bearing of B from home is 270° + 11° = 281°.
The correct answer is C.
10 DAB  90  70
 20
Now find ADB.
sin D sin 20

25
19
25 sin 20
sin D 
19
 0.450...
D  sin 1  0.450...
 26.745...
 27
ADB  27
ADC  70
(alternate angles).
The angle of elevation from D to B is BDC.
BDC  70  27
 43
Exercise 5-06: The cosine rule
1 a d 2  72  32  2  7  3 cos 50
 31.002...
d  31.002...
 5.568...
 5.57 m
b r 2  82  102  2  8  10 cos 37
 36.218...
r  36.218...
 6.018...
 6.02 cm
c k 2  62  72  2  6  7 cos 71
 57.652...
k  57.652...
 7.592...
 7.59 m
d h2  6.12  4.52  2  6.1  4.5 cos 108.4
 74.789...
h  74.789...
 8.648...
 8.65 m
e w2  172  132  2  17  13 cos 1239
 26.729...
w  26.729...
 5.170...
 5.17 m
f
y 2  152  152  2  15  15 cos 13014
 740.655...
y  740.655...
 27.214...
 27.21 mm
2 Let d be the distance between the ball and the jack.
d 2  8.42  7.92  2  8.4  7.9 cos 215
 0.352...
d  0.352...
 0.593...
 0.6 m
3 Let d be the distance between planes.
d 2  16.42  7.52  2  16.4  7.5 cos 136
 502.167...
d  502.167...
 22.409...
 22.4 km
4 a X = 360° – 230°
= 130°
b YZ 2  422  232  2  42  23 cos 130
 3534.856...
YZ  3534.856...
 59.454...
 59 km
5 Let d be the distance between the peaks.
d 2  52  82  2  5  8 cos 88.2
 86.487...
d  86.487...
 9.299...
 9.300 km or 9300 m
6 d 2  182  202  2  8  20 cos 8
 11.006...
d  11.006...
 3.317...
 3.3 m
The correct answer is C.
7 a
b The missing angle to the left of north at Y is:
180° – 130° = 50°
(co-interior angles between parallel lines)
XYZ  50  25
 75
c XZ 2  4.22  2.92  2  4.2  2.9 cos 75
 19.745...
XZ  19.745...
 4.443...
 4.4 km
8 a There are 360° in a revolution.
b θ
360
5
 72
c
8 cm
o
72
8 cm
x
x 2  82  82  2  8  8 cos 72
 88.445...
x  88.445...
 9.404...
Perimeter  5  9.404...
 47.022...
 47.02 cm
9 a
b J = 180° – 148°
= 32°
c KL2  842  602  2  84  60 cos 32
 2107.675...
KL  2107.675...
 45.909...
 46 km
18
 2  π  10
360
 3.141...
10 a Arc AB 
 3.14 cm
b AB2  102  102  2  10  10 cos 18
 9.788...
AB  9.788...
 3.128...
 3.13 cm
c The arc AB is longer by 0.01 cm.
Exercise 5-07: Using the cosine rule to find an unknown angle
202  162  92
1 a cos θ 
2  20  16

575
640
 575 
θ  cos 1 

 640 
 26.046...
 26
b cos θ 
9.42  10.12  8.62
2  9.4  10.1
116.41
189.88
 116.41 
θ  cos 1 

 189.88 
 52.188...
 52

32  7 2  92
c cos θ 
2 3 7

23
42
 23 
θ  cos 1 

 42 
 123.203...
 123
d cos θ 

102  7 2  62
2  10  7
113
140
 113 
θ  cos 1 

 140 
 36.182...
 36
362  422  562
e cos θ 
2  36  42

76
3024
 76 
θ  cos 1 

 3024 
 91.440...
 91
5.02  11.32  8.82
f cos θ 
2  5.0  11.3
75.25
113
 75.25 
θ  cos 1 

 113 
 48.246...
 48

2
S
8.4
8.4
G
12.7
T
8.42  8.42  12.7 2
a cos S 
2  8.4  8.4
20.17
141.12
 20.17 
S  cos 1 

 141.12 
 98.217...
 98

b cos G 
12.7 2  8.42  8.42
2  12.7  8.4
161.29
213.36
 161.29 
G  cos 1 

 213.36 
 40.891...
 41

c The triangle has exactly two sides of equal length, so it is isosceles.
d The triangle is isosceles so G  T .
T  41
3 cos θ 
152  202  82
2  15  20
561
600

 561 
θ  cos 1 

 600 
 20.771...
 21
The correct answer is B.
18.52  162  3.62
4 cos θ 
2  18.5  16
585.29
592
 585.29 
θ  cos 1 

 592 
 8.634...
 9

5 a cos θ 

2.12  2.12  0.92
2  2.1  2.1
8.01
8.82
 8.01 
θ  cos 1 

 8.82 
 24.747...
 2445
b The angle that each leg makes with the ground will be the same because the legs have
the same length. Therefore the triangle is isosceles.
180  24.747...
2
 77.626...
 7738
Angle 
6 The smallest angle in the triangle is opposite the shortest side.
152  17 2  82
cos θ 
2  15  17
450

510
 450 
θ  cos 1 

 510 
 28.072...
 284
7 cos θ 

2002  3502  1552
2  200  350
138 475
140 000
 138 475 
θ  cos 1 

 140 000 
 8.464...
 828
8 cos θ 

852  852  422
2  85  85
12 686
14 450
 12 686 
θ  cos 1 

 14 450 
 28.607...
 29
9 The largest angle in the triangle is opposite the longest side.
82  102  162
cos θ 
2  8  10
92

160
 92 
θ  cos 1 

 160 
 125.099...
 125
The correct answer is B.
10 The smallest angle in the triangle is opposite the shortest side.
cos θ 

82  9 2  52
289
120
144
 120 
θ  cos 1 

 144 
 33.557...
 3333
Exercise 5-08: Area of a triangle
1
 3  7 sin 50
2
 8.043...
1 a A
 8.04 m 2
1
 8  10 sin 37
2
 24.072...
b A
 24.07 cm 2
1
 6  7  sin (71)
2
 19.855...
c A
 19.86 m 2
1
 6.1  4.5  sin (108.4)
2
 13.023...
d A
 13.02 m 2
1
 13  17  sin 1239
2
 24.198...
e A
 24.20 cm 2
f
1
 15  15  sin 13014
2
 85.884...
A
 85.88 mm 2
2
7 cm
130 o
5.2 cm
1
 5.2  7  sin130
2
 27.884...
A  2
 27.9 cm 2
3
4 cm
111.2
o
7 cm
The angle between the sides is 180°  45° – 23.8° = 111.2°
1
 4  7  sin111.2
2
 26.105...
A  2
 26.1 cm 2
The correct answer is D.
1
 17  28 sin 40
2
 152.983...
4 A
 153 m 2
The correct answer is B.
5 A
1
 3  8  sin 30
2
 6 cm 2
6 a θ
360
 60
6
1
 5  5  sin 60
2
 64.951...
b A 6
 64.95 cm 2
c The bases of all the triangles making up the hexagon are equal chords of the circle so
all subtend the same angle (θ) at the centre of the circle. Angle θ was found in part a to
be 60°.
The two sides of each triangle that include angle θ are equal to each other (radii of the
circle). So we know each triangle has at least two equal sides and, therefore, two equal
base angles.
From the sum of the angles of a triangle we know that the two remaining angles (the
base angles of the triangle) will be:
180 – 60 120

 60
2
2
If all three angles of each triangle equal 60°, the triangles must be equilateral, so they
have three equal sides. The length of one side is given as 5 cm. Therefore, all sides of
the triangles have length 5 cm. The sides of this hexagon have length 5 cm.
1
 7  7  sin 60
2
 21.217...
7 A
 21.22 cm 2
8 A
3s 2
4
3  72

4
 21.217
 21.22 cm 2
The answer agrees with the answer for question 7.
9 a Find the angle between the sides measuring 3.8 m and 4.0 m. Call this angle θ.
3.82  4.02  4.52
cos θ 
2  3.8  4.0
10.19

30.4
 10.19 
θ  cos 1 

 30.4 
 70.4154...
 7025
Using a similar method, the other two angles in the triangle are 56°52ʹ and 52°43ʹ.
b A
1
 3.8  4.0  sin (7025)
2
 7.2 m2
100
 π  82
360
 55.8505...
10 a A 
 55.85 cm 2
1
 8  8  sin 100
2
 31.513...
b A
 31.51 cm 2
c Ashaded  55.85  31.51
 24.34 cm2
Exercise 5-09: Applications of the sine and cosine rules
1 a x 2  3.52  4.32  2  3.5  4.3 cos 131
 50.487...
x  50.487...
 7.105...
 7.11 m
b
p
12

sin 7418 sin 4020
12 sin 7418
sin 4020
 17.848...
 17.85 cm
p
c The third angle in the triangle is 180° – 78° – 70° = 32°
r
24

sin 78 sin 32
24 sin 78
r
sin 32
 44.300...
 44.30 mm
d a 2  112  152  2  11  15 cos 49
 129.500...
a  129.500...
 11.379...
 11.38 cm
e
y
8.8

sin 55 sin 38
8.8 sin 55
sin 38
 11.708...
 11.71 m
y
f k 2  452  192  2  45  19 cos 965
 2567.216...
k  2567.216...8...
 50.667...
 50.67 mm
2 a
sin θ sin 66

4.0
5.7
4.0 sin 66
sin θ 
5.7
 0.641...
θ  sin 1  0.641...
 39.872...
 40
92  192  152
b cos θ 
2  9  19

217
342
 217 
θ  cos 1 

 342 
 50.616...
 51
662  1082  1292
c cos θ 
2  66  108

621
14 256
 621 
θ  cos 1 

 14 256 
 92.496...
 92
d
sin θ sin 24.5

38
23
38 sin 24.5
sin θ 
23
 0.685...
θ  sin 1  0.685...
 43.247...
 43
e Let the unmarked side of this triangle be .
sin  sin 111

6.2
15.3
6.2 sin 111
sin  
15.3
 0.378...
  sin 1  0.378...
 22.229...
 22
  = 180° – 111° – 22° (angle sum of a triangle)
= 47°
47°
f cos θ 

262  192  322
2  26  19
13
988
 13 
θ  cos 1 

 988 
 89.246...
 89
3
sin  sin 2429

16
9
16 sin 2429
9
 0.736...
sin  
  sin 1  0.736...
 47.456...
For obtuse ϕ:
  180  47.456...
 132.543...
 13233
The correct answer is B.
4 a
b
h
400

sin 14 sin 33
400 sin 14
sin 33
 177.675...
 180 m
h
5 Let x be the distance Megan walked.
x 2  1002  1602  2  100  160 cos 140
 60 113.422...
x  60 113.422...
 245.180...
 245 m
Distance saved  100  160  245
 15m
6
sin θ sin 3

320
38
320sin 3
38
 0.440...
sin θ 
θ  sin 1  0.440...
 26.150...
For obtuse θ:
θ  180  26.150...
 153.849...
 15351
7 DEF = 180° – 30° = 150°
DF 2  6702  8002  2  670  800 cos 150
 2 017 279.233...
DF  2 017 279.233...
 1420.309...
 1420 km
8 a IFH + 42° = 61°
IFH = 19°
or: FHI = 180° – 61°
= 119°
IFH = 180° – 119° – 42°
= 19°
FH
120
b
=
sin 42 sin 19
FH =
120 sin 42
sin 19
(exterior angle of ΔFIH)
(angles on a straight line)
(angle sum of ΔFIH)
h
FH
h = FH × sin 61°
120 sin 42
=
× sin 61°
sin 19
c sin 61° =
= 216 m
9 a RHA = 180° – 60° = 120°
RAH = 180° – 44° – 120° = 16°
AH
85
=
sin 44
sin 16
AH =
b sin 60° =
85 sin 44 
sin 16
h
AH
h  HA sin 60
85 sin 44
 sin 60
sin 16
 185.516...

 190 m
10 Let l be the length of the longer diagonal. Note that the longer diagonal is opposite the
obtuse angle.
l 2  72  5.22  2  7  5.2 cos 130
 122.834...
l  122.834...
 11.083...
 11.1 cm
The correct answer is B.
11 a
C
124
o
A
10 km
7 km
B
x
The 124° angle and the angle marked x° are co-interior angles.
x  180  124
 56
The bearing of B from C is 116° – 56° = 060°.
b AC 2  72  102  2  7  10 cos 116
 210.371...
AC  210.371...
 14.504...
 15 km
c
sin A sin 116

10
14.504
10 sin 116
14.504
 0.619...
sin A 
A  sin 1  0.619...
A  38.293...
 38
CAB  38
The bearing of C from A is 124° – 38° = 86°.
y
C
124
o
A
10 km
7 km
116
o
B
The angle labelled y in the diagram is 180° – 86° = 94°.
The bearing of A from C is 360° – 94° = 266°.
12 a HIJ = 90° + 52°
= 142°
b HJ 2  172  242  2  17  24 cos 142
(co-interior angles)
 1508.016...
HJ  1508.016...
 38.833...
 38.8 km
J
c
MIJ
H  180  142
 38
o
142
I
M
MJI  90  38
 52
The bearing of Innes from Jokerby is 180° + 52° = 232°.
Exercise 5-10: Offset and radial surveys
1
 41  45  922.5 m 2
2
1
Area of GIF   41  105  2152.5 m 2
2
1
Area of FJE   17  48  408 m 2
2
1
Area of trapezium JEDH    43  48   113  5141.5 m 2
2
1
Area of CHD   20  43  430 m 2
2
Area of field  922.5  2152.5  408.5141.5  430
 9054.5
1 Area of CGI 
 9055 m 2
2 a
1
1
1
1
 26  26   26  41   25  22   22  42
2
2
2
2
2
 1608 m
Area 
b AB  422  222
 2248
 47.413...
BC  222  252
 1109
 33.301...
Distance walked  47.413...  47.413...
 80.714...
 80.7 m
3 a WOY  360 141 125
 94
1
 66  57  sin 94
2
 1876.417...
b Area of WOY 
 1876 m 2
c WY 2  662  572  2  66  57 cos 94
 8129.847...
WY  8129.847...
 90.165...
WX 2  662  662  2  66  66  cos 141
 15 482.495...
WX  15 482.495...
 124.428...
XY 2  57 2  662  2  57  66  cos 125
 11 920.589...
XY  11 920.589...
 109.181...
Perimeter  90.165...  124.428...  109.181...
 323.775...
 320 m
4 a ZOW  360  263  50
 147
1
 60  73  sin 147
2
 1192.759...
b A
 1200 m 2
c YOX  206 129
 77
d YX 2  622  572  2  62  57  cos 77
 5503.045...
YX  5503.045...
 74.182...
 74 m
5 C = 180° – 96° – 42° = 42°
Since ΔABC is iscoceles (C = B), AC = 50 m
CB
50

sin 96 sin 42
50 sin 96
CB 
sin 42
 74.134...
 74.31 m
6 a
1
1
1
1
 38  27   35  21   35  36   36  3
2
2
2
2
2
 1564.5 m
Area of triangles 
1
  36  38   27
2
 999 m2
Area of trapezium 
Total area  1564.5  999
 2563.5 m2
b
P
1
27
N
PN  22  27 2
 733
 27.0739...
 27.1 m
7 a
A
B
E
F
D
C
Construct a scale diagram similar to the one above, which is not to scale.
The areas of the four triangles should be approximately 2700 m2.
b
A
B
44 m
58
43 m
o
D
77 m
C
Using a scale diagram similar to the one above, which is not to scale, measure AB and
ABC. Find the area of each triangle and add the areas together. The total area should
be approximately 2700 m2.
8 a EOF = 360° – 80° – 73° – 86°
 121
1
 65  43  sin 73  1336.435...
2
1
Area of GOF   43  58  sin 86  1243.962...
2
1
Area of FOG   58  50  sin 121  1242.892...
2
1
Area of EOD   50  65  sin 80  1600.312...
2
Area of field  1336.435...  1243.962...  1242.892...  1600.312...
 5423.603...
b Area of DOG 
 5424 m 2
9 a DF = 46 + 12 + 62
= 120 m
b Area of field 
1
1
1
1
 46  58   74  58   62  32   58  32
2
2
2
2
 5400 m2
10 A 
30
30
 0  4  54  79    79  4 
3
3
39  0 
 5300 m2
11 a The answers to questions 8, 9 and 10 are similar, but they are not exactly the same due
to the differences in the accuracy of surveying methods and measurements.
b The most accurate methods are radial or offset surveys because Simpson’s rule is only a
method for approximating areas.
c Simpson’s rule is the least accurate method because it uses approximations for
calculating areas.
Sample HSC problem
a MOL = 283° – 180° = 103°
b ML2  642  602  2  64  60  cos 103
 9423.624...
ML  9423.624...
 97.075...
 97 m
1
 64  60  sin 103
2
 1870.790...
c A
 1871 m 2
Revision
1
x
11o
740 m
x
740
x  740 tan 11
tan 11 
 143.841...
740 m
18o
y
y
740
y  740 tan 18
tan 18 
 240.440...
h x y
 143.841...  240.440...
 384.282...
 384 m
2 TA  18 km
TB  15 km
The angle between TA and TB is 145° – 88° = 57°.
The correct answer is A.
3
o
20
4.2 km
Springfield
s
Shelbyville
s
4.2
s  4.2 cos 20
cos 20 
 3.946...
 3.95 km
Shelbyville is 3.95 km south of Springfield.
4 a If cos θ is positive then θ is acute.
b If tan θ is negative then θ is obtuse.
c If sin θ is positive then θ is either acute or obtuse.
5 a cos θ   0.8731
θ  cos 1   0.8731
 150.820...
 151
b sin θ = 0.2454
θ  sin 1  0.2454 
 14.205...
For obtuse θ:
θ  180  14.205...
 165.794...
 166
c tan θ = –0.9765
θ  tan 1   0.9765
  44.318...
For obtuse θ:
θ   44.318...  180
 135.681...
 136
6
SU
37

sin 106.3 sin 29.5
37 sin 106.3
sin 29.5
 72.118...
 72.1 m
SU 
7 a
sin θ sin 22

18
15
18 sin 22
sin θ 
15
 0.449...
θ  sin 1  0.449...
 26.713...
For obtuse θ:
θ  180  26.713...
 153.286...
 15317
b Call the unlabelled angle in the triangle α.
sin α sin 52.6

31
27
31 sin 52.6
sin α 
27
 0.912...
α  sin 1  0.912...
 65.797...
  180  52.6  65.797...
 61.602...
 6136
8
o
60
17 km
14 km
Let x be the distance between ships.
x
x  14  17  2  14  17 cos 60
 247
2
2
2
x  247
 15.716...
 15.7 km
9 The largest angle in the triangle is opposite the 10 m side. Call this angle θ:
cos θ 
42  7 2  102
247
The correct answer is A.
10
5.1 cm
5.1 cm
x
8.3 cm
Let x be one of the equal angles.
cos x 
5.12  8.32  5.12
2  5.1  8.3
 0.813...
x  cos 1  0.813...
 35.538...
 3532
11 a The missing angle in the triangle is 180° – 23° – 128° = 29°.
l
210

sin 128 sin 29
210 sin 128
l
sin 29
 341.334...
 340 km
1
 210  341.334...  sin 23
2
 14 003.852...
b A
 14 004 km 2
12
25 km
L
M
20 km
11 km
N
112  202  252
2  11  20
104

440
 104 
θ  cos 1 

 440 
 103.672...
 104
cos θ 
13 TH 2  2102  722  2  210  72 cos 118
 63 480.820...
TH  63 480.820...
 251.954...
 252 m
14 a POQ = 360° – 334° + 52°
= 78°
b PQ2  242  352  2  24  35 cos 78
 1451.708...
PQ  1451.708...
 38.101...
 38 m
c POS = 334° – 295°
= 129°
1
 24  38  sin 129
2
 354.378...
d A
 354 m 2
© Cengage Learning Australia Pty Ltd 2013 MAT12MMWS10025 Measurement: Applications of
trigonometry
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