06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 26
Home
Quit
Lesson 6.4 Exercises, pages 478–489
A
3. a) For each triangle, write the Sine Law equation you would use to
determine the length of AC.
i)
ii)
A
10 cm
A
40⬚
B
5.0 cm
C 80⬚
110⬚
30⬚
C
B
Use:
26
b
c
sin B
sin C
Use:
b
c
sin B
sin C
Substitute:
∠ B 30°, ∠C 80°, c 5
Substitute:
∠B 40°, ∠ C 110°, c 10
5
b
sin 30°
sin 80°
b
10
sin 40°
sin 110°
6.4 The Sine Law—Solutions
DO NOT COPY.
©P
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 27
Home
Quit
b) For each triangle in part a, determine the length of AC to the
nearest tenth of a centimetre.
i) b 5 sin 30°
sin 80°
ii) b b 2.5385. . .
AC is approximately
2.5 cm.
10 sin 40°
sin 110°
b 6.8404. . .
AC is approximately
6.8 cm.
4. a) For each triangle, write the Sine Law equation you would use to
determine the measure of ∠ E.
i)
D
8 cm
F
70⬚
ii)
10 cm
D
7 cm
E
115⬚
10 cm
F
E
sin E
sin D
e d
sin E
sin F
Substitute:
∠ D 70°, e 8, d 10
Use: e f
Substitute:
∠F 115°, e 7, f 10
sin 70°
sin E
8
10
sin E
sin 115°
7
10
Use:
b) For each triangle in part a, determine the measure of ∠ E to the
nearest degree.
i) sin E 8 sin 70°
10
Since ∠ E is acute:
∠ E sin
1
8 sin 70°
a
b
10
∠E 48.7425. . .°
∠ E ⬟ 49°
©P
DO NOT COPY.
ii) sin E 7 sin 115°
10
Since ∠ E is acute:
∠E sin1 a
7 sin 115°
b
10
∠E 39.3766. . .°
∠ E ⬟ 39°
6.4 The Sine Law—Solutions
27
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 28
Home
Quit
B
5. For each triangle, determine the measure of ∠ J to the nearest
degree.
a)
9 cm
G
50⬚
H
b)
15 cm
G
50⬚
12 cm
7 cm
J
J
Use:
H
sin J
sin H
j
h
Use:
sin J
sin H
j
h
Substitute:
∠ H 50°, j 9, h 7
Substitute:
∠H 50°, j 15, h 12
sin 50°
sin J
9
7
9 sin 50°
sin J 7
1 9 sin 50°
sin a
b ⬟ 80°
7
sin J
sin 50°
15
12
15 sin 50°
sin J 12
1 15 sin 50°
sin a
b ⬟ 73°
12
Since ∠J is obtuse:
∠ J ⬟ 180° 80°, or 100°
Since ∠J is obtuse:
∠J ⬟ 180° 73°, or 107°
6. Given the following information about each possible ⌬ABC,
determine how many triangles can be constructed.
a) c = 10 cm, a = 12 cm, ∠ A = 20°
The ratio of the side opposite the angle to the side adjacent to
a
the angle is: c 12
, which is 1.2
10
a
Since c >1, only 1 triangle can be constructed
b) c = 18 cm, a = 12 cm, ∠ A = 20°
The ratio of the side opposite the angle to the side adjacent to
a
the angle is: c 12
, which is 0.6
18
sin 20° 0.3420. . .
Since sin 20°<0.6<1, two triangles can be constructed
c) c = 18 cm, a = 12 cm, ∠ A = 50°
The ratio of the side opposite the angle to the side adjacent to
a
the angle is: c 12
, which is 0.6
18
sin 50° 0.7660. . .
a
Since c <sin 50°, no triangle can be constructed
28
6.4 The Sine Law—Solutions
DO NOT COPY.
©P
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 29
Home
Quit
7. a) For each triangle, determine the length of MN to the nearest
tenth of a centimetre.
i)
ii)
K
13 cm
9 cm
K
18 cm
40⬚
M
15 cm
100⬚ M
N
N
Determine ∠ N.
Determine ∠N.
sin N
sin M
Use: n m
Substitute: ∠ M 100°,
n 9, m 13
Use: n m
Substitute: ∠M 40°,
n 18, m 15
sin 100°
sin N
9
13
9 sin 100°
sin N 13
sin N
sin 40°
18
15
18 sin 40°
sin N 15
sin N
sin M
Since ∠ N is acute:
Since ∠N is acute:
∠ N sin1 a
∠N sin1 a
∠ N 42.9836. . .°
So, ∠ K
180° (100° 42.9836. . .°)
37.0163. . .°
∠ N ⬟ 50.4748. . .°
So, ∠ K
180° (40° 50.4748. . .°)
89.5251. . .°
9 sin 100°
b
13
Use:
k
m
sin K
sin M
18 sin 40°
b
15
Use:
k
m
sin K
sin M
Substitute: ∠ K 37.0163. . .°
∠ M 100°, m 13
Substitute: ∠K 82.5251. . .°
∠M 40°, m 15
13
k
sin 37.0163. . .°
sin 100°
13 sin 37.0163. . .°
k
sin 100°
15
k
sin 89.5251. . .°
sin 40°
15 sin 89.5251. . .°
k
sin 40°
k 7.9472. . .
MN is approximately
7.9 cm.
k 23.3350. . .
MN is approximately
23.3 cm.
b) Suppose you had been given the measures for the triangles in
part a and not the diagrams. In which triangle would there have
been an ambiguous case? Justify your choice.
i) Angle M is obtuse, so
there is only 1 triangle
ii) The ratio of the side
opposite the angle to the side
adjacent to the angle is:
NK
15
, which is 0.83
KM
18
sin 40° 0.6427. . .
since sin 40°<0.83<1, then
there is an ambiguous case
©P
DO NOT COPY.
6.4 The Sine Law—Solutions
29
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 30
Home
Quit
8. Here is a part of a proof of the Sine Law for ⌬UVW with obtuse ∠ W.
Explain each step.
V
w
h
u
␣
U
v
␤
T
W
In ⌬VUT,
h
sin U = w
h = w sin U
In ⌬VWT,
Using the sine ratio in a right triangle
Solving for h
h
Using the sine ratio in a right triangle
sin b = u
b = 180° - ␣
Supplementary angles on a line
sin b = sin (180° - ␣)
The sine of an angle is equal to
= sin ␣, or sin W
the sine of its supplement.
h
So, sin W = u
h = u sin W
u sin W = w sin U
u
w
=
sin U
sin W
sin UWV sin VWT
Solving for h
Equating expressions for h
Dividing each side by sin W sin U
9. Solve each triangle. Give the angle measures to the nearest degree
and side lengths to the nearest tenth of a centimetre.
a)
24 cm
R
35⬚
S
19 cm
T
Determine ∠T.
sin T
sin S
s
Use:
t
Substitute: ∠ S 35°,
t 24, s 19
sin T
sin 35°
24
19
24 sin 35°
sin T 19
98.5712. . .°
r
s
Use:
sin R
sin S
Substitute: ∠ R 98.5712. . .°
∠S 35°, s 19
r
19
sin 98.5712. . .°
sin 35°
19 sin 98.5712. . .°
sin 35°
Since ∠T is acute:
r
∠ T sin1 a
r 32.7555. . .
24 sin 35°
b
19
∠ T 46.4287. . .°
30
So, ∠R 180° (35° 46.4287. . .°)
6.4 The Sine Law—Solutions
∠T ⬟ 46°, ∠R ⬟ 99°,
ST ⬟ 32.8 cm
DO NOT COPY.
©P
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 31
Home
Quit
b) In ⌬KMN, ∠ M = 70°, KN = 14.1 cm, and MK = 14.5 cm
Check how many triangles can be drawn.
The ratio of the side opposite ∠M
to the side adjacent to ∠M is:
M
70⬚
14.5 cm
KN
14.1
, which is 0.9724. . .
KM
14.5
N2
sin 70° 0.9396. . .
N1
Since sin 70°<0.9724. . .<1,
14.1 cm
K
two triangles can be constructed:
This diagram is not drawn to scale.
≤ MKN1 is acute; ≤ MKN2 is obtuse.
In ≤ MKN2
In ≤ MKN1
∠N2 180° 75.0943. . .°
Determine ∠ N1.
sin N1
104.9056. . .°
sin M
Use: n m
Substitute: ∠ M 70°,
n 14.5, m 14.1
So, ∠ K
180° (70° 104.9056. . .°)
sin N1
sin 70°
14.5
14.1
14.5 sin 70°
sin N1 14.1
5.0943. . .°
∠N1 sin1 a
Substitute: ∠K 5.0943. . .°
∠ N1 75.0943. . .°
∠ M 70°, m 14.1
So, ∠ K
k
14.1
sin 5.0943. . .°
sin 70°
14.1 sin 5.0943. . .°
k
sin 70°
14.5 sin 70°
b
14.1
180° (70° 75.0943. . .°)
34.9056. . .°
Use:
k
m
sin K
sin M
Substitute: ∠K 34.9056. . .°
∠ M 70°, m 14.1
14.1
k
sin 34.9056. . .°
sin 70°
14.1 sin 34.9056. . .°
k
sin 70°
Use:
m
k
sin K
sin M
k 1.3323. . .
∠ N2 ⬟ 105°, ∠K ⬟ 5°,
MN2 ⬟ 1.3 cm
k 8.5862. . .
∠N1 ⬟ 75°, ∠K ⬟ 35°,
MN1 ⬟ 8.6 cm
©P
DO NOT COPY.
6.4 The Sine Law—Solutions
31
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 32
Home
Quit
10. For each triangle below, can you use the Sine Law to determine the
indicated measure?
If your answer is yes, determine the measure to the nearest tenth of
a unit. If your answer is no, explain why.
a)
b)
A
E
10 cm
␪
5 cm
8 cm
b
10 cm
D
F
B
100⬚
No, because no angles are given.
C
12 cm
No, because only one angle
is given and it is a contained
angle.
c)
d)
G
M
␪
10 cm
N
30⬚
h
11 cm
48⬚
K
110⬚
H
10 cm
Yes; in ≤ GHJ,
Yes
∠ H 180° 110°
Use: n
70°
g
h
Use:
sin H
sin G
Substitute: ∠ H 70°,
∠ G 30°, g 10
h
10
sin 70°
sin 30°
10 sin 70°
h
sin 30°
h 18.7938. . .
h ⬟ 18.8 cm
32
K
J
6.4 The Sine Law—Solutions
sin N
sin K
k
Substitute: ∠K 48°,
n 11, k 10
sin N
sin 48°
11
10
11 sin 48°
sin N 10
Since ∠ N is acute:
∠N sin1 a
11 sin 48°
b
10
∠N 54.8312. . .°
U 180° (48° 54.8312. . .°)
77.1687. . .°
U ⬟ 77.2°
DO NOT COPY.
©P
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 33
Home
Quit
11. A surveyor constructed this drawing
M
of a triangular lot.
a) Determine the unknown side lengths.
∠M 180° (25° 110°)
45°
Use:
p
sin P
m
sin M
Substitute: ∠ P 110°,
∠M 45°, m 270
p
270
sin 45°
270 sin 110°
p
sin 45°
sin 110°
p 358.8100. . .
MN is approximately
359 m.
25⬚
270 m
N
Use:
110⬚
P
n
m
sin N
sin M
Substitute: ∠ N 25°,
∠M 45°, m 270
n
270
sin 25°
sin 45°
270 sin 25°
n
sin 45°
n 161.3715. . .
PM is approximately
161 m.
b) Determine the total length of fencing needed to enclose the lot.
Give the answers to the nearest metre.
The total length of fencing is:
359 m 161 m 270 m 790 m
12. a) Solve ⌬PQR by drawing a perpendicular
from P to QR, then use primary
trigonometric ratios in each right
triangle formed.
Give the angle measures to the nearest
degree and the side lengths to the nearest
tenth of a centimetre.
In ≤ PRN,
In ≤ PQN,
RN
cos 74° 5
sin Q RN 5 cos 74°
RN 1.3781. . .
sin 74° PN
5
PN 5 sin 74°
PN 4.8063. . .
P
5.0 cm
R
74⬚
N
6.5 cm
Q
4.8063. . .
6.5
∠ Q 47.6830. . .°
∠ Q ⬟ 48°
cos Q QN
6.5
QN 6.5 cos 47.6830. . .°
QN 4.3760. . .
In ≤ PQR,
∠P ⬟ 180° (48° 74°)
⬟ 58°
RQ (1.3781. . . 4.3760. . .) cm
RQ ⬟ 5.8 cm
©P
DO NOT COPY.
6.4 The Sine Law—Solutions
33
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 34
Home
Quit
b) Use the Sine Law to solve ⌬PQR.
sin Q
In ≤ PQR, use: q
sin R
r
sin Q
sin 74°
5
6.5
5 sin 74°
sin Q 6.5
Since ∠ Q is acute:
∠ Q sin1 a
5 sin 74°
b
6.5
∠ Q 47.6830. . .°
∠ P 180° (47.6830. . .° 74°)
∠ P 58.3169. . .°
Use:
p
sin P
p
r
sin R
6.5
sin 74°
6.5 sin 58.3169. . .°
p
sin 74°
sin 58.3169. . .°
p 5.7541. . .
∠ Q ⬟ 48°, ∠ P ⬟ 58°, RQ ⬟ 5.8 cm
c) Which strategy for solving ⌬PQR was more efficient?
Justify your answer.
The Sine Law needed 3 calculations, while the primary trigonometric
ratios needed 6 calculations; so the Sine Law is more efficient.
13. A sailor made this sketch on her navigation chart. How much closer
is the ship at S to lighthouse A than to lighthouse B?
N
Lighthouse A
20⬚
11 km
35⬚
10⬚
Lighthouse B
E
Ship S
In ≤ SAB,
∠ ASB 90° (10° 20°)
60°
So, ∠SAB 180° (60° 35°)
85°
Use:
s
b
sin B
sin S
Use:
a
11
sin A
sin 60°
Substitute: ∠ B 35°,
Substitute: ∠A 85°
∠S 60°, s 11
a
11
sin 85°
sin 60°
11 sin 85°
a
sin 60°
b
11
sin 35°
sin 60°
11 sin 35°
b
sin 60°
a 12.6533. . .
b 7.2853. . .
The required distance is: 12.6533. . . 7.2853. . . 5.3679. . .
The ship is approximately 5 km closer to lighthouse A.
34
6.4 The Sine Law—Solutions
DO NOT COPY.
©P
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 35
Home
Quit
14. Two ships are 1600 m apart. Each ship detects a wreck on the
ocean floor. The wreck is vertically below the line through
the ships. From the ships, the angles of depression to the
wreck are 40° and 28°.
a) To the nearest metre, how far is the wreck from each ship?
The wreck could be between the ships or on one side of both ships.
Sketch a diagram to show both positions.
A
1600 m
28°
40⬚
W1
B
140⬚
W2
Case 1 ≤ABW1
∠ W1 180° (28° 40°)
112°
Case 2 ≤ABW2
∠ W2 180° (28° 140°)
12°
Use:
Use:
a
w
sin A
sin W1
a
w
sin A
sin W2
Substitute: ∠ A 28°,
∠ W1 112°, w 1600
Substitute: ∠ A 28°,
∠W2 12°, w 1600
a
1600
sin 28°
sin 112°
1600 sin 28°
a
sin 112°
a
1600
sin 28°
sin 12°
1600 sin 28°
a
sin 12°
a 810.1462. . .
a 3612.8535. . .
1600
b
Use:
sin B
sin 112°
Use:
b
1600
sin B
sin 12°
Substitute: ∠ B 40°
Substitute: ∠B 140°
1600 sin 40°
b
sin 112°
b
b 1109.2300. . .
The wreck is approximately
810 m and 1109 m from
the ships.
b 4946.6202. . .
The wreck is approximately
3613 m and 4947 m from
the ships.
1600 sin 140°
sin 12°
b) To the nearest metre, what is the depth of the wreck?
In ≤ABW1, draw the
perpendicular from W1 to AB at N.
Then, W1N is the depth of
the wreck.
In right ≤ANW1:
sin 28° W1N
AW1
In ≤ABW2, draw the perpendicular
from W2 to AB extended to M.
Then, W2M is the depth of the
wreck.
In right ≤AMW2:
sin 28° W2M
AW2
W2M 4946.6202. . . (sin 28°)
W1N 1109.2300. . . (sin 28°)
W2M 2322.2975. . .
W1N 520.7519. . .
The wreck is approximately 521 m deep or 2322 m deep.
©P
DO NOT COPY.
6.4 The Sine Law—Solutions
35
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 36
Home
Quit
15. Two people use transits to sight the top of a pole that is on the line
through the bases of the transits. The distances to the top of the pole
are: 19.8 m at an angle of elevation of 38.0°; and 14.4 m at an angle of
elevation of 57.8°. To the nearest tenth of a metre, determine the
distance between the people.
Sketch a diagram. There are 2 solutions.
In ≤ABC1,
In ≤ABC2,
∠ A 180° (38.0° 57.8°)
∠C 2 180° 57.8°
84.2°
122.2°
a
c
So, ∠ A 180° (38.0° 122.2°)
Use:
sin A
sin C1
19.8°
Substitute: ∠ A 84.2°,
a
c
Use:
∠ C1 57.8°, c 19.8
sin A
sin C
a
19.8
sin 84.2°
sin 57.8°
19.8 sin 84.2°
a
sin 57.8°
a 23.2791. . .
2
Substitute: ∠ A 19.8°,
∠C2 122.2°, c 19.8
a
19.8
sin 19.8°
sin 122.2°
19.8 sin 19.8°
a
sin 122.2°
A
19.8 m
a 7.9260. . .
The people are approximately 23.3 m or 7.9 m apart.
B
57.8°
38.0°
C2
14.4 m
57.8°
C1
C
16. Two angles in a triangle measure 60° and 45°. The longest side is
10 cm longer than the shortest side. Determine the perimeter of the
triangle to the nearest tenth of a centimetre.
Sketch a diagram.
The third angle in the triangle is:
180° (45° 60°) 75°
The shortest side is opposite
the least angle, so let BC a.
The longest side is opposite the
greatest angle, so AB a 10.
A
45°
a ⫹ 10
60°
B
75°
a
C
a
c
Use:
sin A
sin C
Substitute: ∠ A 45°, ∠ C 75°, c a 10
a
a 10
sin 45°
sin 75°
a sin 75° a sin 45° 10 sin 45°
a(sin 75° sin 45°) 10 sin 45°
a
10 sin 45°
sin 75° sin 45°
a 27.3205. . .
b
a
Use:
sin B
sin A
Substitute: ∠ B 60°, ∠ A 45°, a 27.3205. . .
b
27.3205. . .
sin 60°
sin 45°
27.3205. . . sin 60°
b
sin 45°
b 33.4606. . .
The perimeter is:
(27.3205. . . 27.3205. . . 10 33.4606. . .) cm ⬟ 98.1 cm
36
6.4 The Sine Law—Solutions
DO NOT COPY.
©P
06_ch06_pre-calculas11_wncp_solution.qxd
5/28/11
2:12 PM
Page 37
Home
Quit
17. A hiker plans a trip in two sections. Her destination is 15 km away
on a bearing of N70°E from her starting position. The first leg of the
trip is on a bearing of N10°E. The second leg of the trip is 14 km.
How long is the first leg? Give the answer to the nearest tenth of a
kilometre.
Sketch a diagram.
∠ DCB 70° 10°
60°
Check how many triangles can be drawn.
The ratio of the side opposite ∠ C
to the side adjacent to ∠ C is:
N
D
14 km
10⬚
BD
14
, which is 0.93
BC
15
B
70⬚
sin 60° 0.8660. . .
Since sin 60°<0.93<1, two triangles
can be constructed:
≤ D1CB is acute; ≤ D2CB is obtuse
N
D1
In ≤ D1CB
Determine ∠ D1.
Use:
sin D1
d
14 km
sin C
c
∠D1 68.1073. . .°
∠ B 180° (60° 68.1073. . .°)
51.8926. . .°
To determine CD1,
b
c
use:
sin B
sin C
Substitute: ∠ B 51.8926. . .°,
∠ C 60°, c 14
b
14
sin 51.8926. . .°
sin 60°
14 sin 51.8926. . .°
b
sin 60°
B
D2
15 km
Substitute: ∠C 60°, d 15, c 14
sin D1
sin 60°
15
14
15 sin 60°
sin D1 14
15 km
C
C
60⬚
In ≤ D2CB
∠D2 180° 68.1073. . .°
111.8926. . .°
So, ∠B 180° (60° 111.8926. . .°)
8.1073. . .°
To determine CD2,
use:
b
14
sin B
sin 60°
Substitute: ∠B 8.1073. . .°,
∠ C 60°, c 14
b
14
sin 8.1073. . .°
sin 60°
14 sin 8.1073. . .°
b
sin 60°
b 2.2798. . .
b 12.7201. . .
The first leg is approximately 12.7 km
or 2.3 km.
©P
DO NOT COPY.
6.4 The Sine Law—Solutions
37