Chapter 7: Applications of Trigonometry
7.1 Exercises
1.
ambiguous
2.
180º; 1
3.
I; II
4.
Longer, larger
5.
Answers will vary.
6.
a.
b.
7.
8.
9.
10.
12.
sin38° sin B
=
190
125
125sin B = 190 sin38°
190 sin38°
sin B =
125
190
sin 38° ⎞
⎛
B = sin −1 ⎜
⎟ ≈ 69.4°
125
⎝
⎠
13. ∠C = 180° − (38° + 64°) = 78° ;
sin38° sin64°
=
b
75
bsin38° = 75sin64°
75sin 64°
b=
≈ 109.5 cm ;
sin 38°
sin78° sin38°
=
75
c
c sin38° = 75sin78°
75 sin 78°
c=
≈ 119.2 cm
sin 38°
angles add to 159o < 180o
side a + side b < side c
sin32° sin18.5°
=
a
15
a sin32° = 15sin18.5°
15 sin 18.5°
a=
≈ 8.98
sin 32°
sin52° sin30°
=
12
b
b sin30° = 12 sin52°
12 sin 52°
b=
≈ 18.91
sin 30°
14. ∠C = 180° − (47° +108°) = 25° ;
sin 47° sin108°
=
a
385
asin 47° = 385sin108°
385sin108°
a=
≈ 500.7 m ;
sin 47°
sin 47° sin25°
=
c
385
c sin 47° = 385sin25°
385 sin 25°
c=
≈ 222.5 m
sin 47°
sin63° sin C
=
21.9
18.6
21.9 sin C = 18.6 sin63°
18.6 sin63°
sin C =
21.9
⎛ 18.6 sin 63° ⎞
C = sin −1 ⎜
⎟ ≈ 49.2°
21.9
⎠
⎝
15. ∠C = 180° − (30° + 60°) = 90° ;
sin60° sin 30°
=
a
10 3
sin B sin105°
=
3.14
6.28
6.28 sin B = 3.14 sin105°
3.14 sin105°
sin B =
6.28
3
.14 sin 105° ⎞
⎛
B = sin −1 ⎜
⎟ ≈ 28.9°
6.28
⎝
⎠
a sin 60° = 10 3 sin30°
10 3 sin 30°
a=
= 10 in. ;
sin 60°
sin60° sin 90°
=
c
10 3
c sin 60° = 10 3 sin 90°
10 3 sin 90°
c=
= 20 in.
sin 60°
sinC sin19°
11.
=
43.2
48.5
43.2sinC = 48.5sin19°
48.5sin19°
sinC =
43.2
⎛ 48.5 sin 19° ⎞
C = sin −1 ⎜
⎟ ≈ 21.4°
43.2
⎠
⎝
717
7.1 Exercises
16. Let ∠A = 27°, ∠B = 98°, a = 7.2 m
∠C = 180° − (27° + 98°) = 55° ;
sin27° sin 98°
=
7.2
b
b sin 27° = 7.2 sin98°
7.2sin 98°
b=
≈ 15.7 m ;
sin 27°
sin27° sin 55°
=
7.2
c
c sin 27° = 7.2 sin 55°
7.2sin 55°
c=
≈ 13 m
sin 27°
18. Let ∠A = 29°, ∠B = 121°, a = 89 yd
∠C = 180° − (29° + 121°) = 30° ;
sin29° sin121°
=
89
b
b sin 29° = 89 sin121°
89sin121°
a=
≈ 157.4 yd ;
sin 29°
sin29° sin 30°
=
c
89
c sin 29° = 89 sin30°
89sin 30°
c=
≈ 91.8 yd
sin 29°
17. Let ∠A = 33°, ∠B = 102°, b = 19 in.
∠C = 180° − (33° + 102°) = 45° ;
sin102° sin33°
=
19
a
a sin102° = 19sin 33°
19sin 33°
a=
≈ 10.6 in. ;
sin102°
sin102° sin 45°
=
c
19
c sin102° = 19sin 45°
19sin 45°
c=
≈ 13.7 in.
sin102°
19. ∠C = 180°− (45° + 45°) = 90°
sin90° sin 45°
=
a
15 2
a sin 90° = 15 2 sin 45°
15 2 sin 45°
a=
= 15 mi
sin 90°
b = 15 mi
20. ∠C = 180° − (20.4° + 63.4°) = 96.2°
sin96.2° sin 20.4°
=
a
12.9
a sin 96.2° = 12.9 sin 20.4°
12.9 sin 20.4°
a=
≈ 4.5 mi ;
sin 96.2°
sin96.2° sin 63.4°
=
12.9
b
b sin 96.2° = 12.9 sin 63.4°
12.9 sin 63.4°
≈ 11.6 mi
b=
sin 96.2°
718
Chapter 7: Applications of Trigonometry
21. ∠A = 180°− (103.4°+19.6°) = 57°
sin57° sin103.4°
=
b
42.7
b sin 57° = 42.7 sin103.4°
42.7 sin 103.4°
b=
≈ 49.5 km ;
sin 57°
sin57° sin19.6°
=
c
42.7
c sin57° = 42.7 sin19.6°
42.7 sin 19.6°
c=
≈ 17.1 km
sin 57°
24. Let ∠A = 37°, ∠B = 47°, c = 27.5 cm
∠C = 180° − (37° + 47°) = 96° ;
sin96° sin 37°
=
a
27.5
a sin 96° = 27.5sin 37°
27.5sin 37°
a=
≈ 16.6 cm ;
sin 96°
sin96° sin 47°
=
27.5
b
b sin 96° = 27.5sin 47°
27.5sin 47°
b=
≈ 20.2 cm
sin 96°
22. Let ∠A = 13°, ∠B = 22°, c = 19 mi
∠C = 180°− (13°+ 22°) = 145° ;
sin145° sin13°
=
126.2
a
a sin145° = 126.2 sin13°
126.2sin13°
a=
≈ 49.5 mi ;
sin145°
sin145° sin22°
=
b
126.2
b sin145° = 126.2 sin22°
126.2sin 22°
b=
≈ 82.4 mi
sin145°
25. a.
20 cm
10 cm
30°
b.
The right triangle is a 30º-60º-90º
triangle, so the short side is half the
hypotenuse, or 10 cm.
If side a is 8 cm, it won’t reach the
base, so no triangle is possible.
c.
20
30°
12
12
Two triangles possible for a = 12.
23. Let ∠A = 56°, ∠B = 112°, c = 0.8 cm
∠C = 180° − (56° +112°) = 12° ;
sin12° sin56°
=
0.8
a
a sin12° = 0.8 sin 56°
0.8sin 56°
a=
≈ 3.2 cm ;
sin12°
sin12° sin112°
=
0.8
b
b sin12° = 0.8 sin112°
0.8sin112°
b=
≈ 3.6 cm
sin12°
d.
20 cm
25 cm
30°
One triangle possible for a = 25.
719
7.1 Exercises
26. a.
30.
6 3
9
The right triangle is a 30º-60º-90º
triangle, so the short side is half the
hypotenuse, and the longer leg is 3
times the shorter leg, 9m.
If a = 8 m, the vertical side is too short
to finish the triangle, so none is
possible.
31.
c.
10
If a = 10, two triangles are possible.
d.
15
60°
If a = 15, only one triangle is possible.
27.
28.
29.
sin B sin 59°
=
; 58sin B = 67sin 59°
67
58
67 sin 59°
sin B =
≈ 0.9902
58
B = sin−1 0.9902 ≈ 82°
or B = 180º – 82º = 98º.
For B = 82º:
A = 180°− (59°+ 82°) = 39°
sin39° sin59°
=
; asin59° = 58 sin39°
a
58
58sin 39°
a=
≈ 42.6 mi ;
sin 59°
For B = 98º:
A = 180°− (98°+ 59°) = 23°
sin 23° sin 59°
=
; a sin 59° = 58sin 23°
58
a
58 sin 23°
a=
≈ 26.4 mi
sin 59°
10
60°
6 3
sin 60°
; 15sin C = 10 3 sin 60°
15
10 3 sin 60°
= 1 ; C = 90º. (This
15
tells us that there’s only one triangle.)
B = 180° − (60°+ 90°) = 30°
This is a 30º-60º-90º triangle, so side b is
half the hypotenuse, or 5 3 in.
3 3
6 3
10 3
=
sin C =
60°
b.
sin C
sin 67° sin A
=
; 385sin A = 490sin 67°
385
490
490sin 67°
sin A =
= 1.17 . Not possible.
385
32.
sin 67° sin A
=
; 12.9sin A = 36.5sin 67°
12.9
36.5
36.5sin 67°
sin A =
= 2.6 . Not possible.
12.9
sin 30° sin C
=
; 12.9sin C = 25.8sin 30°
12.9
25.8
25.8sin 30°
sin C =
= 1 ; C = 90º. (This tells
12.9
us that there’s only one triangle.)
B = 180° − (30° + 90°) = 60°
This is a 30º-60º-90º triangle, so
b = 12.9 3 mi
sin A sin 45°
=
; 24.9sin A = 32.8sin 45°
32.8
24.9
32.8sin 45°
sin A =
≈ 0.9314
24.9
A = sin −1 0.9314 ≈ 68.7° or
180º – 68.7º = 111.3º.
For A = 68.7º:
C = 180° − (45° + 68.7°) = 66.3°
sin 66.3°
c
=
sin 45°
24.9
; c sin 45° = 24.9 sin 66.3°
24.9sin 66.3°
≈ 32.2 km
sin 45°
For A = 111.3º;
C = 180° − (45° + 111.3°) = 23.7°
c=
sin 23.7°
c
c=
720
=
sin 45°
24.9
; c sin 45° = 24.9sin 23.7°
24.9sin 23.7°
≈ 14.2 km
sin 45°
Chapter 7: Applications of Trigonometry
33.
34.
35.
36.
sin 59° sin B
=
; 58sin B = 67sin 59°
58
67
67 sin 59°
sin B =
≈ 0.9902
58
B = sin −1 0.9902 ≈ 82° or
B = 180° − 82° = 98°
For B = 82º:
A = 180° − (82° + 59°) = 39° ;
sin 39° sin 59°
=
; a sin 59° = 58sin 39°
a
58
58sin 39°
= 42.6 ft ;
a=
sin 59°
For B = 98º:
A = 180° − (98° + 59°) = 23°
sin 23° sin 59°
=
; a sin 59° = 58sin 23°
a
58
58 sin 23°
a=
≈ 26.4 ft
sin 59°
37.
sin 62°
sin A
=
25
2.6 × 10
2.9 × 1025
25
2.6 × 10 sin A = 2.9 × 1025 sin 62°
2.9 × 1025 sin 62°
sin A =
≈ 0.9848
2.6 × 1025
A ≈ 80.0° ;
B ≈ 180° − (80.0° + 62°) = 38°
sin 62°
sin 38°
=
2.6 × 1025
b
b sin 62° = 2.9 × 10 25 sin 38°
2.9 × 1025 sin 38°
b=
≈ 2.0 × 1025 mi
sin 62°
38.
sin 38° sin B
=
; 382sin 38° = 432sin B
382
432
382sin 38°
sin B =
≈ 0.5444
432
B = sin −1 (0.5444) ≈ 33° or
B = 180° − 33° = 147° . But 147º + 38º =
185º > 180º, so there is only one triangle.
C = 180°− (33°+ 38°) = 109°
sin109° sin 38°
=
; c sin 38° = 432sin109°
c
432
432sin109°
c=
≈ 663.5 cm
sin 38°
sin 51°
13
=
sin C
5.9 ×1013
6.8 ×10
6.8 × 1013 sinC = 5.9 × 1013 sin51°
5.9 × 1013 sin 51°
sin C =
≈ 0.6743
6.8 × 1013
C = sin −1 0.6743 ≈ 42.4° or
C = 180° − 42.4° = 137.6°. But 137.6º + 51º
= 188.6º > 180º, so only one triangle is
possible.
B = 180° − (42.4° + 51°) = 86.6°
sin 86.6°
sin 51°
=
b
6.8 ×1013
b sin51° = 6.8 × 1013 sin86.6°
6.8 × 1013 sin 86.6°
b=
≈ 8.7 × 1013 km
sin 51°
39.
sin 38° sin A
=
; 6.7 sin A = 10.9sin 38°
6.7
10.9
10.9sin 38°
sin A =
= 1.002 . Not possible
6.7
sin 59° sin A
=
; 398sin A = 465sin 59°
398
465
465sin 59°
sin A =
≈ 1.0015 Not possible
398
721
sin A sin 48°
=
; 27sin A = 12sin 48°
12
27
⎛ 12sin 48° ⎞
12sin 48°
sin A =
; A = sin−1⎜
⎟
27
⎠
⎝ 27
A ≈ 19.3° ;
Another possible angle: 180º – 19.3º =
160.7; 48º + 160.7º = 208.7º > 180º
No second solution possible
7.1 Exercises
40.
41.
sin 60° sin B
=
; 32sin B = 9sin 60°
32
9
⎛ 9sin 60° ⎞
9sin 60°
sin B =
; B = sin −1⎜
⎟
32
⎝ 32 ⎠
B ≈ 14.1° ;
Another possible angle:
180°−14.1° = 165.9° ; 165.9°+ 60° = 225.9°
> 180º; No second solution possible
44.
45. 135º = 3(45º), so we get
sin135° = 3sin 45° − 4 sin3 45° =
⎛
⎞3
⎛
⎞
2⎟ 3 2
2
⎜
⎜2 2 ⎟
− 4⎜
=
−
4
= 3⋅
⎟
⎜ 8 ⎟
2
2
⎠
⎝ 2 ⎠
⎝
sin 57° sin C
=
; 35.6 sin C = 40.2 sin 57°
35.6
40.2
2
3 2 2 2
−
=
2
2
2
The reference angle for 135º is 45º, and 135º
=
40.2 sin 57°
⎛ 40.2 sin 57° ⎞
; C = sin −1⎜
⎟
35.6
35.6
⎝
⎠
C ≈ 71.3° ;
Another possible angle:
180° − 71.3° = 108.7° ;
108.7° + 57° = 165.7° < 180° ; Two possible
solutions.
sin C =
42.
sin B
5.2
=
sin 65°
4.9
is in QII, so sin135° =
46. R =
2
.
2
15
b
=
≈ 9.65 cm
2 sin B 2 sin51°
47. We are given θ = 20° . Let α be the angle
with vertex at Sorus and β be the angle with
vertex at the Sun.
sin20° sin α
=
; 82 sin20° = 51sin α
51
82
82sin 20°
sin α =
≈ 0.5499
51
; 4.9 sin B = 5.2 sin 65°
5.2 sin 65°
4.9
⎛ 5.2 sin 65° ⎞
B = sin − 1 ⎜
⎟ ≈ 74.1°
4.9
⎝
⎠
Another possible angle:
180° − 74.1° = 105.9° ;
105.9° + 65° < 180° ; Two possible
solutions.
sin B =
43.
sin 29° sin B
=
; 121sin B = 321sin 29°
121
321
321sin 29°
= 1.29 . Not possible
sin B =
121
α = sin−1 0.5499 = 33.4° or
α = 180° − 33.4° = 146.6°
When α = 33.4º, the distance is the further
of the two; let d1 represent this distance.
β = 180° − (33.4° + 20°) = 126.6°
sin126.6° sin20°
=
d1
51
d1 sin20° = 51sin126.6°
51 sin 126.6°
d1 =
≈ 119.7 million miles
sin 20°
When α = 146.6º, the distance is the closer
of the two: let d2 represent this distance.
β = 180° − (146.6° + 20°) = 13.4°
sin13.4° sin20°
=
d2
51
d2 sin20° = 51sin13.4°
51 sin 13.4°
d2 =
≈ 34.6 million miles
sin 20°
sin A sin15°
=
; 52sin A = 280sin15°
280
52
280sin15°
= 1.39 . Not possible
sin A =
52
722
Chapter 7: Applications of Trigonometry
48. We are given θ = 15° . Let α be the angle
with vertex at Cirrus and β be the angle
with vertex at the Sun.
sin15° sin α
=
; 105sin15° = 70sin α
70
105
105sin15°
sin α =
≈ 0.3882
70
50. a.
b.
α = sin−1 0.3882 = 22.8° or
α = 180° − 22.8° = 157.2°
When α = 22.8º, the distance is the further
of the two; let d1 represent this distance.
β = 180° − (22.8° + 15°) = 142.2°
sin142.2° sin15°
=
d1
70
d1 sin15° = 70 sin142.2°
70sin142.2°
d1 =
≈ 165.8 million miles
sin15°
When α = 157.2º, the distance is the closer
of the two: let d2 represent this distance.
β = 180° − (157.2° + 15°) = 7.8°
sin7.8° sin15°
=
d2
70
d2 sin15° = 70sin7.8°
70sin 7.8°
d2 =
≈ 36.7 million miles
sin15°
49. a.
b.
sin29° sin B
=
; 90sin B = 200 sin29°
90
200
200sin29°
= 1.08
sin B =
90
With a range of 90 yd, the pride will not
be detected.
sin29° sin B
=
; 120 sin B = 200 sin29°
120
200
200sin 29°
sin B =
≈ 0.8080
120
B = sin −1 0.8080 ≈ 53.9° or
B = 180° − 53.9° = 126.1° .
The pride is first detected when B =
126.1º, in which case the third angle is
180° − (126.1° + 29°) = 24.9° .
sin24.9° sin29°
=
d
120
d sin29° = 120 sin24.9°
120 sin 24.9°
d=
≈ 104.2 yd
sin 29°
51. Segment SR is 55 km.
sin 40° sin ∠P
=
; 55sin ∠P = 80sin 40°
55
80
80sin 40°
≈ 0.9350; ∠P ≈ 69.2°
sin ∠P =
55
∠VRP = 180° − (40° + 69.2°) = 70.8°
Let d1 = the distance from V to P.
sin70.8° sin 40°
=
; d1 sin 40° = 55sin70.8°
55
d1
55sin 70.8°
V ↔ P d1 =
≈ 80.8 km
sin 40°
Since ∠P = 69.2°,
∠VSR = 180° − 69.2° = 110.8° .
Then ∠VRS = 180° − (110.8° + 40°) = 29.2° .
Let d2 = the distance from V to S.
sin 29.2° sin 40°
=
; d 2 sin 40° = 55sin 29.2°
d2
55
sin35° sin B
=
; 8 sin B = 15sin35°
8
15
15sin35°
= 1.08 Not possible
sin B =
8
A radar with a range of 8 mi. will not
detect the ship.
sin35° sin B
=
; 12sin B = 15sin35°
12
15
15sin 35°
sin B =
≈ 0.7170
12
B = sin −1 0.7170 ≈ 45.8° or
B = 180º – 45.8 = 134.2º.
The closest point of detection will be
when B = 134.2º, in which case the
third angle is
180° − (35° + 134.2°) = 10.8° .
sin10.8° sin35°
=
d
12
d sin35° = 12 sin10.8°
12sin10.8°
≈ 3.9 mi
d=
sin35°
V ↔S
723
d2 =
55sin 29.2°
≈ 41.7 km
sin 40°
7.1 Exercises
c.
52. Segment VR is 75 km.
sin33° sin∠O
=
; 75sin∠O = 100 sin33°
75
100
100sin 33°
sin ∠O =
≈ 0.7262
75
∠O = sin −1 0.7262 ≈ 46.6° . Then
∠MRO = 180° − (46.6° + 33°) = 100.4° .
Let d1 = the distance from M to O.
sin100.4° sin33°
=
75
d1
d1 sin33° = 75sin100.4°
75sin100.4°
M ↔ O d1 =
= 135.4 km
sin33°
Since ∠O = 46.6°,
∠MVR = 180° − 46.6° = 133.4° , and
∠MRV = 180° − (133.4° + 33°) = 13.6° .
Let d2 = the distance from M to V.
sin13.6° sin33°
=
; d2 sin33° = 75sin13.6°
75
d2
75sin13.6°
≈ 32.4 km
M ↔ V d2 =
sin 33°
We first need to find the distance
d2 − d1 in the diagram below:
246
55 °
215
215
d1
d2
Let C = the angle at the archer.
sin55° sin B
=
246
215
215sin B = 246 sin55°
246sin 55°
sin B =
≈ 0.9373
215
B = sin −1 0.9373 ≈ 69.6° or
B = 180° − 69.6° = 110.4°
When B = 69.6º, C = 180º – (69.6º +
55º) = 55.4º.
sin55° sin55.4°
=
d2
215
d2 sin55° = 215sin55.4°
215sin55.4°
≈ 216 ft
d2 =
sin55°
When B = 110.4º, C = 180º – (110.4º +
55º) = 14.6º.
sin55° sin14.6°
=
d1
215
d1 sin 55° = 215sin14.6°
215sin14.6°
≈ 66 ft
d1 =
sin55°
The target is in range for 216 – 66 =
150 ft. Moving at 10 ft/sec, the target
will be in range for about 15 seconds.
53. Let B be the angle at the target.
sin55° sin B
=
a.
246
180
180sin B = 246 sin55°
246sin 55°
sin B =
≈ 1.1, Not possible
180
The arrow won’t reach the target.
sin 55° sin 90o
=
b.
a
246
a sin 90o = 246sin 55°
a ≈ 201.5 ft
724
Chapter 7: Applications of Trigonometry
54. Let B = the angle at the target.
sin 53° sin B
=
a.
35
50
35sin B = 50sin 53°
50sin 53°
sin B =
≈ 1.14, Not possible
35
The ball won’t reach the target.
sin 53° sin 90o
b.
=
a
50
a sin 90o = 50sin 53°
a ≈ 39.9 ft
c. We first need to find the distance
d2 − d1 in the diagram below:
55.
8 sin112.9°
≈ 16.8 cm
sin26°
Angles
Sides
a
=
12 cm
A1 ≈ 41.1°
B = 26°
b = 8 cm
c1 =
C1 ≈ 112.9°
50
45
sin26° sin A
; 8sin A = 12sin 26°
=
8
12
12sin 26°
sin A =
≈ 0.6576
8
A = sin −1 0.6576 ≈ 41.1° or A = 180º –
41.1º = 138.9º. Two triangles are possible.
For A = 41.1º, C1 = 180º – (41.1º + 26º) =
112.9º.
sin112.9° sin26°
=
; c1 sin26° = 8 sin112.9°
c1
8
c1 ≈ 16.8 cm
45
d1
For A2 = 138.9º, C2 = 180º – (138.9º + 26º)
= 15.1º.
sin15.1° sin26°
=
; c2 sin26° = 8 sin15.1°
c2
8
8sin15.1°
c2 =
≈ 4.8 cm
sin 26°
Angles
Sides
a = 12 cm
A2 ≈ 138.9°
53°
d2
Let C = the angle at the quarterback.
sin53° sin B
=
50
45
45sin B = 50 sin53°
50sin 53°
sin B =
≈ 0.8874
45
B = sin −1 0.8874 ≈ 62.5° or
B = 180° − 62.5° = 117.5°
When B = 62.5º, C = 180º – (62.5º +
53º) = 64.5º.
sin64.5° sin53°
=
d2
45
d2 sin 53° = 45sin 64.5°
45sin64.5°
d2 =
≈ 50.9 yd
sin53°
When B = 117.5º, C = 180º – (117.5º +
53º) = 9.5º.
sin9.5° sin53°
=
d1
45
d1 sin53° = 45sin9.5°
45sin9.5°
≈ 9.3 yd
d1 =
sin53°
The distance over which the target is in
range is 50.9 yd – 9.3 yd = 41.6 yd.
41.6 yd
≈ 8.3 sec
5 yd / sec
56.
725
B = 26°
b = 8 cm
C 2 = 15.1°
c 2 ≈ 4.8 cm
sin 26° sin A
; 4sin A = 11sin 26°
=
4
11
11sin 26°
sin A =
= 1.2, Not possible
4
7.1 Exercises
57. From the grid, a = 9, c = 5. Consider the
right triangle in the diagram as drawn: the
shorter leg has length 4.
⎛4⎞
4
tanC = ; C = tan−1⎜ ⎟ ≈ 24°
9
⎝9⎠
Now use the Law of Sines with a = 9, c = 5,
C = 24º.
sin24° sin A
=
; 5sin A = 9 sin24°
9
5
9sin 24°
sin A =
≈ 0.7321; A ≈ 47.0°
5
or A = 180º – 47.0º = 133.0º.
For A = 47.0º, B = 180º – (47.0º + 24º) =
109.0º.
sin109.0° sin24°
=
; b sin24° = 5sin109.0°
5
b
5sin109.0°
b=
≈ 11.6
sin 24°
Angles
Sides
a = 9 cm
A1 ≈ 47.0°
B1 ≈ 109.0°
C ≈ 24°
Now use the Law of Sines with a = 82 ,
c = 10, C = 51.3º.
sin51.3° sin A
=
; 10sin A = 82 sin51.3°
10
82
82 sin 51.3°
≈ 0.7067; A ≈ 45°
10
or A = 180º – 45º = 135º. But this is not
possible since 135º + 51.3º > 180º, so there
is only one triangle.
B = 180º – (45º + 51.3º) = 83.7º.
sin83.7° sin51.3°
=
10
b
bsin51.3° = 10 sin83.7°
10sin 83.7°
b=
≈ 12.7
sin 51.3°
Angles
Sides
A ≈ 45°
a = 82 cm
sin A =
B ≈ 83.7°
C ≈ 51.3°
b1 ≈ 11.6 cm
c = 5 cm
59. First, find ∠B = 180°− (32°+ 53°) = 95° .
sin95° sin32°
=
c
42
c sin95° = 42 sin32°
42 sin 32°
c=
≈ 22.3 ft ;
sin 95°
sin95° sin53°
=
a
42
asin95° = 42sin53°
42 sin 53°
a=
≈ 33.7 ft
sin 95°
For A = 133.0º, B = 180º – (133.0º + 24º) =
23.0º.
sin23.0° sin24°
=
; b sin24° = 5sin23.0°
5
b
5sin 23.0°
b=
≈ 4.8
sin 24°
Angles
Sides
a = 9 cm
A2 ≈ 133.0°
B 2 ≈ 23.0°
C ≈ 24°
b2 ≈ 4.8 cm
c = 5 cm
60. First, find ∠B = 180° − (45° + 29°) = 106°
sin106° sin29°
=
c
50
c sin106° = 50sin29°
50 sin 29°
c=
≈ 25.2 ft ;
sin 106°
sin106° sin 45°
=
a
50
asin106° = 50sin 45°
50 sin 45°
a=
≈ 36.8 ft
sin 106°
58. The endpoints of a are (3, 7) and (12, 6), so
a = (7 − 6)2 + (3−12)2 = 82 . The
endpoints of c are (3, 7) and (11,1), so
c = (3− 11)2 + (7 −1)2 = 10 . If side b is
rotated so that it intersects c at (8,2), the
triangle will be roughly a right triangle with
hypotenuse
82 and bottom leg
c′ = (8 − 3)2 + (2 − 7)2 = 50 . Then
sin C =
50
82
b ≈ 12.7 cm
c = 10 cm
≈ 0.7809 ⇒ C ≈ 51.3° .
726
Chapter 7: Applications of Trigonometry
65. In the diagram provided, we need to find h.
48°
61. The third angle is 180º – (96º + 58º) = 26º.
Rhymes to Tarryson:
sin26° sin96°
=
27.2
RT
a
h
62°
70°
RT sin26° = 27.2sin96°
27.2sin 96°
RT =
≈ 61.7 km ;
sin 26°
Sexton to Tarryson:
sin26° sin58°
=
27.2
ST
145 yd
First, find a:
sin 48° sin70°
=
145
a
a sin 48° = 145sin70°
145sin 70°
a=
≈ 183.35 yd
sin 48°
Now use the right triangle on the right to
find h:
h
sin62° =
183.35
h = 183.35 sin 62° ≈ 161.9 yd
ST sin26° = 27.2sin58°
27.2sin 58°
ST =
≈ 52.6 km
sin 26°
62. The third angle is 180º – (32º + 102º) = 46º.
Let h be the height of the tower, and note
that it’s the side across from the 32º angle.
sin 46° sin32°
=
h
112
hsin 46° = 112sin32°
112 sin 32°
h=
≈ 82.5 ft
sin 46°
66. In the diagram provided, we need to find h.
21.8 °
Blimp
a
h
63. The third angle is 180º – (39º + 58º) = 83º.
The shortest side is across from the 39º
angle, so let d be that distance.
sin83° sin39°
=
d
5
d sin83° = 5sin39°
5sin39°
d=
≈ 3.2 mi
sin83°
26.5° 131.7° 48.3°
110 yd
The 131.7º angle was found by subtracting
48.3º from 180º (supplementary angles) and
the 21.8º was found by subtracting 26.5º and
131.7º from 180º. Find side a:
sin21.8° sin131.7°
=
a
110
a sin 21.8° = 110 sin131.7°
110 sin131.7°
a=
= 221.2 yd
sin 21.8°
Now use the big right triangle to find h:
h
sin26.5° =
221.2
64. The third angle is 180º – (110º + 48º) = 22º.
sin 22° sin 28°
=
1, 000
BC
BC sin 22° = 1, 000sin 28°
1, 000sin 28°
BC =
≈ 1253.2 m
sin 22°
h = 221.1 sin 26.5° ≈ 98.7 yd
727
7.1 Exercises
67. Let a = the side across from the 63º angle
and d = the base of the triangle.
sin27° sin 63°
=
a
5
a sin 27° = 5sin63°
5 sin 63°
a=
≈ 9.8 cm ;
sin 27°
sin27° sin 90°
=
5
d
d sin27° = 5sin 90°
5 sin 90°
d=
≈ 11 cm
sin 27°
The diameter of the circle is 11 cm, the base
of the triangle. It is a right triangle.
b.
70. a.
h
; h = 3187 sin 48° m
3187
h ≈ 2368 m
The east side forms a right triangle with
the vertical height (2368) and a 65º
angle.
2368
2368
; he =
≈ 2613
sin65° =
he
sin 65°
sin 48° =
The angle inside the triangle at
Coffeyville is 108º (supplementary
angles) and the angle at Olathe is then
49º (subract from 180).
sin 49° sin108°
=
298
d1
d1 sin 49° = 298sin108°
68. Let a = the side across from the 38º angle
and b = the side across from the 52º angle.
sin90° sin 38°
=
11
a
a sin 90° = 11sin 38°
11 sin 38°
a=
≈ 6.8 cm
sin 90°
sin90° sin 52°
=
11
b
b sin 90° = 11sin 52°
11 sin 52°
b=
≈ 8.7 cm
sin 90°
The longer chord is 8.7 cm; the triangle is a
right triangle.
298sin108°
≈ 376 . Distance from
sin 49°
Liberal to Olathe is about 376 mi.
sin 49° sin23°
=
298
d2
d2 sin 49° = 298 sin 23°
298 sin 23°
d2 =
≈ 154 . Distance from
sin 49°
Olathe to Coffeyville is about 154 mi.
d
sin72° = 3 ; d3 = 154 sin72° ≈ 146
154
Shortest distance to the border is about
146 mi.
d1 =
b.
69. a.
71. The base of the triangle is 10.2 3 cm, and
the hypotenuse is 20.4 cm. Using opposite
over hypotenuse,
10.2 3
10.2
and sin 30° =
, so
20.4
20.4
10.2 3
sin 60°
= 20.4 = 3
10.2
sin 30°
20.4
1
and sin 90° = 1 ,
We know that sin 45° =
2
sin 90°
1
=
= 2
so
1
sin 45°
sin 60° =
132 °
35°
48°
65°
1250 m
The small angle at the top of the left
hand triangle is 180 – (35+132) = 13º.
Let s be the slant height of the west
side, h the vertical height (dashed line).
sin 13°
1250
s=
=
sin 35°
s
; s sin 13° = 1250 sin 35°
1250 sin 35°
≈ 3187 m
sin13°
2
728
Chapter 7: Applications of Trigonometry
74. The area of a triangle is calculated using
1
A = bh . All three triangles have the same
2
base, and since the lines are parallel, the
height of each triangle is the same as well.
72. The unlabeled angle is 180º – (xº + 2xº) =
(180 – 3x)º.
sin x° sin(180 − 3x)°
=
50
20
20sin x° = 50 sin(180 − 3x)°
To solve this equation, graph both sides in a
graph utility (degree mode!) and use the
intersect feature. Limit the window to x’s
between 0º and 90º since any x bigger than
90º will make the two labeled angles add up
to more than 180º.
The solution works out to be x = 53.7º, in
which case 2x = 107.4º and the remaining
angle is 180º – (53.7º + 107.4º) = 18.9º.
Let b = the missing side.
sin 53.7° sin107.4°
=
50
b
b sin 53.7° = 50sin107.4°
50sin107.4°
b=
= 59.2 m
sin 53.7°
75. A diagram of the first sighting:
UFO
103 °
a
35 °
Batesville
d1
13
42°
Cave
City
We can find side a using the law of sines,
then use it to find d1 , which is the distance
from the UFO’s initial location to Batesville.
sin103° sin 42°
=
13
a
a sin103° = 13sin 42°
13sin 42°
a=
≈ 8.9275 mi
sin103°
d1
cos 35° =
8.9275
d1 = 8.9275cos 35° ≈ 7.3130 mi
It will be helpful later to find the height:
h
sin35° =
; h ≈ 5.1206
8.9275
73. Plug in a = 45, A = 19°, B = 31°
⎡1
⎤
tan ⎢ (19° + 31°) ⎥
45 + b
2
⎣
⎦ = tan 25° ≈ −4.44
=
1
45 − b
⎡
⎤ tan(−6°)
tan ⎢ (19° − 31°) ⎥
⎣2
⎦
45 + b
= −4.44; 45 + b = −4.44(45 − b)
45 − b
45 + b = −199.8 + 4.44b
−3.44b = −244.8; b ≈ 71.2 cm
Note that C = 180º – (19º + 31º) = 130º.
Plug in a = 45, A = 19°, C = 130°
⎡1
⎤
tan ⎢ (19° + 130°)⎥
45 + c
2
⎣
⎦
=
1
⎡
⎤
45 − c
tan ⎢ (19° − 130°)⎥
⎣2
⎦
tan 74.5°
=
≈ −2.48
tan(−55.5°)
45 + c
= −2.48; 45 + c = −2.48(45 − c)
45 − c
45 + c = −111.6 + 2.48c
−1.48c = −156.6; c = 105.8 cm
729
7.1 Exercises
Second sighting:
77. tan 2 x − sin 2 x = tan 2 x sin 2 x
sin 2 x
5.1206
5.1206
; d2 =
≈ 11.5011 mi
tan24° =
d2
tan24°
The UFO’s linear distance traveled is
11.5011 − 7.3130 = 4.1881 mi
4.1881 mi 3,600 sec
mi
⋅
≈ 12,564
1.2 sec
hr
hr
cos 2 x
sin 2 x
cos 2 x
− sin 2 x =
−
sin 2 x cos 2 x
cos 2 x
sin 2 x − sin 2 x cos 2 x
(
cos 2 x
sin 2 x 1 − cos 2 x
76. Rewrite the right side as 1 − 2sin 2 x .
2sin x = 1− 2sin 2 x
2
cos x
sin x sin 2 x
=
=
)=
2
2sin 2 x + 2sin x −1= 0
Use the quadratic formula to solve for sin x.
=
cos 2 x
sin 2 x
sin 2 x =
cos 2 x
tan 2 x sin 2 x = tan 2 x sin 2 x
−2 ± 4 − 4(2)(−1) −2 ± 12
=
4
4
3
1
=− ±
2 2
1
3
< −1, so we can’t have
Note that − −
2 2
1
3
3
1
. sin x = − +
⇒
sin x = − −
2 2
2 2
⎛
⎞
1
3⎟
x = sin −1⎜⎜ − +
⎟ ≈ 0.3747
⎝ 2 2 ⎠
sin x =
78. Since 1 + 2i is a zero, so is 1− 2i . The
polynomial must have factors (x – 2),
(x + 1), x − (1 + 2i) , and (x − (1 − 2i)) .
(
)
p(x) = ( x − 2)(x + 1)(x − 1 − 2i)(x − 1 + 2i)
= (x2 − x − 2)(x2 − 2x + 5)
= x 4 − 3x3 + 5x2 − x − 10
Another solution is π − 0.3747 = 2.7669 .
All solutions: x = 0.3747 + 2πk,
2.7669 + 2π k , k ∈ Z
2 − (−3) 5
=
4 − (−5) 9
5
5
20
y − 2 = (x − 4); y − 2 = x −
9
9
9
5
2
y = x−
9
9
79. a.
m=
b.
d=
(2 − (−3))2 + (4 − (−5))2
= 25 + 81 = 106 units
730
Chapter 7: Applications of Trigonometry
7.2 Exercises
11. Yes
1.
cosines
12. Yes
2.
2ab
3.
Pythagorean
4.
a2 = b2 + c2 − 2bc cos A
5.
a.
13. a2 = b2 + c2 − 2bc cos A
52.42 = 502 + 26.62 − 2(50)(26.6) cos 80°
2745.76 ≈ 2745.7
b2 = a2 + c2 − 2ac cos B
502 = 52.42 + 26.62 − 2(52.4)(26.6) cos 70°
2500 ≈ 2499.9
c2 = a2 + b2 − 2ab cos C
26.62 = 502 + 52.42 − 2(50)(52.4) cos 30°
707.6 ≈ 707.8
With some rounding, all result in equality.
Law of Cosines Only:
a2 = 37 2 + 522 − 2(37)(52) cos17°
a2 ≈ 393.1; a ≈ 19.8 m
522 = 37 2 + 393.1 − 2(37)(19.8) cos C
2704 = 1369 + 393.1 − 1465.2 cos C
2704 = 1369 + 393.1 − 1465.2 cos C
941.9
cos C =
−1465.2
⎛ 941.9 ⎞
C = cos−1⎜
⎟ ≈ 130.0°
⎝ −1465.2 ⎠
B = 180 – (130 + 17) = 33.0º
Law of Sines:
After we know that a = 19.8 m:
sin B sin17°
=
; 19.8 sin B = 37 sin17°
37
19.8
⎛ 37 sin17° ⎞
37 sin17°
sin B =
; B = sin−1⎜
⎟
19.8
⎝ 19.8 ⎠
14. a2 = b2 + c2 − 2bc cos A
5772 = 1114.72 + 8162 − 2(1114.7)(816) cos30°
332,929 ≈ 332,946.99
b2 = a2 + c2 − 2ac cos B
1114.72 = 5772 + 8162 − 2(577)(816) cos105°
1, 242,556.09 ≈ 1, 242,505.58
c2 = a2 + b2 − 2ab cos C
8162 = 5772 + 1114.72 − 2(577)(1114.7) cos 45°
665,856 ≈ 665,888.52
With some rounding, all result in equality.
15. 42 = 52 + 62 − 2(5)(6) cos B
16 = 61 − 60 cos B; − 45 = −60 cos B
⎛ 45 ⎞
45
cos B = ; B = cos−1⎜ ⎟ ≈ 41.4°
60
⎝ 60 ⎠
B ≈ 33.1° , C = 180 − (31.117) = 129.9
The second method is simpler, Law of
Sines.
o
b.
6.
a2 = b2 + c2 − 2bc cos A
a2 − b2 − c2 = −2bc cos A
a2 − b2 − c2
= cos A
−2bc
b2 + c2 − a2
= cos A
2bc
If b2 + c2 − a2 > 2bc , the left side is greater
than 1, and there’s no solution.
7.
Yes
8.
Yes
9.
No; there will be two unknowns in any of
the three forms.
16. 12.92 = 15.22 + 9.82 − 2(15.2)(9.8) cos C
166.41 = 327.08 − 297.92 cos C
−160.67
−160.67 = −297.92 cos C; cos C =
−297.92
⎛
⎞
−160.67
C = cos−1⎜
⎟ ≈ 57.4°
⎝ −297.92 ⎠
17. a2 = 92 + 7 2 − 2(9)(7) cos 52° ≈ 52.43
a = 52.43 ≈ 7.24
18. b2 = 3.92 + 9.52 − 2(3.9)(9.5) cos 30° ≈ 41.29
b = 41.29 ≈ 6.43
10. No; there will be two unknowns in any of
the three forms.
731
7.2 Exercises
19. 102 = 122 + 152 − 2(12)(15) cos A
100 = 369 − 360 cos A ; −269 = −360 cos A
⎛ 269 ⎞
−269
; A = cos−1⎜
cos A =
⎟ ≈ 41.6°
−360
⎝ 360 ⎠
24. b2 = 10.92 + 6.7 2 − 2(10.9)(6.7) cos 98°
b 2 ≈ 184.03; b ≈ 13.6 km
sinC sin98°
=
; 13.6 sinC = 10.9 sin98°
10.9
13.6
⎛ 10.9 sin 98° ⎞
10.9 sin98°
sinC =
; A = sin−1⎜
⎟
13.6
⎝ 13.6
⎠
C ≈ 52.5° ; A = 180 − (52.5 + 98) = 29.5°
20. 2022 = 1822 + 982 − 2(182)(98) cos B
40, 804 = 42, 728 − 35, 672 cos B
1924
−1924 = −35,672cos B; cos B =
35,672
⎛
⎞
1924
B = cos−1⎜
⎟ ≈ 86.9°
⎝ 35,672 ⎠
21. c2 = 752 + 322 − 2(75)(32) cos 38° = 2866.55
c = 2866.55 ≈ 53.5 cm
sin38° sin B
=
; 53.5sin B = 32 sin38°
53.5
32
⎛ 32sin38° ⎞
32sin38°
; B = sin−1⎜
sin B =
⎟
53.5
⎝ 53.5 ⎠
25. c2 = 5382 + 4652 − 2(538)(465) cos 29°
c 2 ≈ 68, 061.78; c ≈ 260.9 mm
sin29° sin B
=
; 260.9 sin B = 465sin29°
260.9
465
⎛ 465sin 29° ⎞
465sin29°
sin B =
; B = sin−1⎜
⎟
260.9
⎝ 260.9 ⎠
B ≈ 59.8° ; A = 180 − (59.8 + 29) = 91.2°
B ≈ 21.6° ; A = 180 − (21.6 + 38) = 120.4°
22. c2 = 4902 + 3852 − 2(490)(385) cos 67°
c 2 ≈ 240,902.15; c ≈ 490.8 m
sin67° sin B
=
; 490.8 sin B = 385sin67°
490.8
385
⎛ 385sin67° ⎞
385sin 67°
sin B =
; B = sin−1⎜
⎟
490.8
⎝ 490.8 ⎠
B ≈ 46.2° ; A = 180 − (67 + 46.2) = 66.8°
26. b2 = 32.52 + 27.52 − 2(32.5)(27.5) cos141°
b2 ≈ 3201.65; b ≈ 56.6 ft
sin141° sinC
=
; 56.6 sin C = 32.5sin141°
56.6
32.5
⎛ 32.5sin 141° ⎞
32.5sin 141°
sin C =
; C = sin −1⎜
⎟
56.6
56.6
⎠
⎝
C ≈ 21.2° ; A = 180 − (21.2 + 141) = 17.8°
23. b2 = 12.92 + 25.82 − 2(12.9)(25.8) cos 30°
b 2 ≈ 255.59; b ≈ 16 mi
sin30° sin A
=
; 16 sin A = 12.9 sin30°
16
12.9
⎛ 12.9sin30° ⎞
12.9sin30°
; A = sin−1⎜
sin A =
⎟
16
16
⎝
⎠
A ≈ 23.8° ; C = 180 − (30 + 23.8) = 126.2°
732
Chapter 7: Applications of Trigonometry
27. a2 = b2 + c2 − 2bc cos A
675 = 108 + 300 − 360 cos A
267 = −360 cos A; cos A =
⎛ 267 ⎞
A = cos−1⎜
⎟ ≈ 137.9°
⎝ −360 ⎠
sin 137.9°
15 3
sin B =
=
sin B
6 3
30. 4322 = 2082 + 3822 − 2(208)(382) cos A
186,624 = 189,188 − 158,912 cos A
2564
−2564 = −158,912cos A; cos A =
158, 912
A ≈ 89.1°
sin89.1° sinC
=
; 432sinC = 382 sin89.1°
382
432
⎛ 382 sin 89.1° ⎞
382 sin 89.1°
sin C =
; C = sin −1 ⎜
⎟
432
432
⎝
⎠
C ≈ 62.1° ; B = 180 − (89.1 + 62.1) = 28.8°
267
−360
; 15 3 sin B = 6 3 sin 137.9°
6 3 sin137.9°
15 3
⎛
⎞
6 3 sin137.9° ⎟
B = sin−1⎜⎜
⎟ ≈ 15.6°
15 3
⎝
⎠
C = 180 − (15.6 + 137.9) = 26.5°
28. 3002 = 2822 + 1292 − 2(282)(129) cos C
90, 000 = 96,165 − 72, 756 cos C
−6165
−6165 = −72, 756cos C; cos C =
−72,756
⎛
⎞
6165
C = cos−1⎜
⎟ ≈ 85.1°
⎝ 72,756 ⎠
sin85.1° sin A
=
; 300 sin A = 282 sin85.1°
300
282
⎛ 282 sin 85.1° ⎞
282 sin 85.1°
sin A =
; A = sin −1 ⎜
⎟
300
300
⎝
⎠
A ≈ 69.5°; B = 180 − (69.5 + 85.1) = 25.4°
(
31. 4.1 × 10 25
32.8
=
sin B
24.9
2
25 2
25
25 2
25
1.7 × 1051 = 1.4 ×1051 −1.3×1051 cos A
3.0 × 1050 = −1.3× 1051 cos A
⎛
50 ⎞
3.0 × 1050
−1 3.0 × 10
⎜⎜
⎟
cos A =
;
A
=
cos
51 ⎟
−1.3 × 1051
⎝ −1.3 × 10 ⎠
A ≈ 103.3°
sin103.3°
sin C
=
25
4.1 × 10
2.9 × 1025
4.1 × 1025 sin C = 2.9 × 1025 sin103.3°
2.9 × 1025 sin103.3°
sin C =
4.1 × 1025
⎛ 2.9 × 1025 sin103.3° ⎞
⎟⎟ ≈ 43.5°
C = sin−1⎜⎜
4.1 × 1025
⎝
⎠
B = 180 − (43.5 + 103.3) = 33.2°
29. 32.82 = 24.92 + 12.42 − 2(24.9)(12.4) cos A
1075.84 = 773.77 − 617.52 cos A
302.07
302.07 = −617.52cos A; cos A =
−617.52
⎛ 302.07 ⎞
A = cos−1⎜
⎟ ≈ 119.3° ;
⎝ −617.52 ⎠
sin 119.3°
) = (2.3 × 10 ) + (2.9 × 10 )
− 2(2.3 × 10 )(2.9 × 10 )cos A
; 32.8 sin B = 24.9 sin 119.3°
24.9 sin119.3°
32.8
⎛
⎞
24.9sin119.3°
B = sin−1⎜
⎟ ≈ 41.5°
32.8
⎝
⎠
C = 180 − (41.5 + 119.3) = 19.2°
sin B =
733
7.2 Exercises
(
) (
2
) (
)
− 2(2.9 × 10 )(4.9 × 10 )cos A
2
32. 6.8 × 1013 = 2.9 × 1013 + 4.9 × 1013
13
34. This triangle is not possible: the sum of the
two shorter sides is less than the longer.
2
13
35. a2 = b2 + c2 − 2bc cos A
a2 − b2 − c2 = −2bc cos A
a2 − b2 − c2
= cos A
−2bc
b2 + c2 − a2
= cos A
2bc
Adapting the new formula to the given
triangle, where we should solve for angle C
first, we get
4.624 × 1027 = 3.242 × 1027 − 2.842 × 1027 cos A
1.382 × 1027 = −2.842 × 1027 cos A
1.382 × 1027
cos A =
−2.842 × 1027
⎛
⎞
1.382 × 1027 ⎟
A = cos−1⎜
⎜
27 ⎟
⎝ −2.842 × 10 ⎠
A ≈ 119.1°
sin119.1°
sin C
=
13
4.9 × 1013
6.8 × 10
6.8 × 1013 sin C = 4.9 × 1013 sin119.1°
4.9 × 1013 sin119.1°
sin C =
6.8 × 1013
⎛ 4.9 × 1013 sin119.1° ⎞
⎟⎟ ≈ 39°
C = sin−1⎜⎜
6.8 × 1013
⎝
⎠
B = 180 − (39 + 119.1) = 21.9°
a2 + b2 − c2
2ab
392 + 37 2 − 522
cos C =
≈ 0.0644
2(39)(37)
cosC =
C = cos−1 0.0644 ≈ 86.3°
36. P = a + b + h(csc α + csc β )
P = 5000 + 7500 + 2000(csc 42o + csc 78o )
P ≈ 17,534 ft
37. m2 = 14352 + 6922 − 2(1435)(692) cos 99°
m2 ≈ 2, 848774; m ≈ 1688 mi
38. v2 = 3112 + 4882 − 2(311)(488) cos 95°
v2 ≈ 361, 320; v ≈ 601 km
39. 1982 = 3542 + 4232 − 2(354)(423) cos P
39,204 = 304,245 − 299, 484 cos P
−265, 041 = −299, 484cos P
−265, 041
cos P =
−299, 484
⎛ −265,041 ⎞
P = cos−1⎜
⎟ ≈ 27.7°
⎝ −299, 484 ⎠
The heading is 27.7º north of west or a
heading of 297.7º.
⎛
⎞2
33. ⎜12 3 ⎟ = 12.92 + 9.22 − 2(12.9)(9.2) cos A
⎝
⎠
432 = 251.05 − 237.36cos A
180.95
180.95 = −237.36 cos A; cos A =
−237.36
⎛ 180.95 ⎞
A = cos−1⎜
⎟ ≈ 139.7°
⎝ −237.36 ⎠
sin139.7° sin B
=
12.9
12 3
40. 102 = 212 + 182 − 2(21)(18) cos M
100 = 765 − 756cos M
−665
−665 = −756 cos M ; cos M =
−756
⎛
⎞
−665
M = cos−1⎜
⎟ ≈ 28.4
⎝ −756 ⎠
The heading is 28.4º south of east or a
heading of 118.4º.
12 3 sin B = 12.9 sin139.7°
12.9 sin139.7°
sin B =
12 3
⎛ 12.9 sin139.7° ⎞
B = sin−1⎜⎜
⎟⎟ ≈ 23.7°
⎝
⎠
12 3
C = 180 − (23.7 + 139.7) = 16.6°
734
Chapter 7: Applications of Trigonometry
46. Call the point at the bottom left corner of the
board P. Triangle PAB is right with legs 3
and 4, and hypotenuse 5 completes the
Pythagorean triple. Side BC is the
hypotenuse of right triangle PBC, which has
41. d 2 = 1.82 + 2.62 − 2(1.8)(2.6) cos 51°
d = 10 − 9.36 cos 51° mi
= 10 − 9.36 cos 51° (5280) = 10, 703.6 ft
It cannot be constructed.
42. a.
b.
2
2
legs 10 and 3: BC 2 = 102 + 32 ; BC = 109
Side AC is 6, so the perimeter is
2
d = 685 + 610 − 2(685)(610) cos 79°
d 2 ≈ 681,865.9226; d ≈ 825.75 yd
3 ft $5000
825.75 yd ⋅
⋅
= $12,386,250
yd
ft
This is the cost. To make at least a 25%
profit, the bid must be 25% more than
this, or 1.25 times as much:
1.25(12,386, 250) ≈ $15, 482,800
5 + 6 + 109 ≈ 21.4 cm
Using right triangle PBC:
⎛3⎞
3
tanC = ; C = tan−1⎜ ⎟ ≈ 16.7°
10
⎝ 10 ⎠
sin16.7° sin B
=
; 5sin B = 6 sin16.7°
6
5
6 sin16.7°
6 sin16.7°
sin B =
; B = sin−1
5
5
B ≈ 20.2° ; A = 180 − (16.7 + 20.2) = 143.1°
43. After 5 hours, the distances are 5(450) =
2250 miles, and 5(425) = 2125 miles. The
angle between paths is 45º.
d 2 = 22502 + 21252 − 2(2250)(2125) cos 45°
d 2 ≈ 2,816, 416.405; d ≈ 1678.2 mi
47. 202 = 122 + 92 − 2(12)(9) cos C
400 = 225 − 216 cos C
175
175 = −216 cos C ; cos C =
−216
⎛
⎞
175
C = cos−1⎜
⎟ ≈ 144.1°
⎝ −216 ⎠
sin144.1° sin B
=
; 20 sin B = 9 sin144.1°
9
20
⎛ 9 sin 144.1° ⎞
9sin144.1°
; B = sin−1⎜
sin B =
⎟
20
20
⎝
⎠
B ≈ 15.3°
A ≈ 180 − (144.1 + 15.3) = 20.6°
44. After 10 hours, the distances are 10(15) =
150 nautical mi and 10(12) = 120 nautical
mi. The angle between paths is 50º.
d 2 = 1502 + 1202 − 2(150)(120) cos 50°
d 2 ≈ 13, 759.64605; d ≈ 117.3 nautical mi
45. Call the point at the bottom left corner of the
board P. Triangle PAB is a 45-45-90
triangle with legs 4, so side AB is 4 2 .
Using the Pythagorean Theorem on triangle
PBC, BC 2 = 102 + 42 ; BC = 116 . Side
AC is 6, so the perimeter is
48. 162 = 62 + 112 − 2(6)(11) cos B
256 = 157 − 132cos B
99
99 = −132cos B; cos B =
−132
⎛
⎞
−1 99
B = cos ⎜
⎟ ≈ 138.6°
⎝ −132 ⎠
6 + 4 2 + 116 ≈ 22.4 cm
Again using right triangle PBC,
4
4
tan C = ; C = tan−1
≈ 21.8°
10
10
sin21.8° sin B
=
; 4 2 sin B = 6 sin21.8°
6
4 2
⎛ 6 sin21.8° ⎞
6 sin21.8°
sin B =
; B = sin−1⎜⎜
⎟⎟
⎝ 4 2 ⎠
4 2
B ≈ 23.2° ; A = 180 − (21.8 + 23.2) = 135°
sin138.6° sin A
=
; 16sin A = 11sin138.6°
11
16
⎛ 11sin138.6° ⎞
11sin138.6°
; A = sin−1⎜
sin A =
⎟
16
16
⎝
⎠
A ≈ 27.0°
C ≈ 180 − (138.6 + 27.0) = 14.4°
735
7.2 Exercises
49. A regular pentagon can be made from five
360
= 72°
triangles, each with an angle of
5
52. Drawing in right triangles with each side as
hypotenuse, as in 51, we find that a = 13,
b = 10, and c = 17 .
2
132 = 102 + 17 − 2(10)( 17 ) cos A
169 = 117 − 20 17 cos A
52 = −20 17 cos A; cos A =
10 72° 10
−20 17
⎛ 52 ⎞
A = cos−1⎜⎜
⎟⎟ ≈ 129.1°
⎝ −20 17 ⎠
sin129.1° sin B
=
; 13sin B = 10sin129.1°
13
10
⎛ 10 sin 129.1° ⎞
10 sin 129.1°
; B = sin−1⎜
sin B =
⎟
13
13
⎝
⎠
B ≈ 36.7° ; C = 180 − (129.1 + 36.7) = 14.2°
x
x 2 = 102 + 102 − 2(10)(10) cos 72°
x 2 ≈ 138.1966; x ≈ 11.756
Perimeter = 5(11.756) = 58.78 cm
50. A regular hexagon can be made from six
360
triangles, each with an angle of
= 60° .
6
53. Diagonal: 202 + 302 = 1300 ≈ 36.06
20
tan α =
30
⎛2⎞
α = tan −1 ⎜ ⎟ ≈ 33.7o ;
⎝3⎠
1
A = bc sin α
2
1
A=
1300 (15 ) sin 33.7o
2
≈ 150 square feet
60°
15
15
x
In this case, the triangle is equilateral! Side
x is 15 cm, and the perimeter is 6(15) = 90
cm.
51. Side AC is the hypotenuse of a right triangle
with right angle at (3, 0) that has legs 3 and
4, so its length is 5. Side CB is the
hypotenuse of a right triangle with right
angle at (12, 0) that has legs 12 and 5 so its
length is 13. The third side is the
hypotenuse of a right triangle with sides 1
and 9, so its length is
1
bc sin A
2
1
A = (110 )( 225 ) sin 38o ≈ 7619 ft 2
2
54. A =
92 + 12 = 82 .
2
132 = 52 + 82 − 2(5)( 82) cos A
55. 42o + 65o + x = 180o
x = 73o ;
c 2 sin A sin B
A=
2sin C
2992 sin 42o sin 65o
A=
≈ 28346.7 ;
2sin 73o
28346.7
≈ 0.65 or 65%
a.
43560
b. 3, 000, 000 ( 0.65 ) = $1,950, 000
169 = 107 − 10 82 cos A
62 = −10 82 cos A; cos A =
52
62
−10 82
⎛ 62 ⎞
⎟⎟ ≈ 133.2°
A = cos −1 ⎜⎜
⎝ − 10 82 ⎠
sin133.2° sin B
=
; 13sin B = 5sin133.2°
13
5
⎛ 5sin133.2° ⎞
5sin133.2°
sin B =
; B = sin−1⎜
⎟
13
13
⎝
⎠
B ≈ 16.3° ; C = 180 − (133.2 + 16.3) = 30.5°
736
Chapter 7: Applications of Trigonometry
62. 135º = 3(45º);
cos(3 ⋅ 45°) = 4 cos3 (45°) − 3cos(45°)
c 2 sin A sin B
2sin C
2182 sin 20o sin 20o
Area of small =
2sin140o
Area of small ≈ 4324.3;
56. c = 218, A =
⎛
⎞3 ⎛
⎞ ⎛
⎞
2
⎟ − 3⎜ 2 ⎟ = 4⎜ 2 2 ⎟ − 3 2
= 4⎜⎜
⎟
⎜ 2 ⎟ ⎜ 8 ⎟
2
⎝ 2 ⎠
⎝
⎠ ⎝
⎠
2182 sin 42o sin 45o
2sin 93o
Area of large ≈ 11258.3;
Area = 4324.3 + 11258.3 = 15582.6
About 15,583 km 2
=
Area of large =
Ref. angle = 45º, QII, so cos(135°) = −
2
2
63. 2 log2 4 + 2 log2 3 − 2 log2 6
= log2 42 + log2 32 − log2 62
⎛ 16 ⋅9 ⎞
2
= log2 ⎜
⎟ = log2 4 = log2 2 = 2
⎝ 36 ⎠
1289 + 1063 + 922
= 1637
57. p =
2
A= 1637(1637 −1289)(1637 − 922)(1637 −1063)
A ≈ 483,529 km 2
1
3
⎛ 7π ⎞
= ; cos ⎜
⎟=−
6 2
2
⎝ 6 ⎠
⎛π ⎞
tan⎜ ⎟ = 3
⎝3⎠
64. sin
58. x 2 = 202 + 152
x 2 = 625
x = 25;
36 + 25 + 15
= 38
p=
2
π
65. y = –5, r = 13; x = 132 − 52 = 12 , positive
since cosine is positive.
13
12
13
csc x = − ; cos x = ; sec x =
5
13
12
A = 38 ( 38 − 36 )( 38 − 25 )( 38 − 15 )
A ≈ 150.7 ft 2
150.7 − 150 = 0.7 ft 2
tan x = −
59. The sum of the two smaller sides is 889,
which is less than the longer one.
5
12
; cot x = −
12
5
66.
60. The related horizontal distance x – c
becomes c – x, but this doesn’t change
anything because it gets squared in the
second step. The verification remains
essentially the same.
−2 − 1 − 1 7
1 −6
2 − 2 − 10 18
−1
1
5 −9
12
Technology Highlight
61. 53.9 = 78cos 25° + 37 cos117° ≈ 53.9
a2 = b2 + c2 − 2bc cos A
b2 = a2 + c2 − 2ac cos B
Substitute the right side of the first
expression in for a2 in the second:
b 2 = b 2 + c 2 − 2bc cos A + c 2 − 2ac cos B
(
2
2 2 3 2
−
=−
2
2
2
Exercise 1:They would be equal; verified
Exercise 2:The component values will “switch
places” compared to θ = 30o ; verified
)
Exercise 3: v = 9.5; v = 9.5 for all values of θ
2
0 = 2c − 2bc cos A − 2ac cos B
(Divide both sides by 2c)
0 = c − b cos A − a cos B
−c = −b cos A − a cos B
c = b cos A + a cos B
737
7.3 Exercises
12.
7.3 Exercises
1.
Scalar
2.
Vector
3.
Directed line
4.
Magnitude; direction
13.
5, 6. Answers will vary.
7.
14.
8.
9.
15.
10.
16.
11.
738
Chapter 7: Applications of Trigonometry
17. Terminal point = (–2+7, –3 + 2) = (5, –1);
23. a.
v = 7 + 2 = 53
2
2
18. Terminal point = (5 – 6, –2 + 1) = (–1, –1);
v = (−6)2 + 12 = 37
19. Terminal point = (2 − 3, 6 − 5) = (−1,1) ;
v = (−3)2 + (−5)2 = 34
20. Terminal point: (−3 + 8,−5 − 2) = (5, −7) ;
v = 82 + (−2)2 = 68 = 2 17
b.
v = (−2)2 + (−5)2 = 29
c.
tan θ =
⎛5⎞
−5
; θ = tan−1⎜ ⎟ ≈ 68.2°
−2
⎝2 ⎠
21. a.
24. a.
b.
v = 82 + 32 = 73
c.
⎛ 3⎞
3
tan θ = ; θ = tan−1⎜ ⎟ ≈ 20.6°
8
⎝8⎠
b.
c.
22. a.
v = 82 + (−6)2 = 10
⎛ −6 ⎞
−6
; θ = tan−1⎜ ⎟ ≈ −36.9 º
8
⎝8⎠
Angle with the nearest x-axis is 36.9º
tan θ =
25. a = 12 cos 25° ≈ 10.9; b = 12 sin 25° ≈ 5.1
−10.9,5.1
26. a = 25cos 32° ≈ 21.2; b = 25sin 32° ≈ 13.2
−21.2,−13.2
b.
c.
27. a = 140.5cos 41° ≈ 106.0
b = 140.5sin 41° ≈ 92.2
106.0, − 92.2
v = (−7)2 + 62 = 85
6
⎛ 6 ⎞
; θ = tan −1 ⎜
⎟ ≈ 40.6°
−7
⎝ −7⎠
Angle with nearest x-axis is 40.6º
tan θ =
28. a = 15cos 65° ≈ 6.3; b = 15sin65° ≈ 13.6
6.3,13.6
29. a = 10 cos15° ≈ 9.7; b = 10 sin15° ≈ 2.6
−9.7,−2.6
30. a = 4.75cos 62° ≈ 2.2; b = 4.75sin62° ≈ 4.2
−2.2, 4.2
739
7.3 Exercises
31. a.
u + v = 2 + (−3), 3 + 6 = −1,9
32. a.
u + v = −3 + 0,−4 + 5 = −3,1
b.
u − v = 2 − (−3), 3 − 6 = 5,−3
b.
u − v = −3 − 0,−4 − 5 = −3,−9
c.
2u + 1.5v = 4,6 + −4.5,9 = −0.5,15
c.
2u + 1.5v = −6,−8 + 0, 7.5
= −6,−0.5
d.
u − 2v = 2,3 − −6,12 = 8, −9
d.
740
u − 2v = −3, −4 − 0,10 = −3, −14
Chapter 7: Applications of Trigonometry
33. a.
u + v = 7 + 1, −2 + 6 = 8, 4
34. a.
u + v = −5 + 6,−3 + (−4) = 1,−7
b.
u − v = 7 − 1,−2 − 6 = 6,−8
b.
u − v = −5 − 6, −3 − (−4) = −11,1
c.
2u + 1.5v = 14,−4 + 1.5, 9 = 15.5, 5
c.
2u + 1.5v = −10, −6 + 9,−6
= −1,−12
d.
u − 2v = 7, −2 − 2,12 = 5, −14
d.
741
u − 2v = −5, −3 − 12, −8 = −17,5
7.3 Exercises
35. a.
u + v = −4 + 1,2 + 4 = −3,6
36. a.
u + v = 7 + (−7), 3 + 3 = 0,6
b.
u − v = −4 − 1,2 − 4 = −5,−2
b.
u − v = 7 − (−7), 3 − 3 = 14, 0
c.
2u + 1.5v = −8, 4 + 1.5, 6 = −6.5,10
c.
2u + 1.5v = 14, 6 + −10.5, 4.5
= 3.5,10.5
d.
u − 2v = −4, 2 − 2,8 = −6,−6
d.
742
u − 2v = 7, 3 − −14, 6 = 21,−3
Chapter 7: Applications of Trigonometry
46. u + v = −4 + 5, 4 + 2 = 1, 6
37. True
u − v = −4 − 5, 4 − 2 = −9,2
38. True
39. False ( c + d = h )
40. False
41. True
42. True
43. u + v = 1 + 7, 4 + 2 = 8,6
47. u + v = −5 + 2,−3 + (−3) = −3, −6
u − v = 1 − 7, 4 − 2 = −6, 2
u − v = −5 − 2,−3 − (−3) = −7, 0
44. u + v = 3 + 4,−6 + (−1) = 7, −7
48. u + v = −5 + 4, −1 + 3 = −1,2
u − v = 3 − 4,−6 − (−1) = −1,−5
u − v = −5 − 4, −1 − 3 = −9, −4
45. u + v = −1 + (−8),−3 + (−3) = −9,−6
49.
u − v = −1 − (−8),−3 − (−3) = 7,0
8,15 = 8i + 15j
u = 82 + 152 = 289 = 17
743
7.3 Exercises
50.
53. a.
−5,12 = −5i + 12j
v = (−5)2 + 122 = 169 = 13
b.
a = 12 cos16° ≈ 11.5
b = 12 sin 16° ≈ 3.3
v = −11.5,−3.3
c.
v = −11.5i − 3.3j
51.
54. a.
−3.2, −5.7 = −3.2i − 5.7j
p = (−3.2) 2 + (−5.7) 2 = 42.73 ≈ 6.54
52.
b.
a = 10.5cos 25° ≈ 9.5
b = 10.5sin25° ≈ 4.4
u = −9.5, 4.4
c.
u = −9.5i + 4.4j
55. a.
7.5, −3.4 = 7.5i − 3.4j
q = 7.52 + (−3.4)2 = 67.81 ≈ 8.23
744
b.
a = 9.5cos 74.5° ≈ 2.5
b = 9.5sin74.5° ≈ 9.2
w = 2.5, 9.2
c.
w = 2.5i + 9.2j
Chapter 7: Applications of Trigonometry
56. a.
58. a.
v1 + v 2 = 7.8i + 4.2j + 5j = 7.8 i + 9.2j
7.82 + 9.22 ≈ 12.1
⎛ 9.2 ⎞
θ = tan−1⎜ ⎟ ≈ 49.7°
⎝ 7.8 ⎠
Mag. =
b.
v1 − v 2 = 7.8i + 4.2j − 5j = 7.8 i − 0.8j
7.82 + 0.82 ≈ 7.8
⎛ − 0.8 ⎞
θ r = tan −1 ⎜
⎟ ≈ −5.9° ; In QIV,
⎝ 7.8 ⎠
θ = 354.1°
2v1 + 1.5v 2 = 15.6i + 8.4 j + 7.5j
= 15.6i + 15.9 j
Mag. =
b.
a = 20cos 32.6° ≈ 16.8
b = 20sin 32.6° ≈ 10.8
v = 16.8,−10.8
c.
v = 16.8 i − 10.8 j
57. a.
b.
c.
c.
15.62 + 15.92 ≈ 22.3
⎛ 15.9 ⎞
θ = tan−1⎜
⎟ ≈ 45.5°
⎝ 15.6 ⎠
Mag. =
v1 + v 2 = (2 − 4)i + (−3 + 5)j = −2i + 2j
d.
Mag. = (−2)2 + 22 = 8 = 2 2
⎛ 2 ⎞
θ r = tan −1 ⎜
⎟ = −45° ; In QII,
⎝ −2⎠
θ = 135°
v1 − v 2 = (2 − (−4))i + (−3 − 5)j = 6i − 8j
7.82 + (−5.8)2 ≈ 9.7
⎛ −5.8 ⎞
θ r = tan −1 ⎜
⎟ = −36.6° ; In QIV,
⎝ 7.8 ⎠
θ = 323.4°
Mag. =
Mag. = 62 + (−8)2 = 10
⎛ −8⎞
θ r = tan −1 ⎜
⎟ ≈ −53.1° ; In QIV,
⎝ 6 ⎠
θ = 306.9°
2v1 + 1.5v 2 = 4i − 6j + −6i + 7.5j
= −2 i + 1.5j
59. a.
v1 + v 2 = 5 2 i + 7 j + (−3 2 i − 5j)
= 2 2i + 2j
(2 2)2 + 22 ≈ 3.5
⎛ 2 ⎞
θ = tan−1⎜⎜
⎟⎟ ≈ 35.3°
⎝2 2 ⎠
Mag. =
Mag. = (−2)2 + (1.5)2 = 6.25 = 2.5
⎛ 1.5 ⎞
θ r = tan −1 ⎜
⎟ ≈ −36.9° ; In QII,
⎝ −2⎠
b.
v1 − v 2 = 5 2i + 7j − (−3 2i − 5j)
= 8 2i +12j
(8 2)2 +122 ≈ 16.5
⎛ 12 ⎞
θ = tan−1⎜⎜
⎟⎟ ≈ 46.7°
⎝8 2 ⎠
Mag. =
θ = 143.1°
d.
v1 − 2v 2 = 7.8i + 4.2j − 10j = 7.8i − 5.8j
v1 − 2v 2 = 2i − 3j − (−8 i + 10j)
= 10i − 13j
Mag. = 102 + (−13)2 ≈ 16.4
⎛ − 13 ⎞
θ r = tan −1 ⎜
⎟ ≈ −52.4° ; In QIV,
⎝ 10 ⎠
θ = 307.6°
c.
2v1 + 1.5v 2 = 10 2i + 14j +
(−4.5 2i − 7.5 j) = 5.5 2i + 6.5 j
(5.5 2)2 + (6.5)2 ≈ 10.1
⎛ 6.5 ⎞
θ = tan−1⎜⎜
⎟⎟ ≈ 39.9°
⎝ 5.5 2 ⎠
Mag. =
d.
v1 − 2v 2 = 5 2i + 7j − (−6 2i −10j)
= 11 2i +17j
(11 2)2 + 17 2 ≈ 23.0
⎛ 17 ⎞
θ = tan−1⎜⎜
⎟⎟ ≈ 47.5°
⎝ 11 2 ⎠
Mag. =
745
7.3 Exercises
60. a.
v1 + v 2 = 6.8i − 9j + (−4i) + 9j = 2.8i
v1 + v 2 = 2 3i − 6j + (−4 3i + 2j)
62. a.
2.8 + 0 = 2.8
⎛
0 ⎞
θ = tan−1⎜ ⎟ = 0°
⎝ 2.8 ⎠
Mag. =
b.
2
2
= −2 3i − 4j
Mag. =
10.82 + (−18)2 ≈ 21.0
⎛ −18 ⎞
θ r = tan −1 ⎜
⎟ ≈ −59.0° ; In QIV,
⎝ 10.8 ⎠
θ = 301.0°
2v1 + 1.5v 2 = 13.6i − 18j + (−6i + 13.5j)
= 7.6i − 4.5j
b.
= 6 3i − 8j
Mag. =
⎛ −8 ⎞
⎟⎟ ≈ −37.6° ; In QIV,
⎝6 3⎠
θ = 322.4°
2v1 + 1.5v 2 = 4 3i − 12j + (−6 3i + 3j)
7.62 + (−4.5)2 ≈ 8.8
⎛ − 4.5 ⎞
θ r = tan −1 ⎜
⎟ ≈ −30.6° ; In QIV,
⎝ 7.6 ⎠
θ = 329.4°
v1 − 2v 2 = 6.8i − 9j − (−8i + 18j)
= 14.8i − 27j
c.
= −2 3i − 9j
Mag. =
14.82 + (−27)2 ≈ 30.8
⎛ − 27 ⎞
θ r = tan −1 ⎜
⎟ ≈ −61.3° ; In QIV,
⎝ 14.8 ⎠
⎛ −9 ⎞
⎟⎟ ≈ 68.9° ; In QIII,
⎝−2 3⎠
θ = 248.9°
v1 − 2v 2 = 2 3i − 6j − (−8 3i + 4j)
d.
= 10 3i −10j
θ = 298.7°
Mag. =
v1 + v 2 = 12i + 4j + (−4i) = 8i + 4j
⎛ − 10 ⎞
⎟⎟ = −30° ; In QIV,
⎝ 10 3 ⎠
θ = 330°
v1 − v 2 = 12i + 4j − (−4i) = 16i + 4j
63. u = 7 2 + 242 = 25; Unit vector =
162 + 42 ≈ 16.5
⎛4⎞
θ = tan−1⎜ ⎟ ≈ 14.0°
⎝ 16 ⎠
Mag. =
c.
7 24
,
. Mag. =
25 25
2v1 + 1.5v 2 = 24i + 8j + (−6i) = 18i + 8j
⎛ 7 ⎞2 ⎛ 24 ⎞2
⎜ ⎟ +⎜ ⎟ =1
⎝ 25 ⎠ ⎝ 25 ⎠
64. v = (−15)2 + 362 = 39 ;
182 + 82 ≈ 19.7
⎛8⎞
θ = tan−1⎜ ⎟ ≈ 24.0°
⎝ 18 ⎠
Mag. =
d.
(10 3)2 + (−10)2 = 20
θ r = tan −1 ⎜⎜
82 + 42 ≈ 8.9
⎛4⎞
θ = tan−1⎜ ⎟ ≈ 26.6°
⎝8 ⎠
Mag. =
b.
(−2 3)2 + (−9)2 ≈ 9.6
θ r = tan −1 ⎜⎜
Mag. =
61. a.
(6 3)2 + (−8)2 ≈ 13.1
θ r = tan −1 ⎜⎜
Mag. =
d.
⎛ −4 ⎞
⎟⎟ ≈ 49.1° ; In QIII,
⎝−2 3⎠
θ = 229.1°
v1 − v 2 = 2 3i − 6 j − (−4 3i + 2 j)
θ r = tan −1 ⎜⎜
v1 − v 2 = 6.8i − 9j − (−4i + 9j)
= 10.8i −18j
Mag. =
c.
(−2 3)2 + (−4)2 ≈ 5.3
u= −
v1 − 2 v 2 = 12i + 4 j − (−8i ) = 20i + 4 j
5 12
15 36
= − ,
,
39 39
13 13
⎛ −15 ⎞2 ⎛ 36 ⎞2
u= ⎜
⎟ +⎜ ⎟ =1
⎝ 39 ⎠ ⎝ 39 ⎠
202 + 42 ≈ 20.4
⎛4⎞
θ = tan−1⎜ ⎟ ≈ 11.3°
⎝ 20 ⎠
Mag. =
65. p = (−20)2 + 212 = 29 ; u = −
⎛ 20 ⎞2 ⎛ 21 ⎞2
u = ⎜− ⎟ + ⎜ ⎟ = 1
⎝ 29 ⎠ ⎝ 29 ⎠
746
20 21
,
29 29
Chapter 7: Applications of Trigonometry
66. p = 122 + (−35)2 = 37 ; u =
73. Mag. = 62 + 112 = 157
6
11
u=
i+
j
157
157
12 35
,−
37 37
⎛ 12 ⎞2 ⎛ 35 ⎞2
u = ⎜ ⎟ + ⎜− ⎟ = 1
⎝ 37 ⎠ ⎝ 37 ⎠
202 + (−21)2 = 29 ; u =
67. Mag. =
⎛ 6 ⎞2 ⎛ 11 ⎞2
u = ⎜⎜
⎟⎟ + ⎜⎜
⎟⎟ = 1
⎝ 157 ⎠ ⎝ 157 ⎠
21
20
i−
j
29
29
74. Mag. = (−2.5)2 + 7.22 ≈ 7.62
−2.5
7.2
u=
i+
j = −0.33i + 0.94 j
7.62
7.62
⎛ 20 ⎞2 ⎛ 21 ⎞2
u = ⎜ ⎟ + ⎜− ⎟ = 1
⎝ 29 ⎠ ⎝ 29 ⎠
u = −0.332 + 0.952 ≈ 1
68. Mag. = (−4)2 + (−7.5)2 = 8.5;
7.5
8
15
4
i−
j=− i− j
u=−
17
17
8.5 8.5
75. p = 22 + 7 2 = 53; cos 52° =
53
⎛ 8 ⎞2 ⎛ 15 ⎞2
u = ⎜− ⎟ + ⎜− ⎟ = 1
⎝ 17 ⎠ ⎝ 17 ⎠
r = 53 cos 52° ≈ 4.48
q = 102 + 42 = 116 = 2 29
69. Mag. = 3.52 + 122 = 12.5;
3.5
12
7
24
u=
i+
j=
i+
j
12.5 12.5
25
25
2
r = 4.48
= 4.48
2
⎛ 7 ⎞ ⎛ 24 ⎞
u = ⎜ ⎟ +⎜ ⎟ =1
⎝ 25 ⎠ ⎝ 25 ⎠
,
2 29 2 29
5
29
2
,
29
≈ 4.16,1.66
q = 92 + 12 = 82
2
r = 7.42
9
82
,
1
82
≈ 7.37, 0.82
77. p = 42 + (−6)2 = 52; cos 36° =
3
r = 52 cos 36° ≈ 5.83
⎛ 13 ⎞2 ⎛ 3 ⎞2
u = ⎜⎜
⎟⎟ + ⎜⎜
⎟⎟ = 1
⎝ 178 ⎠ ⎝ 178 ⎠
q = 82 + (−3)2 = 73
r = 5.83
8
73
72. v 2 = (−4)2 + 7 2 = 65 ;
u= −
4
65
,
r
52
178
178
r
r = 65 cos 23° ≈ 7.42
71. v1 = 132 + 32 = 178 ;
13
4
,
65
⎛ 8 ⎞ ⎛ 15 ⎞
u = ⎜− ⎟ + ⎜ ⎟ = 1
⎝ 17 ⎠ ⎝ 17 ⎠
u=
10
76. p = 7 2 + 42 = 65; cos 23° =
70. Mag. = (−9.6)2 + 182 = 20.4
9.6
18
8
15
u=−
i+
j=− i+
j
20.4
20.4
17
17
2
r
7
65
⎛ 4 ⎞2 ⎛ 7 ⎞2
u = ⎜⎜−
⎟⎟ + ⎜⎜
⎟⎟ = 1
⎝ 65 ⎠ ⎝ 65 ⎠
747
,
−3
73
≈ 5.46, −2.05
7.3 Exercises
85. The plane vector makes a 75º angle with the
positive x –axis, and the wind vector makes
a 10º angle with the positive x-axis. Find
the components of each:
Plane: a = 250 cos 75° ≈ 64.7
b = 250sin 75° ≈ 241.5
Wind: a = 35cos10° ≈ 34.5
b = 35sin10° ≈ 6.1
Resultant: 64.7 + 34.5, 241.5 + 6.1
r
78. p = (−2)2 + 7 2 = 53; cos 48° =
53
r = 53 cos 48° ≈ 4.87
q = (−10)2 + 52 = 125 = 5 5
r = 4.87
= 4.87
−10
,
5
5 5 5 5
−2
5
,
1
5
≈ −4.36, 2.18
= 99.2, 247.6
99.2 2 + 247.6 2 ≈ 266.7 mph
⎛ 99.2 ⎞
θ = tan −1 ⎜
⎟ ≈ 21.8° from positive x⎝ 247.6 ⎠
axis:
Heading = 68.2º
Mag. =
79. v = 52 + 92 + 102 = 206 ≈ 14.4
80. Answers will vary.
81. Find the vertical component of W2 :
a = 700 sin32°
Find the angle that makes the vertical
component of W1 = 700 sin 32° :
700 sin 32°
700 sin 32° = 900 sin θ ; sin θ =
900
700
sin
32
°
⎛
⎞
θ = sin −1 ⎜
⎟ ≈ 24.3°
900
⎝
⎠
86.
60°
Note that the wind is directly opposite the
ship, so the heading remains the same
(300º), and the speed is 16 – 6 = 10 knots.
82. Vertical component of T1: 1250 sin 40°
Find θ so the vert. components are equal:
1250 sin 40° = 1750 sin θ
1250 sin 40°
;
sin θ =
1750
⎛ 1250sin 40° ⎞
θ = sin−1⎜
⎟ ≈ 27.3°
⎝ 1750
⎠
83. Horizontal: a = 100cos 37° ≈ 79.9
Vertical: b = 100sin37° ≈ 60.2
30°
87. x = 85 cos15° ≈ 82.10
y = 85 sin 15° ≈ 22.00
(82.10 cm, 22.00 cm)
88. Think of the first segment as a vector with
length 45 and θ = 75° …
a1 = 45 cos 75° ≈ 11.65
b1 = 45 sin 75° ≈ 43.47
… and the second length 40, θ = −30°
a2 = 40 cos(−30°) ≈ 34.64
b2 = 40 sin(−30°) ≈ −20
Now add the vectors:
Sum =
11.65 + 34.64, 43.47 − 20 = 46.29, 23.47
ft
sec
ft
sec
m
84. Horizontal.: a = 20 cos 42° ≈ 14.9 sec
m
Vertical: b = 20sin 42° ≈ 13.4 sec
Coordinates: (46.29 cm, 23.47 cm)
748
Chapter 7: Applications of Trigonometry
89. 1 ⋅ u = 1 ⋅ a, b = 1 ⋅ a,1 ⋅ b = a, b = u
c.
90. 0 ⋅ u = 0 ⋅ a, b = 0 ⋅ a, 0 ⋅ b = 0,0 = 0
k 0 = k 0, 0 = k ⋅ 0, k ⋅ 0 = 0, 0 = 0
91. u − v = a, b − c, d = a − c, b − d
For a bearing of 315º, θ = 135° (The
diagram is very similar to the one in
question 85.)
aw = 40cos135° ≈ −28.3
bw = 40 sin135° ≈ 28.3
Resultant: 141.4 − 28.3,141.4 + 28.3
= 113.1,169.7
= a + (−c), b + (− d) = a, b + −c, −d
= u + ( −v)
d.
92. (u + v ) + w = ( a, b + c, d ) + e, f
= a + c, b + d + e, f
= a + c + e, b + d + f
= a + ( c + e), b + ( d + f )
e.
= a, b + c + e, d + f = u + ( v + w)
93. ( ck ) u = ck a, b = cka, ckb = c ka, kb
Mag. = 113.12 + 169.7 2 ≈ 204.0 mph
Bearing = 270º; θ = 180°
aw = 40cos180° = −40
bw = 40 sin180° = 0
Resultant: 101.4,141.4
Mag. = 101.42 + 141.42 ≈ 174.0 mph
Bearing 225º: θ = 225°
aw = 40 cos 225° ≈ −28.3
bw = 40sin225° ≈ −28.3
Resultant: 141.4 − 28.3,141.4 − 28.3
= 113.1,113.1
= c( ku) = cka, ckb = kca, kcb
= k ca, cb = k ( cu)
Mag. = 113.12 + 113.12 ≈ 160.0 mph
Answers will vary.
94. u + 0 = a, b + 0, 0 = a + 0, b + 0 = u
99. Find the components of each vector, then
add all horiz. and vert. components.
p = 1,3 ; r = 3, 3 ; s = 4, −1
95. u + (− u) = a, b + − a,−b = a − a, b − b
= 0,0
t = 2, −4 ; u = −4, −3 ; v = −6,2
Horizontal: 1 + 3 + 4 + 2 + −4 + −6 = 0
Vertical: 3 + 3 + −1 + −4 + −3 + 2 = 0
96. k ( u + v) = k a + c, b + d
= k ( a + c), k ( b + d ) = ka + kc, kb + kd
100. v = ai + bj, v = a 2 + b 2
= ka, kb + kc, kd = ku + kv
Let u =
97. ( c + k ) u = ( c + k ) a, b
= ( c + k ) a, ( c + k ) b
=
= ca + ka, cb + kb = ca, cb + ka, kb
c a, b + k a, b = cu + ku
a2 + b2
a
b
i+
j;
2
2
2
a +b
a + b2
⎛
a
u = ⎜
⎜
2
2
⎝ a +b
98. Vector for the plane: Mag. = 200, θ = 45°
a p = 200cos 45° ≈ 141.4
b p = 200sin 45° ≈ 141.4
a. Wind in same direction, so speed = 200 +
40 = 240 mph
b. For a bearing of 0º, θ = 90° .
aw = 40cos 90° = 0
=
bw = 40sin90° = 40
Resultant = 141.4,181.4
Mag. =
ai + bj
141.42 + 181.42 ≈ 230 mph
749
a2
a2 + b2
+
2
⎞ ⎛
b
⎟ +⎜
⎟ ⎜
2
2
⎠ ⎝ a +b
b2
a2 + b2
=
⎞
⎟
⎟
⎠
a2 + b2
a2 + b2
2
=1
7.3 Exercises
101. Answers will vary. One possibility: Place
the first segment at 0º; it will end at (45, 0).
Place the second to reach the point (51,
39.6), and the third to reach (80, 20). In this
case, the second segment has components
⎛ 39.6 ⎞
6,39.6 and θ = tan−1⎜
⎟ ≈ 81.4° , while
⎝ 6 ⎠
the third segment has components
⎛ −19.6 ⎞
29, −19.6 and θ = tan−1⎜
⎟ ≈ −34°.
⎝ 29 ⎠
102.
sin2 x
2
+
cos2 x
=
1
=
1
2
(
g ( x) = x ( x 2 − 7 ) = x x + 7
)( x − 7 )
x = 0, x = ± 7
+
cos2 x
2
Mid-Chapter Check
2
sin x sin x sin x
1 + cot2 x = csc2 x
103. a.
b.
c.
104.
g ( x) = x3 − 7 x
cos x cos x cos2 x
tan2 x + 1 = sec2 x
sin2 x
2
105.
1.
ln(2(3) − 7) = ln(−1) not a real number
5
not possible
3− 3
2.
b2 = a2 + c2 − 2ac cos B
2ac cos B = a2 + c2 − b2
a2 + c2 − b2
cos B =
2ac
3.
a2 = 207 2 + 2502 − 2(250)(207) cos 31°
1
(3) − 5 = −4 not a real number
3
⎡
⎛ 55 ⎞ ⎤
csc ⎢ tan −1 ⎜ ⎟ ⎥
⎝ 48 ⎠ ⎦
⎣
a2 = 16,632.2; a ≈ 129 m
sin31 sin B
=
; 129sin B = 250sin31°
250
129
⎛ 250sin31° ⎞
250sin31°
sin B =
; B = sin−1⎜
⎟
129
⎝ 129
⎠
B ≈ 86.5° ; C = 180 − (31 + 86.5) = 62.5°
4.
csc θ =
sin A sin B
=
; a sin B = b sin A
a
b
b sin A
sin B =
a
73
55
252 = 212 + 17 2 − 2(21)(17) cos B
625 = 730 − 714cos B
−105
−105 = −714cos B; cos B =
−714
⎛
⎞
−1 105
B = cos ⎜
⎟ ≈ 81.5°
⎝ 714 ⎠
sin81.5° sin A
=
; 25sin A = 17 sin81.5°
25
17
⎛ 17 sin 81.5° ⎞
17 sin 81.5°
sin A =
; A = sin−1⎜
⎟
25
25
⎝
⎠
A ≈ 42.3°; C = 180 − (81.5 + 42.3) = 56.2°
750
Chapter 7: Applications of Trigonometry
5.
sin 44° sin C
=
; 2.1sin C = 2.8sin 44°
2.1
2.8
⎛ 2.8sin 44° ⎞
2.8sin 44°
sin C =
; C = sin−1⎜
⎟
2.1
2.1
⎝
⎠
9.
21
21 = 132 +162 − 2(13)(16) cos β
441 = 425 − 416 cos β
16
16 = −416 cos β ; cos β =
−416
⎛
⎞
16
β = cos−1⎜−
⎟ ≈ 92.2°
⎝ 416 ⎠
sin 92.2° sin α
=
; 21 sin α = 16 sin 92.2°
21
16
⎛ 16sin92.2° ⎞
16sin92.2°
; α = sin−1⎜
sin α =
⎟
21
21
⎝
⎠
α ≈ 49.6° ; γ = 180 − (92.2 + 49.6) = 38.2°
2
10. d 2 = 17 2 + 232 − 2(17)(23) cos 21°
d 2 ≈ 87.94; d ≈ 9.4 mi
sin27° sin A
=
; 100sin A = 70sin27°
100
70
⎛ 70 sin27° ⎞
70 sin27°
sin A =
; A = sin−1⎜
⎟
100
⎝ 100 ⎠
A ≈ 18.5° ; B = 180 – (18.5 + 27) = 134.5º
sin27° sin134.5°
=
100
b
b sin27° = 100sin134.5°
100sin134.5°
≈ 157.1 yd
b=
sin27°
Reinforcing Basic Concepts
1.
2.
7.
8.
75
h
; h sin 25° = 75sin 20°
=
sin 25° sin 20°
75sin 20°
h=
≈ 60.7 ft
sin 25°
h
198.87
sin35° sin B
=
; 11.6sin B = 20 sin35°
20
11.6
⎛ 20sin35° ⎞
20sin35°
sin B =
; B = sin−1⎜
⎟
11.6
⎝ 11.6 ⎠
B ≈ 81.5°; C ≈ 180 − (81.5+ 35) = 63.5°
The measurements are very close.
a2 = 182 + 152 − 2(18)(15) cos 35°
a 2 ≈ 106.66; a ≈ 10.3 cm
a2 = 182 + 152 − 2(18)(15) cos 50°
a2 ≈ 201.89; a ≈ 14.2 cm
a2 = 182 + 152 − 2(18)(15) cos 70°
a2 ≈ 364.31; a ≈ 19.1 cm
The measurements are very close.
Consider the left triangle: the obtuse angle
is supplementary with 70º, so is 110º. That
makes the angle at the top 180 – (58 + 110)
= 12º. Let x be the leftmost side.
sin12° sin110°
=
; x sin12° = 44sin110°
x
44
44sin110°
≈ 198.87
x=
sin12°
Now use that and 58º in a right triangle with
the height of the monument opposite the 58º
angle:
sin58° =
16
13
C ≈ 67.9° or 180 − 67.9 = 112.1°
For C = 67.9º:
B = 180 – (44 + 67.9) = 68.1º
sin 44° sin68.1°
=
2.1
b
b sin 44° = 2.1sin68.1°
2.1sin68.1°
≈ 2.8 km
b=
sin 44°
For C = 112.1º:
B = 180 – (112.1 + 44) = 23.9º
sin 44° sin23.9°
=
; b sin 44° = 2.1sin23.9°
2.1
b
2.1sin23.9°
≈ 1.2 km
b=
sin 44°
6.
Adding the appropriate radii to get lengths:
; h = 198.87 sin58° ≈ 169 m
751
7.4 Exercises
16. F1 = 20 cos 115°, 20 sin 115° = −8.45, 18.13
7.4 Exercises
1.
equilibrium, zero
2.
comp v u ; u cos θ
3.
orthogonal
4.
scalar; vector
F2 = 17 cos 161°, 17 sin 161° = −16.07, 5.53
F3 = 18 cos ( −35° ) ,18 sin ( −35° ) = 14.74, −10.32
F = −8.45 − 16.07 + 14.74,18.13 + 5.53 − 10.32
F = −9.78,13.34 ; −1F = 9.78,−13.34
17. F1 + F2 = 19 + 5,10 + 17 = 24,27
F3 = −24, −27
5,6. Answers will vary.
7.
18. F1 + F2 + F3 = − 12 − 6 + 3, 2 + 17 + 15
F = F1 + F2 = −8 + 2,−3 − 5 = −6, −8
= −15,34 ; F4 = 15, −34
−1F = 6,8
8.
19. F1 = 2210 cos 40°,2210 sin 40°
F = F1 + F2 = −2 + 5, 7 + 3 = 3,10
= 1693.0,1420.6
−1F = −3,−10
9.
F2 = 2500 cos130°, 2500 sin130°
= −1607.0,1915.1
F = F1 + F2 + F3 = −2 + 2 + 5, −7 − 7 + 4
F1 + F2 = 86.0,3335.7
F = 5, −10 ; − 1F = − 5,10
F3 = −86, −3335.7
10. F = F1 + F2 + F3 = −3 − 10 − 9,10 + 3 − 2
F3 = (−86)2 + (−3335.7)2 ≈ 3336.8 ;
F = −22,11 ; − 1F = 22,−11
⎛ − 3335.7 ⎞
⎟ ≈ 88.5 ; In QIII,
⎝ − 86 ⎠
θ r = tan −1 ⎜
11. F = F1 + F2 = (5 + 1)i + (−2 + 10)j = 6i + 8j
−1F = −6i − 8j
268.5º
20. F1 = 200cos 75°, 200sin75° = 51.8,193.2
12. F = F1 + F2 = (−7 − 8)i + (6 − 3)j = −15i + 3j
−1F = 15i − 3j
F2 = 170 cos161°,170 sin161°
= −160.7,55.3
13. F = F1 + F2 + F3 = (2.5 − 0.3)i
+(4.7 + 6.9 − 12)j = 2.2i − 0.4j
−1F = −2.2i + 0.4j
F1 + F2 = −108.9, 248.5
F3 = 108.9, −248.5
F3 = 108.9 2 + (−248.5) 2 ≈ 271.31
14. F = F1 + F2 + F3 = (3 2 − 2 + 5)i
⎛ −248.5 ⎞
⎟ ≈ −66.3 or 293.7º
⎝ 108.9 ⎠
+(−2 3 + 7 + 2 3)j = (3 2 + 3)i + 7j
θ = tan −1 ⎜
−1F = (−3 2 − 3)i − 7j
21. comp vu = 50 cos 42° = 37.16 kg
15. F1 = 10 cos104°,10sin104° = −2.42,9.70
F2 = 6cos 25°,6sin25° = 5.44,2.54
22. comp vu = 3.5 cos 128° = −2.15 tons
F3 = 9 cos ( −20° ) ,9 sin ( −20° ) = 8.46, −3.08
23. comp vu = 1525 cos 65° = 644.49 lbs
F = −2.42 + 5.44 + 8.46,9.70 + 2.54 − 3.08
F = 11.48, 9.16 ; − 1F = −11.48, −9.16
24. comp v u = 221 cos 70°8 = 75.59 lbs
25. comp vu = 3010 cos 30° = 2606.74 kg
752
Chapter 7: Applications of Trigonometry
26. comp vu = 2 cos 115° = −0.85 tons
38. 9000 = F cos 25° ⋅150
F=
27. G makes an angle of 55º with the incline (v)
comp vG = 500 cos 55° ≈ 286.79 lb
39. W = 30 cos 20° ⋅100 ≈ 2819.08 N-m
28. G makes an angle of 70º with the incline (v)
comp vG = 1200 cos 70° ≈ 410.42 lb
40. Dist. = 24(50) = 1200 m
W = 40 cos 39° ⋅1200 ≈ 37,303 N-m
29. Let β = the angle between G and the
incline, and let θ = the angle of incline.
225
325cos β = 225; cos β =
325
⎛
⎞
225
β = cos−1⎜
⎟ ≈ 46.2°
⎝ 325 ⎠
θ = 90 − 46.2 = 43.8°
41. F = 152 +102 ≈ 18.0
⎛ 10 ⎞
⎟ ≈ 33.7°
⎝ 15 ⎠
⎛5⎞
θ v = tan−1 ⎜ ⎟ ≈ 5.7°
⎝ 50 ⎠
Angle between vectors: 28.0º
comp v F = F cos θ = 18.0cos 28.0° ≈ 15.9
θ f = tan−1 ⎜
30. G makes an angle of 72º with the incline (v).
1.75
G cos 72° = 1.75; G =
≈ 5.66
cos 72°
About 5.66 tons
v = 502 + 52 ≈ 50.2
W = 15.9(50.2) ≈ 800 ft-lb
31. W = (15 m)(75 N )=1125 N-m
42. F = (−5)2 + 122 = 13
⎛ 12 ⎞
⎟ ≈ −67.4°
⎝ −5 ⎠
⎛ 10 ⎞
θ v = tan−1 ⎜
⎟ ≈ −21.8°
⎝ −25 ⎠
Angle between vectors: 45.6º
comp v F = 13cos 45.6° ≈ 9.1
32. W = (30 m)(185 N) = 5550 N-m
33. R =
θ f = tan−1 ⎜
1752 sin 45°cos 45°
≈ 957.0 ft
16
952 sin 40°cos 40°
≈ 277.7 ft; it will
16
not break the record.
To find the smallest angle that will break the
record, input Y1 = (952 sin X cos X ) / 16 in
your calculator, and set up a table with
TblStart = 40, ΔTbl = 0.1. The smallest
angle where the value exceeds 280 is 41.6º.
952 sin 45°cos 45°
≈ 282.03. At 45º,
R=
16
the record will fall by about 2.03 ft.
34. R =
v = (−25)2 + 102 ≈ 26.9
W = 9.1(26.9) ≈ 245 ft-lb
43. F = 82 + 22 ≈ 8.2
⎛2 ⎞
θ f = tan−1 ⎜ ⎟ ≈ 14.0°
⎝8 ⎠
⎛ −1 ⎞
θ v = tan−1 ⎜ ⎟ ≈ −3.8°
⎝ 15 ⎠
Angle between vectors: 17.8º
comp v F = 8.2 cos17.8° ≈ 7.8
35. The component of force in the direction of
movement is 250 cos 30° lb.
W = (300)(250 cos 30°) = 64, 951.9 ft-lb
v = (−1)2 + 152 ≈ 15.0
W = 7.8(15) ≈ 117 ft-lb
36. W = (55cos 32°)(100) ≈ 4664.26 ft-lb
37. 45,000 = F cos 5° ⋅100
F=
9000
≈ 66.20 lb
150 cos 25°
45,000
= 451.72 lb
100 cos 5°
753
7.4 Exercises
52. a.
44. F = 152 + (−3)2 ≈ 15.3
⎛ −3 ⎞
⎟ ≈ −11.3°
⎝ 15 ⎠
⎛ −20 ⎞
θ v = tan−1 ⎜
⎟ ≈ −39.8°
⎝ 24 ⎠
Angle between vectors: 28.5º
comp v F = 15.3cos 28.5° ≈ 13.4
b.
θ f = tan−1 ⎜
53. a.
⎛
⎞⎛
⎞
p ⋅q = ⎜7 2 ⎟⎜2 2 ⎟ + (−3)(9) = 1
⎝
⎠⎝
⎠
b.
⎛
⎞2 ⎛
⎞2
p = ⎜7 2 ⎟ + ⎜2 2 ⎟ = 106
⎝
⎠ ⎝
⎠
v = 242 + (−20) 2 ≈ 31.2
q = (−3)2 + 92 = 90
W = 13.4(31.2) ≈ 418 ft-lb
⎛
⎞
⎟⎟ ≈ 89.4°
⎝ 106 90 ⎠
Verified
54. a.
46. F ⋅ v = −5,12 ⋅ −25,10
= (−5) ⋅(−25) + 12 ⋅10 = 245
Verified
b.
= 11
−9
56. u ⋅ v = (−3.5)(−6) + (2.1)(−10) = 0 Yes
57. u ⋅ v = (−6)(−8) + (−3)(15) = 3 No
2
p = 5 + 2 = 29
⎛
58. u ⋅ v = (−5)(−9) + 4(−11) = 1 No
59. u ⋅ v = (−2)(9) + (−6)(−3) = 0 Yes
⎞
⎟⎟ = 45°
⎝ 29 58 ⎠
29
⎛
⎞⎛
⎞
60. u ⋅ v = ⎜3 2 ⎟⎜2 2 ⎟ + (−2)(6) = 0 Yes
⎝
⎠⎝
⎠
p⋅q = (−3) ⋅ 2 + 6 ⋅(−5) = −36
61. v = 7 2 + 12 = 50
u ⋅ v = 3(7) + 5(1) = 26
26
≈ 3.68
comp v u =
50
p = (−3)2 + 62 = 45
q = 22 + (−5)2 = 29
⎛
⎞
⎟⎟ ≈ 175.2°
⎝ 45 29 ⎠
θ = cos−1⎜⎜
b.
2
55. u ⋅ v = 7(4) + (−2)(14) = 0 Yes
p⋅q = 5 ⋅ 3 + 2 ⋅7 = 29
θ = cos−1⎜⎜
51. a.
2
⎞
⎟⎟ ≈ 114.4°
⎝ 11 43 ⎠
q = 32 + 7 2 = 58
b.
( 2 ) + (− 3)
⎛
48. F ⋅ v = 15, −3 ⋅ 24,−20
= 15⋅24 + (−3) ⋅(−20) = 420
Verified
50. a.
p=
θ = cos−1⎜⎜
Verified
b.
⎛ ⎞⎛
⎞
p ⋅q = ⎜ 2 ⎟⎜3 2 ⎟ + (−3)(5) = −9
⎝ ⎠⎝
⎠
q = (3 2)2 + 52 = 43
47. F ⋅ v = 8, 2 ⋅ 15, −1 = 8 ⋅15 + 2 ⋅ −1 = 118
2
1
θ = cos−1⎜⎜
45. F ⋅ v = 15,10 ⋅ 50, 5 = 15 ⋅ 50 + 10 ⋅ 5 = 800
49. a.
p⋅q = (−4)(−6) + 3(−8) = 0
⎛
⎞
0 ⎟
θ = cos−1⎜
= cos−1 0 = 90°
⎜ pq ⎟
⎝
⎠
−36
62. v = (−7)2 + 12 = 50
u ⋅ v = 3(−7) + 5(1) = −16
−16
≈ −2.27
comp v u =
50
p⋅q = (−2)(−6) + 3(−4) = 0
⎛
⎞
0 ⎟
θ = cos−1⎜
= cos−1 0 = 90°
⎜ pq ⎟
⎝
⎠
754
Chapter 7: Applications of Trigonometry
63. v = 02 + (−10)2 = 10
u ⋅ v = (−7)(0) + 4(−10) = −40
−40
= −4
comp v u =
10
69. a.
b.
64. v = 102 + 32 = 109
u ⋅ v = 8(10) + 0(3) = 80
80
≈ 7.66
comp v u =
109
comp v u =
111
70. a.
b.
≈ 3.17
67. a.
b.
129
71. a.
≈ 5.16
b.
v = 82 + 32 = 73
u ⋅ v = 2(8) + 6(3) = 34
⎛ 34 ⎞
proj v u = ⎜ ⎟ 8,3 ≈ 3.73,1.40
⎝ 73 ⎠
72. a.
u1 = 3.73,1.40 ; u2 = u − u1
= 2 − 3.73,6 − 1.40 = −1.73, 4.60
68. a.
b.
u1 = 0.21, 0.34 ; u2 = u − u1
= −4.2 − 0.21, 3 − 0.34 = −4.41,2.66
⎞2
⎛
66. v = 2 + ⎜5 5⎟ = 129
⎠
⎝
⎛
⎞
u ⋅ v = −3 2(2) + 6⎜5 5⎟ = −6 2 + 30 5
⎝
⎠
−6 2 + 30 5
v = (−5)2 + (−8.3)2 = 93.89
u ⋅ v = (−4.2)(−5) + 3(−8.3) = −3.9
⎛ −3.9 ⎞
proj v u = ⎜
⎟ −5,−8.3
⎝ 93.89 ⎠
≈ 0.21,0.34
2
comp v u =
u1 = −0.65, 0.11 ; u2 = u − u1
= − 2 + 0.65,−8 − 0.11 = − 1.35,−8.11
⎛
⎞2
65. v = 62 + ⎜5 3⎟ ≈ 111
⎝
⎠
⎛⎜
⎞⎟
⎛
⎞
u ⋅ v = 7 2 6 + (−3)⎜5 3⎟ ≈ 42 2 − 15 3
⎝
⎠
⎝
⎠
42 2 − 15 3
v = (−6)2 + 12 = 37
u ⋅ v = (−2)(−6) + (−8)(1) = 4
⎛ 4⎞
proj v u = ⎜ ⎟ −6,1 ≈ −0.65, 0.11
⎝ 37 ⎠
b.
v = (−12)2 + 32 = 153
u ⋅ v = (−3)(−12) + 8(3) = 60
⎛ 60 ⎞
proj v u = ⎜
⎟ −12,3 ≈ −4.71,1.18
⎝ 153 ⎠
73. a.
b.
u1 = −4.71,1.18 ; u2 = u − u1
= −3 − (−4.71),8 − 1.18 = 1.71,6.82
74. a.
b.
755
v = 122 + 22 = 148
u ⋅ v = 10(12) + 5(2) = 130
⎛ 130 ⎞
projv u = ⎜
⎟(12i + 2 j)
⎝ 148 ⎠
≈ 10.54i + 1.76j
u1 = 10.54i + 1.76j; u2 = u − u1
= (10 − 10.54)i + (5 − 1.76) j
= −0.54i + 3.24j
v = 52 + (−3)2 = 34
u ⋅ v = (−3)(5) + (−9)(−3) = 12
⎛ 12 ⎞
projv u = ⎜ ⎟(5i − 3 j) = 1.76i − 1.06 j
⎝ 34 ⎠
u1 = 1.76i − 1.06j; u2 = u − u1
= (−3 − 1.76)i + (−9 − (−1.06))j
= −4.76 i − 7.94j
x = (250 cos 60°)(3) = 375 ft
y = (250 sin60°)(3) − 16(3)2 ≈ 505.52 ft
y = (250 sin60°) t − 16 t 2 = 250
−16 t 2 + 216.51t − 250 = 0
Solve using quadratic formula:
t ≈ 1.27 sec, 12.26 sec
x = (300 cos 55°)(3) ≈ 516.22 ft
y = (300 sin55°)(3) − 16(3)2 ≈ 593.24 ft
y = (300 sin55°) t − 16 t 2 = 250
−16 t 2 + 245.75t − 250 = 0
Solve using quadratic formula:
t ≈ 1.10 sec, 14.26 sec
7.4 Exercises
75. a.
b.
76. a.
b.
x = (200 cos 45°)(3) ≈ 424.26 ft
82.
y = (200 sin 45°)(3) − 16(3)2 ≈ 280.26 ft
y = (200 sin 45°) t − 16 t 2 = 250
−16 t 2 + 141.42t − 250 = 0
Solve using quadratic formula:
t ≈ 2.44 sec, 6.40 sec
u v
⋅ =
u v
=
ac + bd
uv
a b
c d
,
⋅
,
u u
v v
=
=
ac
bd
+
uv uv
u⋅v
uv
83. u ⋅ v = 1(5) + 5(2) = 15
x = (500 cos 70°)(3) ≈ 513.03 ft
u = 12 + 52 = 26; v = 52 + 22 = 29
y = (500sin70°)(3) − 16(3)2 = 1265.54 ft
y = (500 sin70°) t − 16 t 2 = 250
−16t 2 + 469.85t − 250 = 0
Solve using quadratic formula:
t ≈ 0.54 sec, 28.82 sec
cos θ =
15
26 29
⎛ 15 ⎞
θ = cos−1⎜⎜
⎟⎟ ≈ 56.9°
⎝ 26 29 ⎠
Δy 5
= =5
Slope of 1i + 5j =
Δx 1
Δy 2
=
Slope of 5i + 2j =
Δx 5
2
−5
−23 / 5
23
=−
tan θ = 5 2 =
3
15
1 + ⋅5
77. y = (90sin65°)(1.2) − 16(1.2)2 ≈ 74.84 ft
To find another time, set height equal to
74.84: y = (90sin65°) t − 16 t 2 = 74.84
−16t 2 + 81.57t − 74.84 = 0
Solve using quadratic formula:
t ≈ 1.2, 3.9; After 3.9 seconds, which is
about 2.7 seconds later.
5
⎛ 23 ⎞
θ = tan−1⎜− ⎟ = −56.9
⎝ 15 ⎠
The angle between is 56.9º. Answers to last
part of question will vary.
78. y = (130 sin30°)(1.5) − 16(1.5)2 ≈ 61.5 ft
To find another time, set height equal to
61.5: y = (130 sin30°) t − 16 t 2 = 61.5
−16t 2 + 65t − 61.5 = 0
Solve using quadratic formula:
t ≈ 1.5, 2.6; After 2.6 seconds, which is
about 1.1 seconds later.
84. x = v cos θ ⋅ t ; t =
x
v cos θ
y = v sin θ ⋅ t − 16t 2
⎛ x ⎞
x2
= v sin θ ⎜
⎟ − 16 2
⎝ v cos θ ⎠
v cos2 θ
79. w ⋅ (u + v) = e, f ⋅ a + c, b + d
= e( a + c) + f ( b + d) = ea + ec + fb + fd
= ( ea + fb) + ( ec + fd )
= e, f ⋅ a, b + e, f ⋅ c, d
= (tan θ )x −
16
2
2
x2
v cos θ
This is a parabola that opens down. It’s
concave down and has two x-intercepts as
long as 0 < θ ≤ 90° . The maximum height
occurs at the vertex. Other answers will
vary.
= w⋅u+ w⋅v
80. k ( u ⋅ v) = k ( ac + bd) = kac + kbd
= ka, kb ⋅ c, d = ku ⋅ v
Also, kac + kbd = a( kc) + b( kd)
= a, b ⋅ kc, kd = u ⋅ kv
81. 0 ⋅ u = 0, 0 ⋅ a, b = 0( a) + 0( b) = 0
u ⋅ 0 = a, b ⋅ 0,0 = a(0) + b(0) = 0
756
Chapter 7: Applications of Trigonometry
88. Vector representing wind:
40 1
= as long
200 5
Horizontal component:
a = 40 cos 0°
a = 40
Vertical component:
b = 40 sin 0°
b=0;
Vector in QI
w = 40i + 0 j ;
angle between x-axis and vector is 60°
Horizontal component:
a = 200 cos 60°
a = 100
Vertical component:
b = 200 sin 60°
85. 2.9e −0.25t + 7.6 = 438
2.9e −0.25t = 430.4
430.4
e −0.25t =
2.9
⎛ 430.4 ⎞
ln e −0.25t = ln ⎜
⎟
⎝ 2.9 ⎠
⎛ 430.4 ⎞
−0.25t = ln ⎜
⎟
⎝ 2.9 ⎠
⎛ 430.4 ⎞
ln ⎜
⎟
2.9 ⎠
t= ⎝
≈ −20
−0.25
86. y = 3 cos⎛⎜ 2θ − π ⎞⎟ = 3 cos ⎡2⎛⎜θ − π ⎞⎟⎤
⎝
4⎠
⎢
⎣ ⎝
8 ⎠⎥⎦
Amplitude: 3
2π
Period:
=π
2
Phase shift: −
b = 100 3 ;
p = 100i + 100 3 j ;
π
w + p = 40i + 0 j + 100i + 100 3 j
8
Reference Rectangle: 2A=6 by P = π units
π π 3π
, and π .
Rule of fourths: t = 0, , ,
4 2 4
= 140i + 100 3 j ;
w +p =
(140)2 + (100
3
)
2
= 222.7 mph ;
Bearing:
⎛ 100 3 ⎞
⎟⎟ ≈ 51°
⎝ 140 ⎠
θr = tan −1 ⎜⎜
From x: 90° − 51° = 39°
7.5 Exercises
87. a 2 = 1722 + 2502 − 2 (172 )( 250 ) cos 32°
1.
modulus; argument
2.
⎛π ⎞
2 cis ⎜ ⎟
⎝4⎠
3.
multiply; add
4.
divide; subtract
5.
z = (−1)2 + (− 3)2 = 4 = 2
⎛
⎞
− 3⎟
θ r = tan−1 ⎜⎜
⎟ = 60° ; In QIII, θ = 240°
⎝ −1 ⎠
a ≈ 138.4 m ;
sin C sin 32°
=
172
138.4
172 sin 32°
sin C =
138.4
⎛ 172 sin 32° ⎞
C = sin −1 ⎜
⎟
⎝ 138.4 ⎠
C ≈ 41.2°
180° − 41.2° − 32° = 106.8°
Angles
Sides
A = 32°
138.4 m
B ≈ 106.8°
250 m
C ≈ 41.2°
172 m
−1 − 3i = 2(cos 240° + i sin240°)
6.
P = 138.4 + 250 + 172 = 560.4 m
A=
1
(250)(172) sin 32° ≈ 11393.3 m 2
2
757
Answers will vary.
7.5 Exercises
11.
7.
z = −2 − 2 i ; QIII
z1 + z3 = (7 + 2i) + (1 + 4i) = 8 + 6i = z2
r = (−2)2 + (−2)2 = 8 = 2 2
⎛ −2 ⎞
θ r = tan−1 ⎜ ⎟ = 45°. In QIII, θ = 225°
⎝ −2 ⎠
8.
z = 2 2 (cos 225° + i sin 225°)
12.
z2 + z3 = (3 + 4i) + (−1 + 3i) = 2 + 7i
= z1
z = 7 − 7i ; QIV
9.
r = 7 2 + (−7)2 = 98 = 7 2
⎛ −7 ⎞
θ r = tan−1 ⎜ ⎟ = −45°. In QIV, θ = 315°
⎝7 ⎠
z = 7 2 (cos 315° + i sin 315°)
13.
z1 + z3 = (−2 − 5i) + (3 − 2 i) = 1 − 7 i = z2
10.
z = −5 3 − 5i ; QIII
⎛
⎞2
r = ⎜5 3⎟ + (−5)2 = 100 = 10
⎝
⎠
⎛ −5 ⎞
θ r = tan−1 ⎜⎜
⎟⎟ = 30° . In QIII, θ = 210°
⎝ −5 3 ⎠
z = 10(cos 210° + i sin 210°)
z1 + z2 = (−2 + 6i) + (7 − 2 i) = 5 + 4i = z3
758
Chapter 7: Applications of Trigonometry
16.
14.
z = 5 7 − 5 7i ; QIV
z = 2 − 2 3i; QIV
2
⎛
⎞
r = 2 + ⎜−2 3 ⎟ = 16 = 4
⎝
⎠
⎛
⎞
−2 3 ⎟
θ r = tan−1 ⎜⎜
⎟ = −60° . In QIV, θ = 300°
⎝ 2 ⎠
z = 4(cos 300° + i sin 300°)
2
r=
(5 7 ) + ( −5 7 )
2
2
= 350 = 5 14
⎛
⎞
7π
−5 7 ⎟
π
⎟ = − 4 . In QIV, θ = 4
⎝5 7 ⎠
⎡ ⎛ 7π ⎞
⎛ 7 π ⎞⎤
z = 5 14⎢cos⎜
⎟ + isin⎜
⎟⎥
⎝ 4 ⎠⎦
⎣ ⎝ 4 ⎠
θ r = tan−1 ⎜⎜
15.
17.
z = −3 2 + 3 2i ; QII
z = 4 3 − 4i ; QIV
⎛
⎞2 ⎛
⎞2
r = ⎜−3 2 ⎟ + ⎜3 2 ⎟ = 36 = 6
⎝
⎠ ⎝
⎠
⎛
⎞
3π
3 2 ⎟
π
θ r = tan−1 ⎜⎜
⎟ = − 4 . In QII, θ = 4
⎝ −3 2 ⎠
⎡ ⎛ 3π ⎞
⎛ 3π ⎞⎤
z = 6⎢cos⎜ ⎟ + isin⎜ ⎟⎥
⎝ 4 ⎠⎦
⎣ ⎝ 4 ⎠
⎛
⎞2
r = ⎜4 3⎟ + (−4)2 = 64 = 8
⎝
⎠
⎛ −4 ⎞
π
11π
θ r = tan−1 ⎜⎜
⎟⎟ = − . In QIV, θ =
6
6
⎝4 3 ⎠
⎡ ⎛ 11π ⎞
⎤
⎛ 11π ⎞
z = 8⎢cos⎜
⎟ + isin⎜
⎟⎥
⎝ 6 ⎠⎦
⎣ ⎝ 6 ⎠
759
7.5 Exercises
21.
18.
z = −5−12 i ; QIII
z = −6 + 6 3i ; QII
r = (−5)2 + (−12)2 = 13
⎛ −12 ⎞
θ r = tan−1 ⎜
⎟ ≈ 67.4° . QIII, θ = 247.4°
⎝ −5 ⎠
⎛
⎞2
r = (−6) + ⎜6 3⎟ = 144 = 12
⎝
⎠
⎛
⎞
2π
6 3⎟
π
= − . In QII, θ =
θ r = tan−1 ⎜⎜
⎟
3
3
⎝ −6 ⎠
⎡ ⎛ 2π ⎞
⎛ 2π ⎞⎤
z = 12⎢cos⎜ ⎟ + i sin⎜ ⎟⎥
⎝ 3 ⎠⎦
⎣ ⎝ 3 ⎠
2
⎡
⎛ 12 ⎞ ⎤
z = 13 cis ⎢180+tan -1 ⎜ ⎟ ⎥ ≈ 13 cis 247.4°
⎝ 5 ⎠⎦
⎣
22.
19.
z = −8 +15i ; QII
r = (−8)2 + 152 = 17
⎛ 15 ⎞
θ r = tan−1 ⎜ ⎟ ≈ −61.9° . QII, θ = 118.1°
⎝ −8 ⎠
z = 8 + 6i ; QI
r = 82 + 62 = 10
⎛6 ⎞
θ = tan−1⎜ ⎟ ≈ 36.9°
⎝8 ⎠
⎡
⎛ 15 ⎞ ⎤
z = 17 cis ⎢180+tan -1 ⎜ − ⎟ ⎥ ≈ 17 cis118.1°
⎝ 8 ⎠⎦
⎣
⎡
⎛ 6 ⎞⎤
z = 10cis ⎢ tan −1 ⎜ ⎟ ⎥ ≈ 10cis36.9°
⎝ 8 ⎠⎦
⎣
23.
20.
z = 6 + 17.5i ; QI
r = 62 +17.52 = 342.25 = 18.5
⎛ 17.5 ⎞
θ = tan−1⎜
⎟ ≈ 1.2405
⎝ 6 ⎠
⎡
⎛ 17.5 ⎞⎤
18.5cis ⎢ tan-1 ⎜
⎟⎥ ≈ 18.5cis1.2405
⎝ 6 ⎠⎦
⎣
z = −9 + 12 i ; QII
r = (−9)2 +122 = 225 = 15
⎛ 12 ⎞
θ r = tan−1 ⎜ ⎟ ≈ −53.1°. QII, θ = 126.9°
⎝ −9 ⎠
⎡
⎛ 12 ⎞ ⎤
z = 15 cis ⎢180+tan -1 ⎜ − ⎟ ⎥ ≈ 15 cis126.9°
⎝ 9 ⎠⎦
⎣
760
Chapter 7: Applications of Trigonometry
26.
24.
z = 30 − 5.5i ; QIV
2
z = 12 − 4i ; QIV
2
r = 30 + (−5.5) = 930.25 = 30.5
⎛ −5.5 ⎞
θ r = tan−1 ⎜
⎟ ≈ −0.1813. In QIV,
⎝ 30 ⎠
θ = 6.1019
⎡
⎛ 5.5 ⎞ ⎤
z = 30.5cis ⎢ 2π + tan -1 ⎜ −
⎟ ⎥ ≈ 30.5cis 6.1019
⎝ 30 ⎠ ⎦
⎣
r = 122 + (−4)2 = 160 = 4 10
⎛ −4 ⎞
θ r = tan−1 ⎜ ⎟ ≈ −0.3218. In QIV,
⎝ 12 ⎠
θ = 5.9614
⎡
⎛ 1 ⎞⎤
z = 4 10cis ⎢2π + tan−1 ⎜ − ⎟⎥ ≈ 4 10cis5.9614
⎝ 3 ⎠⎦
⎣
27. r = 2, θ =
π
25.
4
⎡ ⎛π ⎞
⎛π ⎞
⎛ π ⎞⎤
z = 2cis⎜ ⎟ = 2 ⎢cos⎜ ⎟ + i sin ⎜ ⎟⎥
4
4
⎝ ⎠
⎝ 4 ⎠⎦
⎣ ⎝ ⎠
z = −6 + 10 i ; QII
r = (−6)2 + 102 = 136 = 2 34
⎛ 10 ⎞
θ r = tan−1 ⎜ ⎟ ≈ −1.0304. In QII,
⎝ −6 ⎠
θ = 2.1112
⎡
⎛ 5 ⎞⎤
z = 2 34 cis ⎢π + tan−1 ⎜ − ⎟⎥ ≈ 2 34 cis2.112
⎝ 3 ⎠⎦
⎣
⎡ 2
2⎤
= 2⎢
+i
⎥ = 2 + 2i
2 ⎥⎦
⎢⎣ 2
28. r = 12, θ =
π
6
⎡ ⎛π ⎞
⎛π ⎞
⎛ π ⎞⎤
z = 12cis⎜ ⎟ = 12 ⎢cos⎜ ⎟ + i sin ⎜ ⎟⎥
6
6
⎝ ⎠
⎝ 6 ⎠⎦
⎣ ⎝ ⎠
⎡ 3
1⎤
= 12 ⎢
+ i ⎥ = 6 3 + 6i
2 ⎥⎦
⎢⎣ 2
761
7.5 Exercises
29. r = 4 3 , θ =
π
⎛ 15 ⎞
31. r = 17, θ = tan −1 ⎜ ⎟
⎝8⎠
3
⎡ ⎛π ⎞
⎛π ⎞
⎛ π ⎞⎤
z = 4 3cis⎜ ⎟ = 4 3 ⎢cos⎜ ⎟ + i sin ⎜ ⎟⎥
3
3
⎝ ⎠
⎝ 3 ⎠⎦
⎣ ⎝ ⎠
⎡
⎛ 15 ⎞⎤
z = 17cis ⎢ tan −1 ⎜ ⎟⎥
⎝ 8 ⎠⎦
⎣
⎡1
3⎤
= 4 3⎢ + i
⎥ = 2 3 + 6i
2 ⎥⎦
⎣⎢ 2
30. r = 5 3 , θ =
⎡ ⎛
⎛
⎛ 15 ⎞ ⎞
⎛ 15 ⎞ ⎞⎤
= 17 ⎢cos⎜⎜ tan −1 ⎜ ⎟ ⎟⎟ + i sin ⎜⎜ tan −1 ⎜ ⎟ ⎟⎟⎥
⎝ 8 ⎠⎠
⎝ 8 ⎠ ⎠⎦⎥
⎝
⎣⎢ ⎝
⎡8
15 ⎤
= 17⎢ + i ⎥ = 8 +15i
⎣ 17 17 ⎦
7π
6
⎛3⎞
32. r = 10, θ = tan −1 ⎜ ⎟
⎝4⎠
⎡ ⎛ 7π ⎞
⎛ 7π ⎞
⎛ 7π ⎞⎤
z = 5 3cis⎜
⎟ = 5 3 ⎢cos⎜
⎟ + i sin ⎜
⎟⎥
6
6
⎝
⎠
⎠
⎝ 6 ⎠⎦
⎣ ⎝
⎡ 3
1⎤
15 5 3
= 5 3 ⎢−
−i ⎥ = − −
i
2 ⎥⎦
2
2
⎣⎢ 2
⎡
⎛ 3 ⎞⎤
z = 10cis ⎢ tan −1 ⎜ ⎟⎥
⎝ 4 ⎠⎦
⎣
⎡ ⎛
⎛
⎛ 3 ⎞⎞
⎛ 3 ⎞ ⎞⎤
= 10 ⎢cos⎜⎜ tan −1 ⎜ ⎟ ⎟⎟ + i sin ⎜⎜ tan −1 ⎜ ⎟ ⎟⎟⎥
⎝ 4 ⎠⎠
⎝ 4 ⎠ ⎠⎦⎥
⎝
⎣⎢ ⎝
⎡ 4 3⎤
= 10⎢ + i ⎥ = 8 + 6i
⎣ 5 5⎦
762
Chapter 7: Applications of Trigonometry
⎛ 5 ⎞
⎟
33. r = 6, θ = π − tan −1 ⎜⎜
⎟
⎝ 11 ⎠
⎛ 7⎞
⎟
34. r = 4, θ = π + tan −1 ⎜
⎜ 3 ⎟
⎠
⎝
First, pretend that the “π –“ is not there:
6
First, pretend that the “π +” is not there:
4
5
θ
3
11
⎡
⎛ 5 ⎞⎤
11
cos⎢ tan−1⎜⎜
⎟⎟⎥ =
6
⎢⎣
⎝ 11 ⎠⎥⎦
⎡
⎛ 5 ⎞⎤ 5
sin⎢ tan−1⎜⎜
⎟⎟⎥ =
⎢⎣
⎝ 11 ⎠⎥⎦ 6
Now use subtraction identities: For
⎛ 5 ⎞
convenience, let θ = tan−1⎜⎜
⎟⎟ .
⎝ 11 ⎠
cos(π − θ ) = cos π cos θ + sin π sin θ
⎛
⎞
11
11 ⎟
+0 = −
= −1⎜⎜
⎟
6
⎝ 6 ⎠
sin(π − θ ) = sin π cos θ − cos π sin θ
= 0 − (−1)
7
θ
⎡
⎛ 7 ⎞⎤ 3
⎟ =
cos ⎢ tan −1 ⎜
⎜ 3 ⎟⎥⎥ 4
⎢⎣
⎝
⎠⎦
⎡
⎛ 7 ⎞⎤
⎟⎥ = 7
sin ⎢ tan −1 ⎜
⎜
⎟⎥
3
4
⎝
⎠⎦
⎣⎢
Now use addition identities. For
⎛
⎞
7⎟
convenience, let θ = tan−1⎜⎜
⎟
⎝ 3 ⎠
cos(π + θ ) = cos π cos θ − sin π sin θ
3
3
+0 = −
4
4
sin(π + θ ) = sin π cos θ + cos π sin θ
= (−1)
7
7
=−
4
4
⎡
7⎤
z = 4cis ⎢π − tan −1
⎥
3
⎣
⎦
= 4 ⎡⎣ cos (π + θ ) + i sin (π + θ ) ⎤⎦
5 5
=
6 6
= 0 + (−1)
5 ⎤
⎡
z = 6cis ⎢π − tan −1
⎥
11 ⎦
⎣
= 6 ⎣⎡cos (π − θ ) + i sin (π − θ ) ⎦⎤
⎡ 3
7⎤
= = 4 ⎢− − i
⎥ = −3 − 7i
4 ⎥⎦
⎢⎣ 4
⎡ 11 5 ⎤
= 6⎢−
+ i ⎥ = − 11 + 5i
6 ⎥⎦
⎣⎢ 6
763
7.5 Exercises
⎛
⎞2
38. r1 = ⎜− 3⎟ + 12 = 4 = 2
⎝
⎠
⎛ 1 ⎞
θ1r = tan−1⎜⎜
⎟⎟ = −30° . QII, θ1 = 150°
⎝− 3 ⎠
35. r1 = (−2) 2 + 22 = 8 = 2 2
⎛2⎞
θ1r = tan−1⎜ ⎟ = −45°. In QII, θ1 = 135°
⎝ −2 ⎠
r2 = 32 + (−3)2 = 18 = 3 2
⎛3⎞
θ 2 = tan−1 ⎜ ⎟ = 45°
⎝3⎠
r2 = 32 + 02 = 3
θ 2 = tan−1 0 = 0
(− 2 + 2i )(3 + 3i ) = −6 − 6i + 6i + 6i 2
3 1
− 3+i
=−
+ i;
3 3
3
r1 2
= ; θ1 − θ 2 = 150°
r2 3
= −12 + 0i
⎛⎜
⎞⎟⎜⎛
⎞⎟
r1r2 = 2 2 3 2 = 12 ;
⎝
⎠⎝
⎠
θ1 + θ 2 = 135 + 45 = 180°
12 (cos 180º + isin 180º) = –12 + 0i
⎛
⎞
2
(cos 150° + i sin 150°) = 2 ⎜⎜ − 3 + i 1 ⎟⎟
3⎝ 2
2⎠
3
36. r1 = 12 + ( 3)2 = 4 = 2
⎛ ⎞
3
θ1 = tan−1 ⎜⎜ ⎟⎟ = 60°
1
⎝ ⎠
=−
5π π
+ =π
6
6
z1 z2 = 24 cis π = −24 + 0i ;
r1 8
5π π 2π
= ; θ1 − θ 2 =
− =
r2 3
6
6
3
⎛
⎞
z1 8
2π 8 ⎜ 1
3⎟
4 4 3
= cis
= ⎜− + i
i
=− +
z2 3
3
2 ⎟⎠
3
3⎝ 2
3
39. r1r2 = 24; θ1 + θ 2 =
2
r2 = 32 + 3 = 12 = 2 3
⎛ ⎞
3
θ 2 = tan−1 ⎜⎜ ⎟⎟ = 30°
⎝ 3 ⎠
⎜⎛1 + 3i⎟⎞⎜⎛3 + 3i⎟⎞ = 3 + 4 3i + 3i2 = 0 + 4 3i
⎝
⎠⎝
⎠
⎛⎜
⎞⎟
r1r2 = 2 2 3 = 4 3 ;
⎝
⎠
θ1 + θ 2 = 60 + 30 = 90°
40. r1r2 = 30; θ1 + θ 2 =
4 3 (cos 90° + i sin 90°) = 0 + 4 3i
2
2
3 + i 1 − 3i
3 − 3i + i − 3i2
⋅
=
1 + 3i 1 − 3i 1 − 3i + 3i − 3i2
3 1
2 3 − 2i
=
− i;
4
2 2
r1 2
= = 1; θ1 − θ 2 = 30 − 60 = −30°
r2 2
1(cos (–30º) + isin (–30º)) =
+
π
=
2π
3
= −15 + 15 3i ;
r1 5
π π
π
= ; θ1 − θ 2 = − = −
r2 6
6 2
3
⎛
⎞
z1 5 ⎛ π ⎞ 5 ⎜ 1
3⎟ 5 5 3
= cis ⎜− ⎟ = ⎜ − i
−
i
=
z2 6 ⎝ 3 ⎠ 6 ⎝ 2
2 ⎟⎠ 12 12
r2 = 12 + 3 = 4 = 2
⎛ ⎞
3
θ 2 = tan−1 ⎜⎜ ⎟⎟ = 60°
⎝ 1 ⎠
=
π
6 2
⎛
⎞
2π
1
3⎟
z1 z2 = 30 cis
= 30⎜⎜− + i
3
2 ⎟⎠
⎝ 2
3 + 12 = 4 = 2
⎛ 1 ⎞
θ1 = tan−1 ⎜⎜ ⎟⎟ = 30°
⎝ 3⎠
37. r1 =
3 1
+ i
3 3
3 1
− i
2 2
764
Chapter 7: Applications of Trigonometry
44. r1r2 = 2(8.4) = 16.8
⎛
⎞⎛
⎞
41. r1r2 = ⎜2 3 ⎟⎜7 3⎟ = 42
⎝
⎠⎝
⎠
5π 11π
θ1 + θ 2 = π +
=
6
6
⎛
⎞
11π
3
1
z1 z2 = 42 cis
= 42⎜⎜
− i ⎟⎟
6
2⎠
⎝ 2
3π π 4 π
+ =
5
5
5
⎡ ⎛ 4π ⎞
⎛ 4π ⎞⎤
z1 z 2 = 16.8⎢cos⎜
⎟ + i sin ⎜
⎟⎥
5
⎠
⎝ 5 ⎠⎦
⎣ ⎝
≈ −13.59 + 9.87i ;
r1
2
3π π 2π
=
= 0.23; θ1 − θ 2 =
− =
r2 8.4
5
5
5
θ1 + θ 2 =
= 21 3 − 21i ;
5π π
r1 2 3 2
=
= ; θ1 − θ 2 = π −
=
r2 7 3 7
6
6
π 2 ⎛ 3 1 ⎟⎞
z1 2
3 1
= cis = ⎜⎜
+i ⎟=
+ i
z2 7
6 7⎝ 2
2⎠ 7 7
z1
⎡ ⎛ 2π
= 0.23⎢cos⎜
z2
⎣ ⎝ 5
⎞
⎛ 2π ⎞⎤
⎟ + i sin ⎜
⎟⎥
⎠
⎝ 5 ⎠⎦
≈ 0.07 + 0.22i
45. r1r2 = 40; θ1 + θ 2 = 60 + 30 = 90°
z1 z2 = 40 cis 90° = 0 + 40i ;
r1 10 5
=
= ; θ1 − θ 2 = 60 − 30 = 30°
r2
4 2
⎛
⎞
z1 5
5 3
1
5 3 5
= cis 30° = ⎜⎜
+ i ⎟⎟ =
+ i
z2 2
4
2⎝ 2
2⎠
4
⎛
⎞⎛
⎞
42. r1r2 = ⎜6 2 ⎟⎜3 2 ⎟ = 36
⎝
⎠⎝
⎠
3π π 11π
θ1 + θ 2 =
+ =
2
3
6
⎛
⎞
11π
3
1
z1 z2 = 36 cis
= 36⎜⎜
− i ⎟⎟
6
2⎠
⎝ 2
= 18 3 − 18i ;
46. r1r2 = 14; θ1 + θ 2 = 120 + 300 = 420°
⎛
⎞
1
3⎟
z1 z2 = 14 cis 420° = 14⎜⎜ + i
2 ⎟⎠
⎝2
r1 6 2
3π π 7 π
=
= 2; θ1 − θ 2 =
− =
r2 3 2
2
3
6
⎛
⎞
z1
7π
3
1
= 2 cis
= 2⎜⎜−
+ i ⎟⎟ = − 3 − i
z2
6
2⎠
⎝ 2
= 7 + 7 3i ;
r1 7
= ; θ1 − θ 2 = 120 − 300 = −180°
r2 2
43. r1r2 = 9(1.8) = 16.2
π 2π 11π
+
=
15 3
15
⎡ ⎛ 11π ⎞
⎛ 11π ⎞⎤
z1 z 2 = 16.2 ⎢cos⎜
⎟ + i sin ⎜
⎟⎥
⎝ 15 ⎠⎦
⎣ ⎝ 15 ⎠
≈ −10.84 + 12.04i ;
r1
π 2π
9
3π
=
= 5; θ1 − θ 2 = −
=−
r2 1.8
15 3
5
z1 7
7
= cis (−180)° = − + 0i
z2 2
2
θ1 + θ 2 =
⎛
⎞⎛
⎞
47. r1r2 = ⎜5 2 ⎟⎜2 2 ⎟ = 20
⎝
⎠⎝
⎠
θ1 + θ 2 = 210 + 30 = 240°
⎛
⎞
1
3⎟
z1 z2 = 20 cis 240° = 20⎜⎜− − i
2 ⎟⎠
⎝ 2
z1
⎡ ⎛ 3π ⎞
⎛ 3π ⎞⎤
= 5⎢cos⎜ −
⎟ + i sin ⎜ −
⎟⎥
5
z2
⎠
⎝ 5 ⎠⎦
⎣ ⎝
≈ −1.55 − 4.76i
= −10 − 10 3i ;
r1 5 2 5
=
= ; θ1 − θ 2 = 210 − 30 = 180°
r2 2 2 2
z1 5
5
= cis180° = − + 0i
z2 2
2
765
7.5 Exercises
⎛
⎞⎛ ⎞
48. r1r2 = ⎜5 3 ⎟⎜ 3 ⎟ = 15
⎝
⎠⎝ ⎠
θ1 + θ 2 = 240 + 90 = 330°
⎛
⎞
3
1
z1 z2 = 15cis 330° = 15⎜⎜
− i ⎟⎟
2⎠
⎝ 2
51. Distance from u to v:
d=
(10 − 2)2 + (
3− 3
From v to w:
d=
(6 − 10)2 + (5
3− 3
d=
(2 − 6)2 + (
2
)
2
From w to u:
15 3 15
− i;
2
2
r1 5 3
=
= 5; θ1 − θ 2 = 240 − 90 = 150°
r2
3
⎛
⎞
z1
3
1
5 3 5
= 5cis150° = 5⎜⎜−
+ i ⎟⎟ = −
+ i
z2
2
2
2
2
⎝
⎠
=
)
3 −5 3
=8
= 16 + 48 = 8
)
2
= 16 + 48 = 8
All sides have length 8.
⎛
⎞⎛
⎞
u2 = ⎜2 + 3i⎟⎜2 + 3i⎟ = 4 + 4 3i + 3i2
⎝
⎠⎝
⎠
u2 = 1 + 4 3i ;
⎛
⎞⎛
⎞
v2 = ⎜10 + 3i⎟⎜10 + 3i⎟ = 100 + 20 3i + 3i2
⎝
⎠⎝
⎠
49. r1r2 = 6(1.5) = 9; θ 1 + θ 2 = 82 + 27 = 109°
z1 z 2 = 9(cos109° + i sin 109°) ≈ −2.93 + 8.5i ;
v2 = 97 + 20 3i ;
⎛
⎞⎛
⎞
w 2 =⎜6 + 5 3i⎟⎜6 + 5 3i⎟=36+60 3i+75i2
⎝
⎠⎝
⎠
r1
6
=
= 4; θ1 − θ 2 = 82 − 27 = 55°
r2 1.5
z1
= 4(cos55° + i sin 55°) ≈ 2.29 + 3.28i
z2
w 2 = −39 + 60 3i ;
⎛
⎞⎛
⎞
uv = ⎜2 + 3i⎟⎜10 + 3i⎟ = 20 + 12 3i + 3i2
⎝
⎠⎝
⎠
uv = 17 + 12 3i ;
⎛
⎞⎛
⎞
uw = ⎜2 + 3i⎟⎜6 + 5 3i⎟ = 12 + 16 3i + 15i2
⎝
⎠⎝
⎠
50. r1r2 = 1.6(8) = 12.8 ;
θ1 + θ 2 = 59 + 275 = 334°
z1 z 2 = 12.8(cos334° + i sin 334°)
uw = −3 + 16 3i ;
⎛
⎞⎛
⎞
vw = ⎜10 + 3i⎟⎜6 + 5 3i⎟ = 60 + 56 3i + 15i2
⎝
⎠⎝
⎠
≈ 11.50 − 5.61i ;
r1 1.6
=
= 0.2
r2
8
θ1 − θ 2 = 59 − 275 = −216° or 144º
z1
= 0.2(cos144° + i sin 144°)
z2
vw = 45 + 56 3i;
⎛
⎞ ⎛
⎞
u2 + v2 + w 2 = ⎜1 + 4 3i⎟ + ⎜97 + 20 3i⎟
⎠
⎝
⎠ ⎝
⎛⎜
⎞⎟
+ −39 + 60 3i = 59 + 84 3i ;
⎝
⎠
⎛⎜
⎞ ⎛
⎞
uv + uw + vw = 17 + 12 3i⎟ + ⎜−3 + 16 3i⎟
⎠
⎝
⎠ ⎝
≈ −0.16 + 0.12i
+45 + 56 3i = 59 + 84 3i
52. (1 − 2i )3
= 13 + 3(1)2 (−2i) + 3(1)(−2i)2 + (−2i)3
= 1− 6i + 12i2 − 8i3 = 1− 6 i −12 + 8i
= −11 + 2i
766
Chapter 7: Applications of Trigonometry
53. a.
b.
A = 170; V (t) = 170 sin( f (2πt))
V (t) = 170 sin(60(2πt)) = 170 sin(120 πt)
The graph of V is at height 140 at about
t = 0.00257 sec.
54. a.
A = 325; V (t ) = 325 sin( f (2πt ))
V (t ) = 325 sin(50(2πt ) ) = 325 sin(100πt )
55. a.
b.
57. a.
Z = 7 + j(6 − 11) = 7 − 5 j (QIV)
Z = 7 2 + (−5)2 = 74 ≈ 8.60 ;
⎛ −5 ⎞
⎟ ≈ −35.5° . In QIV,
⎝7 ⎠
θ = 324.5°
Z = 8.60 cis 324.5º
VRLC = I Z = 1.8(8.60) = 15.48 V
θ r = tan−1 ⎜
b.
58. a.
Z = 9.2 + j(5.6 − 8.3) = 9.2 − 2.7 j QIV
Z = 9.22 + (−2.7)2 = 91.93 ≈ 9.59 ;
One cycle is 1/50 or 0.02 seconds, so
our table should go up to 0.01.
t
V
t
V
0.006 309.1
0
0
0.001 100.4 0.007 262.9
0.008 191.0
0.002 191.0
0.009
100.4
0.003 262.9
0
0.004 309.1 0.010
⎛ −2.7 ⎞
⎟ ≈ −16.4° . In QIV,
⎝ 9.2 ⎠
θ = 343.6°
Z = 9.59 cis 343.6º
VRLC = I Z = 2.0(9.59) = 19.18 V
θ r = tan−1 ⎜
b.
59. a.
Z = 12 + j(5 − 0) = 12 + 5 j (QI)
Z = 122 + 52 = 13;
⎛5⎞
⎟ ≈ 22.6°
⎝ 12 ⎠
Z = 13 cis 22.6º
VRLC = I Z = 1.7(13) = 22.1 V
θ = tan−1⎜
The graph of V is at height 215 at about
t = 0.0023 sec.
b.
Z = 15 + j(12 − 4) = 15 + 8 j (QI)
Z = 152 + 82 = 17
b.
⎛7 ⎞
⎟ ≈ 16.3°
⎝ 24 ⎠
Z = 25 cis 16.3º
VRLC = I Z = 2.5(25) = 62.5 V
θ = tan−1⎜
0.005 325.0
c.
Z = 24 + j(12 − 5) = 24 + 7 j (QI)
Z = 242 + 7 2 = 25
One cycle is 1/6 = 0.0167 seconds, so
our table should go up to 0.008.
t
V
t
V
0
0
0.005 161.7
0.001 62.6
0.006 131.0
0.002 116.4 0.007 81.9
0.003 153.8
0.008 21.3
0.004 169.7
c.
b.
56. a.
60. a.
⎛8⎞
θ = tan−1⎜ ⎟ ≈ 28.1°
⎝ 15 ⎠
Z = 17 cis 28.1º
VRLC = I Z = 3(17) = 51 V
Z = 35 + j(12 − 0) = 35 + 12 j (QI)
Z = 352 + 122 = 37 ;
⎛ 12 ⎞
⎟ ≈ 18.9°
⎝ 35 ⎠
Z = 37 cis 18.9º
VRLC = I Z = 4(37) = 148 V
θ = tan−1⎜
b.
767
7.5 Exercises
61. Both are in QI.
64. rI = 42 + 32 = 5
⎛3⎞
θ I = tan−1 ⎜ ⎟ = 36.9°
⎝4⎠
2
3 + 12 = 4 = 2
⎛ 1 ⎞
θ I = tan−1 ⎜⎜ ⎟⎟ = 30°
⎝ 3⎠
rI =
rZ = 22 + 42 = 20 = 2 5
⎛ −4 ⎞
θ Z = tan−1 ⎜ ⎟ = −63.4° or 296.6º
⎝2 ⎠
I = 5cis36.9o ;
rZ = 52 + 52 = 50 = 5 2
⎛5⎞
θ Z = tan−1 ⎜ ⎟ = 45°
⎝5⎠
I = 2cis30o ;
Z = 2 5cis296.6o
( )
V = IZ = 5 2 5 cis(36.9° + 296.6°)
Z = 5 2cis45o
( )
V = IZ = 2 5 2 cis(30° + 45°)
= 10 5 cis 333.5°
= 10 2 cis 75°
2
65. rV = 22 + 2 3 = 16 = 4
⎛
⎞
2 3⎟
θV = tan−1 ⎜⎜
⎟ = 60°
⎝ 2 ⎠
2
3 + 12 = 4 = 2
⎛ 1 ⎞
θ I = tan−1 ⎜⎜− ⎟⎟ = −30° or 330°
⎝ 3⎠
62. rI =
rZ = 42 + (−4)2 = 32 = 4 2
⎛ −4 ⎞
θ Z = tan−1 ⎜ ⎟ = −45° or 315º
⎝4⎠
V = 4cis60o ;
rZ = 22 + 22 = 8 = 2 2
⎛2 ⎞
θ Z = tan−1 ⎜ ⎟ = 45°
⎝2 ⎠
I = 2cis330o ;
Z = 4 2cis315o
4
V
cis(60° − (−45°) )
I= =
Z 4 2
Z = 2 2cis45
o
( )
V = IZ = 2 2 2 cis(330° + 45°)
= 4 2 cis 375° = 4 2 cis15°
=
2
63. rI = 32 + 22 = 13
⎛ 2⎞
θ I = tan−1 ⎜− ⎟ ≈ −33.7° or 326.3°
⎝ 3⎠
17
rZ = 22 + 3.752 = 4.25 =
4
⎛
⎞
3.75
θ Z = tan−1 ⎜
⎟ = 61.9°
⎝ 2 ⎠
66. rV = 4 3 + (−4)2 = 64 = 8
⎛ −4 ⎞
θV = tan−1 ⎜⎜
⎟⎟ = −30° or 330º
⎝4 3 ⎠
rZ = 12 + (−1)2 = 2
⎛ −1 ⎞
θ Z = tan−1 ⎜ ⎟ = −45° or 315º
⎝1⎠
V = 8cis330o ;
I = 13cis326.3o ;
17
Z = cis61.9o
4
⎛ 17 ⎞
V = IZ = 13 ⎜ ⎟cis(326.3° + 61.9°)
⎝ 4⎠
=
2
cis105°
2
Z = 2cis315o
V
8
I= =
cis(330° − 315°)
Z
2
= 4 2 cis15°
17 13
17 13
cis 388.2° =
cis 28.2°
4
4
768
Chapter 7: Applications of Trigonometry
67. rV = 32 + (−4)2 = 25 = 5
⎛ −4 ⎞
θV = tan−1 ⎜ ⎟ = −53.1° or 306.9º
⎝3⎠
69. r1 = 12 + 22 = 5
⎛2 ⎞
θ1 = tan−1 ⎜ ⎟ ≈ 63.4° (Q1)
⎝1 ⎠
rZ = 42 + 7.52 = 8.5
⎛ 7.5 ⎞
θ Z = tan−1 ⎜ ⎟ = 61.9°
⎝ 4 ⎠
r2 = 32 + 22 = 13
⎛ −2 ⎞
θ 2 r = tan−1⎜ ⎟ ≈ −33.7° . In QIV,
⎝3⎠
θ 2 = 326.3°
V = 5cis306.9o ;
Z1Z2 = 5 13 cis (63.4° + 326.3°)
Z = 8.5cis61.9o
V
5
I= =
cis(306.9° − 61.9°)
Z 8.5
10
=
cis 245°
17
= 65 cis 389.7° = 65 cis 29.7°
Z1 + Z 2 = (1 + 2 j ) + (3 − 2 j ) = 4
Z=
68. rV = 2.82 + 9.62 = 100 = 10
⎛ 9.6 ⎞
θV = tan−1 ⎜ ⎟ = 73.7°
⎝ 2.8 ⎠
Z1Z2
65 cis 29.7°
=
Z1 + Z2
4
70. r1 = 32 + (−1)2 = 10
⎛ −1 ⎞
θ1r = tan−1⎜ ⎟ ≈ −18.4° In QIV,
⎝3⎠
θ = 341.6°
rZ = 1.42 + (−4.8)2 = 25 = 5
⎛ −4.8 ⎞
θ Z = tan−1 ⎜
⎟ = −73.7° or 286.3º
⎝ 1.4 ⎠
r2 = 22 + 12 = 5
⎛1 ⎞
θ 2 = tan−1 ⎜ ⎟ ≈ 26.6° (QI)
⎝2 ⎠
V = 10cis73.7o ;
Z = 5cis286.3o
V 10
I = = cis(73.7° − 286.3°)
Z 5
= 2 cis (−212.6°) = 2 cis147.4°
Z1Z2 = 5 10 cis (341.6° + 26.6°)
= 50 cis 368.2° = 5 2 cis 8.2°
Z1 + Z 2 = (3 − j ) + (2 + j ) = 5
Z=
769
5 2 cis 8.2°
Z1Z2
=
= 2 cis 8.2°
Z1 + Z2
5
7.5 Exercises
71.
74. If X C = X L and R is as small as possible
r1
[cos(α − β ) + i sin(α − β )]
r2
=
⎛
π⎞
75. 350 = 750 sin⎜2x − ⎟ − 25
4⎠
⎝
⎛
π⎞
375 = 750sin⎜2 x − ⎟
4⎠
⎝
⎛
π⎞
1
= sin⎜2x − ⎟
4⎠
2
⎝
For x ∈ [0,2π ) , 2x ∈ [0, 4 π ) so we need all
numbers in [0, 4π) for which sine is ½.
π π 5π 13π 17 π
,
,
2x − = ,
4 6 6
6
6
r1
(cos α cos β + sin α sin β )
r2
+i
r1
(sin α cos β + cos α sin β )
r2
cos α + isin α cos β − isin β
⋅
cos β + isin β cos β − isin β
=
=
=
2
cos α cos β − i cos α sin β + i cos β sin α − i sin α sin β
2
2
2
cos β − i sin β
cos α cos β + sin α sin β
sin α cos β + cos α sin β
+i
2
2
2
2
cos β + sin β
cos β + sin β
cos α cos β + sin α sin β
+i
sin α cos β + cos α sin β
2x =
1
1
r1
times this expression is equal
Note that
r2
to each side)
4
π
+
π 5π
,
+
π 13π
,
6 4 6
4 6
5π 13π 29 π 37 π
,
,
,
2x =
12 12 12 12
5π 13π 29π 37 π
,
,
,
x=
24 24 24 24
z1
, and is also equal to the expanded
z2
version of the right side, and we’re done.
to
+
π 17 π
4
,
6
+
π
4
76. Vertical asymptote at x = 0; no horizontal
asymptote; oblique asymptote y = x since
72. The voltage is 100 rms, and the two
frequencies are 50 Hz and 60 Hz. The true
1+ x3
voltage is 100 2 ≈ 141 V and the
equations are V (t) = 141sin(100πt) for 50 Hz
and V (t) = 141sin(120πt) for 60 Hz.
73. The slope for segment 1 is
π
(Add
x
2
=
1
x2
+x
77.
Δy 24
=
, so we
Δx 7
7
. We can accomplish this
24
with y = –7, x = 24, or y = 7, x = –24. This
gives us –24 + 7i and 24 – 7i. The
magnitude of each is the same as the
magnitude of z1 , so we need to divide each
by 5 to get magnitude one-fifth as great:
24 7
24 7
z2 =
− i , z3 = − + i
5 5
5 5
need slope −
78. The angle at the ship is 180 – (41 + 63) =
76º. Use the law of sines:
sin 76° sin 41°
4 sin 41°
=
;d =
≈ 2.70 mi
4
d
sin 76°
770
Chapter 7: Applications of Trigonometry
7.6 Exercises
1.
r 5 [cos(5θ ) + i sin(5θ )] ; DeMoivre’s
2.
even; odd; imaginary; 90º
3.
complex
4.
360° 2π
;
n
n
5.
z 5 = 2cis366° = 2cis6°
2
⎛ 1 ⎞2 ⎛ 3 ⎞
⎟ = 1; n = 5
11. r = ⎜ ⎟ + ⎜−
⎝ 2 ⎠ ⎜⎝ 2 ⎟⎠
⎛
⎞
− 3 / 2⎟
θ = tan−1⎜⎜
⎟ = −60° (QIV)
⎝ 1/ 2 ⎠
5
⎞
⎛1
⎜ − 3 i ⎟ = 15 [cos(−300°) + i sin(−300°)]
⎜2
2 ⎟⎠
⎝
⎡ 1
3⎤
3
1
i
= 1⎢− + i ⎥ = − +
2 ⎥⎦
2 2
⎢⎣ 2
z 6 = 2cis438° = 2cis78°
⎛
⎞2 ⎛ ⎞2
3
⎟ + 1 = 1; n = 6
12. r = ⎜−
⎜ 2 ⎟ ⎜⎝ 2 ⎟⎠
⎝
⎠
⎛ 1/ 2 ⎞
θ r = tan−1 ⎜⎜
⎟⎟ = −30° ; θ = 150° (QII)
⎝ 3 /2⎠
6(150º) = 900º; θ is coterminal with 180º.
z 7 = 2cis510° = 2cis150°
These are equal to z0 , z1, and z2 .
6.
Answers will vary.
(1− 3i) 4 = 28 + 96i; 4 28 + 96i = 3+ i
It appears the calculator found the principal
fourth root of 28 + 96i. 1 – 3i is another.
7.
6
⎞
⎛
⎜ − 3 + 1 i ⎟ = 16 [cos180° + i sin 180°]
⎜ 2
2 ⎟⎠
⎝
= 1[− 1 + 0] = −1
r = 32 + 32 = 18 = 3 2
⎛3⎞
θ = tan−1⎜ ⎟ = 45° (QI); n = 4
⎝3⎠
(3 + 3i )
4
⎛
⎞2 ⎛
⎞2
2
2
⎟ = 1; n = 6
13. r = ⎜ ⎟ + ⎜−
⎜ 2 ⎟ ⎜ 2 ⎟
⎝
⎠ ⎝
⎠
⎛
⎞
− 2 / 2⎟
θ = tan−1⎜⎜
⎟ = −45° (QIV)
⎝ 2 /2 ⎠
( ) [cos(180°) + i sin180°)]
= 3 2
4
= 324[− 1 + 0] = −324
8.
r = (−2)2 + 22 = 8 = 2 2 ; n = 6
⎛2⎞
θ r = tan−1 ⎜ ⎟ = −45° ; θ = 135° (QII)
⎝ −2 ⎠
6(135º) = 810º, which is coterminal with 90º
⎛
⎞6
2
2
6
⎜
⎟
⎜ 2 − 2 i⎟ = 1 [cos(−270°) + isin(−270°)]
⎝
⎠
= 1[0 + i ] = i
(−2 + 2i) 6 = (2 2 ) 6 [cos 90° + i sin 90°]
= 512[0 + 1i] = 512i
9.
⎛
⎞2 ⎛ ⎞2
2⎟ ⎜ 2⎟
⎜
+
= 1; n = 5
14. r = −
⎜ 2 ⎟ ⎜ 2 ⎟
⎝
⎠ ⎝ ⎠
r = (−1)2 + ( 3)2 = 4 = 2; n = 3
⎛ ⎞
3
θ r = tan−1 ⎜⎜ ⎟⎟ = −60° ; θ = 120° (QII)
−1
⎝ ⎠
θ r = tan −1 (− 1) = −45° ; θ = 135º (QII)
5(135º) = 675º; θ coterminal with 315º
5
⎞3
⎛⎜
−1 + 3i⎟ = 23 [cos 360° + isin360°]
⎠
⎝
= 8[1 + 0] = 8
⎛
⎞
⎜ − 2 + 2 i ⎟ = 15 [cos 315° + i sin 315°]
⎜ 2
2 ⎟⎠
⎝
⎡ 2
2⎤
2
2
= 1⎢
−i
i
−
⎥=
2
2
2
2
⎣⎢
⎦⎥
10. r = ( 3)2 + 12 = 4 = 2; n = 3
⎛ −1 ⎞
θ = tan−1⎜⎜ ⎟⎟ = −30° (QIV)
⎝ 3⎠
( 3 − i)
3
= 23 [cos(−90°) + i sin(−90°)]
= 8[0 − i ] = −8i
771
7.6 Exercises
20. r = 3, θ = −90°
z 4 = 34 [cos(−360°) + i sin(−360°)] = 81
⎞2
⎛
15. r = ⎜2 3⎟ + (−2)2 = 16 = 4; n = 3
⎠
⎝
z 3 = 33 [cos( −270°) + i sin(−270°)] = 27i
⎛ −2 ⎞
o
⎟ = −30° or 330 (QIV)
⎝2 3⎠
θ = tan −1 ⎜
(2
z 2 = (−3i)2 = 9i2 = −9
z 4 − z 3 + 7 z 2 − 9z − 18
= 81− 27i + 7(−9) − 9(−3i) −18
= 81− 27i − 63+ 27i − 18 = 0
Verified
)
3 − 2i = 43 [cos(−90°) + i sin(−90°)]
= 64[0 − i ] = −64i
3
⎞2
⎛
16. r = 22 + ⎜−2 3 ⎟ = 16 = 4; n = 3
⎠
⎝
21. r = (−3)2 + (−3)2 = 18 = 3 2
⎛ −3 ⎞
θ r = tan−1 ⎜ ⎟ = 45°; θ = 225° (QIII)
⎝ −3 ⎠
4(225º) = 900º, coterminal with 180º
3(225º) = 675º, coterminal with 315º
2(225º) = 450º, coterminal with 90º
⎛ −2 3 ⎞
o
⎟⎟ = −60° or 300 (QIV)
2
⎝
⎠
θ = tan −1 ⎜⎜
⎛⎜
⎞3
2 − 2 3i⎟ = 43 [cos(−180°) + i sin(−180°)]
⎝
⎠
= 64[− 1 + 0i ] = −64
( )
z 4 = 3 2 [cos180° + i sin 180°]
= 324[− 1 + 0i ] = −324
4
⎛ 1 ⎞2 ⎛ 1 ⎞2
1
2
=
; n=5
17. r = ⎜− ⎟ + ⎜ ⎟ =
2
2
⎝ 2⎠ ⎝2⎠
( )
z 3 = 3 2 [cos 315° + i sin 315°]
3
θ r = tan −1 (− 1) = −45° ; θ = 135° in QII
5(135º)=675º; θ coterminal with 315º
⎡ 2
2⎤
= 54 2 ⎢
−i
⎥ = 54 − 54i
2 ⎥⎦
⎣⎢ 2
( )
5
5
⎛ 1 1 ⎞ ⎛⎜ 2 ⎞⎟
[cos 315° + i sin 315°]
⎜− + i⎟ = ⎜
⎝ 2 2 ⎠ ⎝ 2 ⎟⎠
2⎡ 2
2⎤ 1 1
=
−i
⎢
⎥= − i
8 ⎢⎣ 2
2 ⎥⎦ 8 8
z 2 = 3 2 [cos 90° + i sin 90°]
= 18[0 + i ] = 18i
2
z 4 + 6 z 3 + 19 z 2 + 6 z + 18
= −324 + 6(54 − 54i) + 19(18i) + 6(−3 − 3i) + 18
= −324 + 324 − 324i + 342i −18 −18i + 18 = 0
Verified
⎛ 1 ⎞2 ⎛ 1 ⎞2
1
2
18. r = ⎜− ⎟ + ⎜ ⎟ =
=
; n=8
2
2
2
2
⎝ ⎠ ⎝ ⎠
22. r = 12 + (−1)2 = 2; θ = tan−1(−1) = −45°
z 4 = 2 [cos(−180°) + i sin(−180°)] = −4
4
θ = tan −1 (− 1) = −45° ; θ = 135° in QII
8(135) = 1080º; θ coterminal with 0º
z 3 = 2 [cos(−135°) + i sin(−135°)]
3
8
8
⎛ 1 1 ⎞ ⎛⎜ 2 ⎞⎟
[cos 0° + i sin 0°]
⎜− + i⎟ = ⎜
⎝ 2 2 ⎠ ⎝ 2 ⎟⎠
1
1
= [1 + 0i ] =
16
16
⎡
2
2⎤
= 2 2 ⎢−
−i
⎥ = −2 − 2i
2 ⎥⎦
⎢⎣ 2
z 2 = 2 [cos(−90°) + i sin(−90°)] = −2i
2
2z 4 + 3z 3 − 4z 2 + 2z +12
= 2(−4) + 3(−2 − 2i) − 4(−2i) + 2(1− i) + 12
= −8 − 6 − 6i + 8i + 2 − 2i +12 = 0
Verified
19. r = 2, θ = 90°
z 4 = 24 [cos 360° + i sin 360°] = 16
z 3 = 23 [cos 270° + i sin 270°] = 8[0 − i ] = −8i
z 2 = 4i2 = −4
z 4 + 3z 3 − 6z 2 +12z − 40
= 16 + 3(−8i) − 6(−4) + 12(2i) − 40
= 16 − 24i + 24 + 24i − 40 = 0
772
Chapter 7: Applications of Trigonometry
23. r =
⎛ −1 ⎞
2
3 + (−1)2 = 2; θ = tan−1⎜⎜ ⎟⎟ = −30°
⎝ 3⎠
()
25. r = 12 + (2)2 = 5; θ = tan−1 2
4
[ (
)
(
)]
)
(
)]
)
(
)]
z 4 = 5 cos 4 tan −1 (2) + i sin 4 tan −1 (2)
z 5 = 25 [cos(−150°) + i sin(−150°)]
= −7 − 24i
⎡
3
1⎤
= 32 ⎢ −
− i ⎥ = −16 3 − 16i
2 ⎥⎦
⎢⎣ 2
3
[ (
z = 5 cos 3 tan −1 (2) + i sin 3 tan −1 (2)
3
= −11 − 2i
z 4 = 24 [cos(−120°) + i sin(−120°)]
2
[ (
z 2 = 5 cos 2 tan −1 (2) + i sin 2 tan −1 (2)
⎡ 1
3⎤
= 16 ⎢ − − i
⎥ = −8 − 8 3i
2 ⎥⎦
⎢⎣ 2
= −3+ 4i
z − 4z 3 + 7 z 2 − 6z − 10
= −7 − 24i − 4(−11− 2i) + 7(−3+ 4i)
− 6(1 + 2i) − 10
= −7 − 24i + 44 + 8i − 21+ 28i − 6 − 12i − 10
=0
Verified
4
z 3 = 23 [cos( −90°) + i sin(−90°)] = −8i
z 2 = 2 2 [cos(−60°) + i sin(− 60°) )]
⎡1
3⎤
= 4⎢ − i
⎥ = 2 − 2 3i
2 ⎥⎦
⎣⎢ 2
⎛ 2⎞
26. r = 32 + (−2)2 = 13; θ = tan−1⎜− ⎟
⎝ 3⎠
z 5 + z 4 − 4z 3 − 4z 2 +16z +16
= −16 3 − 16i − 8 − 8 3i − 4(−8i)
( ( )) = −119 − 120i
13 cis(3 tan (− )) = −9 − 46i
13 cis(2 tan (− )) = 5 − 12i
4
z 4 = 13 cis 4 tan −1 −
−4(2 − 2 3i) +16( 3 − i) + 16
= −16 3 − 16i − 8 − 8 3i + 32i − 8
+ 8 3i + 16 3 −16i +16 = 0
Verified
z3 =
z2 =
2
3
3
−1
2
3
2
−1
2
3
z 4 − 2z 3 − 7 z 2 + 28z + 52
= −119 −120i − 2(−9 − 46i) − 7(5− 12i)
+28(3 − 2i) + 52
= −119 −120i +18 + 92i − 35+ 84i
+ 84 − 56i + 52
=0
Verified
2
⎛
⎞
24. r = ⎜2 3⎟ + 22 = 16 = 4
⎝
⎠
⎛ 2 ⎞
θ = tan−1⎜⎜
⎟⎟ = 30°
⎝2 3 ⎠
z 5 = 45 [cos150° + i sin 150°]
⎡
3
1⎤
= 1024 ⎢ −
+ i ⎥ = −512 3 + 512i
2 ⎥⎦
⎣⎢ 2
27. r = 1, θ = 0°, n = 5
0° 360°k
5
1 = 1,
+
= 72°k
5
5
z0 = cis 0° = 1
z1 = cis 72° ≈ 0.3090 + 0.9511i
z2 = cis144° ≈ −0.8090 + 0.5878i
z3 = cis 216° ≈ −0.8090 − 0.5878i
z 4 = cis 288° ≈ 0.3090 − 0.9511i
z 4 = 44 [cos120° + i sin 120°]
⎡ 1
3⎤
= 256 ⎢ − + i
⎥ = −128 + 128 3i
2 ⎥⎦
⎣⎢ 2
z 3 = 43 [cos 90° + i sin 90°] = 64i
z 2 = 42 [cos 60° + i sin 60°]
⎡1
3⎤
= 16 ⎢ + i
⎥ = 8 + 8 3i
2 ⎥⎦
⎢⎣ 2
z 5 + z 4 − 16z 3 −16z 2 + 256z + 256
= −512 3 + 512i − 128 +128 3i −16(64i)
−16(8 + 8 3i) + 256(2 3 + 2i) + 256
= −512 3 + 512i − 128 +128 3i −1024i
−128 −128 3i + 512 3 + 512i + 256 = 0
Verified
773
7.6 Exercises
28. r = 1, θ = 0°, n = 6
0° 360°k
6
1 = 1,
+
= 60°k
6
6
z0 = cis 0° = 1
32. r = 32, θ = 90°, n = 5
90° 360°k
5
32 = 2;
+
= 18° + 72°k
5
5
z0 = 2 cis18° ≈ 1.9021 + 0.6180i
z1 = 2cis 90° = 2i
z2 = 2cis162° ≈ −1.9021 + 0.6180i
z3 = 2 cis 234° ≈ −1.1756 − 1.6180i
z 4 = 2cis 306° ≈ 1.1756 −1.6180i
1
3
+
i
2 2
1
3
z2 = cis120° = − +
i
2 2
z3 = cis180° = −1
z1 = cis 60° =
33. x5 − 32 = 0; x5 = 32 ; For z = 32, r = 32,
0° 360°k
θ = 0° ; 5 32 = 2,
+
= 72°k
5
5
z0 = 2cis[0°] = 2
z1 = 2cis[72°] ≈ 0.6180 + 1.9021i
z2 = 2cis[144°] ≈ −1.6180 + 1.1756i
z3 = 2cis[216°] ≈ −1.6780 − 1.1756i
z4 = 2cis[288°] ≈ 0.6180 − 1.9021i
1
3
z 4 = cis 240° = − −
i
2 2
1
3
z5 = cis 300° = −
i
2 2
29. r = 243, θ = 0°, n = 5
0° 360°k
3
243 = 3 ;
+
= 72°k
5
5
z0 = 3cis 0° = 3
z1 = 3cis 72° ≈ 0.9271+ 2.8532i
z2 = 3cis144° ≈ −2.4271+1.7634i
z3 = 3cis 216° ≈ −2.4271 − 1.7634i
z 4 = 3cis 288° ≈ 0.9271− 2.8532i
34. x5 − 243 = 0; x5 = 243; For z = 243, r =
0° 360°k
243, θ = 0°; 5 243 = 3,
+
= 72°k
5
5
z0 = 3cis[0°] = 3
z1 = 3cis[72°] ≈ 0.9271 + 2.8535i
z2 = 3cis[144°] ≈ −2.4271 + 1.7634i
z3 = 3cis[216°] ≈ −2.4271 − 1.7634i
z4 = 3cis[288°] ≈ 0.9271 − 2.8532i
30. r = 8, θ = 0°, n = 3
0° 360°k
3
8 = 2;
+
= 120°k
3
3
z0 = 2cis 0° = 2
⎡ 1
3 ⎤
z1 = 2cis120° = 2⎢ − +
i ⎥ = −1 + 3i
2 ⎥⎦
⎢⎣ 2
35. x3 − 27i = 0; x3 = 27i; For z = 27i, r = 27,
θ = 90° ,
3
27 = 3 ,
90° 360°k
+
= 30° + 120°k
3
3
⎡ 3
1⎤ 3 3 3
z0 = 3cis30° = 3⎢
+i ⎥=
+ i
2 ⎦⎥
2
2
⎣⎢ 2
⎡ 1
3 ⎤
z2 = 2cis240° = 2 ⎢ − −
i ⎥ = −1 − 3i
2 ⎥⎦
⎢⎣ 2
31. r = 27, θ = 270°, n = 3
270° 360°k
3
27 = 3;
+
= 90° + 120°k
3
3
z0 = 3cis90° = 3[0 + i ] = 3i
⎛
⎞
3 1 ⎟
3 3 3
− i⎟ = −
− i
z1 = 3cis 210° = 3⎜⎜−
2
2
⎝ 2 2 ⎠
⎛
⎞
3 1 ⎟ 3 3 3
− i⎟ =
− i
z2 = 3cis 330° = 3⎜⎜
2
2
⎝ 2 2 ⎠
⎡
3
1⎤
3 3 3
z1 = 3cis150° = 3⎢ −
+i ⎥=−
+ i
2 ⎥⎦
2
2
⎢⎣ 2
z2 = 3cis270° = 3[0 − i ] = −3i
774
Chapter 7: Applications of Trigonometry
36. x3 + 64i = 0; x3 = −64i; For z = –64i,
40. r = 1, θ = 0°;
3
r = 64, θ = 270° ; 64 = 4 ,
270° 360°k
+
= 90° + 120°k
3
3
z0 = 4cis90° = 4[0 + i ] = 4i
0° 360°k
+
= 90°k
4
4
z 2 = 1 cis 180° = 1[−1 + 0i ] = −1
z 3 = 1 cis 270° = 1[0 − i ] = −i
(
)(
)
(
)
x 4 − 1 = x 2 − 1 x 2 + 1 = (x − 1)(x + 1) x 2 + 1
x −1 = 0 ⇒ x = 1
x + 1 = 0 ⇒ x = −1
x2 + 1 = 0 ⇒ x2 = −1 ⇒
The results are the same.
37. x5 − 2 − 2i = 0; x5 = 2 + 2i . For
x = ±i
⎛
⎞2
41. r = (−8)2 + ⎜8 3⎟ = 256 = 16 ; n = 4
⎝
⎠
⎛
⎞
8 3⎟
(QII)
θ r = tan−1 ⎜⎜
⎟ = −60°; θ = 120°
⎝ −8 ⎠
120° 360°k
4
+
= 30° + 90°k
16 = 2;
4
4
⎡ 3
1⎤
+i ⎥
z 0 = 2[cos 30° + i sin 30°] = 2 ⎢
2 ⎦⎥
⎣⎢ 2
z = 2 + 2i, r = 2, θ = 45°
45° 360°k
5
2;
+
= 9° + 72°k
5
5
z0 = 5 2 cis 9° ≈ 1.1346 +1.1797i
z1 = 5 2 cis 81° ≈ 0.1797 + 1.1346i
z2 = 5 2 cis153° ≈ −1.0235 + 0.5215i
z3 = 5 2 cis 225° ≈ −0.8123− 0.8123i
z 4 = 5 2 cis 297° ≈ 0.5215 − 1.0235i
= 3 +i
⎡ 1
3⎤
z1 = 2[cos 120° + i sin 120°] = 2⎢ − + i
⎥
2 ⎥⎦
⎣⎢ 2
5
38. x −1+ 3i = 0; x = 1− 3i . For
z = 1− 3i , r = 2, θ = −60°
−60° 360°k
2;
+
= −12° + 72°k
5
5
z0 = 5 2 cis (−12°) ≈ 1.1236 − 0.2388i
= −1 + 3i
5
⎡
3
1⎤
−i ⎥
z2 = 2[cos 210° + i sin 210°] = 2⎢ −
2 ⎥⎦
⎢⎣ 2
z1 = 5 2 cis 60° ≈ 0.5743 + .09948i
=− 3−i
z2 = 5 2 cis132° ≈ −0.7686 + 0.8536i
⎡1
3⎤
z3 = 2[cos 300° + i sin 300°] = 2⎢ − i
⎥
2 ⎦⎥
⎣⎢ 2
5
z3 = 2 cis204° ≈ −1.0494 − 0.4672i
z4 = 5 2 cis276° ≈ 0.1201 − 1.1424i
(
1 = 1,
z0 = 1 cis 0° = 1
z1 = 1 cis 90° = 1[0 + i ] = i
⎡
3
1⎤
z1 = 4cis210° = 4⎢ −
− i ⎥ = −2 3 − 2i
2 ⎥⎦
⎣⎢ 2
⎡ 3
1⎤
z2 = 4cis330° = 4⎢
− i ⎥ = 2 3 − 2i
2 ⎦⎥
⎣⎢ 2
5
4
= 1 − 3i
)
39. x 3 − 1 = (x − 1) x 2 + x + 1 = 0
x −1 = 0 ⇒ x = 1
x2 + x +1 = 0 ⇒
−1 ± 1 − 4(1)(1)
1
3
=− ±
i
x=
2
2 2
These are the same results as in Ex. 3.
775
7.6 Exercises
45. z 3 − 6 z + 4 = 0 , p = –6, q = 4
⎛
⎞2
42. r = 162 + ⎜−16 3 ⎟ = 32; n = 5
⎝
⎠
⎛
⎞
−16 3 ⎟
θ = tan−1⎜⎜
⎟ = −60°
⎝ 16 ⎠
−60° 360°k
5
+
= −12° + 72°k
32 = 2;
5
5
z0 = 2 cis (−12°) ≈ 1.9563 − 0.4158i
D=
−
3
⎡1
3⎤
z1 = 2cis(60°) = 2 ⎢ + i
⎥ = 1 + 3i
2 ⎥⎦
⎢⎣ 2
z2 = 2 cis (132°) ≈ −1.3383 + 1.4863i
z3 = 2 cis (204°) ≈ −1.8271 − 0.8135i
z4 = 2cis(276°) ≈ 0.2091 − 1.9890i
2
4(−6)3 + 27(4)2 −864 + 432
=
= −4
108
108
q
+ D = −2 + 2i; r = 8, θ = 135° (QII)
2
1
135°
= 86 ;
r
z0
3
1
= 86
+
360°k
= 45° + 120°k
3
cis 45° ; z1 = 81/ 6 cis 165°
1
z 2 = 8 6 cis 285°
−
q
− D = −2 − 2i; r = 8, θ = 225°
2
3
r = 86 ;
1
2
43. r = (−7) + (−7) = 98 = 7 2; n = 4
⎛ −7 ⎞
θ r = tan−1 ⎜ ⎟ = 45°; θ = 225° (QIII)
⎝ −7 ⎠
(QIV)
225° 360°k
+
= 75° + 120°k
3
3
1
1
z 0 = 8 6 cis 75° ; z1 = 8 6 cis 195°
z2 =
4
7 2 ≈ 1.7738 ;
225° 360°k
+
= 56.25° + 90°k
4
4
z0 = 1.7738 cis 56.25° ≈ 0.9855 + 1.4749i
z1 = 1.7738 cis146.25° ≈ −1.4749 + 0.9855i
z2 = 1.7738 cis 236.25° ≈ −0.9855 − 1.4749i
z3 = 1.7738cis326.25° ≈ 1.4749 − 0.9855i
1
86
cis 315°
46. z 3 − 12 z − 8 = 0 , p = –12, q = –8
D=
3
2
4(−12) + 27(−8)
108
=
−6912 + 1728
108
= −48
q
+ D = 4 + 4 3i; r = 8, θ = 60° (QI)
2
60° 360°k
3
r = 2;
+
= 20° + 120°k
3
3
z0 = 2 cis 20°; z1 = 2 cis 140°
z2 = 2 cis 260°
q
− − D = 4 − 4 3i; r = 8, θ = −60° (QIV)
2
60° 360°k
3
r = 2; −
+
= −20° + 120°k
3
3
z0 = 2 cis (−20)°; z1 = 2 cis 100°
z2 = 2 cis 220°
−
44. r = 92 + 92 = 162 = 9 2 ; n = 3
⎛9 ⎞
θ = tan−1⎜ ⎟ = 45°
⎝9 ⎠
3
9 2 ≈ 2.3348
45° 360°k
+
= 15° + 120°k
3
3
z0 = 2.3348 cis15° ≈ 2.2553 + 0.6043i
z1 = 2.3348 cis135° ≈ −1.6510 + 1.6510i
z2 = 2.3348 cis 255° ≈ −0.6043 − 2.2553i
47. We’ll need four times each of the angles
given to use DeMoivre’s Theorem:
4(15) = 60º; 4(105) = 420º, coterminal with
60º; 4(195) = 780º, coterminal with 60º;
4(285) = 1140º, coterminal with 60º. So
raising all four given z’s to the fourth power
results in
⎡1
3⎤
24 cis 60° = 16 ⎢ + i
⎥ = 8 + 8 3i .
2 ⎦⎥
⎣⎢ 2
Verified
776
Chapter 7: Applications of Trigonometry
48. We’ll need five times each of the angles
given to use DeMoivre’s Theorem:
5(6) = 30º; 5(78) = 390º, coterminal with
30º; 5(150) = 750º, coterminal with 30º;
5(222) = 1110º, coterminal with 30º; 5(294)
= 1470, coterminal with 30º. So raising all
five given z’s to the fifth power results in
⎡ 3
1⎤
25 cis 30° = 32 ⎢
+ i ⎥ = 16 3 + 16i .
2 ⎥⎦
⎢⎣ 2
Verified
49. a.
b.
( (
)
(
))
= −117 +44j;
)
(
)
( (
Z = 53 cos 2 tan −1 4 + j sin 2 tan −1 4
2
3
3
Z3
3Z
2
=
=
c.
50. a.
1000 j
⋅
150 − 150 3 j
150 + 150 3 j 150 − 150 3 j
150, 000 j − 150, 000 3 j 2
22,500 − 67,500 j 2
150, 000 j + 150, 000 3
90, 000
5 3 5
+ j
3
3
Z 5 3 +5j 5 3 5
=
=
+ j ; Verified
3
3
3
3
=
51, 52. Answers will vary.
⎛2⎞
53. For 1 + 2i, r = 5, θ = tan−1⎜ ⎟ . The
⎝1⎠
related right triangle is:
= –7 + 24j
3Z 2 = −21 + 72 j
b.
=
=
−1
Z 3 = 53 cos 3 tan −1 43 + j sin 3 tan −1 43
3Z
2
=
c.
⎛4⎞
Z = 3 + 4 j; r = 5, θ = tan ⎜ ⎟
⎝3⎠
Z3
θ
5
−117 + 44 j −21 − 72 j
⋅
−21 + 72 j −21 − 72 j
1
2
2457 + 8424 j − 924 j − 3168 j 2
sin θ =
2
, cos θ =
1
5
5
sin(4θ ) = sin(2(2θ ) ) = 2 sin(2θ )cos(2θ )
2
441 − 5184 j
4
5625 + 7500i
= 1+ j
=
3
5625
4
Z 3+ 4 j
=
= 1 + j ; Verified
3
3
3
(
)
= 2(2 sin θ cos θ ) cos 2 θ − sin 2 θ
⎛ ⎛ 2 ⎞⎛ 1 ⎞⎞⎛⎛ 1 ⎞2 ⎛ 2 ⎞2 ⎞
= 2⎜⎜2⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎟⎟⎜⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ ⎟
⎜
⎟
⎝ ⎝ 5 ⎠⎝ 5 ⎠⎠⎝⎝ 5 ⎠ ⎝ 5 ⎠ ⎠
24
8 ⎛ 3⎞
= ⎜− ⎟ = −
25
5 ⎝ 5⎠
Z = 5 3 + 5 j; r = 10, θ = 30°
Z 3 = 103 [cos 90° + j sin 90°] = 1000 j ;
Z 2 = 10 2 [cos 60° + j sin 60°]
2
⎛ 24 ⎞
cos 2 (4θ ) = 1 − sin 2 (4θ ) = 1 − ⎜ − ⎟
⎝ 25 ⎠
7
576 49
= 1−
=
; cos(4θ ) = −
(Note
625 625
25
that a quick approximation on the calculator
shows us that θ = tan−1(2) ≈ 63° , so
4θ ≈252º, and is in QIII).
⎡1
3⎤
= 100 ⎢ + j
⎥ = 50 + 50 3 j
2 ⎥⎦
⎢⎣ 2
3Z 2 = 150 +150 3 j
4
24 ⎤
⎡ 7
z 4 = 5 [cos(4θ ) + i sin(4θ )] = 25⎢ − − i ⎥
25
25 ⎦
⎣
= −7 − 24 i
777
7.6 Exercises
≈ (0.3660 − 1.3660i ) + (0.3660 + 1.3660i )
= 0.7320 ;
Note: Using sum and difference identities,
all three solutions can actually be found in
exact form. The latter two are −1 − 3 and
⎛ ⎞
5
54. For 2 + 5i , r = 3, θ = tan−1⎜⎜ ⎟⎟ . The
2
⎝ ⎠
related right triangle is:
θ
3
2
−1 + 3 .
5
56. Add the roots from question 46 whose
angles add to 360º:
2 cis 20° + 2 cis − 20°
= (1.8794 + 0.6840i ) + (1.8794 − 0.6840i )
5
2
, cos θ =
sin θ =
3
3
sin(4θ ) = sin(2(2θ ) ) = 2 sin(2θ )cos(2θ )
(
= 2(2 sin θ cos θ ) cos 2 θ − sin 2 θ
⎛ ⎛ ⎞⎛ ⎞⎞⎛⎛ ⎞2 ⎛ ⎞2 ⎞
5 ⎟
5 2 ⎜ 2
= 2⎜2⎜⎜ ⎟⎟⎜ ⎟⎟⎜⎜ ⎟ − ⎜⎜ ⎟⎟ ⎟
⎜ 3 ⎝ 3 ⎠⎟⎜⎝ 3 ⎠
3
⎝ ⎠ ⎟⎠
⎝ ⎝ ⎠ ⎠⎝
⎛
⎞
8 5
8 5 ⎟⎛ 1 ⎞
= ⎜⎜
⎟⎜⎝− 9 ⎟⎠ = − 81
9
⎝
⎠
)
= 3.7588 ;
2 cis 140° + 2 cis 220°
= (−1.5321 + 1.2856i ) + (−1.5321 − 1.2856i )
= −3.0642 ;
2 cis 100° + 2 cis 260°
= (−0.3473 + 1.9696i ) + (−0.3473 − 1.9696i )
= −0.6946
2
⎛ 8 5⎞
⎟
cos 2 (4θ ) = 1 − sin 2 (4θ ) = 1 − ⎜ −
⎜ 81 ⎟
⎝
⎠
79
320 6241
= 1−
=
; cos(4θ ) = −
(Note
6561 6561
81
that a quick approximation on the calculator
⎛
⎞
shows us that θ = tan−1⎜ 5 / 2⎟ ≈ 48° , so
⎝
⎠
4θ ≈ 192º, and is in QIII).
z 4 = 34 [cos(4θ ) + i sin(4θ )]
57.
58. f (− 1) = 2(−1) 2 − 3(−1) = 2 + 3 = 5
⎛ 1 ⎞ ⎛ 1 ⎞2 ⎛ 1 ⎞ 2
7
f ⎜ ⎟ = 2⎜ ⎟ − 3⎜ ⎟ = − 1 = −
9
⎝3⎠ ⎝3⎠
⎝ 3⎠ 9
⎡ 79 8 5 ⎤
= 81⎢ −
−i
⎥ = −79 − 8 5i
81 ⎥⎦
⎢⎣ 81
f (a ) = 2a 2 − 3a
f (a + h ) = 2(a + h )2 − 3(a + h )
55. Look at the solutions to 45, and note that
1
( )
1
(
1
1
cis 45° + 8 6
1
86
1
cis 165° + 8 6
2
2
= 2a + 4ah + 2h − 3a − 3h
59. Goes through (–2, 4) and (3, 0):
0− 4
4
m=
=−
3− (−2)
5
4
4
12
y − 0 = − (x − 3); y = − x +
5
5
5
cis 315°
⎡ 2
⎡ 2
2⎤
2⎤
= 2⎢
+i
−i
⎥ + 2⎢
⎥
2 ⎥⎦
2 ⎦⎥
⎣⎢ 2
⎣⎢ 2
2 2 2 2
= + i+ − i = 2;
2 2 2 2
60. A = 90 − 52 = 38°
b
sin52° =
; b = 213sin52° ≈ 167.85 m
213
a
cos 52° =
; a = 213cos 52° ≈ 131.14 m
213
cis 195°
≈ (−1.3660 + 0.3660i ) + (−1.3660 − 03.660i )
= −2.7320
1
)
= 2 a 2 + 2ah + h 2 − 3a − 3h
8 4 = 23 4 = 2 4 = 2
Add the roots from 45 whose angles add to
360º:
1
86
tan 2 x
sec 2 x − 1 ( sec x + 1)( sec x − 1)
=
=
sec x + 1 sec x + 1
sec x + 1
1
1
cos x
= sec x − 1 =
−1 =
−
cos x
cos x cos x
1 − cos x
=
cos x
1
8 6 cis 285° + 8 6 cis 75°
778
Chapter 7: Applications of Trigonometry
Summary and Concept Review
1.
2.
3.
4.
5.
A = 180 − (21 + 123) = 36°
sin123° sin21°
=
; bsin123° = 293sin21°
293
b
293sin21°
b=
≈ 125.20 cm
sin123°
sin123° sin36°
=
; asin123° = 293sin36°
293
a
293sin36°
a=
≈ 205.35 cm
sin123°
sin35° sin B
=
; 67 sin B = 105sin35°
67
105
⎛ 105sin35° ⎞
105sin35°
sin B =
; B = sin−1⎜
⎟
67
67
⎠
⎝
B ≈ 64.0° or 180 – 64.0 = 116.0º
For B1 = 64.0º:
C1 = 180 − (35+ 64.0) = 81.0°
sin35° sin81.0°
=
; c1 sin35° = 67 sin81.0°
67
c1
67 sin81.0°
≈ 115.37 cm
sin35°
For B2 = 116.0º:
C2 = 180 − (35 + 116.0) = 29.0°
sin35° sin29°
=
; c2 sin35° = 67 sin29°
67
c2
67 sin29°
≈ 56.63 cm
c2 =
sin35°
c1 =
B = 180 − (142 + 28) = 10°
sin10° sin142°
=
; c sin10° = 52sin142°
c
52
52 sin142°
c=
≈ 184.36 yd
sin10°
sin10° sin28°
=
; asin10° = 52sin28°
52
a
52sin28°
a=
≈ 140.59 yd
sin10°
6.
The third angle is 180 – (110 + 25) = 45º.
sin 45° sin25°
=
; hsin 45° = 70 sin25°
70
h
70sin25°
h=
≈ 41.84 ft
sin 45°
sin θ sin50°
14sin50°
=
; sin θ =
14
31
31
⎛
⎞
14 sin50°
θ = sin−1⎜
⎟ ≈ 20.2° or
⎝ 31 ⎠
180 – 20.2 = 159.8º.
sin 40° sin θ
=
; 35sin θ = 60 sin 40°
35
60
60 sin 40°
≈ 1.1 Not possible, so the
sin θ =
35
bystander will not get hit.
We need the smallest angle α for which the
60sin α
expression
is less than one.
35
60sin α
35
= 1 ⇒ sin α =
⇒ α ≈ 35.7°
35
60
The smallest safe angle is about 36º.
7.
81 = 369 − 360 cos B; − 360cos B = −288
−288
; B = cos−1 0.8 ≈ 36.9°
cos B =
−360
8.
x2 = 167 2 + 3252 − 2(167)(325) cos 98°
x2 ≈ 148,621.24; x ≈ 385.5 m
779
Summary and Concept Review
9.
Let A, B and C be the angles from largest to
smallest.
18202 = 12502 + 7202 − 2(720)(1250) cos C
3,312,400 = 2,080,900 – 1,800,000cosC
1, 231, 500 = −1, 800, 000 cos C
1, 231, 500
; C ≈ 133.2°
cos C =
−1, 800,000
sin B sin133.2°
=
1250
1820
1820 sin B = 1250 sin133.2°
1250 sin133.2°
; B ≈ 30.1°
sin B =
1820
A = 180 − (133.2 + 30.1) = 16.7°
15. u = 7 2 + 122 = 193
7
193
i+
12
193
j
16. Quadrant II – the x-component will be
negative.
17. Karl’s velocity in the direction across the
stream is 3 mi/hr, so he’ll make it ½ mile in
1
1/6 hours. The current will carry him
6
mile downstream.
18. The vertical components must be equal.
928sin18°
928 sin18° = 850 sin θ ; sin θ =
850
⎛
⎞
928 sin18°
θ = sin−1⎜
⎟ ≈ 19.7 º
⎝ 850 ⎠
⎛1
⎞
10. A = 4 ⎜ ( 230 ) 218.7 2 − 1152 ⎟
⎝2
⎠
2
A ≈ 85,570.7 m
11.
19. Resultant: −20 + 45, 70 + 53 = 25,123
Additional:
−25, −123
20. v = 192 + (−13)2 = 530
u ⋅ v = −12(19) + (−16)(−13) = −20
−20
≈ −0.87
comp v u =
530
v = 92 + 52 = 106 ≈ 10.30
21. 2(−18) + 9d = 0; 9d = 36; d = 4
⎛5⎞
θ = tan−1⎜ ⎟ ≈ 29.1°
22. p ⋅ q = −5(4) + (−2)(−7) = −20 + 14 = −6
⎝9 ⎠
p = (−5)2 + 22 = 29
q = 42 + (−7)2 = 65
12. u = −8 i + 3j; u = (−8)2 + 32 = 73 ≈ 8.54
⎛ 3 ⎞
⎟ ≈ −20.6° or 159.4°
⎝ −8 ⎠
cos θ =
θ = tan −1 ⎜
−6
29 65
⎛ −6 ⎞
θ = cos−1⎜⎜
⎟⎟ ≈ 97.9°
⎝ 29 65 ⎠
13. Horiz. component = 18cos 52° ≈ 11.08
Vert. component = 18sin52° ≈ 14.18
14. 2 u + v = −6, −10 + 2, 8 = −4, −2
(−4) 2 + (−2) 2 = 20 = 2 5 ≈ 4.47
⎛ −2 ⎞
θ r = tan−1 ⎜ ⎟ ≈ 26.6° . θ = 206.6° (QIII)
⎝ −4 ⎠
Mag. =
780
Chapter 7: Applications of Trigonometry
23. W = F ⋅ V = 50(85) + 15(6) = 4340 ft-lb
31. I =
24. F = 650cos 50° ≈ 417.81 lb
=
25. F = 75cos 25°
W = ( 75cos 25° )(120 ) = 8156.77 ft-lb
=
26. a.
b.
x = (280 cos 50°)(1.5) ≈ 269.97 ft
1 − 3 j2
8 3 +8j
= 2 3 +2j
4
Z = 102 + 32 = 109 ≈ 10.44
⎛3⎞
⎟ ≈ 16.7°
⎝ 10 ⎠
Z = 10.44cis16.7°
−16t 2 + 214.49t − 150 = 0
Quadratic formula: t ≈ 0.74, 12.67 sec
It reaches height 150 ft after 0.74 sec
θ = tan−1⎜
27. r = (−1)2 + (− 3)2 = 4 = 2
⎛
⎞
− 3⎟
θ r = tan−1 ⎜⎜
⎟ = 60°; θ = 240° (QIII)
⎝ −1 ⎠
z = 2[cos 240 ° + i sin 240°]
π
4 3 + 12 j − 4 j − 4 3 j 2
32. Z = R + j ( X L − X C ) = 10 + j (8 − 5) = 10 + 3 j
y = (280sin50°)(1.5) − 16(1.5)2
≈285.74 ft
y = 280 sin50°t − 16t 2 = 150
2
33. r = (−1)2 + 3 = 4 = 2
⎛ ⎞
3
θ r = tan−1 ⎜⎜ ⎟⎟ = −60°; θ = 120° (QII)
⎝ −1 ⎠
(− 1 + 3i ) = 2 [cos(5(120°) + i sin(5(120°)]
5
5
= 32[cos 240° + i sin 240°] (240º coterminal
with 600º)
⎡ 1
3⎤
32 ⎢ − − i
⎥ = −16 − 16 3i
2 ⎥⎦
⎢⎣ 2
2
=3
4
2
π
2
=3
y = 3 2 sin = 3 2
4
2
z = 3 + 3i
28. x = 3 2 cos
V 4 3 − 4 j 1+ 3 j
=
⋅
Z
1− 3 j 1+ 3 j
=3 2
29.
34. r = 12 + (−1)2 = 2
θ = tan −1 (− 1) = −45° (QIV)
z 4 = 2 [cos(−180°) + i sin(−180°)]
= 4[−1 + 0i ] = −4
4
z 3 = 2 [cos(−135°) + i sin(−135°)]
3
⎡
2
2⎤
= 2 2 ⎢−
−i
⎥ = −2 − 2i
2 ⎦⎥
⎣⎢ 2
z 2 = 2 [cos(−90°) + i sin(−90°)]
= 2[0 − i ] = −2i
2
⎛π π ⎞
⎛ 5π ⎞
30. z1 z2 = 8(2) cis ⎜ + ⎟ = 16 cis ⎜ ⎟
⎝4 6⎠
⎝ 12 ⎠
⎛
⎞
⎛
⎞
π π
π
z1 8
= cis ⎜ − ⎟ = 4 cis ⎜ ⎟
z2 2 ⎝ 4 6 ⎠
12
⎝ ⎠
z 4 + z 3 − 2z 2 + 2z + 4
= −4 + (−2 − 2i) − 2(−2i) + 2(1 − i) + 4
= −4 − 2 − 2i + 4i + 2 − 2i + 4 = 0
781
Summary and Concept Review
38. u = z 2 ; u2 + 6u + 25 = 0
35. r = 125; θ = 90°; 3 125 = 5
90° 360°k
+
= 30° + 120°k
3
3
⎡ 3
1⎤
+i ⎥
z0 = 5[cos 30° + i sin 30°] = 5⎢
2 ⎥⎦
⎢⎣ 2
−6 ± 36 − 4(1)(25) −6 ± 8i
=
= −3 ± 4i
2
2
z 2 = −3± 4i so we need the square roots of
u=
−3 + 4i, − 3 − 4i . For each, r = 32 + 42 = 5 .
For –3 + 4i:
⎛ 4⎞
θ r = tan−1 ⎜− ⎟ ≈ −53.1301°; θ = 126.8699°
⎝ 3⎠
126.8699° 360°
+
= 63.43495° + 180°
(QII)
2
2
z0 = 5 cis 63.43495° = 1 + 2i
5 3 5
+ i or 4.3301 + 2.5i
2
2
⎡
3 1⎤
_i ⎥
z1 = 5[cos 150° + i sin 250°] = 5⎢−
2 ⎥⎦
⎢⎣ 2
=
5 3 5
+ i or − 4.3301 + 2.5i
2
2
z2 = 5[cos 270° + i sin 270°] = 5[0 − i ] = −5i
=−
z1 = 5 cis183.43495° = −1 − 2i
For –3 –4i:
⎛4⎞
θ r = tan−1 ⎜ ⎟ ≈ 53.1301°; θ = 233.1301°
⎝3⎠
(QIII)
233.1301° 360°
+
= 116.56505° + 180°
2
2
z0 = 5 cis116.5605° = −1 + 2i
36. x3 − 216 = 0; x3 = 216 ; r = 216, θ = 0°
0° 360°k
3
216 = 6;
+
= 120°k
3
3
z0 = 6[cos 0° + i sin 0°] = 6
⎡ 1
3⎤
z1 = 6[cos 120° + i sin 120°] = 6 ⎢ − + i
⎥
2 ⎦⎥
⎣⎢ 2
z1 = 5 cis 296.5605° = 1 − 2i
= −3+ 3 3i
⎡ 1
3⎤
z2 = 6[cos 240° + i sin 240°] = 6 ⎢ − − i
⎥
2 ⎥⎦
⎢⎣ 2
⎛
⎞2 ⎛ ⎞2
5 3 5
5
3
⎟ + ⎜5⎟ = 5
39.
+ i; r = ⎜
⎜
⎟
2
2
⎝ 2 ⎠ ⎝2⎠
= −3− 3 3i
⎛ 5
⎜
−1 ⎜ 2
θ = tan ⎜
5 3
⎜⎜
⎝ 2
37. r = 22 + 22 = 8; θ = tan−1 (1) = 45°
The other roots will have the same r and will
be spaced 90º apart.
⎡
2
2⎤
+i
2 2 cis135° = 2 2 ⎢ −
⎥ = −2 + 2i
2 ⎦⎥
⎣⎢ 2
The remaining two must be conjugates of
the two we have: 2 – 2i and –2 – 2i
⎞
⎟
⎟ = 30°
⎟
⎟⎟
⎠
⎛
⎞3
⎜ 5 3 + 5 i⎟ = 53 cis 90° = 125i
⎜ 2
2 ⎟⎠
⎝
⎛
⎞2 ⎛ ⎞2
5 3 5
5
3
⎟ + ⎜5⎟ = 5
−
+ i; r = ⎜
⎜
⎟
2
2
⎝ 2 ⎠ ⎝2 ⎠
⎛ 5 ⎞
⎜
⎟
θ r = tan −1 ⎜⎜ 2 ⎟⎟ = −30°; θ = 150° (QII)
−5 3 ⎟
⎜⎜
⎟
⎝ 2 ⎠
⎛
⎞3
⎜− 5 3 + 5 i⎟ = 53 cis450° = 125i
⎜ 2
2 ⎟⎠
⎝
(− 5i )3 = −125i 3 = 125i
782
Chapter 7: Applications of Trigonometry
Mixed Review
1.
A = 180 − (27 + 112 ) = 41°
sin112° sin 27°
=
; b sin112° = 19sin27°
19
b
19 sin27°
b=
≈ 9.30 in.
sin112°
sin112° sin 41°
=
; a sin112° = 19 sin 41°
19
a
19 sin 41°
a=
≈ 13.44 in.
sin112°
Angles
Sides
A = 41°
a ≈ 13.44in.
B = 27°
b ≈ 9.30in.
C = 112°
x = 21cos 40° ≈ 16.09
y = 21sin 40 ≈ 13.50
4.
12, −6 + −2, 8 = 10, 2
102 + 22 = 104 ≈ 10.20
⎛2⎞
θ = tan−1⎜ ⎟ ≈ 11.3°
⎝ 10 ⎠
Mag. =
c = 19in.
192 sin 27o sin 41o
2sin112o
Area ≈ 58 in.2
A=
2.
3.
5.
Missing angle = 180 – (35 + 122) = 23º
sin23° sin 35°
=
; h sin23° = 120 sin35°
120
h
120 sin35°
h=
≈ 176.15 ft
sin23°
6.
Complement of 28º = 62º.
F = 900cos 62° ≈ 422.52 lb
7.
The plane’s velocity vector makes an angle
of 60º with the pos. x-axis, so it’s
components are x = 750cos 60° = 375 ,
y = 750 sin 60° ≈ 649.52 . The wind vector
is 0,50 . The resultant is
a2 = 312 + 522 − 2(31)(52) cos 37° = 1090.20
a ≈ 33.02 cm
sin 37° sin B
=
; 31sin 37° = 33.02sin B
33.02
31
⎛ 31sin 37° ⎞
31sin37°
; B = sin−1⎜
sin B =
⎟
33.02
⎝ 33.02 ⎠
v = 375, 699.52 .
3752 + 699.522 ≈ 793.70 mph;
⎛ 699.52 ⎞
θ = tan−1⎜
⎟ ≈ 61.8° . An angle of
⎝ 375 ⎠
61.8º with the x-axis is a heading of 28.2º
Mag. =
B ≈ 34.4°
C = 180 − ( 37 + 34.4 ) = 108.6°
Angles
A = 37°
B = 34.4°
C = 108.6°
Sides
a ≈ 33.02 cm
8.
b = 31cm
c = 52 cm
1
( 31)( 52 ) sin 37o
2
Area = ≈ 485.1 in 2
A=
v = (−8)2 + 52 = 89 ≈ 9.43 ;
⎛5⎞
⎟ ≈ −32.0°; θ = 148.0° (QII)
⎝ −8 ⎠
θ r = tan−1 ⎜
783
Mixed Review
9.
sin31° sin B
=
; 36sin B = 24 sin31°
36
24
⎛ 24 sin31° ⎞
24sin 31°
sin B =
; B = sin−1⎜
⎟
36
⎠
⎝ 36
B ≈ 20.1° or 180 − 20.1 = 159.9°
Second one not possible since its sum with
31º is over 180º. So B ≈ 20.1° , and
C = 180 − (31+ 20.1) = 128.9°
sin128.9° sin 31°
=
36
c
c sin 31° = 36 sin128.9°
36 sin128.9°
c=
≈ 54.4 m
sin31°
Angles
Sides
A = 31°
a = 36 m
B ≈ 20.1°
b = 24 m
C ≈ 128.9°
12.
sin35° sin θ
=
; 12sin θ = 20sin35°
12
20
20 sin35°
= 1.15; No.
sin θ =
10
To find the smallest possible angle, we need
20sin α
1
= 1; sin α = , and the
α so that
10
2
smallest angle is 30º.
13. a.
c ≈ 54.4 m
10. 1.22 = 0.92 + 0.7 2 − 2(0.9)(0.7) cos C
1.44 = 1.3 − 1.26cos C ; − 1.26 cos C = 0.14
⎛ 0.14 ⎞
0.14
cos C =
; C = cos−1⎜
⎟ ≈ 96.4° ;
−1.26;
⎝ −1.26; ⎠
r = 42 + (−4)2 = 32 = 4 2
⎛ −4 ⎞
θ r = tan−1 ⎜ ⎟ = −45°; θ = 315° (QIV)
⎝4⎠
z = 4 2 (cos 315° + i sin 315°)
b.
sin96.4° sin A
=
; 1.2 sin A = 0.9 sin96.4°
1.2
0.9
⎛ 0.9 sin 96.4° ⎞
0.9 sin 96.4°
sin A =
; A = sin−1⎜
⎟
1.2
1.2
⎝
⎠
A ≈ 48.2°; B = 180 − (96.4 + 48.2) = 35.4°
11. p ⋅ q = −5(4) + (2)(7) = −6
p = (−5)2 + 22 = 29
q = 42 + 7 2 = 65
cos θ =
−6
⎛ 1⎞
x = 6cos120° = 6⎜− ⎟ = −3
⎝ 2⎠
⎛ ⎞
3
y = 6 sin120° = 6⎜⎜ ⎟⎟ = 3 3
⎝ 2 ⎠
29 65
⎛ −6 ⎞
θ = cos−1⎜⎜
⎟⎟ ≈ 97.9°
⎝ 29 65 ⎠
z = −3 + 3 3i
784
Chapter 7: Applications of Trigonometry
14. a.
z = 4 − 5i ;
17. u ⋅ v = −12(19) + (−16)(−13) = −20
2
z − 8 z + 41 = 0
(4 − 5i )
2
v = 192 + (−13)2 = 530
− 8(4 − 5i ) + 41 = 0
comp v u =
16 − 40i + 25i − 32 + 40i + 41 = 0
0=0;
(4 + 5i )2 − 8(4 + 5i ) + 41 = 0
b.
16 + 40i + 25i 2 − 32 − 40i + 41 = 0
0=0;
z 2 − 6iz + 7 = 0
a = 1, b = −6i, c = 7
z=
− (− 6i ) ±
−20
≈ −0.87
530
−20
38 26
proj v u =
19,−13 = − i +
j
530
53
53
2
18. r = (2 3)2 + (−2)2 = 4
⎛ −2 ⎞
θ = tan−1⎜⎜
⎟⎟ = −30°
⎝2 3 ⎠
(− 6i )2 − 4(1)(7 )
2(1)
⎛⎜
⎞6
2 3 − 2 i⎟ = 46 cis (6 ⋅30°) = 4096 cis180°
⎝
⎠
= 4096 (−1 + 0i ) = −4096
6i ± 36i 2 − 28
2
6i ± − 64
z=
2
6i ± 8i
z=
2
14i
−2i
z=
= 7i or z =
= −i
2
2
z=
19. r = (−2)2 + (2 3)2 = 4
⎛
⎞
2 3⎟
θ r = tan−1 ⎜⎜
⎟ = −60°; θ = 120° (QII)
⎝ −2 ⎠
120° 360° k
4
4 = 2;
+
= 30° + 90°k
4
4
⎛
⎞
2
3
1
6
+ i ⎟⎟ =
+
i
z0 = 2 cis 30° = 2 ⎜⎜
2⎠ 2
2
⎝ 2
⎛
⎞
1
3⎟
z1 = 2 cis120° = 2 ⎜⎜− + i
2 ⎟⎠
⎝ 2
15. Set vertical components equal:
418 sin10° = 320sin θ ; sin θ =
⎛ 418sin10° ⎞
⎟ ≈ 13.1°
⎝ 320
⎠
418sin10°
320
θ = sin−1⎜
16. a.
b.
=−
2
6
+
i
2
2
⎛
⎞
3
1
− i ⎟⎟
z2 = 2 cis 210° = 2 ⎜⎜−
2⎠
⎝ 2
z1 z2 = 8(4) cis (45° + 15°) = 32cis 60°
⎛
⎞
1
3⎟
= 32⎜⎜ + i
= 16 + 16 3i
2 ⎟⎠
⎝2
z 1 8(cos 45 + i sin 45)
=
z 2 4(cos 15 + i sin 15)
=−
6
2
−
i
2
2
⎛
⎞
1
3⎟
z3 = 2 cis 300° = 2 ⎜⎜ − i
2 ⎟⎠
⎝2
= 2[cos(45 − 15) + i sin (45 − 15)]
= 2[cos(30 ) + i sin (30 )]
=
⎡ 3 ⎛ 1 ⎞⎤
= 2⎢
+ i⎜ ⎟⎥ = 3 + i
⎝ 2 ⎠⎥⎦
⎢⎣ 2
2
6
−
i
2
2
20. Z = 12 + j (15.2 − 9.4) = 12 + 5.8 j
r = 122 + 5.82 ≈ 13.33
⎛ 5.8 ⎞
θ = tan−1⎜ ⎟ ≈ 25.8° ;
⎝ 12 ⎠
Z = 13.33(cos 25.8° + i sin 25.8°)
VRLC = 6.5(13.33) ≈ 86.6 V
785
Practice Test
For A2 = 121.2º:
Practice Test
1.
C2 = 180 − (20 + 121.2) = 38.8°
The angle at the fire is 73º.
sin39° sin73°
=
; d sin73° = 10sin39°
10
d
10sin 39°
d=
≈ 6.58 mi
sin73°
2.
sin38.8° sin 20°
=
; c2 sin 20° = 6 sin 38.8°
6
c2
6 sin38.8°
≈ 10.99 in
sin 20°
Angles
Sides (in.)
a = 15
A1 ≈ 58.8°
B = 20°
b=6
C1 ≈ 101.2° c1 ≈ 17.21
c2 =
40°
x
Angles
A2 ≈ 121.2°
B = 20°
C 2 ≈ 38.8°
h
68°
72°
100
sin68° sin 40°
=
; x sin 40° = 100 sin68°
100
x
100 sin68°
x=
≈ 144.24
sin 40°
h
sin72° =
144.24
h = 144.24 sin72° ≈ 137.18 ft
3.
4.
b.
Let b = 6, a = 15, B = 20º
sin20° sin A
=
; 6 sin A = 15sin20°
6
15
⎛ 15sin 20° ⎞
15sin 20°
sin A =
; A = sin−1⎜
⎟
6
6
⎠
⎝
A ≈ 58.8° or 121.2°
For A1 = 58.8º:
C1 = 180 − (20 + 58.8) = 101.2°
sin101.2° sin 20°
=
; c1 sin 20° = 6 sin101.2°
6
c1
c1 =
a.
6 sin101.2°
≈ 17.2 in
sin 20°
786
Sides
a = 15
b=6
c 2 ≈ 11.0
sin32° sin θ
=
; 3sin θ = 6 sin32°
3
6
6 sin 32°
≈ 1.06 ; No
sin θ =
3
sin32° sin θ
=
; 4 sin θ = 6 sin32°
4
6
⎛ 6 sin 32° ⎞
6 sin 32°
sin θ =
; θ = sin−1⎜
⎟
4
⎝ 4 ⎠
θ = 52.6° or 127.4°
First contact occurs when θ = 127.4°
180 − (127.4 + 32) = 20.6°
sin20.6° sin32°
=
4
d
d sin32° = 4 sin20.6°
4 sin 20.6°
d=
≈ 2.66 mi
sin32°
Chapter 7: Applications of Trigonometry
5.
a.
b.
c.
6.
a.
b.
7.
sin53° sin θ
=
; 25sin θ = 35sin53°
25
35
35sin 53°
= 1.11 ; No
sin θ =
25
sin53° sin θ
=
; 28sin θ = 35sin53°
28
35
35sin 53°
≈ 1 ; Only 1 throw
sin θ =
25
With a range of 35 yd, the target is in
range from the bottom left corner of the
triangle, and we’ll get an isosceles
triangle with angles 53º, 53º and 74º.
sin74° sin53°
=
; d sin53° = 35sin74°
35
d
35sin74°
d=
≈ 42.13 yd
sin53°
1 sec
42.13 yd ⋅
≈ 8.43 sec
5 yd
8.
50°
100°
v plane = 90 cos 50°, 90 sin 50°
≈ 57.85, 68.94
v wind = 20 cos ( −100° ) , 20sin ( −100° )
≈ −3.47, −19.70
Resultant = 54.38, 49.24
54.382 + 49.242 ≈ 73.36 mph;
⎛ 49.24 ⎞
θ = tan−1⎜
⎟ ≈ 42.2° ;
⎝ 54.38 ⎠
Heading ≈ 90º – 42.2º = 47.8º
Speed =
l2 = 3.52 + 52 − 2(3.5)(5) cos 24° ≈ 5.28
l ≈ 2.30 mi
h
sin24° =
; h = 3.5sin24° mi
3.5
5280 ft
= 7516.5 ft
3.5sin24° mi ⋅
mi
9.
Set vert. components equal:
250 sin30° = 210sin θ
250 sin 30°
sin θ =
210
⎛
⎞
−1 250 sin 30°
θ = sin ⎜
⎟ ≈ 36.5°
⎝ 210
⎠
10. Think of each segment as a vector with
initial point at the origin, and find
components.
v1 = 100 cos ( −22° ) ,100sin ( −22° )
10252 = 10202 + 977 2 − 2(1020)(977) cos P
1,050,625 = 1,994,929 − 1, 993, 080cos P
−944, 304 = −1, 993, 080 cos P
944,304
cos P =
1, 993, 080
⎛ 944, 304 ⎞
P = cos−1⎜
⎟ ≈ 61.7° ;
⎝ 1, 993, 080 ⎠
≈ 92.72, −37.46
v 2 = 75cos ( −138° ) , 75sin ( −138° )
≈ −55.74, −50.18
sin61.7° sin B
=
1025
1020
1025sin B = 1020 sin 61.7°
1020 sin61.7°
sin B =
1025
⎛
⎞
−1 1020sin61.7°
B = sin ⎜
⎟ ≈ 61.2°;
1025
⎝
⎠
M = 180 − (61.7 + 61.2) = 57.1° ;
1020 + 1025 + 977
p=
= 1511
2
v 3 = 50 cos ( −58° ) ,50sin ( −58° )
≈ 26.50, −42.40
The components of the resultant (add all 3
components) will be the coordinates of the
last joint.
Resultant = 63.48, −130.04
The joint is 63.48 cm to the right and 130.04
cm below the pivot point at the ceiling.
A = 1511(1511 −1020)(1511 −1025)(1511 − 977)
Area about 438, 795 mi 2
787
Practice Test
11. F1 = 150 cos 42°,150 sin 42°
15. z1 = (−6)2 + 62 = 72 = 6 2
= 111.47,100.37
⎞2
⎛
z2 = 42 + ⎜−4 3 ⎟ = 8
⎠
⎝
⎛6 ⎞
θ1r = tan−1⎜ ⎟ = −45°; θ1 = 135° (QII)
⎝ −6 ⎠
⎛
⎞
−4 3 ⎟
θ 2 = tan−1 ⎜⎜
⎟ = −60° (QIV)
⎝ 4 ⎠
F2 = 110 cos113°,110 sin113°
= −42.98,101.26
Resultant = 68.49, 201.63
F = −68.49, −201.63
F = 68.492 + 201.632 ≈ 212.94 N
z = z1 z2 = 6 2(8) cis (135° + (−60°))
⎛ 201.63 ⎞
⎟ ≈ 71.2° ; θ = 251.2°
⎝ 68.49 ⎠
θ r = tan−1 ⎜
12. a.
= 48 2 cis 75° ≈ 17.57 + 65.57i
z = 17.57 2 + 65.57 2 ≈ 67.88
u ⋅ v = (−9)(−2) + 5(6) = 48
z1 z2 = 48 2 ≈ 67.88
u = (−9)2 + 52 = 106
2
⎛ 65.57 ⎞
⎟ ≈ 75° = θ1 + θ 2
⎝ 17.57 ⎠
θ = tan−1⎜
2
v = (−2) + 6 = 40
cos θ =
b.
c.
48
106 40
⎛
⎞
48
θ = cos−1⎜⎜
⎟⎟ ≈ 42.5°
⎝ 106 40 ⎠
u ⋅ v 48
=
−2, 6 = −2.4, 7.2
proj v u =
2
40
v
⎞4
⎛⎜
3 − i⎟ = 2 4 cis (4 ⋅ (−30°)) = 16cis − 120°
⎠
⎝
⎛
⎞
1
3⎟
= 16⎜⎜− − i
= −8 − 8 3i
2 ⎟⎠
⎝ 2
u = u 1 + u 2 where
u1 = −2.4,7.2 , u2 = u− u1 = −6.6,−2.2
13. y = 110 sin50°(2) − 16(2)2 ≈ 104.53 ft;
−16 t 2 + 110 sin50°t = 104.53
−16t 2 +110sin50°t − 104.53 = 0
Quadratic formula: t ≈ 1.2, 3.27 . It will be
at that height again after 3.27 sec.
14.
2
3 + 12 = 2
⎛ 1 ⎞
θ = tan−1⎜⎜− ⎟⎟ = −30° (QIV)
⎝ 3⎠
16. r =
2
17. r = 22 + 2 3 = 16 = 4
⎛
⎞
2 3⎟
(QI)
θ = tan−1⎜⎜
⎟ = 60°
⎝ 2 ⎠
⎛
⎞
1
3⎟
z 5 = 45 cis 300° = 1024⎜⎜ − i
2 ⎟⎠
⎝2
⎛π ⎞
z1 6 5 ⎛ π π ⎞
=
cis ⎜ − ⎟ = 2 cis ⎜ ⎟
z2 3 5 ⎝ 8 12 ⎠
⎝ 24 ⎠
= 512 − 512 3i
z 3 = 43 cis180° = 64(−1 + 0 i) = −64
⎛
⎞
1
3⎟
z 2 = 42 cis120° = 16⎜⎜− + i
2 ⎟⎠
⎝ 2
= −8 + 8 3i
z + 3z 3 + 64 z 2 + 192
= 512 − 512 3i + 3(−64) + 64(−8 + 8 3i)
= 192
= 512 − 512 3i − 192 − 512 + 512 3i − 192
=0
Verified
5
788
Chapter 7: Applications of Trigonometry
18. x3 − 125i = 0; x3 = 125i
90° 360°k
r = 125, θ = 90°;
+
3
3
3
= 30° + 120°k ; r = 5
⎛
⎞
3
1
5 3 5
+ i ⎟⎟ =
+ i;
z0 = 5cis 30° = 5⎜⎜
2⎠
2
2
⎝ 2
⎛
⎞
3
1
5 3 5
+ i ⎟⎟ = −
+ i;
z1 = 5cis150° = 5⎜⎜−
2
2
2
2
⎝
⎠
z2 = 5cis270° = 5(0 − i ) = −5i
Calculator Exploration
and Discovery
19. u = z 2 ; u2 − 6u + 58 = 0
1.
a.
b.
c.
d.
Approx. 50.5 ft
Approx. 50.5 ft
Approx. 224.54 ft
Approx. 3.55 sec
2.
a.
b.
c.
d.
931 ft
4900 ft
33,948.20 ft
35 sec
3.
a. 111.87 ft
b. 132.04 ft
c. 443.16 ft
d. 5.75 sec
It will clear the fence.
4.
a. 16.24 ft
b. 21.72 ft
c. 124.07 ft
d. 2.33 sec
No, it will not clear the crossbar. At a
distance of 40 yd, or 120 ft, the height is
only about 2.75 feet.
2
6 ± 6 − 4(1)(58) 6 ± 14i
=
= 3 ± 7i
2
2
So z 2 = 3 ± 7i, and we need the square roots
of 3 + 7i and 3 – 7i. For both,
u=
r = 32 + 7 2 = 58 , and r = 581/ 4
7
For 3 + 7i: θ = tan−1 ≈ 66.8°
3
66.8° 360°k
+
= 33.4° + 180°k
2
2
z0 = 581/ 4 cis 33.4° ≈ 2.3039 + 1.5192 i ;
z1 = 581/ 4 cis 213.4° ≈ −2.3039 − 1.5192 i ;
For 3 – 7i, θ = −66.8° (QIV)
z0 = 581/ 4 cis − 33.4° ≈ 2.3039 − 1.5192 i ;
Strengthening Core Skills
1.
z1 = 581/ 4 cis146.6° ≈ −2.3039 + 1.5192 i
20. a 2 = 2390 2 + 15902 − 2 ( 2390 )(1590 ) cos 23o
≈ −0.9063 u i + 0.4226 u j
a = 23902 + 15902 − 2 ( 2390 )(1590 ) cos 23o
A=
v = v cos 20°i + v sin 20°j
a ≈ 1115.43
2010 + 1275 + 1115.43 + 1590
p1 =
= 2995.215
2
2995.215 ( 2995.215 − 2010 )( 2995.215 − 2390.43)( 2995.215 − 1590 )
A1 ≈ 1583621.549
2390 + 1590 + 1115.43
= 2547.715
2
2547.715 ( 2547.715 − 2390 )( 2547.715 − 1590 )( 2547.715 − 1115.43)
A2 ≈ 574515.2611
Area = 1583621.549 + 574515.2611 =
2158136.81
About 2.2 million miles
p2 =
A=
Let u and v be the force vectors for the
ropes, and w for the weight.
u = − u cos 25°i + u sin 25°j
789
≈ 0.9397 v i + 0.3420 v j
w = −500 j
The sum of all first components and all
second components must be zero.
⎧−0.9063 u + 0.9397 v = 0
⎪
⎨
⎪⎩0.4226 u + 0.3429 v − 500 = 0
Solving this system, we get
u = 664.46 lb, v = 640.86 lb
Cumulative Review
2.
5.
Let u and v be the force vectors for the
ropes, and w for the weight.
u = − u cos 45°i + u sin 45°j
3
≈ −0.7071 u i + 0.7071 u j
α
v = v cos 45°i + v sin 45° j
4
The third side is length 5 (Pythagorean
triple), and since cosine is positive, sine is
negative.
3
5
4
sin α = − ; csc α = − ; cos α =
5
3
5
5
3
4
sec α = ; tan α = − ; cot α = −
4
4
3
≈ 0.7071 v i + 0.7071 v j
w = −150 j
The sum of all first components and all
second components must be zero.
⎧−0.7071 u + 0.7071 v = 0
⎪
⎨
⎪⎩0.7071 u + 0.7071 v − 150 = 0
Solving this system, we get
u = 106.07 lb, v = 106.07 lb
3.
6.
The system remains the same as in the
example, except the 180 in the second
equation is replaced with 200. The solution
is now x = 537.49 lb, y = 547.13 lb. The
rope will hold.
Cumulative Review
1.
This is a 30-60-90 triangle: β = 60° , c is
twice a, or 40 m, and b is
3 times a, or
7.
x=
8.
(
3(x
20 3 m.
2.
α = 90 − 63 = 27°
b
; b = 82sin 63° ≈ 73.06°
82
a
sin27° =
; a = 82sin 27° ≈ 37.23
82
γ = 90°
sin63° =
3.
(
)
A = π 2 R 2 − r 2 = π 2 R 2 − π 2r 2
1
π 2 R 2 = A + π 2r 2 ; R 2 = 2 A + (πr )2
π
R=
4.
1
π
(
−8 ± 64 − 4(5)(2) −8 ± 24
=
10
10
8 2 6
4
6
=− ±
=− ±
10 10
5 5
)
3 x 2 − 24 x = −427
2
)
− 24 x + 144 = −427 + 3(144)
5
3(x − 12 )2 = 5; (x − 12)2 =
3
5
15
=±
x − 12 = ±
3
3
)
x = 12 ±
A + (π r ) 2
9.
(a + bi ) + (a − bi ) = 2a
a.
which has no
imaginary part, and is real.
b.
(a + bi )(a − bi ) = a 2 + abi − abi − bi 2
15
3
Using a right triangle, or the Pythagorean
Identity, we find that if
cos 53° ≈ 0.6, sin53° ≈ 0.8 and if
cos 72° ≈ 0.3, sin72° ≈ 0.95
19° = 72° − 53° , so cos19° = cos(72° − 53°)
= cos 72° cos 53° + sin72° sin 53°
≈ (0.3)(0.60) + (0.95)(0.8) = 0.94
Similarly, cos125° = cos ( 72° + 53° )
= a2 + b2 which is also real.
= cos 72° cos 53° − sin 72° sin 53° = −0.58
790
Chapter 7: Applications of Trigonometry
3 + 2 sin(2 x) = 2 3; 2 sin(2 x) = 3
10.
13. a.
3
; 2 x = 60° + 360°k
2
or 2 x = 120° + 360°k . Then
x = 30° + 180°k or x = 60° + 180°k
m=
sin(2 x ) =
11. a.
b.
1
(1475 )( 2008) sin 25.9o
2
A = 646859.7684 ft 2
A
≈ 14.85 acres
43560
(1485 acres )( $4500 / acre ) = $66,825
A=
Pythagorean triple 5, 12, 13.
13 + 13 + 7 2
7 2
= 13 +
2
2
⎛
7 2 ⎞ ⎛ 7 2 ⎞⎛ 7 2 ⎞ ⎛
7 2⎞
A = ⎜⎜13 +
⎟⎜
⎟⎜
⎟⎟ ⎜⎜ 13 −
⎟
2 ⎟⎠ ⎜⎝ 2 ⎟⎜
2 ⎟⎠
⎝
⎠⎝ 2 ⎠ ⎝
3
or
2
⎡3 ⎞
x ∈ ⎢ ,∞⎟
⎣2 ⎠
b.
x + 3 > 0; x > −3 or x ∈ (−3, ∞ )
c.
x2 − 5 ≠ 0; x2 ≠ 5; x ≠ ± 5
⎛
⎞ ⎛
⎞ ⎛
⎞
x ∈ ⎜−∞, − 5 ⎟ ∪ ⎜− 5, 5⎟ ∪ ⎜ 5, ∞⎟
⎝
⎠ ⎝
⎠ ⎝
⎠
d.
v( x) = x 2 − x − 6
c.
x=
d.
d=
e.
A = Pe rt
−b ± b2 − 4ac
2a
(x2 − x1 )2 + ( y2 − y1 )2
15. a2 = 312 + 522 − 2(31)(52) cos 37°
a2 ≈ 1090.1991; a ≈ 33 cm
sin37° sin B
=
; 33sin B = 31sin37°
33
31
⎛ 31sin37° ⎞
31sin37°
; B = sin−1⎜
sin B =
⎟
33
⎝ 33 ⎠
A ≈ 59.5 mi 2
2 x − 3 ≥ 0; 2 x ≥ 3; x ≥
⎛x + x y + y ⎞
⎜ 2 1 , 2 1⎟
2 ⎠
⎝ 2
sin112° sin27°
=
; a sin112° = 19 sin27°
19
a
19 sin27°
a=
≈ 9.3 in.
sin112°
sin112° sin 41°
=
; b sin112° = 19sin 41°
19
b
19 sin 41°
b=
≈ 13.4 in.
sin112°
⎛
⎞⎛
⎞⎛
⎞
7 2 ⎞⎛
7 2
7 2
7 2
A = ⎜⎜13 +
− 13 ⎟⎟ ⎜⎜ 13 +
− 13 ⎟⎟ ⎜⎜ 13 +
− 7 2 ⎟⎟
⎟⎜ 13 +
2 ⎟⎜
2
2
2
⎝
⎠⎝
⎠⎝
⎠⎝
⎠
12. a.
b.
14. B = 180 − (112 + 27) = 41°
d = 72 + 72 = 7 2 ;
From origin to (5, 12), 13 units
From origin to (12,5), 13 units
From (5, 12) to (12, 5), 7 2 units;
p=
y2 − y1
x2 − x1
B ≈ 34.4° ; C = 180 − (37 + 34.4) = 108.6°
16. a.
x2 − x − 6 ≥ 0
( x − 3)( x + 2 ) ≥ 0
7500 =
1 + 25e−0.2t
7500 + 187,500e−0.2t = 12,000
187,500e−0.2t = 4500; e−0.2t = 0.024
ln(0.024)
−0.2 t = ln(0.024); t =
≈ 18.6
−0.2
It will take about 18.6 months.
Boundary Numbers 3 and −2 .
b.
12, 000
P(2) =
12,000
1 + 25e−0.2(24)
≈ 9952
⎛3⎞
⎜ ⎟ ( 9952 ) = 7464 fish
⎝4⎠
x ∈ ( −∞, −2] ∪ [3, ∞ )
791
Cumulative Review
17. Complement of 28º is 62º.
F = 900 cos 62° ≈ 422.5 lb
⎞2
⎛
21. r = 12 + ⎜− 3⎟ = 4 = 2
⎠
⎝
⎛
⎞
− 3⎟
θ = tan−1⎜⎜
⎟ = −60° (QIV)
⎝ 1 ⎠
18. The plane’s path makes an angle of 60º with
the positive x-axis.
v p = 750 cos 60°, 750 sin 60° ≈ 375, 649.5
v w = 0,50
Resultant:
(1 − 3 ) = 2 cis(8(−60°)) = 256cis(−480°)
8
375,699.5
8
⎛
⎞
1
3⎟
= 256 cis 240° = 256⎜⎜− − i
2 ⎟⎠
⎝ 2
Speed = 3752 + 699.52 ≈ 794 mph
⎛ 699.5 ⎞
θ = tan−1⎜
⎟ ≈ 61.8° ; Bearing is 28.2º.
⎝ 375 ⎠
= −128 − 128 i 3
22. ln(x + 2) + ln (x − 3) = ln(4 x )
ln(( x + 2)( x − 3)) = ln(4 x )
19.
x2 − x − 6 = 4 x; x2 − 5x − 6 = 0
(x − 6)(x + 1) = 0; x = 6, − 1
x = –1 makes the equation undefined, so the
only solution is x = 6.
23. R = 0.08 / 12 ≈ 0.00667
.08
10,000 012
AR
P=
=
nt
12t
(1 + R ) − 1 (1.00667) − 1
66.67
200 =
(1.00667)12t − 1
( )
( )
f ( x) < 0 for x ∈ (−∞,−1) and x ∈ 2,3 .
20.
(
)
200(1.00667 ) − 200 = 66.67
200 (1.00667 ) = 266.67
200 1.0066712t − 1 = 66.67
12t
12t
1.0066712t = 1.33; 12t ln1.00667 = ln(1.33)
ln(1.33)
t=
≈ 3.6 yr
12 ln1.00667
h
; h = 5 tan 38° mi
5
5280 ft
≈ 20, 626 ft
5 tan 38° mi ⋅
1 mi
24. tan38° =
g( x) ↑ on (0,1) ∪ (1, ∞ )
g( x) ↓ on (−∞,−1) ∪ (−1,0)
25. A = 2; P = 2π, so B = 1; the graph is shifted
π
4
792
units left, so C =
π
4
.