# User manual | C H A P T E R 8 Additional Topics in Trigonometry

```C H A P T E R 8
Section 8.1
Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . 731
Section 8.2
Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . 738
Section 8.3
Vectors in the Plane
Section 8.4
Vectors and Dot Products
Section 8.5
Trigonometric Form of a Complex Number . . . . . . . . 771
. . . . . . . . . . . . . . . . . . . . 749
. . . . . . . . . . . . . . . . . 762
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806
Practice Test
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809
C H A P T E R 8
Section 8.1
■
Law of Sines
If ABC is any oblique triangle with sides a, b, and c, then
a
b
c
.
sin A sin B sin C
■
You should be able to use the Law of Sines to solve an oblique triangle for the remaining three parts, given:
(a) Two angles and any side (AAS or ASA)
(b) Two sides and an angle opposite one of them (SSA)
1. If A is acute and h b sin A:
(a) a < h, no triangle is possible.
(b) a h or a > b, one triangle is possible.
(c) h < a < b, two triangles are possible.
2. If A is obtuse and h b sin A:
(a) a ≤ b, no triangle is possible.
(b) a > b, one triangle is possible.
■
The area of any triangle equals one-half the product of the lengths of two sides and the sine of their included angle.
1
1
1
A 2ab sin C 2ac sin B 2bc sin A
Vocabulary Check
1. oblique
2.
1.
b
sin B
C
b
2.
105°
C 180 A B 105
a
20 sin 45
b
sin B 202 28.28
sin A
sin 30
a
20 sin 105
38.64
sin C sin A
sin 30
a
40°
B
Given: A 30, B 45, a 20
c
b
45°
c
1
ac sin B
2
C
a = 20
30°
A
3.
A
c = 20
B
Given: B 40, C 105, c 20
A 180 B C 35
a
c
20 sin 35
sin A 11.88
sin C
sin 105
b
c
20 sin 40
sin B 13.31
sin C
sin 105
731
732
Chapter 8
3.
4.
C
25°
A
C
b
a = 3.5
b
10°
35°
c
a
135°
B
A
c = 45
Given: A 25, B 35, a 3.5
Given: B 10, C 135, c 45
C 180 A B 120
A 180 B C 35
b
a
3.5
sin B sin 35 4.75
sin A
sin 25
a
c
45 sin 35
sin A 36.50
sin C
sin 135
c
3.5
a
sin C sin 120 7.17
sin A
sin 25
b
c
45 sin 10
sin B 11.05
sin C
sin 135
5. Given: A 36, a 8, b 5
sin B b sin A 5 sin 36
0.36737 ⇒ B 21.55
a
8
C 180 A B 180 36 21.55 122.45
c
a
8
sin C sin 122.45 11.49
sin A
sin 36
6. Given: A 60, a 9, c 10
sin C c sin A 10 sin 60
0.9623 ⇒ C 74.21 or C 105.79
a
9
Case 1
Case 2
C 74.21
C 105.79
B 180 A C 45.79
B 180 A C 14.21
b
a
9 sin 45.79
sin B 7.45
sin A
sin 60
7. Given: A 102.4, C 16.7, a 21.6
B 180 A C 60.9
b
a
9 sin 14.21
sin B 2.55
sin A
sin 60
8. Given: A 24.3, C 54.6, c 2.68
B 180 A C 101.1
b
a
21.6
sin B sin 60.9 19.32
sin A
sin 102.4
a
c
2.68 sin 24.3
sin A 1.35
sin C
sin 54.6
c
a
21.6
sin C sin 16.7 6.36
sin A
sin 102.4
b
c
2.68 sin 101.1
sin B 3.23
sin C
sin 54.6
9. Given: A 83 20, C 54.6, c 18.1
B 180 A C 180 83 20 54 36 42 4
10. Given: A 5 40, B 8 15, b 4.8
C 180 A B 166 5
a
c
18.1
sin A sin 83 20 22.05
sin C
sin 54.6
a
b
4.8 sin 5 40
sin A 3.30
sin B
sin 8 15
b
c
18.1
sin B sin 42 4 14.88
sin C
sin 54.6
c
b
4.8 sin 166 5
sin C 8.05
sin B
sin 8 15
B
Section 8.1
Law of Sines
733
11. Given: B 15 30 , a 4.5, b 6.8
sin A a sin B 4.5 sin 15 30
0.17685 ⇒ A 10 11
b
6.8
C 180 A B 180 10 11 15 30 154 19
c
6.8
b
sin C sin 154 19 11.03
sin B
sin 15 30
12. Given: B 2 45, b 6.2, c 5.8
sin C c sin B 5.8 sin 2 45
0.04488 ⇒ C 2.57 or 2 34
b
6.2
A 180 B C 174.68, or 174 41
b
6.2 sin 174.68
sin A 11.99
sin B
sin 2 45
a
13. Given: C 145, b 4, c 14
14. Given: A 100, a 125, c 10
sin C b sin C 4 sin 145
0.16388 ⇒ B 9.43
sin B c
14
B 180 A C 75.48
A 180 B C 180 9.43 145 25.57
a
c sin A 10 sin 100
0.07878 ⇒ C 4.52
a
125
b
c
14
sin A sin 25.57 10.53
sin C
sin 145
a
125 sin 75.48
sin B 122.87
sin A
sin 100
15. Given: A 110 15 , a 48, b 16
sin B b sin A 16 sin 110 15
0.31273 ⇒ B 18 13
a
48
C 180 A B 180 110 15 18 13 51 32
c
a
48
sin C sin 51 32 40.06
sin A
sin 110 15
16. Given: C 85 20, a 35, c 50
sin A a sin C 35 sin 85 20
0.6977 ⇒ A 44.24, or 44 14
c
50
B 180 A C 50.43, or 50 26
b
C sin B 50 sin 50.43
38.67
sin C
sin 85 20
17. Given: A 55, B 42, c 3
4
C 180 A B 83
18. Given: B 28, C 104, a 3
A 180 B C 48
a
c
0.75
sin A sin 55 0.62
sin C
sin 83
b
5
a sin B 38 sin 28
2.29
sin A
sin 48
b
0.75
c
sin B sin 42 0.51
sin C
sin 83
c
5
a sin C 38 sin 104
4.73
sin A
sin 48
5
8
734
Chapter 8
19. Given: A 110, a 125, b 100
20. Given: a 125, b 200, A 110
b sin A 100 sin 110
0.75175 ⇒ B 48.74
sin B a
125
No triangle is formed because A is obtuse
and a < b.
C 180 A B 21.26
c
a sin C 125 sin 21.26
48.23
sin A
sin 110
22. Given: A 76, a 34, b 21
21. Given: a 18, b 20, A 76
h 20 sin 76 19.41
sin B Since a < h, no triangle is formed.
b sin A 21 sin 76
0.5993 ⇒ B 36.82
a
34
C 180 A B 67.18
c
a sin C 34 sin 67.18
32.30
sin A
sin 76
23. Given: A 58, a 11.4, c 12.8
sin B b sin A 12.8 sin 58
0.9522 ⇒ B 72.21 or B 107.79
a
11.4
Case 1
Case 2
B 72.21
B 107.79
C 180 A B 49.79
C 180 A B 14.21
c
a
11.4 sin 49.79
sin C 10.27
sin A
sin 58
24. Given: a 4.5, b 12.8, A 58
c
a
11.4 sin 14.21
sin C 3.30
sin A
sin 58
25. Given: A 36, a 5
h 12.8 sin 58 10.86
(a) One solution if b ≤ 5 or b Since a < h, no triangle is formed.
(b) Two solutions if 5 < b <
(c) No solution if b >
(a) One solution if b ≤ 10 or b (c) No solutions if b >
10
.
sin 60
10
.
sin 60
10
.
sin 60
(b) Two solutions if 315.6 < b <
(b) Two solutions if 10.8 < b <
10.8
sin 10
10.8
sin 10
10.8
sin 10
1
1
29. Area ab sin C 46 sin 120 10.4
2
2
(a) One solution if b ≤ 315.6 or b 315.6
sin 88
5
sin 36
(a) One solution if b ≤ 10.8 or b (c) No solution if b >
28. Given: A 88, a 315.6
(c) No solutions if b >
5
sin 36
27. Given: A 10, a 10.8
26. Given: A 60, a 10
(b) Two solutions if 10 < b <
5
sin 36
315.6
sin 88
315.6
sin 88
Section 8.1
Law of Sines
30. Area 12ac sin B 126220 sin 130 474.9
31. Area 12bc sin A 125785 sin 43 45 1675.2
32. A 5 15, b 4.5, c 22
33. Area 2ac sin B 210564sin7230 3204.5
1
735
1
Area 12bc sin A
24.522 sin 5.25 4.5
1
34. C 84 30, a 16, b 20
35. C 180 23 94 63
1
Area ab sin C
2
h
21620 sin 84.5 159.3
1
36. (a)
37.
20°
h
32°
16
38.
16 sin 32
9.0 meters
sin 70
39. Given: c 100
N
W
A 74 28 46,
E
S
B 180 41 74 65,
Elgin
C
a = 720 km
b = 500 km
44°
46°
A
Naples
B
Canton
Given: A 46, a 720, b 500
sin B b sin A 500 sin 46
0.50 ⇒ B 30
a
720
The bearing from C to B is 240.
40.
3000 ft
r
40°
s
r
(b) r 42 25.9
16.1
h
16
sin 32 sin 70
(c) h sin42 sin 48
10
17
sin42 0.43714
70°
12°
(b)
35
sin 23 15.3 meters
sin 63
3000 sin12180 40
4385.71 feet
sin 40
(c) s 40
1804385.71 3061.80 feet
C 180 46 65 69
A
100
46°
65°
69°
C
100
c
sin A sin 46 77 meters
a
sin C
sin 69
B
736
Chapter 8
41. (a)
18.8°
42. Given: A 15, B 135, c 30
17.5°
C 180 A B 30
z
x
From Pine Knob:
y
9000 ft
Not drawn to scale
b
x
9000
sin 17.5 sin 1.3
(b)
c sin B 30 sin 135
42.4 kilometers
sin C
sin 30
From Colt Station:
x 119,289.1261 feet 22.6 miles
a
c sin A 30 sin 15
15.5 kilometers
sin C
sin 30
y
x
sin 71.2 sin 90
(c)
B
y x sin 71.2 119,289.1261 sin 71.2
a
30
80°
112,924.963 feet 21.4 miles
65°
(d) z x sin 18.8 119,289.1261 sin 18.8
70°
c = 65°
15°
b
C
A
38,443 feet 7.3 miles
43.
In 15 minutes the boat has traveled
10
mi
4
70°
63°
20°
10 mph14 hr 104 miles.
27°
y
d
θ
180 20 90 63
7
104
y
sin 7 sin 20
y 7.0161
sin 27 d
7.0161
d 3.2 miles
44. (a) sin (c)
5.45
0.0934 ⇒ 5.36
58.36
(b)
d
58.36
or
sin84.64 sin d
sin sin d
58.36
sin 58.36 sin84.64 sin sin1
(d)
45. True. If one angle of a triangle is obtuse, then there is less
than 90 left for the other two angles, so it cannot contain
a right angle. It must be oblique.
d sin 58.36
sin d58.36
10
20
30
40
50
60
d
324.1
154.2
95.19
63.80
43.30
28.10
46. False. Two sides and one opposite angle do not
necessarily determine a unique triangle.
Section 8.1
47. (a)
Law of Sines
737
sin sin 9
18
sin 0.5 sin arcsin0.5 sin (b)
Domain: 0 < < 1
Range: 0 < ≤
6
␲
0
0
arcsin0.5 sin (c)
c
18
sin sin c
(d)
18 sin 18 sin arcsin0.5 sin sin sin 27
Domain: 0 < < Range: 9 < c < 27
␲
0
0
(e)
0.4
0.8
1.2
1.6
2.0
2.4
2.8
As → 0, c → 27
0.1960
0.3669
0.4848
0.5234
0.4720
0.3445
0.1683
As → , c → 9
c
25.95
23.07
19.19
15.33
12.29
10.31
9.27
48. (a) A 1
1
1
3020 sin 820 sin 830 sin 2
2
2
2 2
3
80 sin 120 sin 2
2
(a)
300 sin
(a)
3
20 15 sin
4 sin 6 sin 2
2
(b)
20 cm
θ
2
θ
8 cm
30 cm
(c) Domain: 0 ≤ ≤ 1.6690
170
The domain would increase in length and the area would
have a greater maximum value if the 8-centimeter line
segment were decreased.
0
1.7
0
49. sin x cot x sin x
51. 1 sin2
cos x
cos x
sin x
2 x 1 cos
2
x sin2 x
53. 6 sin 8 cos 3 612 sin8 3 sin8 3
3sin 11 sin 5
50. tan x cos x sec x tan x cos x
52. 1 cot2
2 x 1 tan
2
1
tan x
cos x
x sec2 x
54. 2 cos 5 sin 2 2 12sin5 2 sin5 2
sin 7 sin 3
738
Chapter 8
Section 8.2
■
■
Law of Cosines
If ABC is any oblique triangle with sides a, b, and c, the following equations are valid.
(a) a2 b2 c2 2bc cos A or
cos A b2 c2 a2
2bc
(b) b2 a2 c2 2ac cos B or
cos B a2 c2 b2
2ac
(c) c2 a2 b2 2ab cos C
cos C a2 b2 c2
2ab
or
You should be able to use the Law of Cosines to solve an oblique triangle for the remaining three parts, given:
(a) Three sides (SSS)
(b) Two sides and their included angle (SAS)
■
Given any triangle with sides of length a, b, and c, the area of the triangle is
Area ss as bs c, where s abc
.
2
(Heron’s Formula)
Vocabulary Check
1. Cosines
2. b2 a2 c2 2ac cos B
3. Heron’s Area
1. Given: a 7, b 10, c 15
cos C a2 b2 c2 49 100 225
0.5429 ⇒ C 122.88
2ab
2710
sin B b sin C 10 sin 122.88
0.5599 ⇒ B 34.05
c
15
A 180 34.05 122.88 23.07
2. Given: a 8, b 3, c 9
cos C a2 b2 c2 82 32 92
0.16667 ⇒ C 99.59
2ab
283
sin A a sin C 8 sin 99.59
0.8765 ⇒ A 61.22
c
9
B 180 61.22 99.59 19.19
3. Given: A 30, b 15, c 30
a2 b2 c2 2bc cos A
225 900 21530 cos 30 345.5771
a 18.59
cos C a2 b2 c2 18.592 152 302
0.5907 ⇒ C 126.21
2ab
218.5915
B 180 30 126.21 13.79
Section 8.2
4. Given: C 105, a 10, b 4.5
c2 a2 b2 2ab cos C 102 4.52 2104.5 cos 105 143.5437 ⇒ c 11.98
a2 c2 b2 102 12.02 4.52
0.93187 ⇒ B 21.27
2ac
21012.0
cos B A 180 105 21.27 53.73
5. a 11, b 14, c 20
cos C a2 b2 c2 121 196 400
0.2695 ⇒ C 105.63
2ab
21114
sin B b sin C 14 sin 105.63
0.6741 ⇒ B 42.38
c
20
A 180 42.38 105.63 31.99
6. Given: a 55, b 25, c 72
cos C a2 b2 c2 552 252 722
0.5578 ⇒ C 123.91
2ab
25525
cos A b2 c2 a2 252 722 552
0.7733 ⇒ A 39.35
2bc
22572
B 180 123.91 39.35 16.74
7. Given: a 75.4, b 52, c 52
cos A b2 c2 a2 522 522 75.42
0.05125 ⇒ A 92.94
2bc
25252
sin B b sin A 520.9987
0.68876 ⇒ B 43.53
a
75.4
C B 43.53
8. Given: a 1.42, b 0.75, c 1.25
cos A b2 c a2 0.752 1.252 1.422
0.05792 ⇒ A 86.68
2bc
20.751.25
cos B a2 c2 b2 1.422 1.252 0.752
0.84969 ⇒ B 31.82
2ac
21.421.25
2
C 180 86.68 31.82 61.50
9. Given: A 135, b 4, c 9
a2 b2 c2 2bc cos A 16 81 249cos 135 147.9117 ⇒ a 12.16
sin B b sin A 4 sin 135
0.2326 ⇒ B 13.45
a
12.16
C 180 135 13.45 31.55
10. Given: A 55, b 3, c 10
a2 b2 c2 2bc cos A 32 102 2310 cos 55 74.585 ⇒ a 8.64
sin B b sin A 3 sin 55
0.2846 ⇒ A 16.53
a
8.636
C 180 16.53 55 108.47
Law of Cosines
739
740
Chapter 8
11. Given: B 10 35, a 40, c 30
b2 a2 c2 2ac cos B 1600 900 24030cos 10 35 140.8268 ⇒ b 11.87
sin C c sin B 30 sin 10 35
0.4642 ⇒ C 27.66 27 40
b
11.87
A 180 10 35 27 40 141 45
12. Given: B 75 20, a 6.2, c 9.5
b2 a2 c2 2ac cos B 6.22 9.52 26.29.5 cos 75 20 98.8636 ⇒ b 9.94
sin A a sin B 6.2 sin 75 20
0.6034 ⇒ A 37.1, or 37 6
b
9.94
C 180 75 20 37 6 67 34
13. Given: B 125 40 , a 32, c 32
b2 a2 c2 2ac cos B 322 322 23232 cos 125 40 3242.1888 ⇒ b 56.94
A C ⇒ 2A 180 125 40 54 20 ⇒ A C 27 10
14. Given: C 15 15, a 6.25, b 2.15
c2 a2 b2 2ab cos C 6.252 2.152 26.252.15 cos 15 15 17.7563 ⇒ c 4.21
cos A b2 c2 a2 2.152 4.21382 6.252
0.9208 ⇒ A 157.04 or 157 2
2bc
22.154.2138
B 180 15 15 157.04 7.7 or 7 43
4
7
15. C 43, a , b 9
9
c2 a2 b2 2ab cos C sin A 49 79
2
2
2
4979cos 43 0.2968 ⇒ c 0.54
a sin C 49 sin 43
0.5564 ⇒ A 33.80
c
0.5448
B 180 43 33.8 103.20
3
3
16. Given: C 103, a , b 8
4
c2 a2 b2 2ab cos C 3
4
b2 c2 a2
cos A 2bc
38 34
2
2
2
2
0.912 2
340.91
3834 cos 103 0.8297 ⇒ c 0.91
38
2
0.9160 ⇒ A 23.65
B 180 23.65 103 53.35
17. d2 52 82 258cos 45 32.4315 ⇒ d 5.69
2 360 245 270 ⇒ 135
c2 52 82 258cos 135 145.5685 ⇒ c 12.07
8
φ
5
c
5
d
45°
8
Section 8.2
18.
19.
35
120°
14
c
φ
25
25
20
10
d
θ
10
d
θ
14
35
c2 252 352 22535 cos 120
cos 2725 ⇒ c 52.20
102 142 202
21014
111.8º
1
360 2120 60
2
2 360 2111.8
68.2
d2 252 352 22535 cos 60
d 2 102 142 21014 cos 68.2
975 ⇒ d 31.22
d 13.86
402 602 802
1
⇒ 104.5
24060
4
1
360 2104.5 75.5
2
20. cos 60
c
φ
40
40
80
c2 402 602 24060 cos 75.5 3998
θ
c 63.23
21. cos 60
12.52 152 102
0.75 ⇒ 41.41
212.515
102 152 12.52
0.5625 ⇒ 55.77
21015
α
cos 12.52
102
θ
α
.5
12
15
u 180 z 97.18
b2
φ
β
15
z
.5 u
12
10
z 180 82.82
b
10
cos δ
β
x
b
212.510cos 97.18 287.4967 ⇒ b 16.96
12.52 16.962 102
0.8111 ⇒ 35.80
212.516.96
41.41 35.80 77.2
2 360 2 ⇒ 22. cos 360 277.21
102.8
2
252 17.52 252
22517.5
ω
69.512
a
a2 17.52 252 217.525 cos 110.488
a 35.18
z 180 2 40.975
252 35.182 17.52
22535.18
27.775
z 68.7
180 41.738
111.3
25
α
17.5
180 110.488
cos Law of Cosines
β
α
25
α
a
25
17.5
µ
z
α
25
741
742
Chapter 8
23. a 5, b 7, c 10 ⇒ s abc
11
2
24. a 12, b 15, c 9 ⇒ s Area ss as bs c 11641 16.25
25. a 2.5, b 10.2, c 9 ⇒ s Area 18639 54
abc
10.85
2
Area ss as bs c 10.858.350.651.85 10.4
26. Given: a 75.4, b 52, c 52
s
75.4 52 52
89.7
2
Area ss as bs c 89.714.337.737.7 1350.2
27. a 12.32, b 8.46, c 15.05 ⇒ s abc
17.915
2
Area ss as bs c 17.9155.5959.4552.865 52.11
28. Given: a 3.05, b 0.75, c 2.45
s
3.05 0.75 2.45
3.125
2
Area ss as bs c 3.1250.0752.3750.675 0.61
29. cos 17002 37002 30002
⇒ 52.9
217003700
N
W
Bearing: 90 52.9 N 37.1 E
Bearing: 90 26.9 S 63.1 E
N 75 E
cos A
00
3700 m
N
W
E
Rosemount
S
1357.8 miles
Bearing from Franklin to Rosemount:
0m
B
30. Distance from Franklin to Rosemount:
d 8102 6482 2810648 cos137
300
m
17002 30002 37002
⇒ C 100.2
217003000
S
17
cos C E
C
32° 648 miles
75°
Centerville
810 miles
Franklin
1357.82 8102 6482
21357.8810
0.9456
19.0
Bearing from Franklin to Rosemont: N 56.0 E
31. b2 2202 2502 2220250cos 105 ⇒ b 373.3 meters
220 m
C
B
75°
105°
b
250 m
A
12 15 9
18
2
Section 8.2
32.
33.
B
A
575
650
115
C
cos A 1152 762 922
0.6028 ⇒ A 52.9
211576
cos C 1152 922 762
0.75203 ⇒ c 41.2
211592
34. cos C
92
76
22 32 4.52
0.60417
223
127.2
B
A
725
The largest angle is across from the largest side.
cos C 6502 5752 7252
2650575
cos C 72.3
35. C 180 53 67 60
c2 a2 b2 2ab cos C
362 482 236480.5
1872
c 43.3 mi
N
36 mi
53°
c
60°
W
E
67°
48 mi
S
36. The angles at the base of the tower are 96 and 84. The longer guy wire g1 is given by:
g12 752 1002 275100 cos 96 17,192.9 ⇒ g1 131.1 feet
The shorter guy wire g2 is given by:
g22 752 1002 275100 cos 84 14,057.1 ⇒ g2 118.6 feet
37. (a) cos (a)
2732 1782 2352
2273178
58.4
(a) Bearing: N 58.4 W
(b) cos (b)
2352 1782 2732
2235178
81.5
(b) Bearing: S 81.5 W
Law of Cosines
N
W
Niagara Falls
235
E
φ
Denver
178
273
θ
Orlando
S
743
744
Chapter 8
38.
N
W
E
S
216 miles
C 165 miles
17.2°
B
72.8°
59.7°
368 miles
13.1°
A
a 165, b 216, c 368
cos B (a) Bearing of Minneapolis (C) from Phoenix (A)
N 90 17.2 13.1 E
1652 3682 2162
0.9551
2165368
B 17.2
N 59.7 E
(b) Bearing of Albany (B) from Phoenix (A)
2162 3682 1652
0.9741
cos A 2216368
N 90 17.2 E
N 72.8 E
A 13.1
39. d 2 60.52 902 260.590 cos 45 4059.8572 ⇒ d 63.7 ft
S
d
T
90
P
F
60.5
45°
H
40. d 3302 4202 2330420 cos 8
41. a2 352 202 23520cos 42 ⇒ a 24.2 miles
103.9 feet
42. a 202 202 22020 cos 11
43. RS 82 102 164 241 12.8 ft
1
1
PQ 2162 102 2356 89 9.4 ft
3.8 miles
tan P 72 1.52 x2 21.5 x cos 44. (a)
(c)
10 QS QS
⇒ QS 5
16
PS
8
10
49 2.25 x2 3x cos x2 3x cos 46.75 0
(b) x x
3 cos ± 3 cos 2 4146.75
21
1
3 cos 9 cos2 187 2
2␲
0
0
(d) Maximum: 8.5 inches
Section 8.2
Law of Cosines
745
45. d 2 102 72 2107 cos arccos
102 72 d 2
2107
360 360 2r 360
45
s
d (inches)
9
10
12
13
14
15
16
(degrees)
60.9
69.5
88.0
98.2
109.6
122.9
139.8
s (inches)
20.88
20.28
18.99
18.28
17.48
16.55
15.37
x
10
sin 20 sin 120
46.
x
40°
x
120°
10
10 sin 20
3.95 feet
sin 120
20°
47. a 200
b 500
c 600 ⇒ s Area 65045015050 46,837.5 square feet
12 70100 sin 70
48. area 2
6577.8 square meters
(The area of the parallelogram is the
sum of the areas of two triangles.)
49. s 510 840 1120
1235
2
Area 12351235 5101235 8401235 1120
201,674 square yards
Cost 50. area ss as bs c
s
200 500 600
650
2
a b c 2490 1860 1350
2850
2
2
2000 \$83,336.36
201,674
4840 51. False. The average of the three sides of a triangle is
abc
abc
, not
s.
3
2
area 28503609901500 1234346.0 ft2
1234346.0 ft2
28.33669 acre
43560 ft2acre
28.33669 acre\$2200acre \$62,340.71
52. False. To solve an SSA triangle, the Law of Sines is needed.
53. False. If a 10, b 16, and c 5, then by the Law of Cosines, we would have:
cos A 162 5 2 10 2
1.13125 > 1
2165
This is not possible. In general, if the sum of any two sides is less than the third side, then they cannot form a triangle. Here
10 5 is less than 16.
746
Chapter 8
a
2
54. (a) Working with ODC, we have cos .
R
a
This implies that 2R .
cos Since we know that
a
b
c
,
sin A sin B sin C
we can complete the proof by showing that
cos sin A. The solution of the system
(b) By Heron’s Formula, the area of the triangle is
Area ss as bs c.
We can also find the area by dividing the area into six
triangles and using the fact that the area is 12 the base
times the height. Using the figure as given, we have
1
1
1
1
1
1
Area xr xr yr yr zr zr
2
2
2
2
2
2
rx y z
A B C 180
rs.
CA
rs ss as bs c ⇒
Therefore:
B
r
is 90 A. Therefore:
a
a
a
.
cos cos90 A sin A
2R s as s bs c.
A
z
z
B
y
r
β α
A
D
α
β
R
O
α−C
x
y
C
B
x
R
C
55. a 25, b 55, c 72
1
(a) area of triangle: s 25 55 72 76
2
area 7651214 570.60
(b) area of circumscribed circle:
cos C R
252 552 722
0.5578 ⇒ C 123.9
22555
1 c
43.37 (see #54)
2 sin C
area R2 5909.2
(c) area of inscribed circle:
r
s as s bs c 5176214 7.51 (see #54)
area r 2 177.09
56. Given: a 200 ft, b 250 ft, c 325 ft
s
200 250 325
387.5
2
Radius of the inscribed circle: r 137.562.5
64.5 ft (see #54)
s as s bs c 187.5387.5
Circumference of an inscribed circle: C 2r 264.5 405.2 ft
Section 8.2
57.
1
1
b2 c2 a2
bc1 cos A bc 1 2
2
2bc
58.
1
2bc b2 c2 a2
bc
2
2bc
3
2
a2 b2 2bc c2
4
1
b c a b c a
4
a2 b c2
4
bca
2
abc
2
bca
2
a b c
2
abc
2
2
60. arccos 0 arccos
3
2
2
3
63. arcsin 5
6
6
3
2
66. Let u arccos 3x
3x
.
1
1
1 − 9x 2
abc
2
3
3
2x
and
1
1
1 4x2
tanarccos 3x tan u .
1
2x
θ
1 − 4x 2
x2
and
1
1
.
x2
1 9x
2
3x
x1
2
69. 5 25 x2, x 5 sin 5 25 5 sin 2
x1
.
2
2
4 − (x − 1) 2
4 x 12
2
x−1
5 251 sin2 5 5 cos u
x1
cos arcsin
cos u
2
tan x 2 cot 3x
a b c
2
67. Let arctanx 2, then
u
sin u 65. Let arcsin 2x, then
sec 68. Let u arcsin
61. arctan3 sin 2x cos u 3x a b c
2
747
2bc a2 b2 c2
1
bc
2
2bc
62. arctan 3 arctan 3 1
1
a2 b2 c2
bc1 cos A bc 1 2
2
2bc
1
b c2 a2
4
59. arcsin1 64. arccos Law of Cosines
cos 1
sec 1
1
cos csc is undefined.
x−2
θ
1
748
Chapter 8
70. x 2 cos , Additional Topics in Trigonometry
< <
2
2
71. 3 x2 9, x 3 sec 3 3 sec 2 9
2 4 x2
3 9sec2 1
2 4 2 cos 2
3 3 tan 2 4 4 cos2 2 41 cos2 tan 2 4 sin2 2
2
2
sin ⇒ cos 1
1
2
cos 2
2
csc 1
sin 1
2
cot 2
sec 3
sec 1 tan2 2 2 sin 3
5
73. cos
cos 2 sin
6
3
2 sin
12 361 tan2 12 36 sec2 12 6 sec 2 sec 1
2
1
sin2 1 sin ±
csc 1 3
4 4
34 ± 23
1
sin 1
±
3
±
2
3
±
23
3
2
74. sin x sin x 2 cos
2
2
23
3
2
12 36 36 tan2 2
2
12 36 6 tan 2
12
2
1
3
tan 12 36 x2
sin2 csc 1 cot2 1 3 2 2
72. x 6 tan , < <
2
2
cos 1 33
2 cos
x
x
2
2
2
2x2 sin2
2 2 cos x sin sin
x
x
2
2
2
5 5 6
3
6
3
sin
2
2
7
sin
12
4
Section 8.3
Section 8.3
Vectors in the Plane
Vectors in the Plane
\
■
A vector v is the collection of all directed line segments that are equivalent to a given directed line segment PQ .
■
You should be able to geometrically perform the operations of vector addition and scalar multiplication.
■
The component form of the vector with initial point P p1, p2 and terminal point Q q1, q2 is
\
PQ q1 p1, q2 p2 v1, v2 v.
■
The magnitude of v v1, v2 is given by v v12 v22.
■
If v 1, v is a unit vector.
■
You should be able to perform the operations of scalar multiplication and vector addition in component form.
(a) u v u1 v1, u2 v2
■
(b) ku ku1, ku2 You should know the following properties of vector addition and scalar multiplication.
(a) u v v u
(b) u v w u v w
(c) u 0 u
(d) u u 0
(e) cdu cdu
(f) c du cu du
(g) cu v cu cv
(h) 1u u, 0u 0
(i) cv c v
v
.
v
■
A unit vector in the direction of v is u ■
The standard unit vectors are i 1, 0 and j 0, 1. v v1, v2 can be written as v v1i v2 j.
■
A vector v with magnitude v and direction can be written as v ai bj vcos i vsin j,
where tan b a.
Vocabulary Check
1. directed line segment
2. initial; terminal
3. magnitude
4. vector
5. standard position
6. unit vector
8. resultant
9. linear combination; horizontal; vertical
1. v 4 0, 1 0 4, 1
uv
2. u 3 0, 4 4 3, 8
v 0 3, 5 3 3, 8
uv
749
750
Chapter 8
3. Initial point: 0, 0
4. Initial point: 0, 0
Terminal point: 3, 2
Terminal point: 4, 2
v 3 0, 2 0 3, 2
v 4 0, 2 0 4, 2
v 32 22 13
v 42 22 20 25
5. Initial point: 2, 2
6. Initial point: 1, 1
Terminal point: 1, 4
Terminal point: 3, 5
v 1 2, 4 2 3, 2
v 3 1, 5 1 4, 6
v 32 22 13
v 42 62 52 213
7. Initial point: 3, 2
8. Initial point: 4, 1
Terminal point: 3, 3
Terminal point: 3, 1
v 3 3, 3 2 0, 5
v 3 4, 1 1 7, 0
v v 72 02 7
02
52
25 5
9. Initial point: 1, 5
10. Initial point: 1, 11
Terminal point: 15, 12
Terminal point: 9, 3
v 15 1, 12 5 16, 7
v 9 1, 3 11 8, 8
v 162
72
305
v 82 82 82
11. Initial point: 3, 5
12. Initial point: 3, 11
Terminal point: 5, 1
Terminal point: 9, 40
v 5 3, 1 5 8, 6
v 82
62
v 9 3, 40 11 12, 29
100 10
v 122 292 985
13. Initial point: 1, 3
14. Initial point: 2, 7
Terminal point: 8, 9
Terminal point: 5, 17
v 8 1, 9 3 9, 12
v 5 2, 17 7 7, 24
v 9 12 225 15
2
15.
v 72 242 25
2
y
17.
16. 5v
y
y
u+v
5v
v
v
x
−v
u
x
v
x
Section 8.3
18. u v
Vectors in the Plane
751
1
20. v 2 u
19. u 2v
y
y
y
u
v
u + 2v
x
v − 12 u
2v
u−v
x
−v
− 12 u
u
x
21. u 2, 1, v 1, 3
(a) u v 3, 4
(c) 2u 3v 4, 2 3, 9 1, 7
(b) u v 1, 2
y
y
y
5
u+v
4
2
v
3
−6
u
1
2
−3
1
−1
−2
−1
−4
3
4
5
2
3
4
6
−6
2u − 3v
−3v
u−v
−v
−1
2
x
1
x
2
x
−2
u
1
2u
2
3
− 10
22. u 2, 3, v 4, 0
(b) u v 2, 3
(a) u v 6, 3
y
(c) 2u 3v 4, 6 12, 0
8, 6
y
6
8
y
5
6
10
4
8
2u − 3v
u−v
u
4
u+v
2
2
x
4
6
2
−3v
−v
v
2u
4
2
u
x
−5 −4 −3 −2 −1
1
2
− 10 − 8 − 6 − 4 − 2
x
2
4
6
−4
8
−6
−8
23. u 5, 3, v 0, 0
(a) u v 5, 3 u
u=u+v
y
y
7
7
12
6
6
10
5
5
4
4
u=u−v
3
2
1
1
−7 −6 −5 −4 −3 −2 −1
8
6
4
v
x
1
2u = 2u − 3v
3
2
v
−7 −6 −5 −4 −3 −2 −1
(c) 2u 3v 2u 10, 6
(b) u v 5, 3 u
y
x
1
2
−3v
− 12 − 10 − 8 − 6 − 4 − 2
−2
2
x
752
Chapter 8
24. u 0, 0, v 2, 1
(c) 2u 3v 0, 0 6, 3
(b) u v 2, 1
(a) u v 2, 1
6, 3
y
y
y
1
3
u
2
−3
v=u+v
1
−2
1
2
x
1
−1
−3
− 3v = 2u − 3v
3
−4
−3
−1
2u
−2
−2
x
−1
−7 −6 −5 −4 −3 −2
−1
−v = u − v
u
1
x
1
−5
−6
−7
25. u i j, v 2i 3j
3
5
2
u−v 4
−1
y
2u − 3v 12
u
1
−2
4i 11j
y
y
−3
(c) 2u 3v 2i 2j 6i 9j
(b) u v i 4j
(a) u v 3i 2j
10
−v
x
u
u+v
−2
−3
v
−3
8
−3v
3
−1
−2
x
−1
1
2
3
−1
2u
x
−8 −6 −4 −2
−2
2
4
6
26. u 2i j, v i 2j
(b) u v i j
(a) u v 3i 3j
2u 3v 4i 2j 3i 6j
i 4j
y
y
4
u+v
(c)
y
2
3
u
1
3
2
1
2u
2
v
−2
u
−4
−3
−2
1
− 6 − 5 −4 − 3
2
x
−1
1 2 3 4
−1
u−v
x
−1
x
−1
1
−v
−2
−1
2u − 3v
− 3v
−5
−6
−7
27. u 2i, v j
(c) 2u 3v 4i 3j
(b) u v 2i j
(a) u v 2i j
y
y
y
1
3
1
2u
u
2
1
−1
u+v
v
−1
u
−1
1
−1
x
2
x
3
−2
3
−3
−v
u−v
−1
1
2
3
−1
−2
−3
−4
−3v
2u − 2v
x
Section 8.3
Vectors in the Plane
753
28. u 3j, v 2i
y
6i 6j
y
u−v
u+v
3
(c) 2u 3v 6j 6i
(b) u v 2i 3j
(a) u v 2i 3j
y
u
u
2
8
2
1
2u − 3v
1
v
−1
1
2
2u
−v
x
−3
3
−2
−1
4
x
−1
1
2
−3v
−1
−8
−6
−4
x
−2
2
−2
29. v 1
1
1
u
3, 0 3, 0 1, 0
u
3
32 02
30. u 0, 2
v
1
1
0, 2
u
u
02 22
1
0, 2 0, 1
2
31. u 1
1
1
v
2, 2 2, 2
v
22 22
22
1
1
,
2 2
2 2
33. u 35. u 2
,
2
32. v 5, 12
u
1
1
v
5, 12
52 122
v
1
5, 12
13
135 , 1312
1
1
1
6i 2j
v
6i 2j 40
v
62 22
34. v i j
1
1
3
6i 2j j
i
10
10
210
u
10
310
i
j
10
10
1
1
w 4j j
w
4
1
v
v
1
12 12
i j 1
2
i j 2
2
i
36. w 6i
v
1
1
w
6i
w
62 02
1
6i i
6
37. u 1
1
1
i 2j
w
i 2j w
12 22
5
25
5
2
1
j
i
i
j
5
5
5
5
38. w 7j 3i
v
1
1
3i 7j
w
w
32 72
3
58
i
7
58
j
358
758
i
j
58
58
2
2
j
754
Chapter 8
39. 5
u1 u 53 1 3 3, 3 35 2 3, 3
2
41. 9
40. v 6
2
52, 52
52 2, 52 2 6
u1 u 92 1 5 2, 5 929 2, 5
2
u1 u 631
2
1829, 4529
182929, 452929 10
43. u 4 3, 5 1
uu 100
1
2
u 3i 8j
46. u 0 6, 1 4
45. u 2 1, 3 5
3, 8
u 6, 3
3i 8j
u 6i 3j
48. v 34 w 34 i 2j
47. v 32u
j
3i 3,
32
3
4i
3
2j
49. v u 2w
2i j 2i 2j
3 3
4, 2
4i 3j 4, 3
y
y
w
2
y
2w
4
1
1
3
w
4
−1
3
2
x
−1
u + 2w
3
x
1
1
2
2
1
u
−1
x
3u
2
−2
3
−1
51. v 123u w
50. v u w
2i j i 2j
i 3j 1, 3
1
2j
7
2,
12
2i j 2i 2j
5j 0, 5
y
−2
2
1
−2
u
−2
4
1 (3u + w)
2
−1
−1
1
−2
2
1
w
2
x
x
x
−1
w
−u
5
u
y
3
2
7
2i
4
52. v u 2w
126i 3j i 2j
y
−u + w
44. u 3 0, 6 2
7i 4j
3
2j
1
10, 0
102
1
u 3, 8
1010, 0 10, 0
7, 4
3
2 2i
3, 3
31 2 3, 3 62, 62
32, 32
42. v 10
2
32
3u
2
−3
−2w
−4
u − 2w
3
Section 8.3
53. v 3cos 60i sin 60ºj
54.
v 8cos 135 i sin 135 j
755
55. v 6i 6j
v 8, 135
v 3, 60
Vectors in the Plane
v 62 62 72
62
tan 6
1
6
Since v lies in Quadrant IV,
315.
v 5i 4j
56.
57. v 3 cos 0, 3 sin 0
v 5 4 41
2
tan 58. v cos 45, sin 45
3, 0
2
4
5
y
22, 22
y
2
Since v lies in Quadrant II,
141.3.
1
1
x
1
2
3
45°
−1
x
1
59. v 72 cos 150, 27 sin 150
73 7
,
4 4
60. v y
52 cos 45, 25 sin 45
52 52
,
4
4
61. v 32 cos 150, 32 sin 150
36 32
,
2
2
y
y
5
4
3
4
3
3
2
2
2
150°
−4
−3
−2
1
1
x
−1
45°
1
1
62. v 43 cos 90, 43 sin 90
0, 43
−5
x
−1
63. v 2
y
2
1
12 32
2
10
6
10
5
4
i 3j
64. v 3
−6
−4
−2
2
4
x
−1
1
1
3i 4j
42
2
3
3i 4j
5
10 310
310
j
,
5
5
5
9
12
9 12
i j ,
5
5
5 5
y
90°
3
3
x
−2
3
y
2
−2
3
i 3j
i
−3
−1
10
8
−4
150°
6
2
2
1
1
−1
x
−1
x
1
2
1
−1
2
3
756
Chapter 8
65. u 5 cos 0, 5 sin 0 5, 0
u 4 cos 60, 4 sin 60 2, 23 66.
v 4 cos 90, 4 sin 90 0, 4
v 5 cos 90, 5 sin 90 0, 5
u v 2, 4 23 u v 5, 5
67. u 20 cos 45, 20 sin 45 102, 102 u 50 cos 30, 50 sin 30 253, 25
68.
43.301, 25
v 50 cos 180, 50 sin 180 50, 0
u v 102 50, 102 v 30 cos 110, 30 sin 110 10.261, 28.191
u v 33.04, 53.19
69. v i j
y
w 2i 2j
1
v
u v w i 3j
x
−1
w 22
1
−1
v w 10
cos u
α
v 2
v2
2
w
−2
w2
v 2v w
w2
2 8 10
0
22 22
90
70. v i 2j
y
w 2i j
2
u v w i 3j
cos v2
w2
v 2v w
1
w2
5 5 10
0
255
90
Force Two: v 60 cos i 60 sin j
Resultant Force: u v 45 60 cos i 60 sin j
u v 45 60 cos 2 60 sin 2 90
2025 5400 cos 3600 8100
5400 cos 2475
cos 2475
0.4583
5400
62.7
u
θ
−2
−1
x
2
−1
−2
71. Force One: u 45i
v
w
Section 8.3
72. Force One: u 3000i
Force Two: v 1000 cos i 1000 sin j
Resultant Force: u v 3000 1000 cos i 1000 sin j
u v 3000 1000 cos 2 1000 sin 2 3750
9,000,000 6,000,000 cos 1,000,000 14,062, 500
6,000,000 cos 4,062, 500
cos 4,062,500
0.6771
6,000,000
47.4
73.
u 300i
v 125 cos 45i 125 sin 45j R u v 300 R 2
74.
125
125
i
j
2
2
300 1252 1252
125
2
tan 125
300 2
125
125
i
j
2
2
2
398.32 newtons
⇒ 12.8
u 2000 cos 30 i 2000 sin 30j
y
1732.05i 1000j
v 900 cos45i 900 sin45j
2000
636.4i 636.4j
u v 2368.4 i 363.6j
u v 2368.42 363.62 2396.19
tan 363.6
0.1535 ⇒ 8.7
2368.4
75. u 75 cos 30i 75 sin 30ºj 64.95i 37.5j
v 100 cos 45i 100 sin 45j 70.71i 70.71j
w 125 cos 120i 125 sin 120j 62.5i 108.3j
u v w 73.16i 216.5j
u v w 228.5 pounds
tan 216.5
2.9593
73.16
71.3º
u+v
x
900
Vectors in the Plane
757
758
Chapter 8
u 70 cos 30 i 70 sin 30j 60.62i 35j
76.
v 40 cos 45i 40 sin 45j 28.28i 28.28j
w 60 cos 135i 60 sin 135j 42.43i 42.43j
u v w 46.48i 35.71j
u v w 58.61 pounds
tan 35.71
0.7683
46.47
37.5
77. Horizontal component of velocity: 70 cos 35 57.34 feet per second
Vertical component of velocity: 70 sin 35 40.15 feet per second
78. Horizontal component of velocity: 1200 cos 6 1193.4 ftsec
Vertical component of velocity: 1200 sin 6 125.4 ftsec
\
79. Cable AC : u ucos 50i sin 50j
\
Cable BC : v vcos 30i sin 30j
Resultant: u v 2000j
u cos 50 v cos 30 0
u sin 50 vsin 30 2000
Solving this system of equations yields:
TAC u 1758.8 pounds
TBC v 1305.4 pounds
\
80. Rope AC : u 10i 24j
The vector lies in Quadrant IV and its reference angle is arctan 5 .
12
u u cosarctan
12
5
i sinarctan j
12
5
\
Rope BC : v 20i 24j
The vector lies in Quadrant III and its reference angle is arctan5 .
6
v v cosarctan 65 i sinarctan 65 j
Resultant: u v 5000j
6
u cosarctan 12
5 v cosarctan 5 0
6
u sinarctan 12
5 v sinarctan 5 5000
Solving this system of equations yields: TAC u 3611.1 pounds
TBC v 2169.5 pounds
81. Towline 1: u ucos 18i sin 18j
Towline 2: v ucos 18i sin 18j
Resultant: u v 6000i
u cos 18 u cos 18 6000
u 3154.4
Therefore, the tension on each towline is u 3154.4 pounds.
Section 8.3
82. Rope 1: u u cos 70i sin 70j
70°
Rope 2: v u cos 70i sin 70j
Vectors in the Plane
70°
20° 20°
Resultant: u v 100j
u sin 70 u sin 70 100
u 53.2
100 lb
Therefore, the tension of each rope is u 53.2 pounds.
y
83. Airspeed: u 875 cos 58i 875 sin 58j
N
140° W
Groundspeed: v 800 cos 50i 800 sin 50j
148°
Wind: w v u 800 cos 50 875 cos 58i 800 sin 50 875 sin 58j
2
138.7 kilometers per hour
Wind speed:
Wind direction: tan Wind direction:
129.2065
50.5507
68.6; 90 21.4
Bearing: N 21.4 E
y
84. (a)
N
W
E
S
28°
580 mph
45°
x
60 mph
(b) The velocity vector vw of the wind has a magnitude of 60 and a direction angle of 45.
vw vwcos i vwsin j
60cos 45i 60sin 45j
60cos 45i sin 45j
60cos 45, sin 45, or 302, 302
(c) The velocity vector vj of the jet has a magnitude of 580 and a direction angle of 118.
vj vjcos i vjsin j
580cos 118i 580sin 118j
580cos 118i sin 118j
580cos 118, sin 118
—CONTINUED—
v
40°
Wind speed: w 50.5507 129.2065
2
S
x
32°
50.5507i 129.2065j
Wind:
E
u
w
759
760
Chapter 8
84. —CONTINUED—
(d) The velocity of the jet (in the wind) is
v vw vj
60cos 45, sin 45 580cos 118, sin 118
60 cos 45 580 cos 118, 60 sin 45 580 sin 118
229.87, 554.54
The resultant speed of the jet is
v 229.872 554.542
600.3 miles per hour
(e) If is the direction of the flight path, then
tan 554.54
2.4124
229.87
Because lies in the Quadrant II, 180 arctan2.4124 180 67.5 112.5. The true bearing of the jet is
112.5 90 22.5 west of north, or 360 22.5 337.5.
85. W FD 100 cos 5030 1928.4 foot–pounds
86. Horizontal force: u ui
Weight: w j
Rope: t t cos 135i sin 135j
100 lb
u w t 0 ⇒ u t cos 135 0
1 t sin 135 0
50°
t 2 pounds
30 ft
u 1 pound
87. True. See Example 1.
88. True.
u a2 b2 1 ⇒ a2 b2 1
89. (a) The angle between them is 0.
(b) The angle between them is 180.
(c) No. At most it can be equal to the sum when the angle between them is 0.
90. F1 10, 0, F2 5cos , sin (a)
F1 F2 10 5 cos , 5 sin (b)
15
F1 F2 10 5 cos 2 5 sin 2
100 100 cos 25 cos2 25 sin2 54 4 cos cos2 sin2 2␲
0
0
54 4 cos 1
55 4 cos (c) Range: 5, 15
Maximum is 15 when 0.
Minimum is 5 when .
(d) The magnitude of the resultant is never 0 because the magnitudes of F1 and F2 are not the same.
Section 8.3
Vectors in the Plane
91. Let v cos i sin j.
v cos2 sin2 1 1
Therefore, v is a unit vector for any value of .
92. The following program is written for a TI-82 or TI-83 or
TI-83 Plus graphing calculator. The program sketches
two vectors u ai bj and v ci dj in standard
position, and then sketches the vector difference u v
using the parallelogram law.
93. u 5 1, 2 6 4, 4
PROGRAM: SUBVECT
:Input “ENTER A”, A
:Input “ENTER B”, B
:Input “ENTER C”, C
:Input “ENTER D”, D
:Line (0, 0, A, B)
:Line (0, 0, C, D)
:Pause
:A – C→E
:B – D→F
:Line (A, B, C, D)
:Line (A, B, E, F)
:Line (0, 0, E, F)
:Pause
:ClrDraw
:Stop
94.
v 9 4, 4 5 5, 1
u 80 10, 80 60 70, 20
v 20 100, 70 0 80, 70
u v 1, 3 or v u 1, 3
u v 70 80, 20 70 10, 50
v u 80 70, 70 20 10, 50
96. x 8 sin 95. x2 64 8 sec 2 64
64 x2 64 8 sin2 64sec2 1
64 64 sin2 8tan2 8 tan for 0 < <
2
81 sin2 8cos2 8 cos for 0 < <
2
98. x 5 sec 97. x2 36 6 tan 2 36
x2 253 5 sec 2 253
36tan2 1
25 sec2 253
6sec2 6 sec for 0 < <
2
25sec2 13
25 tan2 3
15,625 tan6 125 tan3 for 0 < <
99. cos xcos x 1 0
cos x 0
x
or
n
2
cos x 1 0
cos x 1
x 2n
2
761
762
Chapter 8
100. sin x2 sin x 2 0
101. 3 sec x sin x 23 sin x 0
sin x3 sec x 23 0
2 sin x 2 0
sin x 0
x 0 n
sin x sin x 0
2
or 3 sec x 23 0
2
x n
7
5
2 n,
2 n
x
4
4
5
7
x n,
2 n,
2 n
4
4
sec x 23
3
cos x 3
3
23
2
x
2n
6
x
11
2n
6
102. cos x csc x cos x2 0
cos x csc x 2 0
cos x 0
x
csc x 2 0
n
2
csc x 2
x
x
5
7
2 n,
2 n
4
4
5
7
n ,
2 n,
2 n
2
4
4
Section 8.4
Vectors and Dot Products
■
Know the definition of the dot product of u u1, u2 and v v1, v2 .
■
Know the following properties of the dot product:
u v u1v1 u2v2
■
1.
uvvu
2.
0v0
3.
u v w u v u w
4.
v v v2
5.
cu v cu v u cv
If is the angle between two nonzero vectors u and v, then
cos uv
.
u v
■
The vectors u and v are orthogonal if u v 0.
■
Know the definition of vector components.
u w1 w2 where w1 and w2 are orthogonal, and w1 is parallel to v. w1 is called the projection of u onto v
and is denoted by w1 projvu ■
v v. Then we have w
u
v
2
2
Know the definition of work.
\
1.
Projection form: w projPQ F PQ 2.
Dot product form: w F PQ
\
\
u w1.
Section 8.4
Vocabulary Check
1. dot product
2.
uv
u v
3. orthogonal
4.
uv vv
Vectors and Dot Products
763
2
5. projPQ F PQ ; F PQ
\
\
\
1. u 6, 1, v 2, 3
u v 62 13 9
4. u 2, 5, v 1, 2
u v 21 52
3. u 4, 1, v 2, 3
2. u 5, 12, v 3, 2
u v 42 13 11
u v 53 122 9
6. u 3i 4j, v 7i 2j
5. u 4i 2j, v i j
u v 41 21 6
u v 37 42 13
2 10 8
7. u 3i 2j, v 2i 3j
u v 32 23 12
8. u i 2j, v 2i j
9. u 2, 2
u u 22 22 8
u v 12 21 4
The result is a scalar.
10. u 2, 2, v 3, 4
3u v 323 24
32 6
The result is a scalar.
11. u 2, 2, v 3, 4
u vv 23 24
3, 4
23, 4 6, 8
The result is a vector.
12. u 2, 2, v 3, 4, w 1, 2
v uw 32 42
1, 2
13. u 2, 2, v 3, 4, w 1, 2
3w vu 313 324
2, 2
21, 2
332, 2
2, 4 vector
66, 66
The result is a vector.
14. u 2, 2, v 3, 4, w 1, 2
15. w 1, 2
2v 6, 8
w 1 1222 1 5 1
u 2vw 26 28
1, 2
The result is a scalar.
41, 2
4, 8 vector
16. u 2, 2
2 u 2 u u
17. u 2, 2, v 3, 4, w 1, 2
u v u w 23 24
21 22
2 22 22
2 2
2 8
4
2 22 scalar
The result is a scalar.
764
Chapter 8
18. u 2, 2, v 3, 4, w 1, 2
v u w v 32 42
13 24
2 11
13 scalar
20. u 2, 4
19. u 5, 12
u u u 52 122 13
u u u
22 44
20 25
22. u 12i 16j
21. u 20i 25j
u u u 1212 1616
u u u 20 25 1025 541
2
2
400 20
24. u 21i
23. u 6j
u u u 2121 00
u u u 0 6 36 6
2
2
212 21
25. u 1, 0, v 0, 2
cos uv
0
0
u v 12
27. u 3i 4j, v 2j
26. u 3, 2, v 4, 0
cos 90
uv
34 20
u v
13 4
3
13
cos 5
0.83205
arccos 33.69
28. u 2i 3j, v i 2j
cos uv
u v
21 32
22 3212 22
8
65
143.13
29. u 2i j, v 6i 4j
cos uv
8
0.4961
u v 552
60.26
0.992278
7.13
30. u 6i 3j, v 8i 4j
cos u uv
68 34 36
0.6
u v
60
4580
53.13
32. u 2i 3j, v 4i 3j
cos uv
24 33
0.0555
1325
u v
93.18
uv
8
u v
52
31. u 5i 5j, v 6i 6j
cos uv
0
u v
90
4
Section 8.4
3
1
i sin j i j
3
3
2
2
2
2
3
3
i sin
j
i
j
4
4
2
2
33. u cos
v cos
34. u cos
v cos
uv
uv
u v
2 2 2 2
1
2
arccos
3
2 6
4
2
2
i
2
2
j
2 i s in 2 j j
4
u v u v cos 8
64 34 62 32 42 42 cos 6
uv
u v
v
4
u
−8
−6
−4
−2
12 1210 cos x
2
32 cos 12 45
2
37 45
374
574
2
36. u 6i 3j, v 4i 4j
y
1
75 512
v 7i 5j
2 6
4
35. u 3i 4j
cos 4 i sin 4 j 765
2
2
0 1
2
uv
2
2
cos u v
11
2
u v 1
cos Vectors and Dot Products
4
−2
1
−4
10
0.0232
cos1
91.3
cos 110 ⇒ 108.4
y
8
6
4
v
u
2
−6
−4
−2
x
2
4
6
−2
−4
37. u 5i 5j
v 8i 8j
cos uv
u v
58 58
50128
0
90
38. u 2i 3j, v 8i 3j
y
u v u v cos 10
8
v
28 33 22 32 82 32 cos 6
u
4
7 13
2
−8 − 6 − 4 − 2
−2
7
x
2
4
6
13
−4
cos1
73
cos 13 7 73 ⇒ 76.9
y
6
4
v
2
−2
x
2
−2
−4
−6
73 cos u
4
6
8
10
766
Chapter 8
39. P 1, 2, Q 3, 4, R 2, 5
\
\
\
PQ 2, 2, PR 1, 3, QR 1, 1
\
cos PQ
\
\
\
PQ PR \
cos PR
PQ
QR
2
8
⇒ arccos
26.57
5
2210
\
\
\
PQ QR 0 ⇒ 90. Thus, 180 26.57 90 63.43.
40. P 3, 4, Q 1, 7, R 8, 2
\
41. P 3, 0, Q 2, 2, R 0, 6
\
PQ 4, 11, QR 7, 5,
\
\
\
\
PR 11, 6, QP 4, 11
\
cos PQ PR
110
cos ⇒ 41.41
PQ PR 137 157 \
\
\
\
\
\
\
PQ
\
PR
\
\
\
PQ PR \
cos QR QP
27
cos ⇒ 74.45
QR QP 74 137 \
\
QP 5, 2, PR 3, 6, QR 2, 4, PQ 5, 2
QP
\
QR
27
2945
⇒ 41.63
\
\
QP PR 2
2920
⇒ 85.24
180 41.63 85.24 53.13
180 41.41 74.45 64.14
43. u v u v cos 42. P 3, 5, Q 1, 9, R 7, 9
\
\
PQ 2, 4, QR 8, 0,
\
410 cos
\
PR 10, 4, QP 2, 4
cos 2
40 PQ PR
36
⇒ 41.6
PQ PR 20 116 \
\
\
\
\
\
\
1
20
QR QP
16
cos ⇒ 116.6
QR QP 820 \
2
3
180 41.6 116.6 21.8
44. u 100, v 250, u
6
45. u v u v cos v u v cos 936 cos
100250 cos
25,000 6
324 3
2
2
3
4
1622 229.1
2
46. u 4
v 12
3
u v u v cos 412 cos
12,5003
412
47. u 12, 30, v 12, 45
48. u 3, 15, v 1, 5
u kv ⇒ Not parallel
u 24v ⇒ u and v are parallel.
u
v0
Neither
⇒ Not orthogonal
3
2 24
1
1
49. u 3i j, v 5i 6j
4
u kv ⇒ Not parallel
u v 0 ⇒ Not orthogonal
Neither
Section 8.4
50. u 1, v 2i 2j
51. u 2i 2j, v i j
u kv ⇒ Not parallel
u
v0
u
v0
Vectors and Dot Products
767
52. u cos , sin ⇒ u and v are orthogonal.
v sin , cos ⇒ Not orthogonal
u v 0 ⇒ u and v are
orthogonal.
Neither
53. u 2, 2, v 6, 1
w1 projvu uv vv 37146, 1 371 84, 14
2
w2 u w1 2, 2 u
1
14
10 60
10
6, 1 ,
1, 6 10, 60
37
37 37
37
37
1
1
84, 14 10, 60 2, 2
37
37
54. u 4, 2, v 1, 2
w1 projvu 55. u 0, 3, v 2, 15
v v 01, 2 0, 0
u
v
2
w2 u w1 4, 2 0, 0 4, 2
u 4, 2 0, 0 4, 2
w1 projvu 45
2, 15
uv v v 229
w2 u w1 0, 3 u
56. u 3, 2, v 4, 1
w1 projvu 45
90 12
2, 15 ,
229
229 229
6
15, 2
229
45
6
2, 15 15, 2 0, 3
229
229
57. projvu 0 since they are perpendicular.
4, 1
uv vv 14
17 2
14
5
w2 u w1 3, 2 4, 1 1, 4
17
17
u
2
Since u and v are orthogonal, u v 0 and projv u 0.
projvu uv
v 0, since u v 0.
v2
14
5
4, 1 1, 4 3, 2
17
17
58. Because u and v are orthogonal, the projection of u onto v is 0.
projw u uv
v 0 since u
v2
v 0.
59. u 3, 5
60. u 8, 3
For v to be orthogonal to u, u v must equal 0.
For v to be orthogonal to u, u v must be equal to 0.
Two possibilities: 5, 3 and 5, 3
Two possibilities: 3, 8, 3, 8
1
2
61. u 2i 3j
5
62. u 2i 3j
For u and v to be orthogonal, u v must equal 0.
For v to be orthogonal to u, u v must be equal to 0.
2
1
2
1
Two possibilities: v 3i 2j and v 3i 2j
5
Two possibilities: v 3i 2j and v 3i 2j
5
768
Chapter 8
\
64. P 1, 3, Q 3, 5, v 2i 3j
\
63. w projPQ v PQ where PQ 4, 7 and v 1, 4.
\
PQ
PQ \
proj PQ v \
v
\
2
work v PQ
\
32
PQ 4, 7
65
\
326565
\
w proj PQ v PQ \
2i 3j 4i 2j
24 32 14
65 32
65. (a) u 1650, 3200, v 15.25, 10.50
u
(b) Increase prices by 5%: 1.05v The operation is scalar
multiplication.
v 165015.25 320010.50 \$58,762.50
u
This gives the total revenue that can be earned by
selling all of the pans.
1.05v 1.05u v
1.05165015.25 320010.50
1.0558,762.50
61,700.63
66. (a) u 3240, 2450, v 1.75, 1.25
(b) Increase prices by 2.5%:
u v 32401.75 24501.25 8732.5
1.025v scalar multiplication
The fast food stand sold \$8732.50 of hamburgers and
hot dogs in one month.
67. (a) Force due to gravity:
F 30,000j
Unit vector along hill:
v cos di sin dj
Projection of F onto v:
w1 projv F Fv vv F vv 30,000 sin d v
2
The magnitude of the force is 30,000 sin d.
(b)
d
0
1
2
3
4
5
6
7
8
9
10
Force
0
523.6
1047.0
1570.1
2092.7
2614.7
3135.9
3656.1
4175.2
4693.0
5209.4
(c) Force perpendicular to the hill when d 5:
Force 30,0002 2614.72 29,885.8 pounds
68. Force due to gravity: F 5400j
Unit vector along hill: v cos 10i sin 10j
Projection of F onto v:
w1 projvF
Fv vv
2
F vv because v is a unit vector, v 1
0cos 10 5400sin 10
v
5400sin 10v 937.7v
The magnitude of the force is 937.7, so a force of 937.7 pounds is required to keep the vehicle
from rolling down the hill.
Force perpendicular to the hill: Force 54002 9372 5318.0 pounds
Section 8.4
Vectors and Dot Products
69. w 2453 735 newton-meters
70. work 24005 12,000 foot-pounds
71. w cos 304520 779.4 foot-pounds
72. work cos 3515,691800
10,282,652 newton-meters
73. w cos 30250100 21,650.64 foot-pounds
\
74. work cos F PQ cos 2025 pounds50 feet
1174.62 foot-pounds
75. False. Work is represented by a scalar.
76. True.
W F PQ 0 if F and PQ are orthogonal.
\
v0
77. (a) u
⇒ u and v are orthogonal and (b) u v > 0 ⇒ cos > 0 ⇒ 0 ≤ <
(c) u v < 0 ⇒ cos < 0 ⇒
.
2
\
78. (a) projvu u ⇒ u and v are parallel.
(b) projvu 0 ⇒ u and v are orthogonal.
2
< ≤ 2
80. Let u u1, u2 and v v1, v2 .
79. In a rhombus, u v. The diagonals are
u v and u v.
u v u1 v1, u2 v2 u v u v u v u u v v
u v2 u1 v12 u2 v22
uuvuuvvv
u12 2u1v1 v12 u22 2u2v2 v22
u2 v2 0
u12 u22 v12 v22 2u1v1 2u2v2
Therefore, the diagonals are orthogonal.
u 2 v 2 2u1v1 u2v2
u 2 v 2 2u v
u−v
u
u+v
v
81. 42
24 1008
144
82. 18
7
2
3
127
83. 3 8 3i22i
112 18 112
32 24 7
26i 2
222 7
26
1214
84. 12 96 i12
i 212
22
2
23
i96
96
3
25
26 32
32
242
85.
sin 2x 3 sin x 0
2 sin x cos x 3 sin x 0
3
sin x2 cos x 3 0
sin x 0
or 2 cos x 3 0
x 0, cos x x
3
2
11
,
6 6
769
770
Chapter 8
86. sin 2x 2 cos x 0
2 tan x tan 2x
87.
2 sin x cos x 2 cos x 0
2 tan x cos x2 sin x 2 0
cos x 0
sin x x
x
2 tan x1 tan2 x 2 tan x
2 sin x 2 0
3
x ,
2 2
2 tan x
1 tan2 x
2 tan x1 tan2 x 2 tan x 0
2
2 tan x1 tan2 x 1
0
2
2 tan xtan2 x 0
5 7
,
4 4
2 tan3 x 0
5 3 7
, , ,
2 4 2 4
tan x 0
x 0, 88. cos 2x 3 sin x 2
1 2 sin2 x 3 sin x 2 0
2 sin2 x 3 sin x 1 0
2 sin x 1sin x 1 0
2 sin x 1 0
sin x x
x
sin x 1 0
1
2
sin x 1
7 11
,
6 6
x
3
2
7 3 11
, ,
6 2 6
For Exercises 89–92:
5
sin u 12
13 , u in Quadrant IV ⇒ cos u 13
89. sin u v sin u cos v cos u sin v
24
5
7
12
13 25 13 25 253
325
7
cos v 24
25 , v in Quadrant IV ⇒ sin v 25
12
12
5
90. sin u 13, cos u 1 13 13
2
24
7
cos v 24
25 , sin v 1 25 25
2
sinu v sin u cos v cos u sin v
24
5
7
12
13 25 13 25 323
325
91. cosv u cos v cos u sin v sin u
92. sin u 12
5
12
, cos u , tan u 13
13
5
5
7
12
24
25 13 25 13 cos v 204
325
tanu v 24
7
7
, sin v , tan v 25
25
24
tan u tan v
1 tan u tan v
125 247 253
120
7
17
1 12
5 24 10
253
204
Section 8.5
Section 8.5
Trigonometric Form of a Complex Number
Trigonometric Form of a Complex Number
■
You should be able to graphically represent complex numbers and know the following facts about them.
■
The absolute value of the complex number z a bi is z a2 b2.
■
The trigonometric form of the complex number z a bi is z rcos i sin where
(a) a r cos (b) b r sin (c) r a2 b2; r is called the modulus of z.
b
(d) tan ; is called the argument of z.
a
Given z1 r1cos 1 i sin 1 and z2 r2cos 2 i sin 2:
■
(a) z1z2 r1r2cos1 2 i sin1 2
(b)
z1 r1
cos1 2 i sin1 2, z2 0
z2 r2
You should know DeMoivre’s Theorem: If z rcos i sin , then for any positive integer n,
■
zn rn cos n i sin n.
You should know that for any positive integer n, z rcos i sin has n distinct nth roots given by
■
n r cos
2k
2k
i sin
n
n
where k 0, 1, 2, . . . , n 1.
Vocabulary Check
1. absolute value
2. trigonometric form; modulus; argument
3. DeMoivre’s
4. nth root
1.
7i
02 72
2. 7 72 02 49 7
49 7
−2
4
−8
Imaginary
axis
6
Real
axis
− 4 + 4i
4
−2
−6
32 42
8
2
2
−7
−4
4 4i
42 42
Imaginary
axis
Imaginary
axis
−4
3.
−8
−6
−4
−2
4
3
2
Real
axis
2
−2
−7i
5
1
−5 − 4 −3 −2 −1
−1
1
Real
axis
771
772
Chapter 8
4. 5 12i 52 122
6 7i
62 72
5.
169 13
73
Imaginary
axis
Imaginary
axis
2
4
Real
axis
6
2
4
6
Real
axis
8
6
− 8 + 3i
−2
−4
2
−8
− 10 − 8
−6
− 10
r 22 02 4 2
3
tan , undefined ⇒ 0
2
tan i sin
2
2
0
⇒ 2
1
1
tan , is in Quadrant IV.
3
11. z 3 3i
r 1 3 4 2
3
−2
z 10cos 5.96 i sin 5.96
10. z 1 3i
2
Real
axis
r 32 12 10
z 2cos i sin −2
9. z 3 i
r 02 32 9 3
z 3 cos
−4
−4
8. z 2
7. z 3i
−6
6 − 7i
−8
5 − 12i
− 12
z 2 cos
4
−4
−6
tan 85
Imaginary
axis
−6 −4 −2
−2
6. 8 3i 82 32
2
3 ⇒ 2
2
i sin
3
3
r 32 32 18 32
2
3
tan 3
7
1, is in Quadrant IV ⇒ .
3
4
7
7
i sin
4
4
1
2
z 32 cos
Imaginary
axis
Real
axis
3
−1
−2
−3
12. z 2 2i
13. z 3 i
r 3 12 4 2
r 22 22 8 22
tan 2
2
1 ⇒ 2
4
z 22 cos
3 − 3i
i sin
4
4
tan Imaginary
axis
1
3
z 2 cos
3
3
3
⇒ i sin
6
6
6
Imaginary
axis
2 + 2i
2
2
3+i
1
1
1
2
3
Real
axis
−1
1
−1
2
Real
axis
Section 8.5
15. z 21 3i
14. z 4 43i
r 22 232 16 4
r 42 43 2 8
43
5
3 ⇒ 4
3
tan z 8 cos
Trigonometric Form of a Complex Number
5
5
i sin
3
3
3
tan 1
z 4 cos
3, is in Quadrant III ⇒ 4
4
i sin
3
3
Imaginary
axis
Imaginary
axis
Real
axis
2
−2
2
4
6
−4
Real
axis
8
−3
−2
−1
−2
−2
−4
−3
−6
4−4
−8
16. z r
−2( 1 + 3i)
3i
5
3 i
2
17. z 5i
2
5
3
2
1
tan 3
5
1
2
2
3
11
⇒ 3
6
11
11
z 5 cos
i sin
6
6
100
25 5
4
r 02 52 25 5
5
3
, undefined ⇒ 0
2
tan z 5 cos
Imaginary
axis
3
3
i sin
2
2
Imaginary
axis
−4
−2
2
4
Real
axis
−2
2
1
−1
−1
2
3
4
−4
Real
axis
5
−5i
−6
−2
−8
−3
5
2
−4
(
3 − i)
19. z 7 4i
18. z 4i
r
−4
02
42
162
4
4
tan , undefined ⇒ 0
2
z 4 cos
i sin
2
2
Imaginary
axis
r 72 42 65
tan 4
, is in Quadrant II ⇒ 2.62.
7
z 65 cos 2.62 i sin 2.62
Imaginary
axis
−7 + 4 i
4
5
4
2
4i
3
−8
2
−6
−4
Real
axis
−2
−2
1
−3 − 2 − 1
−1
1
2
3
Real
axis
−4
4
.
3
773
774
Chapter 8
20. z 3 i
21. z 7 0i
r 32 12 10
tan 1
3
z 10 cos 5.96 i sin 5.96
Imaginary
axis
r 72 02 49 7
z 7 cos 0 i sin 0
Imaginary
axis
4
1
2
2
Real
axis
3
7
2
3−i
−1
4
6
Real
axis
8
−2
−4
−2
22. z 4
r
0
0 ⇒ 0
7
tan 23. z 3 3i
42
02
162
4
r 32 3 12
2
23
0
0 ⇒ 0
4
tan tan z 4cos 0 i sin 0
3
3
Imaginary
axis
⇒ z 23 cos
6
i sin
6
6
2
Imaginary
axis
1
4
1
2
3
4
Real
axis
4
3
−1
3+
2
−2
3i
1
−1
1
2
3
4
Real
axis
−1
25. z 3 i
24. z 22 i
r 222 12 9 3
tan 2
1
22
4
z 3cos 5.94 i sin 5.94
r 32 12 10
tan 1 1
, is in Quadrant III ⇒ 3.46.
3 3
z 10 cos 3.46 i sin 3.46
Imaginary
axis
Imaginary
axis
−4
1
2
−1
−2
3
Real
axis
2 2−i
−3
−3 − i
Real
axis
−2
−1
−2
−3
−4
Section 8.5
26. z 1 3i
Trigonometric Form of a Complex Number
27. z 5 2i
r 52 22 29
r 12 32 10
tan 31 3 ⇒ 1.25 radians
tan 25
0.38
z 10 cos 1.25 i sin 1.25
z 29cos 0.38 i sin 0.38
Imaginary
axis
Imaginary
axis
1 + 3i
3
5
2
4
1
3
−1
3
2
1
Real
axis
5 + 2i
2
1
−1
−1
−1
2
3
4
3
8
tan 53
8
3.97
0.36
z 73cos 0.36 i sin 0.36
z 139cos 3.97 i sin 3.97
Imaginary
axis
Imaginary
axis
6
− 10 − 8
−6
−4
Real
axis
−2
−2
8 + 3i
4
−4
2
2
4
6
8
−6
Real
axis
−8
−2
− 8 − 5 3i
−4
30. z 9 210i
r 92 2102 121
− 10
1 3
31. 3cos 120 i sin 120 3 i
2
2
3 33
i
2
2
r 11
tan 210
9
Imaginary
axis
4
3.75
−3 + 3 3 i
2
2
z 11cos 3.75 i sin 3.75
3
2
Imaginary
axis
− 10 − 8
Real
axis
r 82 53 2 139
r 82 32 73
−2
5
29. z 8 53i
28. z 8 3i
tan 1
−6
−4
Real
axis
−2
−2
−4
−6
− 9 − 2 10 i
−8
− 10
−3
−2
−1
1
−1
2
Real
axis
775
776
Chapter 8
32. 5cos 135 i sin 135 5 2
2
i
2
2 33.
3
3 1
3
i cos 300 i sin 300 2
2 2
2
52 52
i
2
2
Imaginary
axis
−
5 2 5 2
+
i
2
2
4
−1
1
−1
1
34.
−2
3 33
i
4
4
1
2
−3
Imaginary
axis
3
−4
Real
axis
−1
Real
axis
2
3 −3 3
i
4
4
−2
2
2
1
1
cos 225 i sin 225 i 4
4
2
2
2
8
i
35. 3.75 cos
3
3
152 152
i sin
i
4
4
8
8
2
8
Imaginary
axis
Imaginary
axis
− 15 2 + 15 2 i
8
8
3
2
−1
2
−4
−1
4
2
2 i
−
8
8
−
1
Real
axis
−1
4
−3
−2
Real
axis
−1
−1
−1
2
36. 6 cos
5
5
i sin
1.5529 5.7956i
12
12
37. 8 cos
i sin
80 i 8i
2
2
Imaginary
axis
Imaginary
axis
10
6
1.5529 + 5.7956i
8
4
6
2
4
−2
2
4
Real
axis
6
−2
8i
2
−2
−2
2
4
6
8
Real
axis
10
39. 3cos1845 i sin1845 2.8408 0.9643i
38. 7cos 0 i sin 0 7
Imaginary
axis
Imaginary
axis
4
2
2
2.8408 + 0.9643i
1
7
2
4
6
8
Real
axis
1
−2
−1
−4
−2
2
3
4
Real
axis
Section 8.5
Trigonometric Form of a Complex Number
777
40. 6cos230 30 i sin230 30 3.8165 4.6297i
Imaginary
axis
1
−5 −4 −3 −2 −1
1
Real
axis
−2
−3
−4
− 3.8165 − 4.6297i − 5
41. 5 cos
i sin
4.6985 1.7101i
9
9
42. 10 cos
43. 3cos 165.5 i sin 165.5 2.9044 0.7511i
45. z 2
2
1 i cos 45 i sin 45
cos 135 i sin 135 46. z 2
2
1 i
r
−2
z4
= −1
2 1
2
z 1 cos
3
2
1
(
3i (
z=
z 3 = −1
−2
z4 =
−1
1
−1 −
2
(
1
1+
2
(
1
3i (
Real
axis
3i (
1 3
i sin
i
3
3
2
2
z2 12 cos
2
(1 + i)
2
1
−1 +
2
−2
z2 = i
1
2
z2 =
3
Imaginary
axis
z=
Imaginary
axis
tan 3
The absolute value of each is 1, and consecutive powers
of z are each 45 apart.
2
z3 =
(−1 + i)
2
1
1 3i
2
zn r ncos n i sin n
z4 cos 180 i sin 180 1
2
44. 9cos 58 i sin 58 4.7693 7.6324i
z2 cos 90 i sin 90 i
z3
2
2
3.0902 9.5106i
i sin
5
5
2
1 3
2
i sin
i
3
3
2
2
z3 13cos i sin 1
Real
axis
z4 14 cos
−1
4
4
1 3
i sin
i
3
3
2
2
The absolute value of each is 1 and consecutive powers of z
47.
2cos 4 i sin 4 6cos 12 i sin 12 26cos4 12 i sin4 12 12 cos
48.
34 cos 3 i sin 3 4cos 34 i sin 34 344cos 3 34 i sin3 34
3 cos
49.
i sin
3
3
13
13
i sin
12
12
53cos 140 i sin 140º 23cos 60 i sin 60 53 23 cos140 60 i sin140 60
10
9 cos 200 i sin 200
778
Chapter 8
50. 0.5cos 100 i sin 1000.8cos 300 i sin 300 0.50.8cos100 300 i sin100 300
0.4cos 400 i sin 400
0.4cos 40 i sin 40
51. 0.45cos 310 i sin 310 0.60cos 200 i sin 200 0.450.60cos310 200 i sin310 200
0.27cos 510 i sin 510
0.27cos 150 i sin 150
52. cos 5 i sin 5cos 20 i sin 20 cos5 20 i sin5 20 cos 25 i sin 25
53.
cos 50 i sin 50
cos50 20 i sin50 20 cos 30 i sin 30
cos 20 i sin 20
54.
2cos 120 i sin 120 2
cos120 40 i sin120 40
4cos 40 i sin 40
4
1
cos 80 i sin 80
2
5
5
i sin
3
3
5
5
2
2
cos
i sin
cos
i sin
cos i sin 3
3
3
3
cos
55.
56.
5cos 4.3 i sin4.3 5
cos4.3 2.1 i sin4.3 2.1
4cos 2.1 i sin2.1 4
5
cos2.2 i sin2.2
4
57.
12cos 52 i sin 52
4cos52 110 i sin52 110
3cos 110 i sin 110
4cos58 i sin(58
4cos 302 i sin 302
58.
6cos 40 i sin 40
6
cos 40 100 i sin40 100
7cos 100 i sin 100 7
6
cos 300 i sin 300
7
59. (a) 2 2i 22 cos
i sin
4
4
4 i sin 4 2cos
(a) 1 i 2 cos (b) 2 2i1 i 22 cos
i sin
4
4
7
7
i sin
4
4
7
7
2cos 4 i sin 4 4cos 2 i sin 2
4cos 0 i sin 0 4
(c) 2 2i1 i 2 2i 2i 2i2 2 2 4
Section 8.5
Trigonometric Form of a Complex Number
60. (a) 3 i 2cos 30 i sin 30
1 i 2cos 45 i sin 45
(b) 3 i1 i 2cos 30 i sin 302cos 45 i sin 45
22cos 75 i sin 75
22
6 2
4
6 2
4
i
3 1 3 1 i 0.732 2.732i
(c) 3 i1 i 3 3 1 i i2 3 1 3 1 i 0.732 2.732i
61. (a)
2 i sin 2 2cos
2i 2 cos (a) 1 i 2 cos
i sin
4
4
3
3
i sin
2
2
3
3
2 i sin 2 2cos 4 i sin 4 (b) 2i1 i 2 cos
7
7
4 i sin 4 (b)
22 cos
(b)
22
2 2i 2 2i
1
1
(c) 2i1 i 2i 2i2 2i 2 2 2i
62. (a)
3 i sin 3 (b) 41 3i 8 cos 4 4cos 0 i sin 0
3 i sin3
1 3i 2 cos 8
(c) 41 3i 4 43i
63. (a)
5
5
i sin
3
3
3 4i
3 4i
(c)
1 3i 1 3i
4 43i
3 4i 5cos 0.93 i sin 0.93
1 3 i 2 cos
12 i 23
(b)
3 4i
5cos 0.93 i sin 0.93
5
1 3 i
5
i sin
2 cos
3
3
2.5cos4.31 i sin4.31
1 3 i
1 3 i
5
cos 1.97 i sin 1.97
2
3 4 33 i 43i2
13
0.982 2.299i
3 43 4 33
i
4
4
0.982 2.299i
64. (a) 1 3i 2 cos
i sin
3
3
6 3i 35 cos0.464 i sin0.464
(b)
1 3i
2
cos
0.464 i sin
0.464
6 3i
3
3
35
(c)
1 3i
6 3i
6 3i
6 3i 2155 cos 1.51 i sin 1.51 0.018 0.298i
6 33 i 3 63 2 3 i 1 23 0.018 0.298i
45
15
15
779
780
Chapter 8
65. (a) 5 5cos 0 i sin 0
(a) 2 3i 13cos 0.98 i sin 0.98
(b)
5
5cos 0 i sin 0
5
5
cos0.98 i sin0.98 cos 5.30 i sin 5.30 0.769 1.154i
13
2 3i 13cos 0.98 i sin 0.98 13
(c)
5
5
2 3i 2 3i
2 3i
2 3i 10 15i 10 15
i 0.769 1.154i
13
13 13
4i 4cos 90 i sin 90
66. (a)
(b)
4 2i 25cos 153.4 i sin 153.4
4i
4i
(c)
4 2i 4 2i
4cos 90 i sin 90
4i
4 2i 25cos 153.4 i sin 153.4
0.400 0.800i
8 16i 2 4
i 0.400 0.800i
20
5 5
67. Let z x iy such that:
z
2
⇒ 2
x2
⇒ 4 circle with radius of 2
x2
25
cos 296.6 i sin 296.6
5
4 i
4 2i
68. z 3
Imaginary
axis
y2
3
Imaginary
axis
y2:
4
1
−1
1
−1
3
Real
axis
2
1
−2 −1
1
2
Real
axis
4
−2
−3
−4
69. Let .
6
Imaginary
axis
Since r ≥ 0, we have the
portion of the line 6
70. 4
2
−4
2
4
Real
axis
5
4
Imaginary
axis
Since r ≥ 0, we have the
portion of the line 54
2
1
−3
−2
−1
−2
−2
−4
71. 1 i5 2 cos
i sin
4
4
2 cos
5
42 4 4i
2
2
i
−3
5
5
5
i sin
4
4
2
2
72. 2 2i6 22 cos
−1
i sin
4
4
22 cos
6
512 cos
512i
6
6
6
i sin
4
4
3
3
i sin
2
2
1
2
Real
axis
Section 8.5
3
3
i sin
4
4
73. 1 i10 2 cos
3
2
32 cos
10
6 i sin
2
3
2
10
6
cos
Trigonometric Form of a Complex Number
30
30
i sin
4
4
32cos 32 i sin 32
320 i1 32i
74. 3 2i8 13 cosarctan23 i sinarctan23 8
13 cos8 arctan23 i sin8 arctan23 8
239 28,560i
75. 23 i 2 2 cos
7
i sin
6
6
2 27 cos
256 7
7
7
i sin
6
6
3
2
1
i
2
76. 41 3i 4 2 cos
3
5
5
i sin
3
3
3
423cos 5 i sin 5
321
32
1283 128i
77. 5cos 20 i sin 203 53cos 60 i sin 60 125 1253
i
2
2
78. 3cos 150 i sin 1504 34cos 600 i sin 600
79.
cos 4 i sin 4 12
cos
12
12
i sin
4
4
81cos 240 i sin 240
cos 3 i sin 3
81cos 60 i sin 60
1
81 813
i
2
2
80.
2cos 2 i sin 2 8
28cos 4 i sin 4
81. 5cos 3.2 i sin 3.24 54cos 12.8 i sin 12.8
608.02 144.69i
256cos 0 i sin 0
256
82. cos 0 i sin 020 cos 0 i sin 0
83. 3 2i5 3.6056cos0.588 i sin0.5885
3.60565cos2.94 i sin2.94
1
597 122i
84. 5 4i3 21cos1.06106 i sin1.06106
3
21 3cos31.06106 i sin31.06106
96.15 4.00i
85. 3cos 15 i sin 154 81cos 60 i sin 60
81 813
i
2
2
86. 2cos 10 i sin 108 256cos 80 i sin 80
44.45 252.11i
781
782
87.
Chapter 8
2 cos
i sin
10
10
5
25 cos
i sin
2
2
88.
2cos 8 i sin 8 6
89. (a) Square roots of 5cos 120 i sin 120:
120 360k
120 360k
i sin
2
2
(b)
, k 0, 1
3
1
−3
(a) k 1: 5cos 240 i sin 240
5
2
15
2
i, 5
2
15
2
−1
(b)
16 cos
6
k 0: 4cos 30 i sin 30
2
−6
k 1: 4cos 210 i sin 210
−2
(c) 23 2i, 23 2i
3 8 cos
2
2
i sin
:
3
3
2
2
k 0: 2 cos
i sin
9
9
8
8
i sin
9
9
14
14
i sin
9
9
k 2: 2 cos
Real
axis
2
6
Real
axis
3
Real
axis
−6
(b)
Imaginary
axis
23 2k
23 2k
i sin
, k 0, 1, 2
3
3
k 1: 2 cos
3
Imaginary
axis
60 2k 360 i sin60 2k 360, k 0, 1
(c)
1
−3
i
90. (a) Square roots of 16cos 60 i sin 60:
91. (a) Cube roots of 8 cos
Imaginary
axis
(a) k 0: 5cos 60 i sin 60
(c)
3
3
i sin
4
4
322 322i
32i
(a) 5 cos
64 cos
3
1
−3
−1
−1
1
−3
1.5321 1.2856i, 1.8794 0.6840i, 0.3473 1.9696i
92. (a) Fifth roots of 32 cos
5
5
i sin
:
6
6
(b)
Imaginary
axis
565 2k i sin565 2k, k 0, 1, 2, 3, 4
3
5 32 cos
i sin
6
6
17
17
i sin
30
30
29
29
i sin
30
30
41
41
i sin
30
30
53
53
i sin
30
30
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
k 3: 2 cos
k 4: 2 cos
−3
3
Real
axis
−3
(c) 3 i, 0.4158 1.9563i,
1.9890 0.2091i, 0.8135 1.8271i,
1.4863 1.3383i
Section 8.5
3
3
:
i sin
2
2
93. (a) Square roots of 25i 25 cos
3
2k
2
(a) 25 cos
2
3
3
i sin
4
4
7
7
i sin
4
4
(a) k 0: 5 cos
(a) k 1: 5 cos
(c) 3
2k
2
i sin
2
Trigonometric Form of a Complex Number
(b)
Imaginary
axis
6
4
2
, k 0, 1
−6
−2
−2
2
4
6
Real
axis
4
6
Real
axis
4
6
Real
axis
−4
−6
52 52 52 52
i,
i
2
2
2
2
94. (a) Fourth roots of 625i 625 cos
i sin :
2
2
(b)
2k
2
4
625 cos
4
Imaginary
axis
2k
2
i sin
2
6
−6
k 0, 1, 2, 3
−4
−6
i sin
8
8
5
5
i sin
8
8
9
9
i sin
8
8
13
13
i sin
8
8
k 0: 5 cos
k 1: 5 cos
k 2: 5 cos
k 3: 5 cos
−2
2
(c) 4.6194 1.9134i, 1.9134 4.6194i,
4.6194 1.9134i, 1.9134 4.6194i
95. (a) Cube roots of 125
4
4
1 3i 125 cos
i sin
:
2
3
3
4
2k
3
3
(a) 125 cos
3
(a) k 0: 5 cos
4
2k
3
i sin
3
4
4
i sin
9
9
2
, k 0, 1, 2
−6
−2
−6
16
16
i sin
9
9
Imaginary
axis
−4
(b)
6
10
10
(a) k 1: 5 cos
i sin
9
9
(a) k 2: 5 cos
(c) 0.8682 4.9240i, 4.6985 1.7101i, 3.8302 3.2140i
783
784
Chapter 8
96. (a) Cube roots of 421 i 8 cos
3
2k
4
3
8 cos
3
k 0: 2 cos
i sin
4
4
(b)
19
19
i sin
12
12
Imaginary
axis
3
, k 0, 1, 2
−3
3
2k
4
i sin
3
11
11
k 1: 2 cos
i sin
12
12
k 2: 2 cos
3
3
i sin
:
4
4
Real
axis
3
−3
(c) 2 2i, 1.9319 0.5176i, 0.5176 1.9319i
97. (a) Fourth roots of 16 16cos 0 i sin 0:
4 16 cos
(b)
Imaginary
axis
0 2k
0 2k
i sin
, k 0, 1, 2, 3
4
4
3
k 0: 2cos 0 i sin 0
k 1: 2 cos i sin
2
2
1
−3
k 2: 2cos i sin k 3: 2 cos
3
3
i sin
2
2
−1
1
3
Real
axis
−1
−3
(c) 2, 2i, 2, 2i
98. (a) Fourth roots of i cos
2k
2
4 1 cos
4
k 0: cos
i sin :
2
2
2k
2
i sin
4
i sin
8
8
5
5
k 1: cos
i sin
8
8
k 2: cos
9
9
i sin
8
8
k 3: cos
13
13
i sin
8
8
(c) 0.9239 0.3827i, 0.3827 0.9239i,
0.9239 0.3827i, 0.3827 0.9239i
(b)
Imaginary
axis
2
, k 0, 1, 2, 3
−2
2
−2
Real
axis
Section 8.5
99. (a) Fifth roots of 1 cos 0 i sin 0:
(a) cos
2k
Trigonometric Form of a Complex Number
(b)
Imaginary
axis
2k
5 i sin 5 , k 0, 1, 2, 3, 4
2
(a) k 0: cos 0 i sin 0
(a) k 1: cos
−2
2
2
i sin
5
5
Real
axis
2
−2
4
4
i sin
(a ) k 2: cos
5
5
( a) k 3: cos
6
6
i sin
5
5
(a) k 4: cos
8
8
i sin
5
5
(c) 1, 0.3090 0.9511i, 0.8090 0.5878i, 0.8090 0.5878i, 0.3090 0.9511i
100. (a) Cube roots of 1000 1000cos 0 i sin 0:
3 1000 cos
2k
2k
i sin
3
3
(b)
Imaginary
axis
8
6
4
k 0, 1, 2
−8 −6 −4 −2
k 0: 10cos 0 i sin 0
2
2
i sin
3
3
4
4
i sin
3
3
k 1: 10 cos
k 2: 10 cos
Real
axis
2 4 6 8
−6
−8
(c) 10, 5 53i, 5 53i
101. (a) Cube roots of 125 125cos i sin :
(b)
Imaginary
axis
32k i sin 32k, k 0, 1, 2
6
3
(a) 125 cos
(a) k 0: 5 cos i sin
3
3
2
−6
(a) k 1: 5cos i sin 5
5
(a) k 2: 5 cos
i sin
3
3
(c)
5 53
5 53
i, 5, i
2
2
2
2
4
−2
2
−4
−6
4
6
Real
axis
785
786
Chapter 8
102. (a) Fourth roots of 4 4cos i sin :
4 4 cos
2k
2k
i sin
4
4
(b)
Imaginary
axis
2
1
k 0, 1, 2, 3
i sin
4
4
3
3
i sin
4
4
5
5
i sin
4
4
7
7
i sin
4
4
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
k 3: 2 cos
−2
3
3
3
3
i sin
2152 cos
i sin
4
4
4
4
11
11
i sin
20
20
19
19
i sin
20
20
(b)
27
27
i sin
(a) k 3: 22 cos
20
20
7
7
i sin
(a) k 4: 22 cos
4
4
Imaginary
axis
1
−2
−1
1
Real
axis
2
−2
(c) 2.5201 1.2841i, 0.4425 2.7936i,
2.7936 0.4425i, 1.2841 2.5201i, 2 2i
i sin :
2
2
104. (a) Sixth roots of 64i 64 cos
2k
2
6
64 cos
6
(a) k 2: 22 cos
3
2k
4
, k 0, 1, 2, 3, 4
i sin
5
3
3
i sin
20
20
(a) k 1: 22 cos
Real
axis
(c) 1 i, 1 i, 1 i, 1 i
(a) k 0: 22 cos
2
−2
232
1
−1
103. (a) Fifth roots of 1281 i 1282 cos
3
2k
4
cos
5
−1
(b)
Imaginary
axis
2k
2
i sin
6
3
1
−3
1
3
Real
axis
k 0, 1, 2, 3, 4, 5
i sin
12
12
5
5
i sin
12
12
3
3
i sin
4
4
13
13
i sin
12
12
17
17
i sin
12
12
7
7
i sin
4
4
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
k 3: 2 cos
k 4: 2 cos
k 5: 2 cos
−3
(c) 1.9319 0.5176i, 0.5176 1.9319i, 2 2 i,
1.9319 0.5176i, 0.5176 1.9319i, 2 2i
Section 8.5
Trigonometric Form of a Complex Number
105. x4 i 0
x4 i
The solutions are the fourth roots of i cos
3
2k
2
4 1 cos
4
3
2k
2
i sin
4
3
3
i sin :
2
2
, k 0, 1, 2, 3
k 0: cos
3
3
i sin
0.3827 0.9239i
8
8
k 1: cos
7
7
i sin
0.9239 0.3827i
8
8
k 2: cos
11
11
i sin
0.3827 0.9239i
8
8
k 3: cos
15
15
i sin
0.9239 0.3827i
8
8
Imaginary
axis
1
2
Real
axis
1
−2
106. x3 1 0
Imaginary
axis
x 1
3
2
The solutions are the cube roots of
1 cos i sin :
2k
2k
cos
i sin
3
3
−2
−2
k 0, 1, 2
k 0: cos
Real
axis
2
1 3
i sin i
3
3
2
2
k 1: cos i sin 1
k 2: cos
5
5 1 3
i sin
i
3
3
2
2
107. x5 243 0
x5 243
The solutions are the fifth roots of 243 243cos i sin :
5 243 cos
2k
2k
i sin
5
5
, k 0, 1, 2, 3, 4
i sin
2.4271 1.7634i
5
5
3
3
i sin
0.9271 2.8532i
5
5
k 0: 3 cos
k 1: 3 cos
7
7
i sin
0.9271 2.8532i
k 3: 3 cos
5
5
k 4: 3 cos
4
k 2: 3cos i sin 3
Imaginary
axis
9
9
i sin
2.4271 1.7634i
5
5
−4
−2
2
−4
4
Real
axis
787
788
Chapter 8
108. x 3 27 0
Imaginary
axis
x 3 27
4
The solutions are the cube roots of 27 27cos 0 i sin 0:
3 27
2
2k
2k
cos
i sin
3
3
−4
−2 −1
1
k 0, 1, 2
−2
k 0: 3cos 0 i sin 0 3
−4
2
3 33
2
i sin
i
3
3
2
2
4
4
3 33
i sin
i
3
3
2
2
k 1: 3 cos
k 2: 3 cos
2
4
Real
axis
109. x4 16i 0
x4 16i
The solutions are the fourth roots of 16i 16 cos
3
3
i sin
:
2
2
3
3
2k
2 k
2
2
4 16 cos
i sin
, k 0, 1, 2, 3
4
4
Imaginary
axis
3
3
i sin
0.7654 1.8478i
8
8
3
7
7
i sin
1.8478 0.7654i
8
8
1
11
11
i sin
0.7654 1.8478i
8
8
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
−3
−1
3
Real
axis
−3
15
15
i sin
1.8478 0.7654i
k 3: 2 cos
8
8
110. x6 64i 0
x6
Imaginary
axis
64i
3
3
3
i sin
:
The solutions are the sixth roots of 64i 64 cos
2
2
3
2 k
2
6
64 cos
6
3
2 k
2
i sin
6
k 0, 1, 2, 3, 4, 5
i sin
2 2 i
4
4
7
7
i sin
0.5176 1.9319i
12
12
11
11
i sin
1.9319 0.5176i
12
12
5
5
i sin
2 2 i
4
4
19
19
i sin
0.5176 1.9319i
12
12
23
23
i sin
1.9319 0.5176i
12
12
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
k 3: 2 cos
k 4: 2 cos
k 5: 2 cos
1
−3
−1
−1
−3
1
3
Real
axis
Section 8.5
Trigonometric Form of a Complex Number
789
111. x3 1 i 0
x3 1 i 2 cos
7
7
i sin
4
4
The solutions are the cube roots of 1 i:
7
2k
4
3 2 cos
3
7
2k
4
i sin
3
7
, k 0, 1, 2
Imaginary
axis
7
12 i sin 12 0.2905 1.0842i
5
5
k 1: 2cos
i sin 0.7937 0.7937i
4
4
6 2 cos
k 0: 2
6
6
k 2: 2 cos
−2
23
23
i sin
1.0842 0.2905i
12
12
2
Real
axis
−2
112. x4 1 i 0
Imaginary
axis
x4 1 i 2cos 225 i sin 225
2
The solutions are the fourth roots of 1 i:
4 2 cos
225 360k
225 360k
i sin
4
4
−2
k 0, 1, 2, 3
2
Real
axis
−2
8 2cos 56.25 i sin 56.25 0.6059 0.9067i
k 0: 8 2cos 146.25 i sin 146.25 0.9067 0.6059i
k 1: 8 2cos 236.25 i sin 236.25 0.6059 0.9067i
k 2: 8 2cos 326.25 i sin 326.25 0.9067 0.6059i
k 3: 113. True, by the definition of the absolute value of
a complex number.
114. False. They are equally spaced along the circle centered
n r.
at the origin with radius 115. True. z1z2 r1r2cos1 2 i sin1 2 andz1z2 0 if and only if r1 0 and/or r2 0.
116. False. The complex number must be converted to trigonometric form before applying DeMoivre’s Theorem.
4 6i 8 22 cosarctan
117.
z1 r1cos 1 i sin 1
z2 r2cos 2 i sin 2
6
4
i sinarctan 46
8
cos i sin cos 2 i sin 2
2
2
r1
cos 1 cos 2 sin 1 sin 2 isin 1 cos 2 sin 2 cos 1
2
r2cos 2 sin2 2
r1
cos1 2 i sin1 2
r2
790
Chapter 8
118. z r cos i sin 119. (a) zz r cos i sin r cos i sin r cos i sin (a)
r2cos i sin r cos ir sin (a)
r2cos 0 i sin 0
(a)
r2
which is the complex conjugate of
r cos i sin r cos ir sin .
z
rcos i sin
z rcos i sin r
(b) cos i sin r
(b)
cos 2 i sin 2
(b)
120.
z r cos i sin 121.
z r cos i sin 1
1 3i cos 43 i sin 43
2
21 1 3i cos 43 i sin 43
6
r cos i sin r cos i sin 6
cos 8 i sin 8
1
2141 i 214 2 cos
122.
214 cos
2 cos 74 i sin 74
4
14
7
7
i sin
4
4
7
7
i sin
4
4
2144cos 7 i sin 7 2cos i sin 2
123. (a) 2cos 30 i sin 30
124. (a) 3cos 45 i sin 45
2cos 150 i sin 150
3cos 135 i sin 135
2cos 270 i sin 270
3cos 225 i sin 225
(b) These are the cube roots of 8i.
3cos 315 i sin 315
(b) These are the fourth roots of 81.
(c) The fourth roots of 81:
Imaginary
axis
4
−4
4
−4
Real
axis
Section 8.5
Trigonometric Form of a Complex Number
126. B 66, a 33.5
125. A 22, a 8
A 90 66 24
B 90 A 68
tan 22 8
8
19.80
⇒ b
b
tan 22
b
a sin B 33.5 sin 66
75.24
sin A
sin 24
sin 22 8
8
⇒ c
21.36
c
sin 22
c
a sin C 33.5 sin 90
82.36
sin A
sin 24
128. B 6, b 211.2
127. A 30, b 112.6
A 90 6 84
B 90 A 60
tan 30 a
⇒ a 112.6 tan 30 65.01
112.6
a
b sin A 211.2 sin 84
2009.43
sin B
sin 6
cos 30 112.6
112.6
⇒ c
130.02
c
cos 30
c
b sin C 211.2 sin 90
2020.50
sin B
sin 6
130. B 81 30, c 6.8
129. A 4215 42.25, c 11.2
A 90 81 30 8 30
B 90 A 4745
sin 42.25 a
⇒ a 11.2 sin 42.25 7.53
11.2
a
c sin A 6.8 sin 8 30
1.01
sin C
1
cos 42.25 b
⇒ b 11.2 cos 42.25 8.29
11.2
b
c sin B 6.8 sin 81 30
6.73
sin C
1
131. d 16 cos t
4
132. d Maximum displacement: 16 16
16 cos
t0 ⇒ t
⇒ t2
4
4
2
1
133. d 16
sin54 t
Maximum displacement:
1
16
sin54 t 0
5
4 t
791
1
16
1
cos 12 t
8
Maximum displacement:
d 0 when 12t 1
8
1
, or t 2
24
1
134. d 12
sin 60 t
1
16
Maximum displacement:
1
12
1
d 0 when 60 t , or t 60
4
t5
135. 6 sin 8 cos 3 612 sin8 3 sin8 3
3sin 11 sin 5
136. 2 cos 5 sin 2 2 12sin5 2 sin5 2
sin 7 sin 3
792
Chapter 8
Review Exercises for Chapter 8
1. Given: A 35, B 71, a 8
C 180 35 71 74
2. Given: A 22, B 121, a 17
C 180 A B 37
b
a sin B 8 sin 71
13.19
sin A
sin 35
b
a sin B 17 sin 121
38.90
sin A
sin 22
c
a sin C 8 sin 74
13.41
sin A
sin 35
c
a sin C 17 sin 37
27.31
sin A
sin 22
3. Given: B 72, C 82, b 54
A 180 72 82 26
4. Given: B 10, C 20, c 33
A 180 B C 150
a
b sin A 54 sin 26
24.89
sin B
sin 72
a
c sin A 33 sin 150
48.24
sin C
sin 20
c
b sin C 54 sin 82
56.23
sin B
sin 72
b
c sin B 33 sin 10
16.75
sin C
sin 20
5. Given: A 16, B 98, c 8.4
C 180 16 98 66
6. Given: A 95, B 45, c 104.8
C 180 A B 40
a
c sin A 8.4 sin 16
2.53
sin C
sin 66
a
c sin A 104.8 sin 95
162.42
sin C
sin 40
b
c sin B 8.4 sin 98
9.11
sin C
sin 66
b
c sin B 104.8 sin 45
115.29
sin C
sin 40
7. Given: A 24, C 48, b 27.5
B 180 24 48 108
8. Given: B 64, C 36, a 367
A 180 B C 80
a
b sin A 27.5 sin 24
11.76
sin B
sin 108
b
a sin B 367 sin 64
334.95
sin A
sin 80
c
b sin C 27.5 sin 48
21.49
sin B
sin 108
c
a sin C 367 sin 36
219.04
sin A
sin 80
9. Given: B 150, b 30, c 10
sin C c sin B 10 sin 150
0.1667 ⇒ C 9.59
b
30
A 180 150 9.59 20.41
b sin A 30 sin 20.41
a
20.92
sin B
sin 150
11. A 75, a 51.2, b 33.7
sin B b sin A 33.7 sin 75
0.6358 ⇒ B 39.48
a
51.2
C 180 75 39.48 65.52
c
a sin C 51.2 sin 65.52
48.24
sin A
sin 75
10. Given: B 150, a 10, b 3
sin A a sin B 10 sin 150
1.67 > 1
b
3
No solution
Review Exercises for Chapter 8
793
12. Given: B 25, a 6.2, b 4
sin A a sin B
0.65506 ⇒ A 40.92 or 139.08
b
Case 1: A 40.92
Case 2: A 139.08
C 180 25 40.92 114.08
C 180 25 139.08 15.92
c 8.64
c 2.60
13. Area 12bc sin A 1257sin 27 7.9
14. B 80º, a 4, c 8
Area 12ac sin B 12480.9848 15.8
1
1
15. Area 2ab sin C 2165sin 123 33.5
16. A 11, b 22, c 21
Area 12 bc sin A 12 22210.1908 44.1
17. tan 17 h
⇒ h x 50 tan 17
x 50
h
h x tan 17 50 tan 17
tan 31 31°
x
17°
50
h
⇒ h x tan 31
x
x tan 17 50 tan 17 x tan 31
50 tan 17 xtan 31 tan 17
50 tan 17
x
tan 31 tan 17
x 51.7959
h x tan 31 51.7959 tan 31 31.1 meters
The height of the building is approximately 31.1 meters.
18. 162 w2 122 2w12 cos 140
w2 24 cos 140w 112 0 ⇒ w 4.83
19.
h
75
sin 17 sin 45
75 sin 17
h
sin 45
45°
118°
h 31.01 feet
ft
75
62°
17°
28°
20. The triangle of base 400 feet formed by the two angles of sight to the tree has base angles
of 90 22 30 67 30, or 67.5, and 90 15 75. The angle at the tree measures
180 67.5 75 37.5.
400 sin 75
b
634.683
sin 37.5
h 634.683 sin 67.5
h 586.4
The width of the river is about 586.4 feet.
45°
Tree
A
N
W
E
S
15°
h
22° 30'
C
400 ft
B
h
794
Chapter 8
21. Given: a 5, b 8, c 10
a2
cos C 2ab
b2
c2
22. Given: a 80, b 60, c 100
0.1375 ⇒ C 97.90
a2 c2 b2
0.61 ⇒ B 52.41
2ac
cos B a2 b2 c2 6400 3600 10,000
2ab
28060
cos C 0 ⇒ C 90
A 180 B C 29.69
sin A 80
0.8 ⇒ A 53.13º
100
sin B 60
0.6 ⇒ B 36.87º
100
23. Given: a 2.5, b 5.0, c 4.5
cos B a2 c2 b2
0.0667 ⇒ B 86.18
2ac
cos C a2 b2 c2
0.44 ⇒ C 63.90
2ab
A 180 B C 29.92
24. Given: a 16.4, b 8.8, c 12.2
cos A b2 c2 a2 8.82 12.22 16.42
0.1988 ⇒ A 101.47
2bc
28.812.2
sin B b sin A 8.8 sin 101.47
0.5259 ⇒ B 31.73
a
16.4
C 180 101.47 31.73 46.80
25. Given: B 110, a 4, c 4
b
a2
c2
26. Given: B 150, a 10, c 20
2ac cos B 6.55
A C 12 180 110 35
b2 102 202 21020cos 150 ⇒ b 29.09
sin A a sin B 10 sin 150
⇒ A 9.90
b
29.09
C 180 150 9.90 20.10
27. Given: C 43, a 22.5, b 31.4
c a2 b2 2ab cos C 21.42
cos B a2 c2 b2
0.02169 ⇒ B 91.24
2ac
A 180 B C 45.76
28. Given: A 62, b 11.34, c 19.52
a2 11.342 19.522 211.3419.52 cos 62 ⇒ a 17.37
sin B b sin A 11.34 sin 62
⇒ B 35.20
a
17.37
C 180 62 35.20 82.80
Review Exercises for Chapter 8
29.
5 ft
8 ft
8 ft
28°
5 ft
152°
30.
795
15 m
a
20 m
b
20 m
34°
15 m
146°
s1
s2
a2 52 82 258cos 28 18.364
a 4.3 feet
b2 82 52 285cos 152 159.636
b 12.6 feet
s12 152 202 2 15
20 cos 34 127.58
s1 11.3 meters
s22 15 2 202 2 15
20 cos 146 1122.42
s2 33.5 meters
31. Length of AC 3002 4252 2300425 cos 115
32. d 2 8502 10602 28501060 cos 72
1,289,251
615.1 meters
d 1135 miles
N
W
d
5°
E
S
850
67°
33. a 4, b 5, c 7
s
1060
34. a 15, b 8, c 10
abc 457
8
2
2
s
15 8 10
16.5
2
Area 16.51.58.56.5 36.979
Area ss as bs c
8431 9.80
35. a 12.3, b 15.8, c 3.7
s
36. a 38.1, b 26.7, c 19.4
a b c 12.3 15.8 3.7
15.9
2
2
Area ss as bs c
s
38.1 26.7 19.4
42.1
2
Area 42.1415.422.7 242.630
15.93.60.112.2 8.36
37. u 4 22 6 12 61
v 6 02 3 22 61
38. u 3 12 2 42 210
v 1 32 4 22 210
u is directed along a line with a slope of
61
5
.
4 2 6
u is directed along a line with a slope of
2 4
3.
31
v is directed along a line with a slope of
3 2 5
.
60
6
v is directed along a line with a slope of
4 2
3.
1 3
Since u and v have identical magnitudes and directions,
u v.
39. Initial point: 5, 4
Since u and v have identical magnitudes and directions,
u v.
40. Initial point: 0, 1
Terminal point: 2, 1
7
Terminal point: 6, 2 v 2 5, 1 4 7, 5
v 6 0, 72 1 6, 52
796
Chapter 8
41. Initial point: 0, 10
42. Initial point: 1, 5
Terminal point: 7, 3
Terminal point: 15, 9
v 7 0, 3 10 7, 7
v 15 1, 9 5 14, 4
43. v 8, 120
8 cos 120, 8 sin 120 4, 43
45. u 1, 3, v 3, 6
1
44. v , 225
2
12 cos 225, 21 sin 225 42, 42
46. u 4, 5, v 0, 1
(a) u v 1, 3 3, 6 4, 3
(a) u v 4 0, 5 1 4, 4
(b) u v 1, 3 3, 6 2, 9
(b) u v 4 0, 5 1 4, 6
(c) 3u 31, 3 3, 9
(c) 3u 34, 35 12, 15
(d) 2v 5u 23, 6 51, 3
(d) 2v 5u 20, 21 54, 55
6, 12 5, 15 11, 3
47. u 5, 2, v 4, 4
0 20, 2 25 20, 23
48. u 1, 8, v 3, 2
(a) u v 5, 2 4, 4 1, 6
(a) u v 1 3, 8 2 4, 10
(b) u v 5, 2 4, 4 9, 2
(b) u v 1 3, 8 2 2, 6
(c) 3u 35, 2 15, 6
(c) 3u 31, 38 3, 24
(d) 2v 5u 24, 4 55, 2
(d) 2v 5u 23, 22 51, 58
8, 8 25, 10 17,18
49. u 2i j, v 5i 3j
6 5, 4 40 11, 44
50. u 7i 3j, v 4i j
(a) u v 2i j 5i 3j 7i 2j
(a) u v 7i 3j 4i j 3i 4j
(b) u v 2i j 5i 3j 3i 4j
(b) u v 7i 3j 4i j 11i 2j
(c) 3u 32i j 6i 3j
(c) 3u 37i 3j 21i 9j
(d) 2v 5u 25i 3j 52i j
(d) 2v 5u 8i 2j 35i 15j
10i 6j 10i 5j 20i j
51. u 4i, v i 6j
27i 17j
52. u 6j, v i j
(a) u v 4i i 6j 3i 6j
(a) u v 6j i j i 5j
(b) u v 4i i 6j 5i 6j
(b) u v 6j i j i 7j
(c) 3u 34i 12i
(c) 3u 18j
(d) 2v 5u 2i 6j 54i
(d) 2v 5u 2i 2j 30j
2i 12j 20i 18i 12j
2i 28j
Review Exercises for Chapter 8
53. u 6i 5j, v 10i 3j
54. u 6i 5j, v 10i 3j
2u v 26i 5j 10i 3j
22i 7j
4u 5v 24i 20j 50i 15j
26i 35j
y
v
2
22, 7
−5
−2
y
26,35
x
10
20
20 25 30
− 60
−4
− 40
x
20
−5v
4u
−6
−8
2u + v
2u
4u − 5v − 40
− 10
− 60
− 12
55.
797
v 10i 3j
56. v 10i 3j
y
3v 310i 3j
1
2v
20
30i 9j
y
5i 32j 5, 32
8
6
10
30, 9
3v
4
v
v
2
x
10
20
30
1
v
2
x
2
− 10
57. u 3, 4 3i 4j
58. u 6, 8 6i 8j
59. Initial point: 3, 4
60. Initial point: 2, 7
6
Terminal point: 5, 9
u 9 3i 8 4j 6i 4j
u 5 2, 9 7 7, 16 7i 16j
62. v 4i j
v 10 10 200 102
2
tan 2
10
1 ⇒ 135 since
10
v 102i cos 135 j sin 135
v 7cos 60 i sin 60 j
v 42 12 17
tan 1
, in Quadrant IV ⇒ 346
4
v 17cos 346 i sin 346 j
64. v 3cos 150i sin 150 j
v 3, 150
v 7
60
65.
66. v 4i 7j
v 5i 4j
v 52 42 41
tan 67.
8
Terminal point: 9, 8
61. v 10i 10j
63.
4
−2
4
⇒ 38.7
5
v 3i 3j
tan tan 7
, in Quadrant II ⇒ 119.7
4
68. v 8i j
v 3 3 32
2
v 42 72 65
2
3
1 ⇒ 225
3
v 82 12 65
tan 1
, in Quadrant IV ⇒ 352.9
8
10
798
Chapter 8
69. Magnitude of resultant:
70. Rope One:
23i 21j
c 852 502 28550 cos 165
u ucos 30i sin 30j u
133.92 pounds
Rope Two:
Let be the angle between the resultant and the 85-pound
force.
cos v u cos 30i sin 30j u 133.92 85 50
2133.9285
2
2
2
2
1
i j
2
Resultant: u v uj 180j
0.9953
u 180
⇒ 5.6
Therefore, the tension on each rope is u 180 lb.
71. Airspeed: u 430cos 45i sin 45j 2152i j
Wind: w 35cos 60 sin 60 j y
u w N
135° W
35
i 3j
2
θ
35
353
i
2152
Groundspeed: u w 2152 2
2
215 2 35
2
2
E
S
x
45°
u
353
2152
2
2
w
422.30 miles per hour
Bearing: tan 17.53 2152
2152 17.5
40.4
90 130.4
72. Airspeed: u 724cos 60i sin 60j
y
362i 3j
Wind: w 32i
Groundspeed u w 394i 3623j
u w 3942 36232 740.5 kmhr
tan 3
3623
⇒ 57.9
394
724
30°
x
32
Bearing: N 32.1 E
73. u 6, 7, v 3, 9
u v 63 79 45
75. u 3i 7j, v 11i 5j
u v 311 75 2
77. u 3, 4
2u 6, 8
2u u 63 84 50
The result is a scalar.
74. u 7, 12, v 4, 14
u v 74 1214 140
76. u 7i 2j, v 16i 12j
u v 716 212 136
78. v 2, 1
v2 v v 22 12 5; scalar
Review Exercises for Chapter 8
79. u 3, 4, v 2, 1
799
80. u 3, 4, v 2, 1
u v 32 41 2
3u v 332 41 32 6; scalar
uu v u2 2u 6, 8
The result is a vector.
81. u cos
7
7
1
1
,
i sin j 2
2
4
4
82. u cos 45 i sin 45 j
v cos 300 i sin 300 j
v cos
3 1
5
5
i sin j ,
6
6
2 2
cos 3 1
uv
11
⇒ u v
22
12
Angle between u and v: 60 45 105
84. u 3, 3 , v 4, 33 83. u 22, 4, v 2, 1
cos uv
8
⇒ 160.5
u v 243
86. u 85. u 3, 8
cos uv
21
⇒ 22.4
u v 1243
14, 12, v 2, 4
87. u i
v 8u ⇒ Parallel
v 8, 3
v i 2j
u v 38 83 0
u
u and v are orthogonal.
v ku ⇒ Not parallel
v0
⇒ Not orthogonal
Neither
88. u 2i j, v 3i 6j
u
v0
⇒ Orthogonal
89. u 4, 3, v 8, 2
w1 projvu v v 688, 2 174, 1
u
v
26
w2 u w1 4, 3 u w1 w2 90. u 5, 6, v 10, 0
w1 projvu 50
10, 0 5, 0
uv vv 100
2
13
2
13
16
4, 1 1, 4
17
17
13
16
4, 1 1, 4
17
17
91. u 2, 7, v 1, 1
w1 projvu v v 2 1, 1
u
v
5
2
w2 u w1 5, 6 5, 0 0, 6
5
1, 1
2
u w1 w2 5, 0 0, 6
w2 u w1 2, 7 521, 1
9
1, 1
2
5
9
u w1 w2 1, 1 1, 1
2
2
800
Chapter 8
\
93. P 5, 3, Q 8, 9 ⇒ PQ 3, 6
92. u 3, 5, v 5, 2
w1 projvu uv
25
v 5, 2
v2
29
\
w2 u w1 3, 5 u w1 w2 W v PQ 2, 7
3, 6 48
25
19
5, 2 2, 5
29
29
19
25
5, 2 2, 5
29
25
95. w 18,00048
12 72,000 foot-pounds
94. work v PQ
\
3i 6j 10i 17j
30 102
132
97. 7i 02 72 7
\
96. W cos F PQ Imaginary
axis
cos 2025 pounds12 ft
10
281.9 foot-pounds
8
7i
6
4
2
−6
98. 6i 6
99. 5 3i 52 32
Imaginary
axis
34
8
−4
−2
2
−2
4
Imaginary
axis
4
5 + 3i
4
3
2
2
4
6
8
Real
axis
2
1
−4
−6
−6i
−1
−1
−8
100. 10 4i 102 42
229
tan 4
2
102.
z 5 12i
z tan 52
2
3
r 52 52 50 52
6
− 10 − 4i
1
101. 5 5i
Imaginary
axis
−12 − 10 −8 − 6
Real
axis
5
7
1 ⇒ since the
5
4
complex number is in Quadrant IV.
−2
−4
5 5i 52 cos
−6
7
7
i sin
4
4
103. 33 3i
122
13
12
⇒ 1.176
5
z 13cos 1.176 i sin 1.176
Real
axis
5
6
−8 −6 −4 −2
6
r 33 2 32 36 6
tan 3
1
5
⇒ 3
33
6
since the complex number is in Quadrant II.
33 3i 6 cos
5
5
i sin
6
6
4
5
Real
axis
Review Exercises for Chapter 8
801
z 7
104.
z 7
tan 0
0 ⇒ 7
z 7cos i sin 105. (a) z1 23 2i 4 cos
(a ) z2 10i 10 cos
3
3
i sin
2
2
11
11
i sin
6
6
10
10
i sin
3
3
(b) z1z2 4 cos
40 cos
z1
z2
11
11
i sin
6
6
z2 23 i 4 cos
5
5
i sin
4
4
i sin
6
6
5
5
i sin
4
4
17
17
i sin
12
12
122 cos
5
(b) z1z2 32 cos
3
3
i sin
2
2
10
cos
107. 5 cos
11
11
i sin
6
6
2
cos i sin
3
3
5
3
3
i sin
10 cos
2
2
4 cos
106. (a) z1 31 i 32 cos
z1
z2
4
cos 6 i sin 6 5
4 i sin 4 32 13
13
cos
i sin
4 12
12 4 cos i sin 6
6
32 cos
i sin
12
12
4
54 cos
4
4
i sin
12
12
625 cos
625
i sin
3
3
108.
4
4
2
cos 15 i sin 15 5
25 cos
32 12 23i
4
4
i sin
3
3
1 3
i
2
2
16 163i
625 6253
i
2
2
109. 2 3i6 13cos 56.3 i sin 56.36
110. 1 i8 2cos 315 i sin 315
8
133cos 337.9 i sin 337.9
16cos 2520 i sin 2520
13 0.9263 0.3769i
16cos 0 i sin 0
2035 828i
16
3
802
Chapter 8
111. Sixth roots of 729i 729 cos
3
3
:
i sin
2
2
(a) and (c)
(b)
6
729 cos
3
2k
2
6
i sin
3
2k
2
6
4
, k 0, 1, 2, 3, 4, 5
32 32
k 0: 3 cos i sin
i
4
4
2
2
7
7
0.776 2.898i
i sin
12
12
11
11
i sin
2.898 0.776i
12
12
5
5
32 32
i sin
i
4
4
2
2
19
19
i sin
0.776 2.898i
12
12
23
23
i sin
2.898 0.776i
12
12
k 1: 3 cos
k 2: 3 cos
k 3: 3 cos
k 4: 3 cos
k 5: 3 cos
Imaginary
axis
−4
−2
4
−2
−4
112. (a) 256i 256 cos
i sin
2
2
(b)
Imaginary
axis
5
Fourth roots of 256i:
3
2k
2k
2
2
4 256 cos
i sin
, k 0, 1, 2, 3
4
4
i sin
8
8
5
5
i sin
8
8
9
9
i sin
8
8
13
13
i sin
8
8
k 0: 4 cos
k 1: 4 cos
k 2: 4 cos
k 3: 4 cos
1
−3
−1
1 2 3
−2
−3
−5
(c) 3.696 1.531i
1.531 3.696i
3.696 1.531i
1.531 3.696i
113. Cube roots of 8 8cos 0 i sin 0, k 0, 1, 2
(a) and (c)
3
8 cos
(b)
0 2k
0 2k
i sin
3
3
Imaginary
axis
3
k 0: 2cos 0 i sin 0 2
2
2
k 1: 2 cos
i sin
1 3i
3
3
4
4
i sin
1 3i
3
3
k 2: 2 cos
Real
axis
−3
−1
1
−3
3
Real
axis
5
Real
axis
Review Exercises for Chapter 8
114. (a) 1024 1024cos i sin (b)
Imaginary
axis
Fifth roots of 1024:
5 1024 cos
5
2k
2k
, k 0, 1, 2, 3, 4
i sin
5
5
k 0: 4 cos i sin
5
5
k 1: 4 cos
3
3
i sin
5
5
7
7
i sin
5
5
9
9
i sin
5
5
k 4: 4 cos
2 3
Real
axis
5
−5
(c) 3.236 2.351i
k 2: 4cos i sin k 3: 4 cos
1
−3 −2 −1
1.236 3.804i
4
1.236 3.804i
3.236 2.351i
115. x4 81 0
x4 81
Solve by finding the fourth roots of 81.
81 81cos i sin 4 81 4 81 cos
2k
2k
i sin
, k 0, 1, 2, 3
4
4
Imaginary
axis
4
32 32
k 0: 3 cos i sin
i
4
4
2
2
3
3
32 32
i sin
i
4
4
2
2
5
5
32 32
i sin
i
4
4
2
2
7
7
32 32
i sin
i
4
4
2
2
k 1: 3 cos
k 2: 3 cos
k 3: 3 cos
2
−4
−2
2
Real
axis
4
−2
−4
116. x5 32 0
Imaginary
axis
x5 32
3
32 32cos 0 i sin 0
3 32 5 32 cos 0 2k
2k
i sin 0 5
5
1
−3
−1
k 0, 1, 2, 3, 4
k 0: 2cos 0 i sin 0 2
2
2
i sin
0.6180 1.9021i
5
5
4
4
i sin
1.6180 1.1756i
5
5
6
6
i sin
1.6180 1.1756i
5
5
8
8
i sin
0.6180 1.9021i
5
5
k 1: 2 cos
k 2: 2 cos
k 3: 2 cos
k 4: 2 cos
−3
1
3
Real
axis
803
804
Chapter 8
117. x3 8i 0
x3
Imaginary
axis
8i
Solve by finding the cube roots of 8i.
3
3
8i 8 cos
i sin
2
2
3 8i 3 8 cos
3
2k
2
3
1
i sin
i sin
2i
2
2
7
7
i sin
3 i
6
6
11
11
i sin
3 i
6
6
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
3
3
2 k
2
3
−3
, k 0, 1, 2
3
−1
Real
axis
−3
118. x3 1x2 1 0
Imaginary
axis
x3 1 0
2
x2 1 0
x3 1
−2
2
Real
axis
1 1cos 0 i sin 0
3
3
1 1 cos
0 2k
0 2k
i sin
3
3
, k 0, 1, 2
−2
1cos 0 i sin 0 1
2
2
1 3
i sin
i
3
3
2
2
4
4
1 3
i sin
i
3
3
2
2
1 cos
1 cos
x2 1 0
x2 1
1 1cos i sin 1 1 cos
2k
2k
i sin
, k 0, 1
2
2
k 0, 1
i sin
i
2
2
3
3
i sin
i
2
2
1 cos
1 cos
119. True. sin 90 is defined in the Law of Sines.
120. False. There may be no solution, one solution, or
two solutions.
121. True, by the definition of a unit vector.
v
so v vu
u
v
122. False, a b 0.
Review Exercises for Chapter 8
123. False. x 3 i is a solution to x3 8i 0, not
x2 8i 0.
124.
a
b
c
sin A sin B sin C
or
805
sin A sin B sin C
a
b
c
Also, 3 i2 8i 2 23 8i 0.
125. a2 b2 c2 2bc cos A
b2 a2 c2 2ac cos B
126. A vector in the plane has both a magnitude and
a direction.
c2 a2 b2 2ab cos C
127. A and C appear to have the same magnitude and direction.
128. u v is larger in figure (a) since the angle between
u and v is acute rather than obtuse.
129. If k > 0, the direction of ku is the same, and the
magnitude is ku.
130. The sum of u and v lies on the diagonal of the
parallelogram with u and v as its adjacent sides.
If k < 0, the direction of ku is the opposite direction of
u, and the magnitude is k u.
131. (a) The trigonometric form of the three roots shown is:
4cos 60 i sin 60
132. (a) The trigonometric forms of the four roots shown are:
4cos 60 i sin 60
4cos 180 i sin 180
4cos 150 i sin 150
4cos 300 i sin 300
(b) Since there are three evenly spaced roots on the circle of radius 4, they are cube roots of a complex
number of modulus 43 64.
Cubing them yields 64.
4cos 60 i sin 603 64
4cos 180 i sin 1803 64
4cos 240 i sin 240
4cos 330 i sin 330
(b) Since there are four evenly spaced roots on the circle
of radius 4, they are fourth roots of a complex
number of modulus 44. In this case, raising them to
the fourth power yields 128 1283i.
4cos 300 i sin 3003 64
133. z1 2cos i sin z2 2cos i sin z1z2 22cos i sin 4cos i sin 4
z1
2cos i sin z2 2cos i sin 1cos i sin cos2 i sin2 cos 2 cos sin 2 sin isin 2 cos cos 2 sin cos 2 i sin 2
134. (a) z has 4 fourth roots. Three are not shown.
(b) The roots are located on the circle at 30 90k, k 0, 1, 2, 3.
The three roots not shown are located at 120, 210, 300.
806
Chapter 8
Problem Solving for Chapter 8
\
1. PQ 2 4.72 62 24.76 cos 25
P
4.7 ft
\
PQ 2.6409 feet
θ
φ
25°
sin sin 25
⇒ 48.78
4.7
2.6409
O
θ
β
6 ft
γ
α
Q
T
180 25 48.78 106.22
180 ⇒ 180 106.22 73.78
106.22 73.78 32.44
180 180 48.78 32.44 98.78
180 180 98.78 81.22
\
PT
4.7
sin 25 sin 81.22
\
PT 2.01 feet
2.
3
mile 1320 yards
4
55°
300 yd
55°
35°
25°
x2 13202 3002 21320300cos 10
θ
1320 yd
x 1025.881 yards 0.58 mile
x
sin sin 10
1320 1025.881
sin 0.2234
180 sin10.2234
167.09
Bearing: 55 90 22.09
S 22.09 E
3. (a)
A
75 mi
30°
15°
135°
x
y
60° Lost party
(c)
B
75°
A
80° 20°
60°
10°
20 mi
Rescue
party
27.452 mi
(b)
x
75
y
75
and
sin 15 sin 135
sin 30 sin 135
x 27.45 miles
y 53.03 miles
z
Lost
party
z2 27.452 202 227.4520 cos 20
z 11.03 miles
sin sin 20
27.45
11.03
sin 0.8511
180 sin10.8511
121.7
To find the bearing, we have 10 90 21.7.
Bearing: S 21.7 E
Problem Solving for Chapter 8
4. (a)
(b)
65°
sin C sin 65
46
52
sin C 46 ft
807
46 sin 65
0.801734
52
C 53.296
52 ft
A 180 B C 61.704
1
(c) Area 4652sin 61.704 1053.09 square feet
2
Number of bags:
a
52
sin 61.704 sin 65
1053.09
21.06
50
a
a 50.52 feet
To entirely cover the courtyard, you would
5. If u 0, v 0, and u v 0, then
(a)
u 1, 1,
v 1, 2,
u 2,
(b)
v 5,
u 0, 1,
(c)
(d)
u v 1
u v 3, 2
v 18 32, u v 13
21,
72
u 1,
u uu vv uu vv 1 since all of these are magnitudes of unit vectors.
u v 0, 1
v 3, 3,
u 1,
52 sin 61.704
sin 65
5
2
v 2, 3, u v 3,
v 13, u v ,
u 2, 4,
v 5, 5,
9 494 85
2
u v 7, 1
u 20 25, v 50 52, u v 50 52
120
(d) tan 40 ⇒ tan1 3 ⇒ 71.565
6. (a) u 120j
v 40i
(e)
Up
(b) s u v 40i 120j
140
120
100
Up
80
140
60
120
s
100
80
v
u s
W
60
40
− 60
−20
−60
− 20
E
20 40 60 80 100
Down
v
20
W
u
E
20 40 60 80 100
Down
(c) s 402 1202 16000 4010
126.49 miles per hour
This represents the actual rate of the skydiver’s fall.
s 30i 120j
s 302 1202
15300
123.69 miles per hour
808
Chapter 8
8. Let u v 0 and u w 0.
7. Initial point: 0, 0
Terminal point:
u
1
v1 u2 v2
,
2
2
Then, u cv dw u cv u dw
cu v du w
u v1 u2 v2
1
,
w 1
u v
2
2
2
c0 d0
0.
Initial point: u1, u2
Terminal point:
w
u
v
Thus for all scalars c and d, u is orthogonal to cv dw.
1
u v1, u2 v2
2 1
1
v1
u v2
u1, 2
u2
2
2
1
u1 v2 u2
1
,
v u
2
2
2
→
9. W cos F PQ and F1 F2
(a)
F1
If 1 2 then the work is the same since cos cos .
θ1
θ2
F2
P
(b)
Q
If 1 60 then W1 F1
60°
F2
If 2 30 then W2 30°
P
Q
→
1
F PQ
2 1
3
2
→
F2 PQ
W2 3 W1
The amount of work done by F2 is 3 times as great as the amount of work done by F1.
10. (a)
100 sin 100 cos 0.5
0.8727
99.9962
1.0
1.7452
99.9848
1.5
2.6177
99.9657
2.0
3.4899
99.9391
2.5
4.3619
99.9048
3.0
5.2336
99.8630
(b) No, the airplane’s speed does not equal the sum of the vertical and horizontal components of its velocity. To find speed:
speed v sin2 v cos2
(c) (i) speed 5.235 2 149.909 2 150 miles per hour
(ii) speed 10.463 2 149.634 2 150 miles per hour
Practice Test for Chapter 8
Chapter 8
Practice Test
For Exercises 1 and 2, use the Law of Sines to find the remaining sides and
angles of the triangle.
1. A 40, B 12, b 100
2.
C 150, a 5, c 20
3. Find the area of the triangle: a 3, b 6, C 130.
4. Determine the number of solutions to the triangle: a 10, b 35, A 22.5.
For Exercises 5 and 6, use the Law of Cosines to find the remaining sides and
angles of the triangle.
5. a 49, b 53, c 38
6.
C 29, a 100, b 300
7. Use Heron’s Formula to find the area of the triangle: a 4.1, b 6.8, c 5.5.
8. A ship travels 40 miles due east, then adjusts its course 12 southward. After traveling
70 miles in that direction, how far is the ship from its point of departure?
9. w 4u 7v where u 3i j and v i 2j. Find w.
10. Find a unit vector in the direction of v 5i 3j.
11. Find the dot product and the angle between u 6i 5j and v 2i 3j.
12. v is a vector of magnitude 4 making an angle of 30 with the positive x-axis.
Find v in component form.
13. Find the projection of u onto v given u 3, 1 and v 2, 4.
14. Give the trigonometric form of z 5 5i.
15. Give the standard form of z 6cos 225 i sin 225.
16. Multiply 7cos 23 i sin 23 4cos 7 i sin 7.
5
5
i sin
4
4
.
3cos i sin 9 cos
17. Divide
19. Find the cube roots of 8 cos
18. Find 2 2i8.
i sin .
3
3
20. Find all the solutions to x4 i 0.
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