Algebra and Trigonometry 8th Edition Aufmann Solutions Manual

Algebra and Trigonometry 8th Edition
Aufmann Solutions Manual

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Section 7.1 513
Chapter 7 Applications of Trigonometry
Section 7.1 Exercises 7. Solve the triangles.

1. No. There is an infinite number of triangles with the
given angles.
2. Answers will vary.
3. The SSA case may yield no solution, one solution or
two solutions.
B = 180 -110 - 32
4. It is an example of AAS case.
B = 38
5. Solve the triangles.
b = a c = b
sin B sin A sin C sin B
12 = a c = 12
sin 38 sin110 sin 32 sin 38
 
a = 12sin110 c = 12sin 32
sin 38 sin 38
a » 18 c » 10
C = 180 - 42 - 61
C = 77
b = a c = a
sin B sin A sin C sin A
b = 12 c = 12
sin 61 sin 42 sin 77 sin 42
 
b = 12sin 61 c = 12sin 77
sin 42 sin 42
A = 180 - 78 - 28
b » 16 c » 17
A = 74
b = c a = c
sin B sin C sin A sin C
b = 44 a = 44
sin 28 sin 78 sin 74 sin 78
 
b = 44sin 28 a = 44sin 74
sin 78 sin 78
A = 180 -125 - 25 b » 21 a » 43
A = 30 9. Solve the triangles.
c = b a = b
sin C sin B sin A sin B
c = 5.0 a = 5.0
sin125 sin 25 sin 30 sin 25
 
c = 5.0sin125 a = 5.0sin 30
sin 25 sin 25
c » 9.7 a » 5.9 a = b
sin A sin B
22 = 16
sin132 sin B
514 Chapter 7 Applications of Trigonometry
 b = a c = b
sin B = 16sin132 sin B sin A sin C sin B
22
sin B » 0.5405 41.4 = a c = 41.4 

sin108.5 sin 31.4 sin 40.1
sin108.5
B » 33
 
   a = 41.4sin 31.4 c = 41.4sin 40.1
C » 180 - 33 -132 sin108.5 sin108.5
C » 15 a » 22.7 c » 28.1
a = c 12. Solve the triangles.

sin A sin C
22 » c
sin132 sin15

c » 22sin15
sin132
c » 7.7
10. Solve the triangles. C = 180 - 68.5 - 39.2
C = 72.3
b = a c = a
b = 154 c = 154
sin 39.2 sin 68.5 sin 72.3 sin 68.5
b = c  
sin B sin C b = 154sin 39.2 c = 154sin 72.3
6.0 = 3.0 sin 68.5 sin 68.5
sin 82 sin C b » 105 c » 158

sin C = 3sin 82.0 13. Solve the triangles.
6.0
sin C » 0.4951
C » 29.7
A » 180.0 - 82.0 - 29.7

A » 68.3 c = a
sin C sin A
a = b 14.5 = 17.2
sin A sin B sin C sin141.1
a » 6.0  
sin 68.3
sin 82.0 sin C = 14.5sin141.1 » 0.5294
17.2

a = 6.0sin 68.3 C » 32.0
sin 82.0
a » 5.6 B » 180 -141.1 - 32.0
11. Solve the triangles. B » 6.9
b = a
sin B sin A
b » 17.2
sin 6.9 sin141.1

C = 180 - 31.4 -108.5 b = 17.2sin 6.9 » 3.3
sin141.1
C = 40.1
Section 7.1 515
14. Solve the triangles. A = 180 - 69.2 - 36.9
A = 73.9
b = a c = a
b = 166 c = 166
A = 180 - 54.25 - 41.50 sin 36.9 sin 73.9 sin 69.2 sin 73.9
 
A = 84.25 b = 166sin 36.9 c = 166sin 69.2
sin 73.9 sin 73.9
b = a b » 104 c » 162
sin B sin A
b 17. Solve the triangles.

= 24.75 
sin 54.25 sin 84.25

b = 24.75sin 54.25
sin 84.25
b » 20.19
c = b
c = a sin C sin B
sin C sin A 87.2 = 12.1
c = 24.75 sin114.2 sin B
sin 41.50 sin 84.25 
 sin B = 12.1sin114.2 » 0.1266
c = 24.75sin 41.50 87.2
sin 84.25 B » 7.3
c » 16.48
A » 180 -114.2 - 7.3
A » 58.5
c = a
sin C sin A
87.2 » a

sin114.2 sin 58.5
B = 180 - 98.5 - 33.8 
 a = 87.2 sin 58.5 » 81.5
B = 47.7 sin114.2
a = c b = c 18. Solve the triangles that exist.
sin A sin C sin B sin C
a = 102 b = 102
sin 33.8 sin 98.5 sin 47.7 sin 98.5
 
a = 102 sin 33.8 b = 102 sin 47.7
sin 98.5 sin 98.5
a = c
a » 57.4 b » 76.3 sin A sin C
16. Solve the triangles. 24.42 = 16.92
sin 54.32 sin C

sin C = 16.92 sin 54.32 » 0.5628
24.42
C » 34.25
B = 180 - 54.32 - 34.25 20. Solve the triangles that exist.
B = 91.43
b = a
sin B sin A
b » 24.42
sin 91.43 sin 54.32

sin 32 = h
14
b = 24.42 sin 91.43 » 30.05
sin 54.32 h = 14sin 32
19. Solve the triangles that exist. h » 7.4
Since h < 9.0, two triangles exist.
b = c
sin B sin C
9.0 = 14
sin 32 sin C

sin C = 14sin 32 = 0.8243
9
C = 56 or 124
sin 64.2 = h
75.5
For C = 56
h = 75.5sin 64.2
h » 68 A = 180 - 32 - 56 = 92
Since h < 75.5, two triangles exist. a = 9.0
sin 92 sin 32
c = a 
sin C sin A a = 9sin 92 » 17
71.6 = 75.5 sin 32
sin 64.2 sin A For C = 124

sin A = 75.5sin 64.2 = 0.9493
71.6 A = 180 -124 - 32 = 24
A » 71.7 or 108.3 a = 9.0
sin 24 sin 32
For A = 71.7 
    a = 9sin 24 » 6.9
B = 180 - 71.7 - 64.2 = 44.1 sin 32
b = 71.6 21. Solve the triangles that exist.
sin 44.1 sin 64.2

b = 71.6sin 44.1 = 55.3
sin 64.2
For A = 108.3
B = 180 -108.3 - 64.2 = 7.5 Since b > c, one triangle exists.
b = 71.6 b = c
sin 7.5 sin 64.2 sin B sin C

b = 71.6sin 7.5 » 10.4 9.25 = 5.44
sin 64.2 sin 82.6 sin C
sin C = 5.44sin 82.6 » 0.5832
9.25
C = 35.7
Section 7.1 517
A = 180- 82.6- 35.7 23. Solve the triangles that exist.
A = 61.7
a = b
sin A sin B
a = 9.25
sin 61.7 sin 82.6
a = 9.25sin 61.7 sin 30 = h
sin 82.6 2.4
a » 8.21 h = 2.4sin 30
h » 1.2
Since h > 1, no triangle is formed.
24. Solve the triangles that exist.
sin 31 = h
12 sin 22.6 = h
13.8
h = 12sin 31
h = 13.8sin 22.6
h » 6.2 h » 5.30
Since h < 11, two triangles exist. Since h < 5.55, two solutions exist.
a = c
a = b
sin A sin C
sin A sin B
11 = 12 13.8 = 5.55
sin 31 sin C sin A sin 22.6
 
sin C = 12sin 31 = 0.5619 sin A = 13.8sin 22.6
11 5.55
C = 34 or 146 sin A = 0.9555
A » 72.9 or 107.1
For C = 34
For A = 72.9
   
B = 180 - 34 - 31 = 115
C = 180 - 72.9 - 22.6
b = 11 
 C = 84.5
sin115 sin 31
c = 5.55 

b = 11sin115 » 19 sin 84.5
sin 22.6
sin 31 
c = 5.55sin 84.5

For C = 146 sin 22.6
c » 14.4
B = 180 -146 - 31 = 3
For A = 107.1
b = 11
C = 180 -107.1 - 22.6
sin 3 sin 31
 C = 50.3
b = 11sin 3 » 1.1
sin 31 c = 5.55 

sin 50.3 sin 22.6

c = 5.55sin 50.3

sin 22.6
c » 11.1
25. Solve the triangles that exist. Since h < 2.84, two solutions exist.
b = c
sin B sin C
3.50 = 2.84
sin B sin 37.9
sin B = 3.50sin 37.9
sin14.8 = h 2.84
6.35
sin B = 0.7570
h = 6.35sin14.8 B » 49.2 or 130.8
h » 1.62
For B = 49.2
Since h < 4.80, two solutions exist.
A = 180- 49.2- 37.9
c = a A = 92.9
sin C sin A
a = 2.84
6.35 = 4.80 sin 92.9 sin 37.9
sin C sin14.8
 a = 2.84 sin 92.9
sin 37.9
sin C = 6.35sin14.8
4.80 a » 4.62
sin C = 0.3379
For B = 130.8
C » 19.8 or 160.2 A = 180-130.8- 37.9
For C = 19.8 A = 11.3
a = 2.84
B = 180 -19.8 -14.8
sin11.3 sin 37.9
= 145.4
a = 2.84 sin11.3
b sin 37.9
= 4.80 
sin145.4 
sin14.8 a » 0.906
 27. Solve the triangles that exist.
b = 4.80sin145.4
sin14.8
b » 10.7
For C = 160.2
B = 180 -160.2 -14.8
= 5.0 sin 47.2 = h
b = 4.80 8.25
sin 5.0 sin14.8 h = 8.25sin 47.2
 h » 6.05
b = 4.80sin 5.0
sin14.8 Since h > 5.80, no triangle is formed.
b » 1.64 28. Solve the triangles that exist.
sin 37.9 = h sin 52.7 = h

3.50 16.3
h = 3.50sin 37.9 h = 16.3sin 57.2
h » 2.15 h » 13.70
Since h >12.3, no triangle is formed.
Section 7.1 519
29. Solve the triangles that exist. a = c
sin A sin C
10.3 = 14.1
sin 20.5 sin C

sin C = 14.1sin 20.5
10.3
sin C = 0.4794
Since b > a, one triangle exists.
C » 28.6 or 151.4
a = b
sin A sin B For C = 28.6
12.10 = 62.75
B = 180 - 28.6 - 20.5
sin A sin103.45
B = 130.9
sin A = 12.10sin103.45 » 0.1875
62.75 b = 10.3 
A = 10.81 sin130.9 
sin 20.5
C = 180-10.81-103.45 
b = 10.3sin130.9
C = 65.74 sin 20.5
b » 22.2
c = b
sin C sin B For C = 151.4
c = 62.75
sin 65.74 sin103.45 B = 180 -151.4 - 20.5
c = 62.75sin 65.74 B = 8.1

sin103.45 b = 10.3
c » 58.82 sin 8.1 sin 20.5
30. Solve the triangles that exist. 
b = 10.3sin 8.1
sin 20.5
b » 4.1
sin 49.22 = h
24.62
h = 24.62 sin 49.22
h » 18.64
Since h > 16.92, no triangle is formed. sin 41.2 = h
31.5
h = 31.5sin 41.2
h » 20.7
Since h < 21.6, two solutions exist.
a = b
sin A sin B
sin 20.5 = h

14.1 31.5 = 21.6
sin A sin 41.2
h = 14.1sin 20.5

h » 4.9 sin A = 31.5sin 41.2
21.6
Since h < 10.3, two solutions exist. sin A = 0.9606
A » 73.9 or 106.1
For A = 73.9 35. a = 155 yd, c = 165 yd, A = 42.0

  
C = 180 - 73.9 - 41.2
c = a
C = 64.9 sin C sin A
c = 21.6  165 = 155

sin 64.9 sin 41.2 sin C sin 42.0

c = 21.6sin 64.9
 sin C = 165sin 42.0

sin 41.2 155
c » 29.7 æ ö
C = sin-1 çç165sin 42.0 ÷÷÷ = 45.4
For A = 106.1 è 155 ø
C = 180 -106.1 - 41.2
B = 180 - 42.0 - 45.4 = 92.6
C = 32.7
c b = a
= 21.6 sin B sin A
sin 32.7 sin 41.2
 b = 155
c = 21.6sin 32.7
 sin 92.6 sin 42.0
sin 41.2

c » 17.7 b = 155sin 92.6 = 231 yd
33. Find the distance. sin 42.0
36. b = 365 yd, A = 11.2 , C = 22.9

B = 180 -11.2 - 22.9 = 145.9
a. c = b
sin C sin B
c = 365
sin 22.9 sin145.9

B = 180- (39.4 + 64.9) c = 365sin 22.9 = 253 yd
B = 75.7 sin145.9
a = b b. a = b
sin A sin B sin A sin B
a a = 365 
= 105 
sin 39.4 sin 75.7 sin11.2 sin145.9

a = 105sin 39.4 a = 365sin11.2 = 126 yd
sin 75.7 sin145.9
a » 68.8 miles
37. Find the distance.
C = 180- (59.0 + 77.2) B = 180- 67- 31 c = b

C = 43.8 sin C sin B
B = 82 c = 220
b = c
sin B sin C sin 31 sin 82
x = 7620 c = 220sin 31
sin 77.2 sin 43.8 sin 82
x = 7620sin 77.2 c » 110 feet
sin 43.8
x » 10, 700 feet
Section 7.1 521
38. Solve Example 6 using Law of Sines. 41. Find the length of the guy wire.
CAB = 180- 27.2 = 152.8 A = 21 c = a

sin C sin A
ACB = 180-152.8- 23.9 = 3.3 B = 180- 90- 32 c = 35
AC = AB CD = AC B = 58 sin101 sin 21
sin ABC sin ACB sin CAD sin ADC C = 180- 58- 21 c = 35sin101
AC = 17.5 CD = 123.2 sin 21
C = 101
sin 23.9 sin 3.3 sin 27.2 sin 90 c » 96 feet
AC = 17.5sin 23.9 CD = 123.2 sin 27.2 42. Find the lengths of the other two sides.
sin 3.3 sin 90
» 123.2 m » 56.3 m
Responses will vary.
39. Find the length of the new runway.
C = 180- 54- 47

C = 79
ABC = 180- 3 b = a c = a
= 177 sin B sin A sin C sin A
b = 320 c = 320
AC = BC sin 47 sin 54 sin 79 sin 54
sin ABC sin CAB
b = 320sin 47 c = 320sin 79
AC = 3550 sin 54 sin 54
sin177 sin 2.2 b » 290 feet c » 390 feet
AC = 3550sin177 » 4840 ft 43. Find the height of the hill.
sin 2.2
40. Find the height of the kite.
A = 5
B = 180- 90- 75
ABC = 180- 78
B = 15
= 102
C = 180-15- 5
ACB = 180-102- 62
C = 160
= 16
b = a sin 70 = h
AC = AB sin 62 = h sin B sin A b
sin ABC sin ACB AC
b = 12 h = b sin 70
AC = 30 h = AC sin 62
sin15 sin 5
sin102 sin16
h = 30sin102 sin 62 b = 12 sin15
(
sin 5 )
h = 12 sin15 sin 70
AC = 30sin102 sin16 sin 5
sin16 h » 94 feet h » 33 feet
44. Find the distance. 46. Find the closest distance.
 = 65 A = 90- 55

B = 65 + 8 A = 35
B = 73 C = 90- 25
A = 180- 50- 65 C = 65
A = 65 B = 180- 35- 65
C = 180- 65- 73 B = 80
C = 42 c = b sin 35 = h
sin C sin B c
b = c
sin B sin C c = 8 h = c sin 35
sin 65 sin 80
b = 20 h = 8sin 65 sin 35
sin 73 sin 42 c = 8sin 65 sin 80
sin 80 h » 4.2 miles
b = 20sin 73
sin 42 47. Find the distance.
b » 29 miles
A = 120- 65
A = 55
 = 360- 332  = 65
 = 28 B = 38 + 65
 = 28 B = 103
C = 180-103- 55
C = 82- 28
C = 22
C = 54
A = 28 + 36 b = c
sin B sin C
A = 64
b = 450
B = 180- 64- 54 sin103 sin 22
B = 62
b = 450sin103
sin 22
a = b
sin A sin B b » 1200 miles
a = 8.0 48. Find the distance.
sin 64 sin 62
a = 8.0sin 64
sin 62
a » 8.1 miles
Section 7.2 523
A = 40-15 52. Use the results of Problems 50 and 51.
A = 25 a -b sin A - sin B
B = 180- 90- 40 b = sin B
a + b sin A + sin B
B = 50 b sin B
C = 180- 25- 50 a - b = sin A - sin B
C = 105 a + b sin A + sin B
d = 12 53. Using the figure below,

sin A sin C
d = 12 sin 25
sin105
d » 5.3 feet
3 = cos  , 5 = sin 
d1 d2
d1 = 3 , d = 5
A = 180- 67- 68 cos  2 sin 
A = 45 L = d1 + d 2
B = 67 + 11 3 + 5
L( ) =
B = 78 cos  sin 
C = 180- 45- 78 The graph of L is shown.
C = 57
The minimum value of L is approximately 11.19 m.
c = b
sin C sin B
c = 300
sin 57 sin 78
c = 300sin 57
sin 78
c » 260 meters
50. a = b
sin A sin B Prepare for Section 7.2
a = sin A P1. Evaluate.
b sin B
a -1 = sin A -1 (10.0) 2 + (15.0) 2 - 2(10.0)(15.0) cos110.0 » 20.7
b sin B
a - b = sin A - sin B P2. Find the area of the triangle.
b sin B
A = 1 bh = 1 (6)(8.5) = 25.5 in.2
a = b 2 2
51.
sin A sin B P3. Solve for C.
a = sin A
b sin B c 2 = a 2 + b 2 - 2ab cos C
a + 1 = sin A + 1 c 2 - a 2 - b 2 = -2ab cos C
b sin B
a + b = sin A + sin B
b sin B
2 2 2 7. Find the third side of the triangle.
cos C = c - a - b
-2ab
c 2 = a 2 + b2 - 2ab cos C
æ 2 2 2ö
C = cos-1 çç c - a - b ÷÷÷ c 2 = 9.02 + 7.02 - 2(9.0)(7.0) cos 72
è -2ab ø
æ 2 2 2ö c 2 = 130 -126 cos 72
C = cos-1 ççç a + b - c ÷÷÷
è 2ab ø c = 130 -126 cos 72
P4. Find the semiperimeter. c » 9.5
P = 6 + 9 + 10 = 25 8. Find the third side of the triangle.
semiperimeter = 1 (25) = 12.5 m a 2 = b2 + c 2 - 2bc cos A

2
a 2 = 122 + 222 - 2(12)(22) cos55
P5. Evaluate.
a 2 = 628 - 528cos55
s = a + b + c = 3+ 4 + 5 = 6 a = 628 - 528cos55
2 2
a » 18
6(6 - 3)(6 - 4)(6 - 5) = 6(3)(2)(1) = 6
9. Find the third side of the triangle.
P6. State the relationship.
c 2 = a 2 + b 2 - 2ab cos C
2 2 2
c = a +b
c 2 = 13782 + 24112 - 2(1378)(2411) cos118.25
Section 7.2 Exercises
c 2 = 7, 711,805 - 6, 644, 716 cos118.25
1. It is an example of SAS case.
c = 7, 711,805 - 6, 644, 716 cos118.25
2. Responses will vary. c » 3295
3. The area of a triangle is the produce of the lengths of
any two sides and the sine of the included angle.
a 2 = b 2 + c 2 - 2bc cos A
4. Yes, the Pythagorean Theorem is a special case of the
a 2 = 64.52 + 71.42 - 2(64.5)(71.4) cos 25.9

Law of Cosines. If C = 90 , then
a 2 = 9258.21- 9210.6 cos 25.9
c 2 = a 2 + b 2 - 2ab cos C simplifies to c 2 = a 2 + b 2 . a = 9258.21- 9210.6 cos 25.9
5. Find the third side of the triangle. a » 31.2
a 2 = b 2 + c 2 - bc cos A
a 2 = 53.42 + 71.42 - 2(53.4)(71.4) cos 25.9 b2 = a 2 + c 2 - 2ac cos B
a 2 = 7949.52 - 7625.52 cos 25.9 b2 = 25.92 + 33.42 - 2(25.9)(33.4) cos84.0
a = 7949.52 - 7625.52 cos 25.9 b2 = 1786.37 -1730.12 cos84.0

a » 33.0 b = 1786.37 -1730.12 cos84.0
6. Find the third side of the triangle. b » 40.1
b 2 = a 2 + c 2 - 2ac cos B
b 2 = 2432 + 2052 - 2(243)(205) cos 52.4 c 2 = a 2 + b2 - 2ab cos C
b 2 = 101, 074 - 99, 630 cos 52.4 c 2 = 14.22 + 9.302 - 2(14.2)(9.30) cos 9.20
b = 101, 074 - 99, 630 cos 52.4 c 2 = 288.13 - 264.12 cos 9.20
b » 201 c = 288.13 - 264.12 cos 9.20
c » 5.24
Section 7.2 525
13. Find the third side of the triangle. 18. Find the specified angle.
2 2 2
b2 = a 2 + c 2 - 2ac cos B cos A = b + c - a
2bc
b2 = 1222 + 55.92 - 2(122)(55.9) cos 44.2 2 2 2
2
b = 18,008.81-13,639.6 cos 44.2 cos A = 132 + 160 -108
2(132)(160)
b = 18,008.81-13,639.6 cos 44.2 31,360
cos A =
b » 90.7 42, 240
14. Find the third side of the triangle. ( )

A = cos-1 31360 » 42.1
42240
2 2 2
a = b + c - 2bc cos A 19. Find the specified angle.
2 2 2
a = 444.8 + 389.6 - 2(444.8)(389.6) cos 78.44 2 2 2
2 cos B = a + c - b
a = 349, 635.2 - 346,588.16 cos 78.44 2ac
2 2 2
a = 349, 635.2 - 346,588.16 cos 78.44 cos B = 80 + 124 - 92
2(80)(124)
a » 529.3
13, 312
15. Find the specified angle. cos B =
19,840
2 2
cos C = a + b - c
2ab
2
( )
B = cos-1 13312 » 47.9
19840
2 2 2
cos C = 23 + 29 - 42 20. Find the specified angle.
2(23)(29)
2 2 2
- 394 cos B = a + c - b
cos C = 2ac
1334
2 2 2
cos B = 166 + 139 -124
(
C = cos-1 -394 » 107
1334 ) 2(166)(139)
31,501
16. Find the specified angle. cos B =
46,148
( )
2 2 2
cos A = b + c - a B = cos-1 31501 » 47.0
2bc 46148
2 2 2
cos A = 4.89 + 7.75 - 5.25 21. Find the specified angle.
2(4.89)(7.75)
2 2 2
cos A = 56.4121 cos B = a + c - b
75.795 2ac
( )
2 2 2
A = cos-1 56.4121 » 41.9 cos B = 19.2 + 29.1 -14.3
75.795 2(19.2)(29.1)
17. Find the specified angle. cos B = 1010.96
1117.44
( )
2 2 2
cos C = a + b - c B = cos-1 1010.96 » 25.2
2ab 1117.44
2 2 2
cos C = 82.15 + 61.45 - 72.84 22. Find the specified angle.
2(82.15)(61.45)
2 2 2
cos C = 5219.0594 cos A = b + c - a
10, 096.235 2bc
2 2 2
æ ö
C = cos-1 ççç 5219.0594 ÷÷÷ » 58.87 cos A = 3.2 + 5.9 - 4.7
è10, 096.235 ø 2(3.2)(5.9)
cos A = 22.96
37.76
( )
A = cos-1 22.96 » 53
37.76
23. Find the specified angle. 27. Solve the triangle.
2 2 2
cos B = a + c - b b 2 = a 2 + c 2 - 2ac cos B
2ac
2 2 2 b = 92 + 52 - 2(9)(5) cos 39°
cos B = 32.5 + 29.6 - 40.1 b»6
2(32.5)(29.6)
b = c
cos B = 324.4
1924 sin B sin C
( )
B = cos-1 324.4 » 80.3
1924
6 = 5
sin 39 sin C
æ ö
24. Find the specified angle. C = sin-1 çç 5sin 39 ÷÷÷
è 6 ø
2 2 2
cos C = a + b - c C » 32
2ab
2 2 2
cos C = 112.4 + 96.80 -129.2 A = 180 - 39 - 32
2(112.40)(96.80)
A = 109
cos C » 0.2441
C » 75.87 28. Solve the triangle.
25. Solve the triangle. a 2 = b 2 + c 2 - 2bc cos A
a 2 = b 2 + c 2 - 2bc cos A a = 1232 + 1522 - 2(123)(152) cos 95.1°

a » 204
a = 27.42 + 41.42 - 2(27.4)(41.4) cos 27.6°
a » 21.3 b = a
sin B sin A
b = a 123 = 204
sin B sin A sin B sin 95.1
27.4 = 21.3
æ ö
sin B sin 27.6 B = sin-1 çç123sin 95.1 ÷÷÷
è 204 ø
æ ö
B = sin-1 çç 27.4sin 27.6 ÷÷÷ B » 36.9
è 21.3 ø
B » 36.6 C = 180 - 95.1 - 36.9
C = 180 - 27.6 - 36.6 C = 48.0
C = 115.8 29. Solve the triangle.

2 2 2
26. Solve the triangle. cos B = a + c - b
2ac
c 2 = a 2 + b 2 - 2ab cos C 2 2 2
2 2 cos B = 205 + 157 - 321
c = 1545 + 1282 - 2(1545)(1282) cos121.22° 2(205)(157)
c » 2467 -36,367
cos B =
67,370
b = c
sin B sin C æ -36,367 ö÷
B = cos-1 ççç ÷ » 124.4
1282 = 2467 è 67,370 ø÷
sin B sin121.22
b = a
æ ö
B = sin-1 çç1282sin121.22 ÷÷÷ sin B sin A
è 2467 ø 321 = 205
B » 26.39 sin124.4 sin A
æ ö
A = 180 - 26.39 -121.22 A = sin-1 çç 205sin124.4 ÷÷÷
è 321 ø
A = 32.39
A » 31.8
Section 7.2 527
C = 180 - 31.8 -124.4 32. Solve the triangle.

 2 2 2
C = 23.8
cos C = a + b - c
30. Solve the triangle. 2ab
2 2 2
2 2 2 cos C = 24 + 45 - 63
cos B = a + c - b 2(24)(45)
2ac
cos C = - 1368
2 2 2
cos B = 15.2 + 18.5 - 21.3 2160
cos B =
2(15.2)(18.5)
119.6 ( )
C = cos-1 -1368 » 129
2160
562.4
a = c
(
B = cos-1 119.6 » 77.7
562.4 ) sin A sin C
24 = 63
b = a sin A sin129
sin B sin A æ ö
21.3 = 15.2 A = sin-1 çç 24sin129 ÷÷÷
è 63 ø
sin 77.7 sin A
A » 17
æ ö
A = sin-1 çç15.2sin 77.7 ÷÷÷
è 21.3 ø B = 180 -17 -129
A » 44.2 B = 34
C = 180 - 44.2 - 77.7 33. Find the area of the triangle. Round using the rules of
C = 58.1 significant digits.
31. Solve the triangle. K = 1 bc sin A
2
2 2 2
cos B = a + c - b 1
K = (12)(24) sin105
2ac 2
2 2 2 K » 140 square units
cos B = 1135 + 1462 -1725
2(1135)(1462)
34. Find the area of the triangle. Round using the rules of
450, 044
cos B = significant digits.
3,318, 740
æ 450, 044 ö÷ K = 1 ac sin B
B = cos-1 ççç ÷ » 82.21
è 3,318, 740 ÷ø 2
1
K = (32)(25) sin127
b = a 2
sin B sin A K » 320 square units
1725 = 1135
sin 82.21 sin A
æ ö significant digits.
A = sin-1 çç1135sin 82.21 ÷÷÷
è 1725 ø A = 180- 47.2- 62.4
A » 40.68 A = 70.4
  
C = 180 - 40.68 - 82.21 2
K = a sin B sin C
C = 57.11  2sin A
2
K = 22.4 sin 47.2 sin 62.4
2sin 70.4
K » 173 square units
36. Find the area of the triangle. Round using the rules of K = 1 ab sin C
2
significant digits.
1
K = (22.4)(26.9) sin 83.2
A = 180-102- 27 2
A = 51 K » 299 square units
2 40. Find the area of the triangle. Round using the rules of
K = a sin B sin C
2 sin A significant digits.
2
K = 8.5 sin102 sin 27 K = 1 ab sin C
2 sin 51 2
K » 21 square units 1
K = (9.84)(13.4) sin18.2
37. Find the area of the triangle. Round using the rules of 2
K » 20.6 square units
significant digits.
s = 1 (a + b + c)
2 significant digits.
1
s = (16 + 12 + 14) C = 180-116- 34
2
s = 21 C = 30
2
K = s( s - a )( s - b)( s - c ) K = c sin A sin B
2sin C
K = 21(21-16)(21-12)(21-14) 2
K = 8.5 sin116 sin 34
K = 21(5)(9)(7) 2sin 30
K » 81 square units K » 36 square units
38. Find the area of the triangle. Round using the rules of 42. Find the area of the triangle. Round using the rules of
significant digits. significant digits.
B = 180- 62.15- 47.25 A = 180- 76.3- 42.8

B = 70.6 A = 60.9
2
2
K = b sin A sin C K = c sin A sin B
2sin B 2sin C
2
2
K = 17.9 sin 60.9 sin 42.8
K = 12.52 sin 62.15 sin 47.25 2sin 76.3
2sin 70.6
39. Find the area of the triangle. Round using the rules of 43. Find the area of the triangle. Round using the rules of
significant digits. significant digits.
a = b s = 1 (a + b + c )
sin A sin B 2
1
s = (3.6 + 4.2 + 4.8)
22.4 = 26.9
2
sin A sin 54.3
s = 6.3
sin A = 22.4 sin 54.3
26.9 K = s( s - a )( s - b)( s - c )
sin A » 0.6762
K = 6.3(6.3 - 3.6)(6.3 - 4.2)(6.3 - 4.8)
A » 42.5
C = 180- 42.5- 54.3 K = 6.3(2.7)(2.1)(1.5)
C = 83.2 K » 7.3 square units
Section 7.2 529
44. Find the area of the triangle. Round using the rules of 47. Find the distance from the ball to first base.
significant digits.
s = 1 (a + b + c )
2
s = 1 (10.2 + 13.3 + 15.4)
2
s = 19.45 a 2 = b2 + c 2 - 2bc cos A
K = s( s - a )( s - b)( s - c ) a 2 = 262 + 902 - 2(26)(90) cos 45
K = 19.45(19.45 -10.2)(19.45 -13.3)(19.45 -15.4) a 2 = 8776 - 4680 cos 45
K = 19.45(9.25)(6.15)(4.05) a = 8776 - 4680 cos 45

K » 66.9 square units a » 74 feet
45. Find the distance. 48. Find the wing span.
b2 = a 2 + c 2 - 2ac cos B
b2 = (105.6)2 + (105.6)2 - 2(105.6)(105.6) cos109.05
 = 32 b2 = 22302.72 - 22302.72 cos109.05

 = 72 b = 22302.72 - 22302.72 cos109.05

B = 72 + 32 b » 172.0 feet
B = 104 49. Find the angle.
b2 = a 2 + c 2 - 2ac cos104
b2 = 3202 + 5602 - 2(320)(560) cos104
b2 = 416,000 - 358, 400 cos104
b = 416,000 - 358, 400 cos104
b » 710 miles
46. Find the length of the third side.
Let a = the length of the diagonal on the front of the
box.
Let b = the length of the diagonal on the right side of
the box.
a 2 = b2 + c 2 - 2bc cos A Let c = the length of the diagonal on the top of the box.
a 2 = 3002 + 4162 - 2(300)(416) cos 72 a 2 = (4.75) 2 + (6.50) 2 = 64.8125

a 2 = 263,056 - 249,600 cos 72 a = 64.8125
a = 263,056 - 249,600 cos 72 b2 = (3.25) 2 + (4.75)2 = 33.125
a » 430 feet b = 33.125
c 2 = (6.50) 2 + (3.25)2 = 52.8125
 =C a 2 = b2 + c 2 - 2bc cos A
2 2 2
cos C = a + b - c a 2 = 1802 + 2202 - 2(180)(220) cos120
2ab
64.8125 + 33.125 - 52.8125 a 2 = 120, 400
cos  =
2 64.8125 33.125 a » 350 miles
cos  = 45.125
52. d 2 = 1362 + 1622 - 2(136)(162) cos 78
2 64.8125 33.125
æ ö÷ d 2 = 44,740 - 44, 064 cos 78
 = cos-1 çç 45.125 ÷
èç 2 64.8125 33.125 ø÷ d = 44,740 - 44, 064 cos 78
 » 60.9 d » 189 miles
50. Find the depth of the submarine. 53. Find the length of one side.
2 2 2 A = 360 a 2 = 402 + 402 - 2(40)(40) cos 60

cos A = b + c - a 6
2bc A = 60 a 2 = 1600
2 2 2
(615) + (499) - (629)
cos A = = 231585 a = 40 cm
2(615)(499) 613770
54. Find the angle.
(
A = cos-1 231585 » 67.8
613770 )
2 2 2
cos B = 224 + 182 -165
2(224)(182)
sin A = x cos B » 0.6877
499
B » 46.5
sin 67.8 = x
499 55. Find the distance.
499 sin 67.8 = x
x » 462 feet
51. Find the distance between ships after 10 hrs.
C = 90 + 14
C = 104
b = (18 mph)(10 hours) = 180 miles
180(5280)
c = (22 mph)(10 hours) = 220 miles a= ⋅10
3600
A = 318-198 a = 2640 feet
A = 120
Section 7.2 531
2 2 2
c 2 = 26402 + 4002 - 2(2640)(400)(cos104) cos A = b + c - a
2bc
c 2 » 7,640,539 2 2 2
c » 2800 feet cos A = 402.046592 + 181 - 225
2(402.046592)(181)
56. Find the distance. æ143, 777.4621ö÷
A = cos-1 çç ÷
è145,540.8663 ø÷
A » 8.9
 = 180 - (108.5 + 8.9)
 = 62.6°
The distance is 402 mi and the bearing is S62.6°E.
58. Use the figure below.
 = 270- 254
 = 16
A = 16 + 90 + 32
A = 138
b = 4 ⋅16 = 64 miles
c = 3 ⋅ 22 = 66 miles a. 162 = 42 + c 2 - 2(4)( c ) cos A
a 2 = b2 + c 2 - 2bc cos A b. 162 = 42 + c 2 - 2(4)( c ) cos A

a 2 = 642 + 662 - 2(64)(66) cos138 0 = c 2 - (8cos A)c - 240
a 2 = 8452 - 8448cos138 8cos A  (-8cos A)2 - 4(1)(-240)
a » 120 miles c=
2(1)
57. Find the distance and bearing. 8cos A + 64 cos2 A + 960
c=
2
8cos(55) + 64 cos2 (55) + 960

c. c =
2
c » 18 cm
d. a = b
sin A sin B
16 = 4
sin 55 sin B
B = 108.5 + (180 -124.6)
B = 163.9°
( 16 )
B = sin-1 4 sin 55 » 11.8°
C = 180 - 55 -11.8 = 113.2°

b2 = a 2 + c 2 - 2ac cos B
a = c
b2 = 2252 + 1812 - 2(225)(181) cos163.9 sin A sin C
b2 = 83, 386 - 81, 450 cos163.9 16 = c
sin 55 sin113.2
b = 83, 386 - 81, 450 cos163.9
c = 16sin113.2 » 18 cm
b » 402.046592 sin 55
b » 402 mi They are the same.
59. Find the area. 62. Find the cost.
s = 1 (324 + 412 + 516)

2
s = 626
K = 626(626 - 324)(626 - 412)(626 - 516)

s = 1 (a + b + c) K » 66,710 ft 2
2
1
s = (236 + 620 + 814) cost = 4.15(66,710)
2 cost » $277, 000
s = 835
63. Find the cost.
K = s( s - a )( s - b)( s - c )
K = 835(835 - 236)(835 - 620)(835 - 814)
K = 835(599)(215)(21)
K = 2, 258, 244,975
K » 47,500 square meters
60. Find the exact area. s = 1 (680 + 800 + 1020)
2
s = 1250
K = 1250(1250 - 680)(1250 - 800)(1250 -1020)

K » 271,558 ft 2
271,558
Acres =
s = 1 (8 + 10 + 12) 43,560
2
Acres » 6.23
s = 15
64. Find the number of acres.
K = 2 s( s - a )( s - b)( s - c )
s = 1 (420 + 500 + 540)
K = 2 15(15 - 8)(15 -10)(15 -12) 2
K = 30 7 square feet s = 730
61. Find the cost. K = 730(730 - 420)(730 - 500)(730 - 540)

K » 99, 445
99, 445
Acres =
4840
Acres » 20.5
65. Determine whether triangles ABC or DEF have
correction dimensions.
s = 1 (185 + 212 + 240)
2 For ABC,
s = 318.5
13.0 -16.1 » -0.2981
K = 318.5(318.5 -185)(318.5 - 212)(318.5 - 240) 10.4
K » 18,854 ft 2 (
sin 53.5 - 86.5
2 )
» -0.3022
cost = 2.20(18,854)
cost » $41,000
cos ( )
40.0
2
Triangle ABC has correct dimensions.
Section 7.3 533
For DEF, P3. Solve.
17.2 - 21.3 » -0.1798
22.8 tan  = - 3
3
(
sin 52.1- 59.9
2 )
» -0.0820 tan  = 3
3
cos ( )
68.0
2 æ ö
 = tan-1 ççè 3 ÷÷ø÷ = 30
3
Triangle DEF has an incorrect dimension.
2 2 2 2 2 2
P4. Solve.
66. cos A = b + c - a = b + 2bc + c - a - 2bc
2bc 2bc cos  = -17
338
(b + c )2 - a 2 2bc
= -
2bc 2bc  = cos-1 ççæ -17 ÷÷ö » 157.6
çè 338 ø÷
(b + c - a )(b + c + a )
= -1
2bc P5. Rationalize the denominator.
67. Find the volume. 1 ⋅ 5= 5
V = 18K , where K = area of triangular base 5 5 5
V = 1 (4)(4)(sin 72)(18) P6. Rationalize the denominator.

2
V » 140 in3 28 = 28 = 14 ⋅ 17 = 14 17
68 2 17 17 17 17
68. Show that the area is K = rs.
1. The dot product of two vectors is a scalar.
2. The projection of v onto w is a scalar.
3. Yes, the vector a, b is perpendicular to the vector
-b, a .
K BOC = 1 ar
2 4. a. Yes, v and u are parallel.
1
K AOC = br
2 b. The unit vector is u.
1
K AOB = cr c. The magnitude of v is greater than the magnitude of
2
u only if v ³ 1.
K = K BOC + K AOC + K AOB
5. Find the amount of work.
K = 1 ar + 1 br + 1 cr
2 2 2 W = F ⋅ s = F s cos 
1
K = r(a + b + c)
2 = 16 3 cos 0
K = rs where s = 1 ( a + b + c )
1 = 48 ft-lb
2 2
6. Explain how to use the dot product to determine
Prepare for Section 7.3
whether v and w are orthogonal.
P1. Evaluate.
v and w are orthogonal provided v ⋅ w = 0.
( 53) + (- 45 )
2 2
= 9 + 16 = 25 = 1 7. Find the components, then write the vector.
25 25 25
a = 5 -1 = 4
P2. Use a calculator to evaluate. b = 4-2 = 2
10 cos 228 » -6.691 A vector equivalent to P1P2 is v = 4, 2 .
8. Find the components, then write the vector. 16. Find the components, then write the vector.
a = 3 - 4 = -1 a = 3- 3 = 0
b = -2 - 2 = -4 b = 0 - (-2) = 2
A vector equivalent to P1P2 is v = -1, -4 . A vector equivalent to P1P2 is v = 0, 2 .
9. Find the components, then write the vector. 17. Find the magnitude, direction, and the unit vector.
a = -3 - 2 = -5 4
v = (-3)2 + 42  = tan-1 = tan-1 4
b = 5 -1 = 4 -3 3
v = 9 + 16  » 53.1
A vector equivalent to P1P2 is v = -5, 4 .
v =5  = 180- 
10. Find the components, then write the vector.  » 180- 53.1
a = 3 - (-1) = 4  » 126.9
b = 3 - 4 = -1
u = -3 , 4
A vector equivalent to P1P2 is v = 4, -1 . 5 5
11. Find the components, then write the vector. A unit vector in the direction of v is u = - 3 , 4 .
5 5
a = 5 - (-7) = 12
b = 8 -11 = -3 18. Find the magnitude, direction, and the unit vector.
A vector equivalent to P1P2 is v = 12, -3 . v = 62 + 102  = tan-1 10 = tan-1 5

6 3
12. Find the components, then write the vector. v = 36 + 100  » 59.0
a = -72 -18 = -90 v = 2 34  = 59.0
b = 31- (-39) = 70 » 11.7
A vector equivalent to P1P2 is v = -90, 70 . 6 , 10 = 3 34 , 5 34

u=
2 34 2 34 34 34
13. Find the components, then write the vector.
A unit vector in the direction of v is
a = 7.9 - 2.5 = 5.4
b = -6.5 - 3.1 = -9.6
u = 3 34 , 5 34 .
34 34
A vector equivalent to P1P2 is v = 5.4, -9.6 .
19. Find the magnitude, direction, and the unit vector.
14. Find the components, then write the vector.
v = 82 + (-15)2  = tan-1 -15 = tan-1 1.875
a= 7-1 = 3 8
8 2 8
v = 64 + 225  » 61.9
6 ( )
b = - 11 - - 3 = - 13
4 12 v = 289 = 17  = 360- 
 » 360- 61.9
A vector equivalent to P1P2 is v = 3 , - 13 .
8 12  » 298.1
15. Find the components, then write the vector. u = 8 , -15
17 17
a = 2-2 = 0
b = 3 - (-5) = 8 A unit vector in the direction of v is u = 8 , -15 .
17 17
A vector equivalent to P1P2 is v = 0, 8 .
Section 7.3 535
20. Find the magnitude, direction, and the unit vector. 24. Find the magnitude, direction, and the unit vector.
v = (-33) 2 + 562  = tan-1

56
= tan-1 56 v = (-6) 2 + (2) 2  = tan-1 2 = tan-1 1
-33 33 -6 3
v = 1089 + 3136 v = 36 + 4  » 18.4
 » 59.5
v = 4225 = 65  = 180-  v = 40  = 180-
 » 180- 59.5 = 2 10  » 180-18.4
 » 120.5 » 6.3  » 161.6
u = -33 , 56
65 65
u= -6 , 2 = - 3 10 , 10
A unit vector in the direction of v is u = - 33 , 56 . 2 10 2 10 10 10
65 65
A unit vector in the direction of v is
21. Find the magnitude, direction, and the unit vector.
v = 362 + 77 2  = tan-1 77 = tan-1 77 u = - 3 10 , 10 .

36 36 10 10
v = 1296 + 5929  » 64.9 25. Perform the indicated operations.
v = 7225 = 85  = 64.9
3u = 3 -2, 4 = -6, 12
u = 36 , 77 26. Perform the indicated operations.

85 85
-4 v = -4 -3, -2 = 12, 8
A unit vector in the direction of v is u = 36 , 77 .
85 85 27. Perform the indicated operations.
22. Find the magnitude, direction, and the unit vector. 2u - v = 2 -2, 4 - -3, -2
v = (-9) 2 + (-40) 2  = tan-1 -40 = tan-1 40 = -4, 8 - -3, -2

-9 9
= -1, 10
v = 81 + 1600  » 77.3
v = 61 = 41  = 180- 28. Perform the indicated operations.
 » 180+ 77.3 4 v - 2u = 4 -3, -2 - 2 -2, 4
 » 257.3
u = -9 , -40 = -12, -8 - -4, 8
41 41
= -8, -16
A unit vector in the direction of v is u = - 9 , - 40 . 29. Perform the indicated operations.
41 41
23. Find the magnitude, direction, and the unit vector. 2 u + 1 v = 2 -2, 4 + 1 -3, -2
3 6 3 6
v = 52 + (-2) 2  = tan-1 -2 = tan-1 2 4
= - , 8 + -1 , -1
5 5 3 3 2 3
v = 25 + 4  » 21.8 11 7
= - ,
v = 29  = 360-  6 3
» 5.4  » 360- 21.8 30. Perform the indicated operations.
 » 338.2 3 u - 2 v = 3 -2, 4 - 2 -3, -2
4 4
u= 5 , -2 = 5 29 , - 2 29
29 29 29 29 = - 3 , 3 - -6, -4
2
A unit vector in the direction of v is = 9, 7
2
u = 5 29 , - 2 29 .
29 29
31. Perform the indicated operations. 39. Perform the indicated operations.
2 v + 3 u = 2 (-2i + 3j) + 3 (3i - 2 j)
u = (-2)2 + 42 = 20 = 2 5
3 4 3 4
32. Perform the indicated operations.
(
3 ) (
= - i + 2j + i - 3 j
4 9
4 )2
u ⋅ v = -3, -2 ⋅ -2, 4
= (-3)(-2) + (-2)(4) (
3 4 ) ( )
= - 4 + 9 i + 2- 3 j
2
= -2 11
= i+ j 1
12 2
u v cos 60 = (-2) 2 + 42 (-3) 2 + (-2) 2 cos 60
v = (-2)2 + 32 = 13
= 20 13 1 = 2 5 13 1
2 2 41. Perform the indicated operations.
= 65 » 8.06
u ⋅ v = (3i - 2 j ) ⋅ (-2i + 3 j )
34. Perform the indicated operations. = (3)(-2) + (-2)(3)
-2u = -2(3i - 2 j) = -12
= -6i + 4 j 42. Perform the indicated operations.
u v cos 30 = 32 + (-2) 2 (-2) 2 + 32 cos 30
4 v = 4(-2i + 3j)
= -8i + 12 j = 13 13 3
2
36. Perform the indicated operations. = 13 3
2
3u + 2 v = 3(3i - 2 j) + 2(-2i + 3j)
= (9i - 6 j) + (-4i + 6 j) 43. Find the horizontal and vertical components and write
= (9 - 4)i + (-6 + 6) j an equivalent vector in the form v = a1i + a2 j .
= 5i + 0 j a1 = 5cos 27 » 4.5
= 5i a2 = 5sin 27 » 2.3
37. Perform the indicated operations. v = a1i + a2 j » 4.5i + 2.3j
6u + 2 v = 6(3i - 2 j) + 2(-2i + 3j)
44. Find the horizontal and vertical components and write
= (18i -12 j) + (-4i + 6 j)
an equivalent vector in the form v = a1i + a2 j .
= (18 - 4)i + (-12 + 6) j
= 14i - 6 j a1 = 4 cos127 » -2.4
a2 = 4sin127 » 3.2
v = a1i + a2 j » -2.4i + 3.2 j
1 u - 3 v = 1 (3i - 2 j) - 3 (-2i + 3j)
2 4 2 4 45. Find the horizontal and vertical components and write
(3
2 ) (
= i- j - - 3 i+ 9 j
2 4 ) an equivalent vector in the form v = a1i + a2 j .
( ) (
= 3 + 3 i + -1- 9 j
2 2 4 ) a1 = 4 cos  » 2.8
4
= 3i - 13 j a2 = 4sin  » 2.8
4 4
v = a1i + a2 j » 2.8i + 2.8 j
Section 7.3 537
46. Find the horizontal and vertical components and write
 = sin-1 0.8 = sin-1 0.8
2.6 2.6
an equivalent vector in the form v = a1i + a2 j .
 » 17.9
a1 = 2 cos 8 » -1.8 heading =  = 90- 
7
8   » 90-17.9
a2 = 2sin » -0.9
7  » 72.1
v = a1i + a2 j » -1.8i - 0.9 j
49. Find the ground speed and course.
47. Find the ground speed. heading = 96  direction angle = -6
heading = 124  direction angle = -34
heading = 37  direction angle = 53

wind from the west  direction angle = 0
AB = 50cos53i + 50sin 53 j

» 30.1i + 39.9 j
AB = 45i AD = 225cos(-6)i + 225sin(-6) j
AD = 340cos(-34)i + 340sin(-34) j » 223.8i - 23.5j
AD » 281.9i -190.1j AC = AB + AD
AC = AB + AD » 30.1i + 39.9 j + 223.8i - 23.5 j
AC = 45i + 281.9i -190.1j » 253.9i + 16.4 j
AC » 327i -190 j
AC = (253.9)2 + (16.4)2 » 250
2 2
AC = 327 + (-190)
AC » 380 mph  = tan-1 16.4
253.9
The ground speed of the plane is = tan -1 16.4
253.9
approximately 380 mph.
 » 4
48. Find the heading of the boat.  = 90- 
 » 90- 4
 » 86
The ground speed of the plane is about 250 mph at a
heading of approximately 86 .
50. Find the course. 52. Find the angle of the ramp.
heading = 327  direction angle = 123
 =
heading = 60  direction angle = 30 sin  = 120
800
 » 8.6
F1
53. a. sin 22.4 =
345
F1 = 345sin 22.4
F1 » 131 lb
F2
b. cos 22.4 =
345
F2 = 345cos 22.4
F2 » 319 lb
F1
54. a. sin 31.8 =
17.1 345
AB = 18cos123i + 18sin123 j  = tan-1
-6.3 F1 = 811sin 31.8
AB » -9.8i + 15.1j
AD = 4 cos 30i + 4sin 30 j = tan-1 17.1 F1 » 427 lb
6.3
AD = 3.5i + 2 j  » 70 F2
b. cos 31.8 =
AC = AB + AD  = 270 + 70 345
= -9.8i + 15.1j + 3.5i + 2 j  = 340 F2 = 811cos 31.8
= -6.3i + 17.1j F2 » 689 lb
AC = (-6.3)2 + (17.1)2 » 18 55. Determine whether the forces are in equilibrium, if not
determine the additional force required for equilibrium.
The course of the boat is about 18 mph at a heading of
F1 + F2 + F3
approximately 340.
= (18.2i + 13.1j) + (-12.4i + 3.8 j) + (-5.8i -16.9 j)
51. Find the magnitude of the force. = (18.2 -12.4 - 5.8)i + (13.1 + 3.8 -16.9) j
=0
The forces are in equilibrium.
56. Determine whether the forces are in equilibrium, if not
sin 5.6 = F
3000 F1 + F2 + F3
F = 3000sin 5.6 = (-4.6i + 5.3j) + (6.2i + 4.9 j) + (-1.6i -10.2 j)
F » 293 lb = (-4.6 + 6.2 -1.6)i + (5.3 + 4.9 -10.2) j
=0
Section 7.3 539
57. Determine whether the forces are in equilibrium, if not 62. Find the dot product of the vectors.
determine the additional force required for equilibrium. v ⋅ w = 2, 4 ⋅ 0, 2
F1 + F2 + F3 = 2(0) + (4)2
= (155i - 257 j) + (-124i + 149 j) + (-31i + 98 j) = 0+8
= (155 -124 - 31)i + (-257 + 149 + 98) j =8
= 0i -10 j 63. Find the dot product of the vectors.
The forces are not in equilibrium. F4 = 0i + 10 j v ⋅ w = ( i + 2 j) ⋅ (-i + j)
= 1(-1) + 2(1)
58. Determine whether the forces are in equilibrium, if not
= -1 + 2
=1
F1 + F2 + F3
64. Find the dot product of the vectors.
= (23.5i + 18.9 j) + (-18.7i + 2.5j) + (-5.6i -15.6 j)
v ⋅ w = (5i + 3j) ⋅ (4i - 2 j)
= (23.5 -18.7 - 5.6)i + (18.9 + 2.5 -15.6) j
= 5(4) + 3(-2)
= -0.8i + 5.8 j
= 20 - 6
The forces are not in equilibrium. F4 = 0.8i - 5.8 j = 14
59. Determine whether the forces are in equilibrium, if not 65. Find the angle. Round to the nearest tenth.
determine the additional force required for equilibrium. cos  = v ⋅ w
v w
F1 + F2 + F3 = (189.3i + 235.7 j) + (45.8i - 205.6 j)
2, -1 ⋅ 3, 4
+ (-175.2i - 37.7 j) + (-59.9i + 7.6 j) cos  =
= (189.3 + 45.8 -175.2 - 59.9)i 2 + (-1)2 32 + 42
2
+ (235.7 - 205.6 - 37.7 + 7.6) j 2(3) + (-1)4

cos  =
=0 5 25
cos  = 2 » 0.1789
5 5
60. F1 = ( F1 cos144) i + ( F1 sin144) j  » 79.7
= (6223cos144) i + (6223sin144) j 66. Find the angle. Round to the nearest tenth.
» -5034i + 3658 j
cos  = v ⋅ w
F1 + F2 + F3 = 0 v w
F2 = -F1 - F3 -1, 7 ⋅ 3, -2
cos  =
= -(-5034i + 3658 j) - (0i - 9450 j) (-1)2 + 72 32 + (-2)2
= 5034i + 5792 j (-1)3 + 7(-2)
cos  =
50 13
F2 = F2 cos 49 + F2 sin 49
F2 = 5034 cos 49 + 5792sin 49 cos  = -17 = -0.6668
5 2 13
» 3300 + 4370  = 131.8
» 7670 lbs
61. Find the dot product of the vectors.
v ⋅ w = 3, -2 ⋅ 1, 3
= 3(1) + (-2)3
= 3- 6
= -3
67. Find the angle. Round to the nearest tenth. 71. Find projw v .
cos  = v ⋅ w projw v = v ⋅ w
v w w
(5i - 2 j) ⋅ (2i + 5j) 6, 7 ⋅ 3, 4
cos  =
projw v = = 18 + 28 = 46
52 + (-2)2 22 + 52 2 2 25 5
3 +4
5(2) + (-2)(5)
cos  = 72. Find projw v .
29 29
0
cos  = =0 projw v = v ⋅ w
29 29 w
 = 90 -7, 5 ⋅ -4, 1
projw v = = 28 + 5 = 33
Thus, the vectors are orthogonal. (-4) + 12 2 17 17
68. Find the angle. Round to the nearest tenth.
= 33 17 » 8.0
17
cos  = v ⋅ w
v w 73. Find projw v .
(8i + j) ⋅ (-i + 8 j)
cos  =
82 + 12 (-1)2 + 82 projw v = v ⋅ w
w
8(-1) + (1)(8) -3, 4 ⋅ 2, 5
cos  = projw v = = -6 + 20 = 14
65 65 2 2 29 29
2 +5
cos  = 0 =0
65 65 = 14 29 » 2.6
29
 = 90
74. Find projw v .
Thus, the vectors are orthogonal.
69. Find the angle. Round to the nearest tenth. projw v = v ⋅ w
w
cos  = v ⋅ w 2, 4 ⋅ -1, 5
v w projw v = = -2 + 20 = 18
(5i + 2 j) ⋅ (-5i - 2 j) (-1) + 52 2 26 26
cos  =
52 + 22 (-5)2 + (-2)2 = 9 26 » 3.5
5(-5) + 2(-2) 13
cos  =
29 29 75. Find projw v .
cos  = -29 = -1
29 29 projw v = v ⋅ w
w
 = 180
(2i + j)⋅(6i + 3j) 12 + 3 5
70. Find the angle. Round to the nearest tenth. projw v = = =
62 + 32 45 5
cos  = v ⋅ w = 5 » 2.2
v w
(3i - 4 j) ⋅ (6i -12 j) 76. Find projw v .
cos  =
32 + (-4)2 62 + (-12)2 projw v = v ⋅ w
3(6) + (-4)(-12) w
cos  =
25 180 (5i + 2 j)⋅(-5i +-2 j)
projw v = = -25 - 4 = -29
cos  = 66 = 0.9839 29 29
2 2
(-5) + (-2)
5 180
= - 29 » -5.4
 = 10.3
Section 7.3 541
77. Find projw v . W = F⋅s
W= F s cos 
projw v = v ⋅ w
w W = (50)(6)(cos 48)
(3i - 4 j)⋅(-6i + 12 j) W » 201 foot-pounds
projw v = = -18 - 48 = - 11
2
(-6) + 12 2 180 5 83. Find the sum geometrically.
= - 11 5 » -4.9
5
78. Find projw v .
projw v = v ⋅ w
w
(2i + 2 j)⋅(-4i - 2 j)
projw v = = -8 - 4 = -6
2
(-4) + (-2)
2 20 5
= - 6 5 » -2.7
5
Thus, the sum is 6, 9 .
79. Find the work done.
W = F⋅s 84. Find the sum geometrically.
W= F s cos 
W = (75)(15)(cos 32)
W » 954 foot-pounds
W = F⋅s
W= F s cos 
W = (100)(25)(cos 42)
Thus, u + v - w = 5, - 4 .
85. Find the vector.
W = F⋅s
W= F s cos 
W = (75)(12)(cos 30)
The vector from P1(3, 1) to P2(5, 4) is equivalent to
2i - 3j.
86. Find the vector. 90. w = 4i + j
4, 1 ⋅ a, b = 0
4a + b = 0
a =-1 b
4
Let b = 4
a = -1
Thus, u = -1, 4 is one example.
91. Let u = ci + bj, v = ci + dj, and w = ei + fj.

The vector from P1(2, 4) to P2(3, 7) is equivalent
(u ⋅ v ) ⋅ w = é(ai + bj)⋅(ci + dj)ù ⋅(ei + fj)
ë û
to -1, 3 .
= ac + bd ⋅(ei + fj)
( )
87. Is v perpendicular to w? Why or why not?
ac + bd is a scalar quantity. The product of a scalar
and a vector is not defined. Therefore, no,
(u ⋅ v ) ⋅ w does not equal u ⋅( v ⋅ w ) .
92. Let v = a, b and w = c, d .

v ⋅ w = a, b ⋅ c, d = ac + bd
w ⋅ v = c, d ⋅ a, b = ca + db = ac + bd
v ⋅ w = (2i - 5j) ⋅ (5i + 2 j) Therefore, v ⋅ w = w ⋅ v.
= 10 -10
=0 93. Let v = a, b and w = d , e
cv = ca, cb
The two vectors are perpendicular.
88. Is v perpendicular to w? Why or why not? c ( v ⋅ w ) = c a, b ⋅ d , e = c (ad + be) = cad + cbe
(cv ⋅ w ) = ca, cb ⋅ d , e = cad + cbe
Therefore, c ( v ⋅ w ) = (cv ) ⋅ w.
94. Let  be the angle between vectors v and w.

v⋅w = v w cos 
v ⋅ w = 5, 6 ⋅ 6, 5
v ⋅ w is positive if cos  is positive.
= 30 + 30
= 60 ¹ 0 cos  is positive when 0 <  < 90.
The vectors are not perpendicular. This is an acute angle.

v ⋅ w is negative if cos  is negative.
89. v = -2, 7
cos  is negative when 90 <  < 180.
-2, 7 ⋅ a, b = 0
This is an obtuse angle.
-2 a + 7 b = 0
95. Neither. If the force and the distance are the same, the
a= 7b
2 work will be the same.
Let b = 2
a=7
Thus, u = 7, 2 is one example.
Section 7.4 543
Mid-Chapter 7 Quiz c. projw u = u ⋅ w
w
1. Find b.
3, 5 ⋅ 4, 1
C = 180 - 71.2 - 62.7 projw u = = 12 + 5 = 17 = 17
2
4 +1 2 17 17
C = 46.1
d. u ⋅ v = 3, 5 ⋅ -2, 7
b = c
sin B sin C = 3(-2) + 5(7)
b = -6 + 35

= 18.5 
sin 62.7 sin 46.1 = 29

b = 18.5sin 62.7 5. Find the area of the triangle.
sin 46.1
b » 22.8 in. K = 1 bc sin A
2
K = 1 (15.4)(16.5) sin 34.5
2
B = 74.2 + (180 - 95.5)
B = 158.7° K » 72 m 2
6. Find the angle between the vectors.
b2 = a 2 + c 2 - 2ac cos B
b2 = 67.22 + 43.52 - 2(67.2)(43.5) cos158.7 cos  = v ⋅ w
v w
b2 = 6408.09 - 5846.4 cos158.7 (3i + 5j) ⋅ (-6i + 5j)
cos  =
b = 6408.09 - 5846.4 cos158.7 32 + 52 (-6)2 + 52
b » 108.8812637 3(-6) + 5(5)
cos  =
b » 109 mi 34 61
3. Find A. cos  = 7 = 0.1537
2074
b2 = a 2 + c 2 - 2ac cos B
 = 81.2
2 2
b = 21.2 + 32.5 - 2(21.2)(32.5) cos105.5° 7. Find the magnitude of the force.
b » 43.2890804
b » 43.3 sin 9.5 = F
725
2 2 2 F = 725sin 9.5
cos A = b + c - a
2bc F » 120 lb
2 2 2
cos A = 43.2890804 + 32.5 - 21.2 8. Find the work.
2(43.2890804)(32.5)
W = F⋅s
A = cos (
-1 2480.754482
2813.790226 ) W= F s cos 
A » 28.2
W = (65)(45)(cos 35)
4. a. 3u + 2 v = 3 3, 5 + 2 -2, 7 W » 2396 foot-pounds
= 9, 15 + -4, 14 Prepare for Section 7.4
= 5, 29
P1. Simplify.
b. u = 32 + 52 = 34 (1 + i )(2 + i ) = 2 + 3i + i 2 = 1 + 3i
P2. Simplify.
2 + i ⋅ 3 + i = 6 + 5i + i 2 = 5 + 5i = 1 + 1 i
3- i 3 + i 9 - i2 10 2 2
P3. Find the conjugate. 8. Graph the complex number and find the absolute value.
2 – 3i
P4. Find the conjugate.
3 + 5i
P5. Find the solutions using the quadratic formula.
-1  12 - 4(1)(1) -1  -3
x= = =-1  3 i z = 42 + (-4)
2
2(1) 2 2 2
= 32
P6. Solve.
=4 2
x2 + 9 = 0
9. Graph the complex number and find the absolute value.
x 2 = -9
x =  3i
1. The absolute value of the product is 1.
2. Describe the graph.
2
The graph is a circle, with center at the origin of the z= ( 3 ) + (-1)2
complex plane and a radius of 5. = 3+1 = 4
=2
3. Write the conjugate of 4(cos 60 + i sin 60 ) .
 
4(cos 60 - i sin 60 )
4. Responses will vary.

5. Explain the relationship between the trigonometric
form of a complex number and the polar form.
They are exactly the same.
2
6. Determine if the method is valid. z = 12 + ( 3 ) = 2
No. cis is not a number. 11. Graph the complex number and find the absolute value.
2
z = 02 + (-2) = 2
2 2
z = (-2) + (-2) 12. Graph the complex number and find the absolute value.
= 8=2 2
2
z = (-5) + 02 = 5
Section 7.4 545
 = tan-1 -1
3
= tan-1 1 = 30
3
 = 360 - 30 = 330
z = 2 cis 330
2 2
z = 3 + (-5)
18. Write the complex number in trigonometric form.
= 34
2
14. Graph the complex number and find the absolute value. r = 12 + ( 3 )
r=2
 = tan-1 3
1
= tan-1 3 = 60
 = 60
2 2
z = (-5) + (-4) z = 2 cis 60
= 41 19. Write the complex number in trigonometric form.
r = 0 2 + 52
r = 12 + (-1)2 r =5
r= 2  = 90
 = tan-1 -1 z = 5 cis 90

1
= tan-1 1
2
= 45 r = 02 + (-7)
 = 360- 45 = 315 r=7
z = 2 cis 315  = 270

16. Write the complex number in trigonometric form. z = 7 cis 270
2
r = (-4) + (-4)
2 21. Write the complex number in trigonometric form.
r = 32 2
r = (-17) + 02
r=4 2 r = 17
 = tan-1 -4  = 180
-4
= tan-1 1 = 45 z = 17 cis 180
 = 180 + 45 = 225 22. Write the complex number in trigonometric form.
z = 4 2 cis 225 r = 112 + 02

17. Write the complex number in trigonometric form. r = 11
2  = 0
r= ( 3 ) + (-1)2
r=2 z = 11 cis 0
23. Write the complex number in trigonometric form. 27. Write the complex number in standard form.
r= (-3 3 ) + (3)2
2 z = 2(cos 45 + i sin 45 )
æ ö
r=6 z = 2 çç 2 + 2 i ÷÷÷
è 2 2 ø
 = tan-1 3 z = 2 +i 2
-3 3
28. Write the complex number in standard form.
= tan-1 3 = 30
3 z = 3(cos 240 + i sin 240 )
 = 180 - 30 = 150 æ ö
z = 3çç- 1 - 3 i ÷÷÷
è 2 2 ø
z = 6 cis 150
z =-3-3 3 i
24. Write the complex number in trigonometric form. 2 2
2 2 29. Write the complex number in standard form.
r= (5 2 ) + (-5 2 )
r = 10 z = cos 315 + i sin 315
z= 2- 2i
 = tan-1 5 2 2 2
-5 2
-1
= tan 1 = 45
z = 5(cos 120 + i sin 120 )
 = 360 - 45 = 315
æ ö
z = 5çç- 1 + 3 i ÷÷÷
z = 10 cis 315 è 2 2 ø
25. Write the complex number in trigonometric form. z =-5 + 5 3 i
2 2
2
r = (-2) + (-2 3 )
2
r=4
z = 6 cis 135
 = tan-1 -2 3 z = 6(cos 135 + i sin 135 )
-2
æ ö
= tan-1 3 = 60 z = 6 çç- 2 + 2 i ÷÷÷
è 2 2 ø
 = 180 + 60 = 240 z = -3 2 + 3i 2
z = 4 cis 240 32. Write the complex number in standard form.
26. Write the complex number in trigonometric form. z = cis 315

z = cos 315 + i sin 315
2 2
r= ( 2 ) + (- 2 )
z= 2- 2i
r=2 2 2
 = tan-1 - 2
2 z = 8 cis 0
-1 
= tan 1 = 45 z = 8(cos 0 + i sin 0 )
 = 360 - 45 = 315 z = 8(1 + 0i )
z =8
z = 2 cis 315
Section 7.4 547
34. Write the complex number in standard form. 41. Write the complex number in standard form.
z = 5 cis 90 z = 9 cis 11
z = 5(cos 90 + i sin 90) 6
z = 5(0 + i ) (
z = 9 cos  + i sin 11
11
6 6 )
z = 5i æ ö
z = 9 çç 3 - 1 i ÷÷÷
35. Write the complex number in standard form. è 2 2 ø
(
z = 2 cos 5 + i sin 5
6 6 ) z = 9 3-9i
2 2
æ ö 42. Write the complex number in standard form.
z = 2 çç- 3 + 1 i ÷÷÷
è 2 2 ø
z = cis 3
z =- 3+i 2
36. Write the complex number in standard form. z = cos  + i sin 3
3
2 2
(
z = 4 cos 5 + i sin 5
3 3 ) z = 0-i
z = -i
æ ö
z = 4 çç 1 - 3 i ÷÷÷ 43. Multiply, write answer in trigonometric form.
è2 2 ø
z = 2 - 2i 3 z1 z2 = 2 cis 30⋅ 3 cis 225
z1 z2 = 6 cis(30 + 225)
z1 z2 = 6 cis 255
(
z = 3 cos 3 + i sin 3
2 2 ) 44. Multiply, write answer in trigonometric form.
z = 3(0 - i ) z1 z2 = 4 cis 120⋅ 6 cis 315
z = -3i z1 z2 = 24 cis(120 + 315)
38. Write the complex number in standard form. z1 z2 = 24 cis 435
z1 z2 = 24 cis 75
z = 5(cos  + i sin  )
z = 5(-1 + 0i ) 45. Multiply, write answer in trigonometric form.
z = -5 z1 z2 = 3(cos 122+ i sin 122) ⋅ 4(cos 213+ i sin 213)
39. Write the complex number in standard form. z1 z2 = 12[cos(122+ 213) + i sin(122+ 213)]
z1 z2 = 12(cos 335+ i sin 335)
z = 8 cis 3
4 z1 z2 = 12 cis 335
(
= 8 cos 3 + i sin 3
4 4 ) 46. Multiply, write answer in trigonometric form.
æ ö z1 z2 = 8(cos 88 + i sin 88) ⋅12(cos 112 + i sin 112)
z = 8 çç- 2 + i 2 ÷÷÷
è 2 2 ø = 96[cos(88 + 112) + i sin(88 + 112)]
z = -4 2 + 4i 2 = 96[cos 200 + i sin 200]
= 96 cis 200
47. Multiply, write answer in trigonometric form.
z = 9 cis 4
3
( ) (
z1 z2 = 5 cos 2 + i sin 2 ⋅ 2 cos 2 + i sin 2 )
(
= 9 cos 4 + i sin 4
3 3 ) é
z1 z2 = 10 ê cos
3
2
(+ 2
3
) (
+ i sin 2 +
5
2 ù
ú)
5
æ ö ëê 3 5 3 5 ûú
z = 9 çç- 1 - i 3 ÷÷÷
è 2 2 ø (
z1 z2 = 10 cos 16 + i sin 16
15 15)
z =-9-9 3 i z1 z2 = 10 cis 16 
2 2 15
48. Multiply, write answer in trigonometric form. 53. Divide, write answer in standard form, round answer to
z1 z2 = 5 cis 11 ⋅ 3 cis 4

3 decimal places.
12 3
z1 27(cos 315 + i sin 315 )
(
z1 z2 = 15 cis 11 + 4
12 3 ) z2
=
9(cos 225 + i sin 225 )
z1
z1 z2 = 15 cis 9 = 3 [cos(315 - 225 ) + i sin(315 - 225 )]
4 z2
z1 z2 = 15 cis 
z1
= 3(cos 90 + i sin 90 ) = 3(0 + i ) = 3i
4 z2
49. Multiply, write answer in trigonometric form. 54. Divide, write answer in standard form, round answer to
z1 z2 = 4 cis 2.4 ⋅ 6 cis 4.1 3 decimal places.
z1 z2 = 24 cis (2.4 + 4.1)
z1 9(cos 25 + i sin 25 )
z1 z2 = 24 cis 6.5 =
z2 3(cos 175 + i sin 175 )
50. Multiply, write answer in trigonometric form. z1
= 3 [cos(25 -175 ) + i sin(25 -175 )]
z2
z1 z2 = 7 cis 0.88 ⋅ 5 cis 1.32
z1
z1 z2 = 35 cis (0.88 + 1.32) = 3(cos 150 - i sin 150 )
z2
z1 z2 = 35 cis 2.2 z1 æ ö
= 3çç- 3 - 1 i÷÷÷ = - 3 3 - 3 i
51. Divide, write answer in standard form, round answer to z2 çè 2 2 ÷ø 2 2
3 decimal places. 55. Divide, write answer in standard form, round answer to
z1 32 cis 30 3 decimal places.
=
z2 4 cis 150
z1 z1
=
(
12 cos 2 + i sin 2
3 3 )
= 8 cis(30-150)
z2
z1
(
z2 4 cos 11 + i sin 11
6 6 )
z2
= 8 cis( -120) z1
z2
é
= 3 ê cos
ëê
2
3( - 11
6

) 3 (ù
+ i sin 2 - 11 ú
6 ûú )
z1
z2
= 8 (cos 120- i sin 120)
z1
z2
= 3 cos( 7
6
- i sin 7
6 )
é ù
z1
z2
=8
æ 1 i 3 ö÷
çç- -
èç 2
÷ = -4 - 4i 3
2 ÷ø
z1
z2
= 3 ê- 3 - - 1 i ú
êë 2 ( )
2 úû
z1
52. Divide, write answer in standard form, round answer to =-3 3 + 3 i
z2 2 2
3 decimal places.
56. Divide, write answer in standard form, round answer to
z1 15 cis 240 3 decimal places.
=
z2 3 cis 135
z1
= 5 cis (240 -135 )
( 
z1 10 cos 3 + i sin 3
=

)
z2
z1
z2
(
5 cos  + i sin 
4 4 )
= 5 cis 105
z2
z1
z1
z2
é
= 2 êcos
êë

( ) ù
- + i sin  -  ú
3 4

3 4 úû ( )
= 5 (cos 105 + i sin 105 )
z2
z1
z2 (
= 2 cos  + i sin 
12 12 )
z1 z1
» 5 ( - 0.2588 + 0.9659i ) » 2(0.9659 + 0.2588i )
z2 z2
z1 z1
» -1.294 + 4.830i » 1.932 + 0.518i
z2 z2
Section 7.4 549
57. Divide, write answer in standard form, round answer to z1 z2 = 2(cos300 + i sin 300) ⋅ 2(cos45 + i sin 45)
3 decimal places. z1 z2 = 2 2[cos(300 + 45) + i sin(300 + 45)]
z1 25 cis 3.5 z1 z2 = 2 2(cos 345 + i sin 345)
=
z2 5 cis 1.5 z1 z2 » 2.732 - 0.732i
z1
= 5 cis (3.5 -1.5) 60. Perform indicated operation, write answer in standard
z2
z1 form, round constants to 4 decimal places.
= 5 cis 2
z2 For z1 = 3 - i
z1
= 5 (cos 2 + i sin 2)
z2 r1 = ( 3)2 + (-1) 2  = tan-1 -1 = 30
z1 r1 = 2 3
» 5 ( - 0.4161 + 0.9093i ) 1 = 330
z2
z1 z1 = 2(cos 330 + i sin 330 )
» -2.081 + 4.546i
z2
For z2 = 1 + i 3
58. Divide, write answer in standard form, round answer to
3 decimal places. 2 3 = 60
r2 = 12 + ( 3 )  = tan-1
1
z1 18 cis 0.56 r2 = 2
=  2 = 60
z2 6 cis 1.22
z1 z2 = 2(cos 60 + i sin 60)
= 3 cis (0.56 -1.22)
z2 z1 z2 = 2(cos 330 + i sin 330) ⋅ 2(cos 60 + i sin 60)
z1 z1 z2 = 4[cos(330 + 60) + i sin(330 + 60)]
= 3 cis ( - 0.66)
z2
z1 z2 = 4(cos 390 + i sin 390)
z1
= 3 (cos 0.66 - i sin 0.66) é ù
z2 z1 z2 = 4 ê 3 + i ú = 2 3 + 2i
ëê 2 2 ûú
z1
» 3 (0.7900 - 0.6131i ) 61. Perform indicated operation, write answer in standard
z2
z1 form, round constants to 4 decimal places.
» 2.370 -1.839i
z2 For z1 = 3 - 3i
59. Perform indicated operation, write answer in standard
r1 = 32 + (-3) 2  = tan-1 -3 = 45
form, round constants to 4 decimal places. 3
r1 = 3 2 1 = 315
For z1 = 1- i 3
z1 = 3 2(cos 315 + i sin 315)
2
r1 = 12 + ( 3 )  = tan-1 - 3 = 60 For z2 = 1 + i
1
r1 = 2
1 = 300 r2 = 12 + 12  = tan-1 1 = 45
1
z1 = 2(cos 300 + i sin 300) r1 = 2 1 = 45
For z2 = 1 + i z2 = 2(cos 45 + i sin 45)
z1 z2 = 3 2(cos315+ i sin315) ⋅ 2(cos45+ isin45)

r2 = 12 + 12  = tan-1 1 = 45
1 z1 z2 = 6[cos(315+ 45) + i sin(315+ 45)]
r2 = 2 z1 z2 = 6(cos 360+ i sin 360)
 2 = 45
z1 z2 = 6 + 0i
z2 = 2(cos 45 + i sin 45) z1 z2 = 6
62. Perform indicated operation, write answer in standard 64. Perform indicated operation, write answer in standard
form, round constants to 4 decimal places. form, round constants to 4 decimal places.
For z1 = 2 + 2i For z1 = 1 + i
r1 = 22 + 22  = tan-1 2 = 45 r1 = 12 + 12 1 = tan-1 1 = 45

2 1
r1 = 2 2 1 = 45 r1 = 2 1 = 45 
z1 = 2 2(cos 45 + i sin 45) z1 = 2(cos 45 + i sin 45)
For z2 = 3 - i For z2 = 1- i
r2 =
2
( 3 ) + (-1)2  = tan-1 -1 = 30 r2 = 12 + (-1)2  2 = tan-1 -1 = 45
1
3 r2 = 2  2 = 315
r2 = 2
 2 = 330
z2 = 2(cos 315 + i sin 315)
z2 = 2(cos 330 + i sin 330)
z1 2(cos 45 + i sin 45)
=
z1 z2 = 2 2(cos45+ i sin45) ⋅ 2(cos330+ i sin330) z2 2(cos 315 + i sin 315)
z1 z2 = 4 2[cos(45+ 330) + i sin(45+ 330)] z1
= cos(45- 315) + i sin(45- 315)
z2
z1 z2 = 4 2(cos 375+ i sin 375)
z1
z1 z2 » 5.4641+1.4641i = cos 270- i sin 270 = 0 - i (-1) = 0 + 1i = i
z2
form, round constants to 4 decimal places.
form, round constants to 4 decimal places.
For z1 = 1 + i 3
For z1 = 2 - i 2
r1 = 12 + ( 3)2 1 = tan-1 3 = 60

1 r1 = ( 2)2 + ( 2)2 1 = tan-1 - 2 = 45
r1 = 2 2
1 = 60 r1 = 2 1 = 315
z1 = 2(cos 60 + i sin 60) z1 = 2(cos 315 + i sin 315)
For z2 = 1- i 3 For z2 = 1 + i
r2 = 12 + ( 3)2  = tan-1 - 3 = 60 r2 = 12 + 12  2 = tan-1 1 = 45

1 1
r2 = 2 r2 = 2  2 = 45
 2 = 300
z2 = 2(cos 300 + i sin 300) z2 = 2(cos 45 + i sin 45)
z1 2(cos 315 + i sin 315)

z1 2(cos 60 + i sin 60) =
= z2 2(cos 45 + i sin 45)
z2 2(cos 300 + i sin 300)
z1 z1
= cos(60- 300) + i sin(60- 300) = 2[cos(315- 45) + i sin(315- 45)]
z2 z2
z1
z1 = 2(cos 270 + i sin 270)
= cos 240- i sin 240 = - 1 + 3 i z2
z2 2 2
z1
= 2[0 + i (-1)] = 2(0 -1i ) = 0 - 2i = - 2i
z2
or - i 2
Section 7.4 551
66. Perform indicated operation, write answer in standard z3 = 4(cos 300 + i sin 300)
form, round constants to 4 decimal places. z1 z2 z3 = 2(cos 330 + i sin 330)
For z1 = 1 + i 3 ⋅ 2 2(cos 45 + i sin 45)
⋅ 4(cos 300 + i sin 300)
r1 = 12 + ( 3)2 1 = tan-1 3 = 60
1 z1 z2 z3 = 16 2[cos(330 + 45 + 300)
r1 = 2 1 = 60 + i sin(330 + 45 + 300)]
z1 = 2(cos 60 + i sin 60) z1 z2 z3 = 16 2(cos 675 + i sin 675)
For z2 = 4 - 4i z1 z2 z3 = 16 2(cos 315 + i sin 315)
æ ö
z1 z2 z3 = 16 2 ç 1 - 1 i ÷÷ = 16 -16i
r2 = 42 + (-4)2  2 = tan-1 -4 = 45 çè 2 2 ø÷
4
r2 = 4 2  2 = 315 68. Perform indicated operation in trigonometric form,
z2 = 4 2(cos 315 + i sin 315) write answer in standard form.

For z1 = 1- i
z1 2(cos 60 + i sin 60)
=
z2 4 2(cos 315 + i sin 315)
r1 = 12 + (-1)2 1 = tan-1 -1 = 45
z1 1
= 2 [cos(60- 315) + i sin(60- 315)] r1 = 2 1 = 315
z2 4
z1
= 2 [cos( - 255) + i sin( - 255)] z1 = 2(cos 315 + i sin 315)
z2 4
z1 For z2 = 1 + i 3
» -0.0915 + 0.3415i
z2
r2 = 12 + ( 3)2  2 = tan-1 3 = 60
67. Perform indicated operation in trigonometric form, 1
r2 = 2  2 = 60
write answer in standard form.
For z1 = 3 -1 z2 = 2(cos 60 + i sin 60)
For z3 = 3 - i
r1 = ( 3)2 + (-1)2 1 = tan-1 -1 = 30
3
r1 = 2
1 = 330 r3 = ( 3)2 + (-1)2  3 = tan-1 -1 = 30
3
r3 = 2
z1 = 2(cos 330 + i sin 330) 3 = 330
For z2 = 2 + 2i z3 = 2(cos 330 + i sin 330)
z1 z2 z3 = 2(cos 315 + i sin 315)

r2 = 22 + 22  2 = tan-1 2 = 45
2 ⋅2(cos 60 + i sin 60)
r2 = 2 2 
 2 = 45 ⋅2(cos 330 + i sin 330)
z2 = 2 2(cos 45 + i sin 45) z1 z2 z3 = 4 2[cos(315 + 60 + 330)

+ i sin(315 + 60 + 330)]
For z3 = 2 - 2i 3
z1 z2 z3 = 4 2(cos 705 + i sin 705)
2
r3 = 2 + (-2 3) 2 -1
 3 = tan -2 3 = 60 z1 z2 z3 = 4 2(cos 345 + i sin 345)
2
r3 = 4 z1 z2 z3 » 5.4641-1.4641i
3 = 300
69. Perform indicated operation in trigonometric form, z1 = 4(cos 300 + i sin 300)
For z2 = 1- i 3
For z1 = 3 + i 3
r2 = 12 + (- 3)2  2 = tan-1 - 3 = 60
2 2 -1 3 = 45 1
r1 = ( 3) + ( 3) 1 = tan r2 = 2
3  2 = 300
r1 = 6
1 = 45
z2 = 2(cos 300 + i sin 300)
z1 = 6(cos 45 + i sin 45)
For z3 = 4 3 + 4i
For z2 = 1- i 3
r3 = (4 3)2 + 42  3 = tan-1 1 = 30
3
r2 = 12 + (- 3)2  2 = tan-1 - 3 = 60 r3 = 8
1 3 = 30
r2 = 2
 2 = 300 z3 = 8(cos 30 + i sin 30)
z2 = 2(cos 300 + i sin 300) z1
For z3 = 2 - 2i z2 z3
4(cos 300+i sin 300)⋅2(cos 300+i sin 300)
=
r3 = 22 + (-2)2  3 = tan-1 -2 = 45 8(cos 30+i sin 30)
2
r3 = 2 2 3 = 315 z1
= 4 ⋅ 2 [cos(300 + 300- 30)
z2 z3 8
z3 = 2 2(cos 315 + i sin 315) + i sin(300 + 300- 30)]
z1 z1
= (cos 210 + i sin 210) = - 3 - i
z2 z3 z2 z3 2 2
6(cos 45+i sin 45) 71. Perform indicated operation in trigonometric form,
=
2(cos 300+i sin 300)⋅2 2(cos 315+i sin 315) write answer in standard form.
z1 6(cos 45 + i sin 45) For z1 = 1- 3i
=
z2 z3 4 2[cos(300 + 315) + i sin(300 + 315)]
r1 = 12 + (-3)2 1 = tan-1 -3 » 71.57
z1 6(cos 45 + i sin 45) 1
=
z2 z3 4 2(cos 255 + i sin 255) r1 = 10 
1 = 288.43
z1
= 3 [cos(45- 255) + i sin(45- 255)] z1 = 10(cos 288.4 + i sin 288.4)
z2 z3 4
z1 æ ö For z2 = 2 + 3i
= 3 (cos 210- i sin 210) = 3 çç- 3 + i ÷÷÷
z2 z3 4 4 è 2 2ø
r2 = 22 + 32  2 = tan-1 3 » 56.31
=-3+ 3 i 2
8 8 r2 = 13  2 = 56.31
70. Perform indicated operation in trigonometric form,
z2 = 13(cos 56.3 + i sin 56.3)
For z3 = 4 + 5i
For z1 = 2 - 2i 3
r3 = 42 + 52  3 = tan1 5 » 51.34
r1 = 22 + (-2 3)2 1 = tan-1 -2 3 = 60 4
2 r3 = 41 3 = 51.34
r1 = 4
1 = 300
z3 = 41(cos51.3 + i sin 51.3)
Section 7.5 553
z1 z2 z3 = 10(cos 288.4 + i sin 288.4) 73. z = r (cos  + i sin  ) z = r (cos  - i sin  )
⋅ 13(cos 56.3 + i sin 56.3) z ⋅ z = r (cos  + i sin  ) ⋅ r (cos  - i sin  )

⋅ 41(cos51.3 + i sin 51.3) z ⋅ z = r (cos  + i sin  ) ⋅ r[cos(- ) + i sin(- )]
z ⋅ z = r 2 [cos( - ) + i sin( -  )]
z1 z2 z3 = 10 ⋅ 13 ⋅ 41[cos(288.43 + 56.31
z ⋅ z = r 2 (cos 0 + i sin 0)
+ 51.34) + i sin(288.43 + 56.31 + 51.34)]
z1 z2 z3 » 73.0(cos 396.08 + i sin 396.08) z ⋅ z = r 2 or a 2 + b2
z1 z2 z3 = 73.0(cos 36.08 + i sin 36.08) 74. z = r (cos  + i sin  ) z = r (cos  - i sin  )
z1 z2 z3 » 59.0 + 43.0i
z = r (cos  + i sin  )
72. Perform indicated operation in trigonometric form, z r (cos  - i sin  )
write answer in standard form. = cos  + i sin 
cos(- ) + i sin(- )
For z1 = 2 - 5i = cos( +  ) + i sin( +  )
= cos 2 + i sin 2
r1 = 22 + (-5)2 1 = tan-1 -5 » 68.1986
2
r1 = 29 75. (2 cis 60 )6 = 26 (cis 60 )6
1 = 291.8014
26 (cis 60 + 60 + 60 + 60 + 60 + 60 )
z1 = 29 cis 291.8014
= 26 éëcis 6(60 )ùû
For z2 = 1- 6i
= 26 (cis 360 )
r2 = 12 + (-6)2  2 = tan-1 -6 » 80.5377 = 64(cos360 + i sin360 )

1
= 64 + 0i
r2 = 37
 2 = 279.4623
76. (3 cis 45 ) 4 = 34 (cis 45 )4
z2 = 37 cis 279.4623
= 34 (cis 45 + 45 + 45 + 45 )
For z3 = 3 + 4i = 34 éë cis 4(45 )ùû
r3 = 32 + 42  3 = tan-1 4 » 53.1301 = 34 (cis 180 )

3
r3 = 5 
= 81(cos180 + i sin180 )
 3 = 53.1301
= -81 + 0i
z3 = 5 cis 53.1301 Prepare for Section 7.5
z1 z2 P1. Simplify.
z3
æ 2 ö2
çç + 2 i ÷÷÷ = 2 + 2 2 i + 2 i 2 = i
= 29 cis 291.8014⋅ 37 cis 279.4623 çè 2 2 ø 4 4 4
5 cis 53.1301
29 ⋅ 37 P2. Find the real root.
= cis (291.8014 + 279.4623- 53.1301)
5
x 3 - 8 = ( x - 2)( x 2 + 2 x + 4)
= 29 ⋅ 37 (cos 518.1336 + i sin 518.1336)
5 ( x 2 + 2 x + 4) yields 2 complex solutions
» -6.0800 + 2.4400i ( x - 2) yields 1 real solution
The real root is 2.
P3. Find the real root. 7. Find the indicated power.
x 5 - 243 = ( x - 3)( x 4 + 3x 3 + 9 x 2 + 27 x + 81) [5(cos10 + i sin10)]3

( x 4 + 3x 3 + 9 x 2 + 27 x + 81) yields 4 = 53 [cos(3 ⋅10) + i sin(3 ⋅10)]
complex solutions = 125[cos 30 + i sin 30 ]
( x - 3) yields 1 real solution æ ö
= 125çç 3 + i 1 ÷÷÷
The real root is 3. è 2 2ø
P4. Write 2 + 2i in trigonometric form. = 125 3 + 125 i
2 2
r = 22 + 22 = 2 2 8. Find the indicated power.
 = tan-1 2 [3(cos 45 + i sin 45)]4

2
= 34 [cos(4 ⋅ 45) + i sin(4 ⋅ 45)]
= tan-1 1 = 45
= 81(cos180 + i sin180)
 = 45
= 81(-1 + 0i )
z = 2 2 cis 45 or 2 2 cis  = -81 + 0i = -81
4
9. Find the indicated power.
P5. Write in standard form.
æ ö [cis(150)]4 = cis(4 ⋅150)
2(cos150 + i sin150 ) = 2 çç- 3 + 1 i ÷÷÷ = - 3 + i
 
= cos 600 + i sin 600
è 2 2 ø
= cos 240 + i sin 240
P6. Find the absolute value.
=-1 - 3 i
æ ö2 æ ö2 2 2
z = çç 2 ÷÷÷ + çç- 2 ÷÷÷ = 2 + 2 = 1
è 2 ø è 2 ø 4 4 10. Find the indicated power.
Section 7.5 Exercises [2cis(330)]4 = 24 cis(4 ⋅ 330)
1. Determine the number of solutions. = 16(cos1320 + i sin1320)
There is just 1 solution, since z is to the power of 1. = 16(cos 240 + i sin 240)
2. Determine the number of solutions. = -8 - 8i 3
There are 6 solutions since z is to the power of 6. 11. Find the indicated power.
3. Yes, the student is correct. [2cis (120)]6 = 26 cis (6 ⋅ 2 3)
4. The statement is true. = 64(cos 720 + i sin 720)
5. Find the indicated power. = 64(cos 0 + i sin 0)
= 64
[2(cos 30 + i sin 30)]8 = 28 [cos(8 ⋅ 30) + i sin(8 ⋅ 30)]
= 256(cos 240 + i sin 240) 12. Find the indicated power.
= -128 -128i 3 [4cis(150)]3 = 43 cis(3 ⋅ 5 6 )
6. Find the indicated power. = 64(cos 450 + i sin 450)
= 64(cos 90 + i sin 90)
(cos 60 + i sin 60)3 = cos(3 ⋅ 60) + i sin(3 ⋅ 60)
= 64(0 + 1i ) = 0 + 64i
= cos180 + i sin180
= 64i
= -1 + 0i
= -1
Section 7.5 555
13. Find the indicated power. 16. Find the indicated power.
z = 3 -i z = 2 - 2i 3
2
r= ( 3 ) + (-1)2  = tan-1 -1 = 30 r = 22 + (-2 3)2  = tan-1 -2 3 = 60
3 2
r=2 r=4
 = 360 - 30 = 330  = 300
z = 2(cos 330 + i sin 330) z = 4(cos 300 + i sin 300)
4 4
( 3 - i ) = [2(cos 330 + i sin 330)] (2 - 2i 3) = [4(cos 300 + i sin 300)]3
3
= 24 [cos(4 ⋅ 330) + i sin(4 ⋅ 330)] = 43[cos(3 ⋅ 300) + i sin(3 ⋅ 300)]

= 16(cos1320 + i sin1320) = 64(cos 900 + i sin 900)
= 16(cos 240 + i sin 240) = 64(cos180 + i sin180)
æ ö = -64 + 0i = -64
= 16 çç- 1 - 3 ÷÷÷
è 2 2 ø
= -8 - 8i 3
z = 2 + 2i
z = 1- i
r = 22 + 22  = tan-1 2 = 45
2
r=2 2 
 = 45
r = 12 + (-1)2  = tan-1 -1 = 45
1
r= 2  z = 2 2(cos 45 + i sin 45)
 = 315
(2 + 2i ) = [2 2(cos 45 + i sin 45)]7
7
z = 2(cos 315 + i sin 315)
= 1024 2[cos(7 ⋅ 45) + i sin(7 ⋅ 45)]
(1- i ) = [ 2(cos 315 + i sin 315)]8
8
= 1024 2(cos 315 + i sin 315)
= ( 2)8 [cos(8 ⋅ 315) + i sin(8 ⋅ 315)] = 1024 -1024i
= 16(cos 2520 + i sin 2520)
= 16(cos 0 + i sin 0)
= 16 - 0i = 16 z = 2 3 - 2i

r = (2 3)2 + (-2)2  = tan-1 -2 = 30
z = 1+ i 2 3
r=4

 = 330
r = 12 + 12  = tan-1 1 = 45
1 z = 4(cos 330 + i sin 330)
r= 2 
 = 45 (2 3 - 2i ) = [4(cos 330 + i sin 330)]5
5
z = 2(cos 45 + i sin 45) = 45 [cos(5 ⋅ 330) + i sin(5 ⋅ 330)]

(1 + i )4 = [ 2(cos 45 + i sin 45)]4 = 1024(cos1650 + i sin1650)
= ( 2)4 [cos(4 ⋅ 45) + i sin(4 ⋅ 45)] = 1024(cos 210 + i sin 210)
= 4(cos180 + i sin180) = -512 3 - 512i
= -4 + 0i = -4
19. Find the indicated power. 22. Find all the indicated roots.
z = 2 +i 2 16 = 16(cos 0 + i sin 0 )
2 2
æ     ö
2 2 wk = 161/2 ççcos 0 + 360 k + i sin 0 + 360 k ÷÷÷ k = 0,1
2
r = ( 2 2) + ( 2 2) 2
 = tan -1
= 45 è 2 2 ø
2 2
r =1 w0 = 4(cos0 + i sin 0 )
 = 45
= 4 + 0i = 4
z = cos 45 + i sin 45 æ    ö
w1 = 4ççcos 0 + 360 + i sin 0 + 360 ÷÷÷
æ 2 6 è 2 2 ø
ç - i 2 ö÷÷ = (cos 45 + i sin 45)6
çè 2 2 ÷ø = 4(cos180 + i sin180 )
= cos(6 ⋅ 45) + i sin(6 ⋅ 45) = -4 + 0i = -4
= cos 270 + i sin 270
23. Find all the indicated roots.
= 0 -1i = -i
64 = 64(cos 0 + i sin 0)
z =- 2 +i 2
(
wk = 641/6 cos 0 + 360k + i sin 0 + 360k
6 6 )
2 2
k = 0, 1, 2, 3, 4, 5
2 2
r = (- 2 2)2 + ( 2 2)2  = tan-1 = 45 w0 = 2(cos 0 + i sin 0)
- 2 2 = 2 + 0i = 2
r =1
 = 135
z = cos135 + i sin135 (
w1 = 2 cos 0 + 360 + i sin 0 + 360
6 6 )
æ ö12 = 2(cos 60 + i sin 60)
ç- 2 - i 2 ÷÷ = (cos135 + i sin135)12
çè 2 2 ÷ø æ ö
= 2 çç 1 + i 3 ÷÷÷
= cos(12 ⋅135) + i sin(12 ⋅135) è2 2 ø
= cos1620 + i sin1620 = 1+ i 3
= cos180 + i sin180
= -1 + 0i = -1 (
w2 = 2 cos 0 + 360⋅ 2 + i sin 0 + 360⋅ 2
6 6 )
21. Find all the indicated roots. = 2(cos120 + i sin120)
æ ö
9 = 9(cos 0 + i sin 0 ) = 2 çç- 1 + i 3 ÷÷÷
è 2 2 ø
æ     ö = -1 + i 3
wk = 91/2 ççcos 0 + 360 k + i sin 0 + 360 k ÷÷÷ k = 0,1
è 2 2 ø
w0 = 3(cos 0 + i sin 0 ) (
w3 = 2 cos 0 + 360⋅ 3 + i sin 0 + 360⋅ 3
6 6 )
= 3 + 0i = 3 = 2(cos180 + i sin180)
æ    ö = 2(-1 + 0i )
w1 = 3ççcos 0 + 360 + i sin 0 + 360 ÷÷÷ = -2 + 0i = -2
è 2 2 ø
 
= 3(cos180 + i sin180 )
= -3 + 0i = -3 (
w4 = 2 cos 0 + 360⋅ 4 + i sin 0 + 360⋅ 4
6 6 )
= 2(cos 240 + i sin 240)
æ ö
= 2 çç- 1 - i 3 ÷÷÷
è 2 2 ø
= -1- i 3
Section 7.5 557
(
w5 = 2 cos 0 + 360⋅ 5 + i sin 0 + 360⋅ 5
6 6 ) w1 = cos 180 + 360 + i sin 180 + 360
5 5
= 2(cos 300 + i sin 300) = cos108 + i sin108
æ ö » -0.309 + 0.951i
= 2 çç 1 - i 3 ÷÷÷
è2 2 ø
w2 = cos 180 + 360⋅ 2 + i sin 180 + 360⋅ 2
= 1- i 3 5 5
= cos180 + i sin180
= -1 + 0i = -1
(
32 = 32 cos 0 + i sin 0 )
w3 = cos 180 + 360⋅ 3 + i sin 180 + 360⋅ 3
5 5
5(
wk = 321/5 cos 0 + 360k + i sin 0 + 360k
5 ) = cos 252 + i sin 252
k = 0, 1, 2, 3, 4 » -0.309 - 0.951i
w0 = 2(cos 0 + i sin 0) w4 = cos 180 + 360⋅ 4 + i sin 180 + 360⋅ 4

5 5
= 2 + 0i = 2 = cos 324 + i sin 324
(
w1 = 2 cos 0 + 360 + i sin 0 + 360
5 5 ) » 0.809 - 0.588i
= 2(cos 72 + i sin 72)
-16 = 16(cos180 + i sin180)
» 2(0.3090 + 0.9511i )
= 0.6180 + 1.9021i
(
wk = 161/4 cos 180 + 360k + i sin 180 + 360k
4 4 )
(
w2 = 2 cos 0 + 360⋅ 2 + i sin 0 + 360⋅ 2
5 5 ) k = 0, 1, 2, 3
= 2(cos144 + i sin144) w0 = 2(cos 45 + i sin 45)

» 2(-0.8090 + 0.5878i ) æ ö
= 2 ççè 2 + i 2 ÷÷ø÷
= -1.6180 + 1.1756i 2 2
= 2 +i 2
(
w3 = 2 cos 0 + 360⋅ 3 + i sin 0 + 360⋅ 3 )
5
= 2(cos 216 + i sin 216)
5
(
w1 = 2 cos 180 + 360 + i sin 180 + 360
4 4 )
» 2(-0.8090 - 0.5878i ) = 2(cos135 + i sin135)
= -1.6180 -1.1756i æ ö
= 2 ççè- 2 + i 2 ÷÷ø
2 2
(
w4 = 2 cos 0 + 360⋅ 4 + i sin 0 + 360⋅ 4
5 5 ) =- 2 +i 2
= 2(cos 288 + i sin 288)
» 2(0.3090 - 0.9511i ) (
w2 = 2 cos 180 + 360⋅ 2 + i sin 180 + 360⋅ 2
4 4 )
= 0.6180 -1.9021i = 2(cos 225 + i sin 225)
æ ö
25. Find all the indicated roots. = 2 ççè- 2 - i 2 ÷÷ø
2 2
-1 = 1(cos180 + i sin180)
= - 2 -i 2
wk = 11/5
( cos 180 + 360k + i sin 180 + 360k
5 5 ) (
w3 = 2 cos 180 + 360⋅ 3 + i sin 180 + 360⋅ 3
4 4 )
k = 0, 1, 2, 3, 4
= 2(cos 315 + i sin 315)
w0 = 1(cos 36 + i sin 36) æ ö
= 2 ççè 2 - i 2 ÷÷ø
» 0.809 + 0.588i 2 2
= 2 -i 2
1 = cos 0 + i sin 0
(
w0 = 21/8 cos 45 + i sin 45
4 4 )
= 21/8 (cos11.25 + i sin11.25)
wk = cos 0 + 360k + i sin 0 + 360k k = 0, 1, 2 » 1.070 + 0.213i
3 3
æ    ö
w0 = cos 0 + i sin 0 w1 = 21/8 ççcos 45 + 360 + i sin 45 + 360 ÷÷÷
3 3 è 4 4 ø
= cos 0 + i sin 0 = 21/8 (cos101.25 + i sin101.25)
= 1 + 0i = 1 » -0.213 -1.070i
w1 = cos 0 + 360 + i sin 0 + 360 æ     ö

w2 = 21/8 ççcos 45 + 360 ⋅ 2 + i sin 45 + 360 ⋅ 2 ÷÷÷
3 3 è 4 4 ø
= cos120 + i sin120
= 21/8 (cos191.25 + i sin191.25)
=-1 + 3 i » -1.070 - 0.213i
2 2
æ     ö
w3 = 21/8 ççcos 45 + 360 ⋅ 3 + i sin 45 + 360 ⋅ 3 ÷÷÷
   
w2 = cos 0 + 360 ⋅ 2 + i sin 0 + 360 ⋅ 2 è 4 4 ø
3 3
= cos 240 + i sin 240 = 21/8 (cos 281.25 + i sin 281.25)
» 0.213 -1.070i
=-1- 3 i
2 2 30. Find all the indicated roots.
-1 + i = 2(cos135 + i sin135)
i = cos 90 + i sin 90
wk = cos 90 + 360k + i sin 90 + 360k k = 0,1,2

wk = ( 2 )
1/5
(cos 135 +5360k + i sin 135 +5360k )
3 3 k = 0, 1, 2, 3, 4
 
w0 = cos 90 + i sin 90 æ  ö
3 3 w0 = 21/10 ççcos 135 + i sin 135 ÷÷÷
è 5 5 ø
= cos 30 + i sin 30
= 21/10 (cos 27 + i sin 27)
= 3+1i » 0.955 + 0.487i
2 2
æ    ö
    w1 = 21/10 çççcos 135 + 360 + i sin 135 + 360 ÷÷÷
w1 = cos 90 + 360 + i sin 90 + 360 è 5 5 ø
3 3
= 21/10 (cos 99 + i sin 99)
= cos150 + i sin150
» -0.168 + 1.059i
=- 3 + 1 i æ     ö
w2 = 21/10 ççcos 135 + 360 ⋅ 2 + i sin 135 + 360 ⋅ 2 ÷÷÷
2 2
è 5 5 ø
   
w2 = cos 90 + 360 ⋅ 2 + i sin 90 + 360 ⋅ 2 = 21/10 (cos171 + i sin171)
3 3
= cos 270 + i sin 270 » -1.059 + 0.168i
= 0 - i = -i
(
w3 = 21/10 cos 135 + 360⋅ 3 + i sin 135 + 360⋅ 3
5 5 )
= 21/10 (cos 243 + i sin 243)
1 + i = 2(cos 45 + i sin 45)
» -0.487 - 0.955i
wk = ( 2 )
1/4
( cos 45 + 360k + i sin 45 + 360k
4 4 ) æ     ö
w4 = 21/10 ççcos 135 + 360 ⋅ 4 + i sin 135 + 360 ⋅ 4 ÷÷÷
è 5 5 ø
k = 0, 1, 2, 3
= 21/10 (cos 315 + i sin 315)
» 0.758 - 0.758i
Section 7.5 559
(
w0 = 4 2 cos 120 + i sin 120
2 2 )
2 - 2i 3 = 4(cos 300 + i sin 300) k = 0, 1, 2
= 4 2(cos 60 + i sin 60)
wk = 4 1/3
( cos 300 + 360k + i sin 300 + 360k
3 3 ) » 2 2 + 2i 6
æ    ö
w1 = 4 2 ççcos 120 + 360 + i sin 120 + 360 ÷÷÷
w0 = 41/3 (cos 3003  + i sin 3003  ) è 2 2 ø
= 4 2(cos 240 + i sin 240)
= 41/3 (cos100 + i sin100)
» -0.276 + 1.563i » -2 2 - 2i 6
æ    ö 34. Find all the indicated roots.

w1 = 41/3 ççcos 300 + 360 + i sin 300 + 360 ÷÷÷
è 3 3 ø -1 + 3i = 2(cos120 + i sin120)
= 41/3 (cos 220 + i sin 220)
» -1.216 -1.020i
(
wk = 21/2 cos 120 + 360k + i sin 120 + 360k
2 2 )
k = 0, 1
(
w2 = 41/3 cos 300 + 360⋅ 2 + i sin 300 + 360⋅ 2
3 3 ) (
w0 = 21/2 cos 120 + i sin 120
2 2 )
1/3
= 4 (cos 340 + i sin 340) = 21/2 (cos 60 + i sin 60)
» 1.492 - 0.543i
= 2+ 6i
32. Find all the indicated roots. 2 2
-2 + 2i 3 = 4(cos120 + i sin120) (
w1 = 21/2 cos 120 + 360 + i sin 120 + 360
2 2 )
( )
1/2
wk = 41/3 cos 120 + 360k + i sin 120 + 360k = 2 (cos 240 + i sin 240)
3 3
k = 0, 1, 2 =- 2 - 6 i
2 2
æ  ö 35. Find all the roots.
w0 = 41/3 ççcos 120 + i sin 120 ÷÷÷
è 3 3 ø
x3 + 8 = 0
1/3
= 4 (cos 40 + i sin 40) x 3 = -8
» 1.216 + 1.020i
Find the three cube roots of –8.
w1 = 4 1/3
( cos 120 + 360 + i sin 120 + 360
3 3 ) æ
è
  ö
xk = 81/3 çççcos 180 + 360k + i sin 180 + 360k ÷÷÷
ø
3 3
= 41/3 (cos160 + i sin160) k = 0, 1, 2
» -1.492 + 0.543i
æ  ö
w0 = 2 ççcos 180 + i sin 180 ÷÷÷
w2 = 4 1/3
( cos 120 + 360⋅ 2 + i sin 120 + 360⋅ 2
3 3 ) è 3
= 2(cos 60 + i sin 60)
3 ø
= 41/3 (cos 280 + i sin 280) = 2 cis 60

» 0.276 -1.563i
(
w1 = 2 cos 180 + 360 + i sin 180 + 360
3 3 )
= 2(cos180 + i sin180)
-16 + 16i 3 = 32(cos120 + i sin120)
= 2 cis 180
(
wk = 321/2 cos 120 + 360k + i sin 120 + 360k
2 2 ) (
w2 = 2 cos 180 + 360⋅ 2 + i sin 180 + 360⋅ 2
3 3 )
k = 0, 1
= 2(cos 300 + i sin 300)
= 2 cis 300
36. Find all the roots.
(
wk = 21/3 cos 90 + 360k + i sin 90 + 360k
3 3 )
x5 - 32 = 0
x 5 = 32 (
= 3 2 cos 90 + 360k + i sin 90 + 360k
3 3 )
Find the five fifth roots of 32. k = 0, 1, 2
32 = 32(cos 0 + i sin 0)
w0 = 3 2 cis 90 = 3 2 cis 30
3
(
wk = 321/5 cos 0 + 360k + i sin 0 + 360k
5 5 ) w1 = 3 2 cis 90 + 360 = 3 2 cis 150
k = 0, 1, 2, 3, 4 3
w2 = 3 2 cis 90+360⋅ 2 = 3 2 cis 270

w0 = 2 cis 0 = 2 cis 0 3
5
 
w1 = 2 cis 0 + 360 = 2 cis 72
5 x 3 - 27 = 0
  x 3 = 27
w2 = 2 cis 0 + 360 ⋅ 2 = 2 cis 144
5 Find the three cube roots of 27.
 
w3 = 2 cis 0 + 360 ⋅ 3 = 2 cis 216 27 = 27(cos 0 + i sin 0)
5
 
w4 = 2 cis 0 + 360 ⋅ 4 = 2 cis 288
(
wk = 3 cos 0 + 360k + i sin 0 + 360k
3 3 ) k = 0, 1, 2
5
37. Find all the roots. w0 = 3 cis 0 = 3 cis 0
3
x4 + i = 0
w1 = 3 cis 0 + 360 = 3 cis 120
4
x = -i 3
Find the four fourth roots of –i. w2 = 3 cis 0 + 360⋅ 2 = 3 cis 240
3
-i = (cos 270 + i sin 270 ) 40. Find all the roots.
wk = cos 270 + 360k + i sin 270 + 360k x 5 + 32i = 0

4 4
k = 0, 1, 2, 3 x 5 = -32i
Find the five fifth roots of –32i.
w0 = cis 270 = cis 67.5
4 -32i = 32(cos 270 + i sin 270)
w1 = cis 270 + 360 = cis 157.5

4 (
wk = 321/5 cos 270 + 360k + i sin 270 + 360k
5 5 )
k = 0, 1, 2, 3, 4
w2 = cis 270 + 360⋅ 2 = cis 247.5
4
w0 = 2 cis 270 = 2 cis 54
5
w3 = cis 270 + 360⋅ 3 = cis 337.5 270  + 360 = 2 cis 126
4 w1 = 2 cis
5
38. Find all the roots. 270  + 360⋅ 2 = 2 cis 198
w2 = 2 cis
5
x 3 - 2i = 0 270  + 360⋅ 3 = 2 cis 270
w3 = 2 cis
x 3 = 2i 5
Find the three cube roots of 2i. w4 = 2 cis 270 + 360⋅ 4 = 2 cis 342
5
2i = 2(cos 90 + i sin 90)
Section 7.5 561
41. Find all the roots. æ     ö
wk = 21/4 ççcos 300 + 360 k + i sin 300 + 360 k ÷÷÷
è 4 4 ø
x 4 + 81 = 0
k = 0, 1, 2, 3
x 4 = -81
Find the four fourth roots of –81. w0 = 4 2 cis 300
4
-81 = 81(cos180 + i sin180) 4
= 2 cis 75
(
wk = 811/4 cos 180 + 360k + i sin 180 + 360k
4 4 ) w1 = 4 2 cis 300 + 360
4
4
k = 0, 1, 2, 3 = 2 cis 165
w0 = 3 cis 180 = 3 cis 45 w2 = 4 2 cis 300 + 360⋅ 2

4 4
= 4 2 cis 255
w1 = 3 cis 180 + 360 = 3 cis 135
4
w3 = 4 2 cis 300 + 360⋅ 3
4
w2 = 3 cis 180 + 360⋅ 2 = 3 cis 225 4
4 = 2 cis 345
w3 = 3 cis 180 + 360⋅ 3 = 3 cis 315 44. Find all the roots.
4
x 3 + (2 3 - 2i ) = 0
x 3 = -2 3 + 2i
3
x - 64i = 0
Find the three cube roots of -2 3 + 2i.
x 3 = 64i
Find the three cube roots of 64i. -2 3 + 2i = 4(cos150 + i sin150)
64i = 64(cos 90 + i sin 90)

(
wk = 41/3 cos 150 + 360k + i sin 150 + 360k
3 3 )
(
wk = 641/3 cos 90 + 360k + i sin 90 + 360k
3 3 ) k = 0, 1, 2
k = 0, 1, 2 w0 = 3 4 cis 150
3
w0 = 4 cis 90 3
= 4 cis 50
3
= 4 cis 30 w1 = 3 4 cis 150 + 360
3
w1 = 4 cis 90 + 360 = 3 4 cis 170
3
= 4 cis 150
w2 = 3 4 cis 150 + 360⋅ 2
3
w2 = 4 cis 90 + 360⋅ 2 3
3 = 4 cis 290
= 4 cis 270 45. Find all the roots.
x 3 + (1 + i 3) = 0
x 4 - (1- i 3) = 0
x 3 = -1- i 3
4
x = 1- i 3
Find the three cube roots of -1- i 3.
Find the four fourth roots of 1- i 3.
-1- i 3 = 2(cos 240 + i sin 240)
1- i 3 = 2(cos 300 + i sin 300)
(
wk = 21/3 cos 240 + 360k + i sin 240 + 360k
3 3 ) 47. Let z = a + bi. Then z = a - bi by definition.
Substitute a = r cos  and b = r sin  .
k = 0, 1, 2
Thus z = rcos - ri sin  = r (cos  - i sin  )
w0 = 2 cis 240
3
3 48. 1= 1 = cos - i sin
3
= 2 cis 80 z r (cos + i sin ) r (cos + i sin )(cos - i sin )
= cos - i sin = cos - i sin
w1 = 3 2 cis 240 + 360 r (cos2  - i 2 sin 2  ) r (cos2  + sin 2  )
3
3
= 2 cis 200 z-1 = r-1 (cos - i sin )
49. z = r (cos + i sin )
w2 = 3 2 cis 240 + 360⋅ 2
3 z 2 = r 2 (cos2 + i sin2 )
3
= 2 cis 320 1 = 1
46. Find all the roots. z 2 r 2 (cos2 + i sin2 )
= 2 cos2 - i sin2
x 6 -(4 - 4i ) = 0 r (cos2 + i sin2 )(cos2 - i sin2 )
x 6 = 4 - 4i = 2 cos2  - i sin2 = cos2 - i sin2
Find the six sixth roots of 4 - 4i. r (cos2 2 - i 2 sin 2 2 ) r 2 (cos2 2 + sin 2 2 )
z-2 = r-2 (cos2 - i sin2 )
4 - 4i = 4 2(cos 315 + i sin 315)
50. Exercises 48 and 49 imply that
wk = (4 2 )
1/6
(
cos 315 + 360k + i sin 315 + 360k
6 6 ) z-n = r-n (cos n - i sin n ).
k = 0, 1, 2, 3, 4, 5
z = 1- i 3 = 2(cos 300 + i sin 300)
1/6
w0 = (4 2 ) cis 315 z -4
= 2-4 [cos 4(300) - i sin 4(300)]
6
1/6 z-4 = 16
1 (cos1200- i sin1200)
= (4 2 ) cis 52.5
z-4 = 16
1 (cos120- i sin120)
1/6
w1 = (4 2 ) cis 315 + 360
6 z-4 = 16
1 [cos(-120) + i sin(-120)]
1/6
= (4 2 ) cis 112.5 z-4 = 16
1 cis ( -120)
1/6
w2 = (4 2 ) cis 315 + 360⋅ 2 51. For n = 2, the two square roots of 1 are
6
1/6 1 and –1.
= (4 2 ) cis 172.5
The sum of these roots is
1/6
w3 = (4 2 ) cis 315 + 360⋅ 3 1 + (–1) = 0.
6
1/6 For n = 3, the three cube roots of 1 are
= (4 2 ) cis 232.5
(from exercise 23)
1/6
w4 = (4 2 ) cis 315 + 360⋅ 4
6 1, - 1 + 3 i and - 1 - 3 i .
1/6
2 2 2 2
= (4 2 ) cis 292.5
1/6
w5 = (4 2 ) cis 315 + 360⋅ 5
6 1- 1 + 3 i - 1 - 3 i = 0 .
2 2 2 2
1/6
= (4 2 ) cis 352.5
Section 7.5 563
For n = 4, the four fourth roots of 1 are The product of these roots is
1, –1, i, and –i. æ öæ öæ öæ ö
1⋅(-1)⋅çèç- 1 + 3 i÷÷÷ø⋅çèç- 1 - 3 i÷÷÷ø⋅çèç 1 + 3 i÷÷÷ø⋅èçç 1 - 3 i÷÷÷ø
The sum of these roots is 2 2 2 2 2 2 2 2
=-1
1-1 + i - i = 0
For n ³ 2 , the product of the nth roots of 1 is –1 if n is
For n = 5, the five fifth roots of 1 are
even and 1 if n is odd.
1, cis 72 , cis 144 , cis 216 , cis 288
Exploring Concepts with Technology
1. Find the force using WolframAlpha.
1 + cis 72 + cis 144 + cis 216 + cis 288 = 0
a = A cos 97.5, sin 97.5
For n = 6, the six sixth roots of 1 are
b = B cos10.1, sin10.1
1, –1, - 1 + 3 i , - 1 - 3 i , 1 + 3 i , 1 - 3 i
2 2 2 2 2 2 2 2 a + b + c = 0, 0
The sum of these roots is 0, 0 = A cos 97.5, sin 97.5
+ B cos10.1, sin10.1 + 0, -58
1-1 - 1 + 3 i - 1 - 3 i + 1 + 3 i + 1 - 3 i = 0
2 2 2 2 2 2 2 2
Using WolframAlpha, A » 57.16 and B » 7.57832 .
For n ³ 2 , the sum of the nth roots of 1 is 0.
The force exerted by cable A is about 57.2 lb.
52. For n = 2, the two square roots of 1 are
The force exerted by cable B is about 7.58 lb.
1 and –1.
2. a. Find the resultant force using WolframAlpha.
The product of these roots is
F1 = 4250 cos 0, sin 0
1⋅ (-1) = -1 .
F2 = 3640 cos 32.4, sin 32.4
For n = 3, the three cube roots of 1 are
Using WolframAlpha, F1 + F2 » 7323.35, 1950.41 .
(from exercise 27)
The resultant force is approximately 7323, 1950 .
1, - 1 + 3 i and - 1 - 3 i .
2 2 2 2 b. Find the magnitude and direction angle of resultant.
The product of these roots is From WolframAlpha, vector length is approximately
æ öæ ö 7578.63 and the angle is 14.1933º.
1⋅çèç- 1 + 3 i ÷÷ø÷⋅çèç- 1 - 3 i ÷÷ø÷ = 1 .
2 2 2 2 The magnitude is approximately 7578 lb, and the
For n = 4, the four fourth roots of 1 are direction angle is approximately 14.9º.
1, –1, i, and –i. c. Find the magnitude and direction angle for all.
The product of these roots is F1 = 4250 cos 0, sin 0
1⋅(-1)⋅(i ) ⋅(-i ) = -1 F2 = 3640 cos 32.4, sin 32.4
For n = 5, the five fifth roots of 1 are F3 = 1420 cos8.0, sin 8.0
1, cis 72 , cis 144 , cis 216 , cis 288 Using WolframAlpha,
The product of these roots is F1 + F2 + F3 » 8929.53, 2148.04
1⋅(cis 72)⋅(cis 144) ⋅(cis 216)⋅(cis 288) = 1 vector length » 8989.93 .
For n = 6, the six sixth roots of 1 are angle » 13.8239
The magnitude is approximately 8989 lb, and the
1, –1, - 1 + 3 i , - 1 - 3 i , 1 + 3 i , 1 - 3 i
2 2 2 2 2 2 2 2 direction angle is approximately 13.8º.
3. Find the force using WolframAlpha. 2 2 2
cos B = 12 + 20 -15
F1 = 1280 cos119, sin119 2 (12)(20)
cos B » 0.6646
F2 = 945 cos153, sin153
B » 48
Using WolframAlpha, F1 + F2 , the magnitude is 2 2 2
cos C = 12 + 15 - 20
approximately 2130 lb, and the direction angle is ( )(
2 12 15)
approximately 133º. cos C » -0.0861

C » 95
4. Find the work using WolframAlpha.
A = 180- 48- 95
Enter 25{cos 35,sin 35}.{425, 0} .
A = 37
The work is approximately 8703 ft-lb.
4. Solve the triangle. [7.2]
Chapter 7 Review Exercises
1. Solve the triangle.
2 2 2
cos C = 24 + 32 - 28
C = 180 - 34.1 -104.5 = 41.4 2(24)(32)
cos C » 0.5313
a = c b = c
sin A sin C sin B sin C C » 58
a = 155 b = 155 2 2 2
sin104.5 sin 41.4 sin 34.1 sin 41.4 cos A = 32 + 28 - 24
2(32)(28)
 
a = 155sin104.5 b = 155sin 34.1 cos A » 0.6875
sin 41.4 sin 41.4
A » 47
a » 227 b » 131
B » 180- 58- 47 » 75
C = 180 - 44.4 - 37.7

C = 97.9
c 2 = 222 + 182 - 2 (22)(18) cos 35
a = b a = c c 2 » 159
sin A sin B sin A sin C
58.5 = b 58.5 = c c = 159
sin 44.4 sin 37.7 sin 44.4 sin 97.9 c » 13
 
b = 58.5sin 37.7

c = 58.5sin 97.9

18 = 159
sin 44.4 sin 44.4 sin A sin 35
b » 51.1 c » 82.8 sin A = 18sin 35
159
sin A » 0.8188
A » 55
B » 180 - 35- 55
B » 90
Chapter Review Exercises 565
6. Solve the triangle. [7.2] 9. Solve the triangle.
c = b
2 2 2 sin C sin B
a = 102 + 150 - 2 (102)(150) cos82
54.4 = 32.5
a 2 » 28645 sin121.5 sin B
a = 28645 æ ö
B = sin-1 çç 32.5sin121.5 ÷÷÷
a » 169 è 54.4 ø
B » 30.6
150 = 28645
sin C sin 82 A = 180 - 30.6 -121.5 = 27.9
sin C = 150sin 82
28645 c = a
sin C sin A
sin C » 0.8776
54.4 = a
C » 61 sin121.5 
sin 27.9
B » 180- 61- 82 
a = 54.4 sin 27.9
B » 37 sin121.5
a » 29.9
10 = 8
sin C sin105
sin C = 10sin105 sin 45.2 = h
8 71.4
sin C » 1.207 h = 71.4sin 45.2
h » 50.7
No triangle is formed.
Since h < 71.4, two triangles exist.
a = b
sin A sin B
61.5 = 71.4
sin 45.2 sin B

sin B = 71.4sin 45.2 = 0.824
61.5
B » 55.5 or 124.5
110 = 80
sin B sin 55 For B = 55.5
sin B = 110sin 55 C = 180 - 45.2 - 55.5 = 79.3
80
sin B » 1.1263 c = 71.4
sin 79.3 sin 55.5
No triangle is formed. 
c = 71.4sin 79.3 = 85.1
sin 55.5
For B = 124.5 K = 1 bc sin A

2
C = 180 - 45.2 -124.5 = 10.3 1
K = (8.0)(12) sin 75
c 2
= 71.4
sin10.3 sin 55.5 K » 46 square units

c = 71.4 sin10.3 = 15.5 15. Find the area of the triangle. [7.2]
sin 55.5
11. Find the area of the triangle. [7.2]
s = 1 (a + b + c)
2
1
s = (24 + 30 + 36) 50 = 75
2 sin B sin15
s = 45
sin B = 50sin15
K = s ( s - a )( s - b)( s - c ) 75
sin B » 0.1725
K = 45(45 - 24)(45 - 30)(45 - 36)
B » 10
K = 127,575
A » 180-10-15
A » 155
K » 1 (50)(75) sin155
2
s = 1 (a + b + c)
2 K » 790 square units
1
s = (9.0 + 7.0 + 12)
2 16. Find the area of the triangle. [7.2]
s = 14
K = s ( s - a )( s - b)( s - c )
K = 14 (14 - 9.0)(14 - 7.0)(14 -12)
K = 980
K » 31 square units 18 = 25
sin B sin 68
sin B = 18sin 68
25
sin B » 0.6676
B » 42
C » 180- 42- 68
C » 70
K = 1 ab sin C
2 K » 1 (18)(25) sin 70
2
1
K = (60)(44) sin 44
2 K » 210 square units
K » 920 square units 17. Find the area of the triangle. [7.2]
15 = 32 22. Find the magnitude and direction. [7.3]
sin B sin110
 » tan-1 -3 = tan-1 1
2
sin B = 15sin110 v = 62 + (-3)
32 6 2
v = 36 + 9  » 26.6
sin B » 0.4405
B » 26 v » 6.7  » 360- 26.6
 » 333.4
C » 180-110- 26
C » 44 23. Find the magnitude and direction. [7.3]
α = tan-1 3 = tan-1 3
2
K » 1 (15)(32) sin 44 u = (-2) + 32
2 -2 2
K » 170 square units u = 4+9 α » 56.3
u » 3.6 θ » 180- 56.3
θ » 123.7
24. Find the magnitude and direction. [7.3]
 = tan-1 -7 = tan-1 7
2 2
u = (-4) + (-7)
-4 4
u = 16 + 49  » 60.3
K = 1 bc sin A u » 8.1  » 180 + 60.3
2
 » 240.3
1
K = (18)(22) sin 45
2 25. Find the unit vector. [7.3]
2
w = (-8) + 52
19. Find the components of each vector. Write an
w = 89
equivalent vector. [7.3]
u = -8 , 5 = - 8 89 , 5 89
Let P1P2 = a1i + a2 j. 89 89 89 89
a1 = 3 -(-2) = 5 A unit vector in the direction of
a2 = 7 - 4 = 3
w is u = - 8 89 , 5 89 .
A vector equivalent to P1P2 is v = 5, 3 . 89 89
26. Find the unit vector. [7.3]
20. Find the components of each vector. Write an
2
equivalent vector. [7.3] w = 7 2 + (-12)
Let P1P2 = a1i + a2 j. w = 193
a1 = -3 - (-4) = 1 u= 7 , -12 = 7 193 , - 12 193

193 193 193 193
a2 = 6 - 0 = 6
A unit vector in the direction of
A vector equivalent to P1P2 is v = 1, 6 .
21. Find the magnitude and direction. [7.3] w is u = 7 193 , - 12 193 .

193 193
 » tan-1 2 = tan-1 1
2
v = (-4) + 22
-4 2
v = 16 + 4  » 26.6
v » 4.5  » 180- 26.6
 » 153.4
27. Find the unit vector. [7.3] 33. Find the distance from A to C. [7.1]
From the drawing, we find A.
v = 52 + 12
v = 26 A = 360- 84.5- 222.1 = 53.4
From the drawing, the angle formed from the north
u = 5 i + 1 j = 5 26 i + 26 j
26 26 26 26 line, vertex B and point C is
A unit vector in the direction of NBC = 360- 332.4 = 27.6
We make a drawing by extending the north lines,
v is u = 5 26 i + 26 j.
26 26 which are parallel, at both A and B. We find the angle,
28. Find the unit vector. [7.3] 42.1º, by subtracting 180º from 222.1º.
2
v = 32 + (-5)
v = 34
u = 3 i - 5 j = 3 34 i - 5 34 j
34 34 34 34
A unit vector in the direction of

Next we see that alternate interior angles of 42.1º are
v is u = 3 34 i - 5 34 j. formed. Therefore, we can find angle B and then C.
34 34
B = 27.6 + 42.1 = 69.7
29. Perform the indicated operation. [7.3]
C = 180- A - B = 180- 53.4- 69.7 = 56.9
v - u = -4, -1 - 3, 2 = -7, -3
b = c
30. Perform the indicated operation. [7.3] sin B sin C
b = 345
2u - 3v = 2 3, 2 - 3 -4, -1 sin 69.7 sin 56.9
= 6, 4 - -12, -3 
b = 345sin 69.7
= 18, 7 sin 56.9
b » 386
31. Perform the indicated operation. [7.3]
The distance between A and C is about 386 miles.
-u + 1 v = -(10i + 6 j) + 1 (8i - 5j)
2 2 34. Find b. [7.1]
(
= (-10i - 6 j) + 4i - 5 j
2 ) C = 180- 85.0- 62.1
C = 32.9
(
= (-10 + 4) i + -6 - 5 j
2 ) b = c
17 sin B sin C
= -6i - j
2 b = 285
32. Perform the indicated operation. [7.3] sin 62.1 sin 32.9

2 v - 3 u = 2 (8i - 5j) - 3 (10i + 6 j) b = 285sin 62.1
3 4 3 4 sin 32.9
b » 464 ft
= (16
3
10
3 ) (
i - j - 15 i + 9 j
2 2 )
= (16 -
3 2
15
) ( 10
i+ - - j
3 2
9
)
= 32 - 45 i+ - 20 - 27 j
6 6
= - 13 i - 47 j
6 6
35. v = 400sin 204i + 400 cos 204 j [7.3] 42. Find the angle between the vectors. [7.3]
v » -162.7i - 365.4 j -5, 2 ⋅ 2, -4
w = -45i cos  =
(-5)2 + 22 22 + (-4)2
R = v+w
R » -162.7i - 365.4 j - 45i cos α = -10 - 8
R » -207.7i - 365.4 j 29 20
cos  » -0.7474
2
R » (-207.7) + (-365.4)
2  = 138
R » 420 mph 43. Find the angle between the vectors. [7.3]
The ground speed is approximately 420 mph. (6i -11j)⋅(2i + 4 j)

cos  =
2
62 + (-11) 22 + 42
36.  = sin-1 40 [7.3]
320
 » 7 cos  = 12 - 44
157 20
37. Find the dot product of the vectors. [7.3] cos  » -0.5711
u ⋅ v = 3, 7 ⋅ -1, 3 cos  » 125
= (3)(-1) + (7)(3) 44. Find the angle between the vectors. [7.3]
= 18 (i - 5j)⋅(i + 5j)
cos  =
38. Find the dot product of the vectors. [7.3] 2
12 + (-5) 12 + 52
v ⋅ u = -8, 5 ⋅ 2, -1
cos  = 1- 25
= (-8)(2) + (5)(-1) 26 26
= -21 cos  » -0.9231
39. Find the dot product of the vectors. [7.3]  » 157
v ⋅ u = (-4i - j)⋅(2i + j) 45. Find projw v . [7.3]
= (-4)(2) + (-1)(1)
projw v = v ⋅ w
= -9 w
40. Find the dot product of the vectors. [7.3] -2, 5 ⋅ 5, 4
projw v =
u ⋅ v = (-3i + 7 j) ⋅(-2i + 2 j) 52 + 42
= (-3)(-2) + (7)(2)
= -10 + 20 = 10
= 20 41 41
41. Find the angle between the vectors. [7.3] = 10 41
41
7, -4 ⋅ 2,3
cos  = 46. Find projw v . [7.3]
2 2
7 + (-4) 22 + 32
projw v = v ⋅ w
( )
cos  = 14 + -12
w
65 13
cos  » 0.0688 (4i - 7 j)⋅(-2i - 5j)
projw v =
 » 86 (-2)2 + (-5)2
= -8 + 35 = 27
29 29
= 27 29
29
47. w = F S cos  [7.3] 51. Write the number in trigonometric form. [7.4]
w = 60 ⋅14 cos 38 z = - 3 + 3i
w » 662 foot-pounds
2
r= (- 3 ) + 32 = 12 = 2 3
48. Find the modulus and the argument, and graph. [7.4]
2
 = tan-1 3 = tan-1 3
r = 22 + (-3) = 13
- 3 3
 = tan-1 -3 = tan-1 3  = 60
2 2
 = 180- 60
 » 56
 = 120
 » 360- 56
 » 304 z = 2 3 cis 120
52. Write the number in standard form. [7.4]
z = 5(cos 315 + i sin 315)
æ ö
z = 5çç 2 - i 2 ÷÷÷
è 2 2 ø
49. Find the modulus and the argument, and graph. [7.4] z = 5 2 -5 2 i
2 2
2 53. Write the number in standard form. [7.4]
r = (-5) + ( 3 )
2
r = 28 » 5.29 (
z = 6 cos 4 + i sin 4
3 3 )
3 = tan-1 3 æ ö
 = tan-1 z = 6 çç- 1 - i 3 ÷÷÷
-5 5 è 2 2 ø
 » 19 z = -3 - 3i 3
 » 180-19 54. Multiply the numbers, write in standard form. [7.4]
 » 161
z1 z2 = 5 cis 162⋅ 2 cis 63
z1 z2 = 10 cis (162 + 63)
z1 z2 = 10 cis 225
z1 z2 = 10 (cos 225 + i sin 225)
æ ö
z1 z2 = 10 ççè- 2 - 2 i ÷÷ø
50. Write the number in trigonometric form. [7.4] 2 2
z = 2 - 2i z1 z2 = -5 2 - 5 2i
or - 5 2 - 5i 2
2
r = 22 + (-2)
55. Multiply the numbers, write in standard form. [7.4]
r= 8=2 2
z1 z2 = 3 cis 12⋅ 4 cis 126
 = tan-1 -2 = tan-1 1 z1 z2 = 12 cis (12 + 126)
2
z1 z2 = 12 cis 138
 = 45
 = 360- 45 z1 z2 = 12 (cos138 + i sin138)
 = 315 z1 z2 » -8.918 + 8.030i
z = 2 2 cis 315
56. Multiply the numbers, write in standard form. [7.4] 62. Find the indicated power, write in standard form. [7.5]
z1 z2 = 7 cis 2 ⋅ 4 cis  (3 cis 45)6 = 36 cis 6 ⋅ 45

3 4
= 729 cis 270
z1 z2 = 28 cis (
2
3
+ 
4 ) = 729 (cos 270 + i sin 270)
z1 z2 = 28 cis 11 = 729 (0 -1i ) = 0 - 729i
12
(
z1 z2 = 28 cos  + i sin 11
11
12 12 ) 63. Find the indicated power, write in standard form. [7.5]
(cis 116 ) = cis 8 ⋅ 116 = cis 443 = cis 23

8
z1 z2 » -27.046 + 7.247i
57. Multiply the numbers, write in standard form. [7.4]

= cos 2 + i sin 2
z1 z2 = (3 cis 1.8) ⋅ (5 cis 2.5) 3 3
=-1 + 3 i or - 1 + i 3
z1 z2 = 15 cis (1.8 + 2.5)
2 2 2 2
z1 z2 = 15 cis 4.3
64. Find the indicated power, write in standard form. [7.5]
z1 z2 = 15(cos 4.3 + i sin 4.3)
7
z1 z2 » -6.012 -13.742i (1- i 3 ) = (2 cis 300)7 = 27 cis 7 ⋅ 300
58. Divide the numbers, write in trigonometric form. [7.4] = 128 cis 2100 = 128 cis 300
=128(cos 300 + i sin 300)
z1
= 6 cis 50 æ ö
z2 2 cis 150 = 128ççè 1 - 3 i ÷÷÷ø
z1 2 2
= 3 cis (50-150)
z2 = 64 - 64i 3
z1 65. Find the indicated power, write in standard form. [7.5]
= 3 cis ( -100) or 3 cis 260
z2
(-2 - 2i )10
59. Divide the numbers, write in trigonometric form. [7.4] 10
= (2 2 cis 225) = 32,768 cis 10 ⋅ 225
z1 30 cis 165
= = 32,768 cis 2250
z2 10 cis 55
z1 = 32,768(cos 2250 + i sin 2250)
= 3 cis (165- 55)
z2 = 0 + 32,768i
z1 = 32,768i
= 3 cis 110
z2
66. Find all the roots, write in trigonometric form. [7.5]
60. Divide the numbers, write in trigonometric form. [7.4]
27i = 27 cis 90
z1 40 cis 66
= = 5 cis (66-125)
z2 8 cis 125 wk = 271/3 cis 90 + 360k k = 0, 1, 2
3
z1
= 5 cis (-59) or 5 cis 301
z2 w0 = 3 cis 90
3
61. Divide the numbers, write in trigonometric form. [7.4] = 3 cis 30
z1
= 3 -i w1 = 3 cis 90 + 360
z2 1+ i 3
2 cis 330 = 3 cis 150
=
2 cis 45
= 2 cis (330- 45) w2 = 3 cis 90 + 300⋅ 2
3
= 2 cis 285 = 3 cis 270
3 - i = 2 cis 330 1 + i = 2 cis 45

w1 = cis 60 + 360
5
8i = 8 cis 90
= cis 84
wk = 81/4 cis 90 + 360k = 4 8 cis 90 + 360k
4 4 w2 = cis 60 + 360⋅ 2
5
k = 0, 1, 2, 3 = cis 156
w0 = 4 8 cis 90 w3 = cis 60 + 360⋅ 3
4 5
4
= 8 cis 22.5 = cis 228
w1 = 4 8 cis 90 + 360 w4 = cis 60+360⋅ 4

4 5
4
= 8 cis 112.5 = cis 300
Chapter 7 Test
w2 = 4 8 cis 90 + 360⋅ 2
4 1. Solve the triangle. [7.1]
= 4 8 cis 202.5
w3 = 4 8 cis 90 + 360⋅ 3

4
4
= 8 cis 292.5
68. Find all the roots, write in trigonometric form. [7.5] B = 180 -11.2 - 65.5
B = 103.3
wk = 2561/4 cis 120 + 360k k = 0, 1, 2, 3
4 a = c b = c
sin A sin C sin B sin C
w0 = 4 cis 120 a
4 = 16.2 b = 16.2
= 4 cis 30 sin 65.5 sin11.2 sin103.3 sin11.2

a = 16.2sin 65.5 b = 16.2sin103.3
w1 = 4 cis 120 + 360 sin11.2 sin11.2
4
a » 75.9 b » 81.2
= 4 cis 120
w2 = 4 cis 120 + 360⋅ 2
4
= 4 cis 210
w3 = 4 cis 120+360⋅ 3
4
= 4 cis 300
sin 58.8 = h
47.5
æ ö2
( 12 ) + çèç 23 ø÷÷÷
2
= 1 + 3 =1 h = 47.5sin 58.8
4 4 h » 40.6
z = 1 cis 60 Since h < 47.5, two triangles exist.

a = b
cis 60 + 360k = 1 cis 60 + 360k
1/5
wk = (1) sin A sin B
5 5
42.5 = 47.5
k = 0, 1, 2, 3, 4
sin 58.8 sin B

w0 = cis 60 sin B = 47.5sin 58.8 = 0.956
5 42.5
= cis 12 B » 72.9 or 107.1
Chapter 7 Test 573
For B = 72.9 6. Find the area. [7.2]
C = 180 - 58.8 - 72.9 = 48.3

c = 47.5
sin 48.3 sin 72.9
s = 1 (a + b + c )

c = 47.5sin 48.3 = 37.1 2
sin 72.9
1
s = (17 + 55 + 42) = 57
For B = 107.1 2
K = s( s - a )( s - b)( s - c )
C = 180 - 58.8 -107.1 = 14.1
c K = 57(57 -17)(57 - 55)(57 - 42)
= 47.5
sin14.1 sin 72.9 K » 260 square units

c = 47.5sin14.1 = 12.1 7. Write an equivalent vector. [7.3]
sin 72.9
a1 = 12 cos 220 » -9.2
a2 = 12 sin 220 » -7.7
b2 = a 2 + c 2 - 2ac cos B v = a1i + a2 j
2 2
b = 36.5 + 42.4 - 2(36.5)(42.4) cos51.5° v = -9.2i - 7.7 j
b » 34.7 8. Find 3u – 5v. [7.3]
2 2 2
3u - 5v = 3(2i - 3j) - 5(5i + 4 j)
cos A = b + c - a
2bc = (6i - 9 j) - (25i + 20 j)
2 2 2
cos A = 34.7 + 42.4 - 36.5 = (6 - 25)i + (-9 - 20) j
2(34.7)(42.4)
= -19i - 29 j
(
A = cos-1 1669.6
2942.56 ) 9. Find the dot product. [7.3]
A » 55.4
u ⋅ v = (-2i + 3j) ⋅ (5i + 3j)
C = 180 - 55.4 - 51.5 = 73.1 = (-2 ⋅ 5) + (3 ⋅ 3) = -10 + 9
4. Find B. [7.2] = -1
2 2 2 10. Find the smallest angle between the vectors. [7.3]
cos B = a + c - b
2ac 3, 5 ⋅ -6, 2
2 2 2 cos  = u ⋅ v =
cos B = 24.75 + 31.45 - 42.25 u v 3 + 52 (-6)2 + 22
2
2(24.75)(31.45)
cos  = -18 + 10 = -8
cos B = -183.3975 34 40 34 40
1556.775
 » 103
(
B = cos-1 -183.3975 » 96.77
1556.775 ) 11. Write in trigonometric form. [7.4]
5. Find the area. [7.2]
z = -3 2 + 3i
z = (-3 2)2 + 32 = 3 3
 = tan-1 3 = tan-1 2
-3 2 2
 » 35
K = 1 ab sin C
2  » 180- 35
1
K = (7)(12)(sin110)  » 145
2
K » 39 square units z » 3 3 cis 145
12. Write in standard form. [7.4] 16. Find the roots, write in standard form. [7.5]
z = 5 cis 315 27i = 27(cos 90 + i sin 90) = 27 cis 90
z = 5(cos 315 + i sin 315)
æ ö wk = 271/3 cis 90 + 360k k = 0, 1, 2
3
z = 5ççè 2 - 2 i ÷÷ø
2 2
w0 = 3 cis 90 = 3 cis 30
z = 5 2 -5 2 i 3
2 2
= 3 3 + 3i
13. Simplify, write in standard form. [7.5] 2 2
w1 = 3 cis 150
z= 1+ 3i
2 2
=-3 3 + 3i
2 2
r = (1 2)2 + ( 3 2)2 = 1
w2 = 3 cis 270
3 2 = 0 - 3i or - 3i
 = tan-1 = 60
12
17. Find the distance. [7.3]
z = cos 60 + i sin 60
3
æ1 3 ö÷
ç 3
çè 2 + i 2 ÷÷ø = (cos 60 + i sin 60)
= cos(3 ⋅ 60) + i sin(3 ⋅ 60)
= cos180 + i sin180
= -1 + 0i = -1 A = 142- 65 = 77
14. Simplify, write in trigonometric form. [7.4] R 2 = 242 + 182 - 2(24)(18) cos 77
z1 25 cis 115 R » 27 miles
=
z2 10 cis 210 18. Find the ground speed and course. [7.3]
z1 From the x-axis, the plane makes the angle
= 2.5 cis (115- 210)
z2
90- 84.5 = 5.5
z1
= 2.5 cis ( - 95) or 2.5 cis 265 plane = 145cos5.5i + 145sin 5.5 j
z2
» 144.3i + 13.9 j
15. Simplify, write in standard form. [7.5]
From the x-axis, the wind makes the angle
z = 2 -i
90- 68.4 = 21.6
2 2
z= 2 + (-1) = 3 wind = 24.6 cos 21.6i + 24.6sin 21.6 j
» 22.9i + 9.1j
 = tan-1 -1 = tan-1 2
2 2 plane + wind » 144.3i + 13.9 j + 22.9i + 9.1j
 » 35.2644 » 167.2i + 23.0 j
 » 360- 35.2644
plane + wind = (167.2)2 + (23.0)2 » 169
 » 324.7356
z » 3 cis 324.7356  = tan-1 23.0
167.2
z 5 » ( 3)5 cis (5 ⋅ 324.7356) -1 23.0
= tan
z 5 » 9 3(cos 1623.678 + i sin1623.678) 167.2
 » 7.8
z 5 » -15.556 -1.000i
Cumulative Review Exercises 575
 = 90-  Cumulative Review Exercises
 » 90- 7.8 1. Find ( f  g )( x ) . [2.6]
 » 82.2
( f  g )( x ) = f [ g ( x ) ] = f éë x 2 + 1ùû = cos( x 2 + 1)
The ground speed of the plane is about 169 mph at a
heading of approximately 82.2º. 2. Find f -1 ( x ) . [4.1]
19. Find the roots, write in trigonometric form. [7.5] f ( x) = 2 x + 8
2 + 2 i = 1(cos 45 + i sin 45) y = 2x + 8
2 2 x = 2y +8
x -8 = 2 y
wk = cos 45 + 360k + i sin 45 + 360k
5 5 x -8 = y
k = 0, 1, 2, 3, 4 2
  f ( x) = 1 x - 4
-1
2
w0 = cos 45 + i sin 45
5 5
3. Convert to degrees. [5.1]
= (cos 9 + i sin 9)
= cis 9 3 æç180 ö÷÷ = 270
2 çè  ÷ø
   
w1 = cos 45 + 360 + i sin 45 + 360 4. Find sin  , cos  , tan  . [5.2]
5 5
= (cos81 + i sin 81)
hyp = 32 + 42 = 25 = 5
= cis 81
    sin  = 3 , cos  = 4 , tan  = 3
w2 = cos 45 + 360 ⋅ 2 + i sin 45 + 360 ⋅ 2 5 5 4
5 5
= cos153 + i sin153 5. Find c. [5.2]
= cis 153 cos 26.0 = 15.0
   
c
w3 = cos 45 + 360 ⋅ 3 + i sin 45 + 360 ⋅ 3 c = 15.0  » 16.7 cm
5 5
cos 26.0
= cos 225 + i sin 225
6. Graph.
= cis 225
   
y = 3sin  x
w4 = cos 45 + 360 ⋅ 4 + i sin 45 + 360 ⋅ 4
5 5
= cos 297 + i sin 297
= cis 297
20. Find the cost. [7.2]
7. Find the amplitude, period and phase shift. [5.7]
(
y = 4 cos 2 x - 
2 )
S = 1 (112 + 165 + 140) = 208.5
0 £ 2 x -  £ 2
2
2
K = 208.5(208.5 -112)(208.5 -165)(208.5 -140)  £ 2 x £ 5
K » 7743 2 2
cost » 8.50(7743)  £ x £ 5
4 4
cost » $66,000
amplitude = 4, period =  , phase shift = 
4
8. Write in the form k sin ( x +  ) . [6.4] 14. Solve. [6.6]
sin x - cos x sin 2 x = 3 sin x

a = 1, b = -1 2 sin x cos x - 3 sin x = 0
sin x(2 cos x - 3) = 0
k = 12 + (-1)2 = 2
sin x = 0 2 cos x - 3 = 0
sin  = 1 , cos  = -1 ,  = 7 or -  x = 0, 
2 2 4 4 cos x = 3
2
4( )
sin x - cos x = 2 sin x + 7 or 2 sin x - 
4 ( ) x =  , 11
6 6
9. Determine whether the function is odd, even, or
The solutions are 0,  ,  , 11 .
neither. [5.4] 6 6
y = sin x is an odd function. 15. Find the magnitude and direction angle. [7.3].
4
10. Verify the identity. [6.1] v = (-3)2 + 42  = tan-1 = tan-1 4
-3 3
1 - sin x = 1- sin 2 x v = 9 + 16  » 53.1
sin x sin x
2
v =5  = 180- 
= cos x  » 180- 53.1
sin x
 » 126.9
= cos x cos x
sin x magnitude: 5, angle: 126.9
= cos x cot x
16. Find the angle between the vectors. [7.3]
11. Evaluate. [6.5]
cos  = v ⋅ w
( 13 ( ))
tan sin-1 12 = tan (67.38 ) = 12
5
v w
2, - 3 ⋅ -3, 4
cos  =
12. Express in terms of a sine function. [6.2]
2 + (-3)2 (-3)2 + 42
2
sin 2 x cos 3 x - cos 2 x sin 3 x = sin(2 x - 3 x ) 2(-3) + (-3)(4)

= sin(-x ) or - sin x cos  =
13 25
13. Solve. [6.6] cos  = - 18 » -0.9985
5 13
2 cos2 x + sin x -1 = 0  = 176.8
2(1- sin 2 x ) + sin x -1 = 0 17. Find the ground speed and course. [7.2]
2
2 sin - sin x + 1 = 0 AB = 415(cos 42i + sin 42 j) » 308.4i + 277.6 j
(2 sin x + 1)(sin x -1) = 0 AD = 55[cos(-25)i + sin(-25) j] » 49.8i - 23.2 j
2 sin x + 1 = 0 sin x -1 = 0 AC = AB + AD
sin x = 1 AC = 308.4i + 277.6 j + 49.8i - 23.2 j
sin x = - 1
2 AC » 358.2i + 254.4 j
 x=
x= 7 , 11 2
6 6 AC = 358.22 + (254.4)2
AC » 439 mph
The solutions are  , 7 , 11 .
2 6 6
( 358.2 )
 = 90 - = 90 - tan-1 254.4 » 54.6
Cumulative Review Exercises 577
18. Find A. [7.1]
a = b
sin A sin B
sin A = a sin B
b

= 42 sin 32 = 0.4364041
51
A = sin-1 (0.4364041) » 26
19. Simplify. [7.5]
z = 1- i
r = 12 + (-1)2  = tan-1 -1 = 45

1
r= 2 
 = 315
z = 2(cos 315 + i sin 315)
(1- i )8 = [ 2(cos 315 + i sin 315)]8
= ( 2)8 [cos(8 ⋅ 315) + i sin(8 ⋅ 315)]
= 16(cos 2520 + i sin 2520)
= 16(cos 0 + i sin 0)
= 16 - 0i = 16
20. Find the square roots. [7.5]
i = cos 90 + i sin 90
wk = cos 90 + 360k + i sin 90 + 360k k = 0, 1

2 2
 
w0 = cos 90 + i sin 90
2 2
= cos 45 + i sin 45
= 2+ 2i
2 2
   
w1 = cos 90 + 360 + i sin 90 + 360
2 2
= cos 225 + i sin 225
=- 2 - 2 i
2 2

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Algebra and Trigonometry 8th Edition

Aufmann Solutions Manual

Section 7.1 Exercises 7. Solve the triangles.

a = c 12. Solve the triangles.

A » 180.0 - 82.0 - 29.7

B = 180 - 54.32 - 34.25 20. Solve the triangles that exist.

24. Solve the triangles that exist.

sin 37.9 = h sin 52.7 = h

c = 62.75sin 65.74 B = 8.1

For A = 73.9 35. a = 155 yd, c = 165 yd, A = 42.0

36. b = 365 yd, A = 11.2 , C = 22.9

C = 180- (59.0 + 77.2) B = 180- 67- 31 c = b

CAB = 180- 27.2 = 152.8 A = 21 c = a

C = 180- 54- 47

 = 65 A = 90- 55

d = 12 53. Using the figure below,

semiperimeter = 1 (25) = 12.5 m a 2 = b2 + c 2 - 2bc cos A

a 2 = 7949.52 - 7625.52 cos 25.9 b2 = 25.92 + 33.42 - 2(25.9)(33.4) cos84.0

a = 7949.52 - 7625.52 cos 25.9 b2 = 1786.37 -1730.12 cos84.0

14. Find the third side of the triangle. ( )

25. Solve the triangle. a 2 = b 2 + c 2 - 2bc cos A

a 2 = b 2 + c 2 - 2bc cos A a = 1232 + 1522 - 2(123)(152) cos 95.1°

C = 180 - 27.6 - 36.6 C = 48.0

C = 115.8 29. Solve the triangle.

C = 180 - 31.8 -124.4 32. Solve the triangle.

significant digits. significant digits.

B = 180- 62.15- 47.25 A = 180- 76.3- 42.8

significant digits. significant digits.

K = 19.45(19.45 -10.2)(19.45 -13.3)(19.45 -15.4) a 2 = 8776 - 4680 cos 45

K = 19.45(9.25)(6.15)(4.05) a = 8776 - 4680 cos 45

45. Find the distance. 48. Find the wing span.

 = 32 b2 = 22302.72 - 22302.72 cos109.05

a 2 = 3002 + 4162 - 2(300)(416) cos 72 a 2 = (4.75) 2 + (6.50) 2 = 64.8125

2 2 2 A = 360 a 2 = 402 + 402 - 2(40)(40) cos 60

a 2 = b2 + c 2 - 2bc cos A b. 162 = 42 + c 2 - 2(4)( c ) cos A

8cos(55) + 64 cos2 (55) + 960

C = 180 - 55 -11.8 = 113.2°

s = 1 (324 + 412 + 516)

K = 626(626 - 324)(626 - 412)(626 - 516)

K = 1250(1250 - 680)(1250 - 800)(1250 -1020)

61. Find the cost. K = 730(730 - 420)(730 - 500)(730 - 540)

V = 1 (4)(4)(sin 72)(18) P6. Rationalize the denominator.

A vector equivalent to P1P2 is v = 12, -3 . v = 62 + 102  = tan-1 10 = tan-1 5

A vector equivalent to P1P2 is v = -90, 70 . 6 , 10 = 3 34 , 5 34

v = (-33) 2 + 562  = tan-1

v = 362 + 77 2  = tan-1 77 = tan-1 77 u = - 3 10 , 10 .

u = 36 , 77 26. Perform the indicated operations.

v = (-9) 2 + (-40) 2  = tan-1 -40 = tan-1 40 = -4, 8 - -3, -2

heading = 37  direction angle = 53

AB = 50cos53i + 50sin 53 j

+ (235.7 - 205.6 - 37.7 + 7.6) j 2(3) + (-1)4

85. Find the vector.

91. Let u = ci + bj, v = ci + dj, and w = ei + fj.

92. Let v = a, b and w = c, d .

94. Let  be the angle between vectors v and w.

The vectors are not perpendicular. This is an acute angle.

4. Responses will vary.

 = tan-1 -1 z = 5 cis 90

z = 2 cis 315  = 270