section 7.3 the law of cosines sin sin121 sin 7.04 3 3sin121 sin 0.36527016 21.4 ... 7.23 8.94 2...

Section 7.3 The Law of Cosines 353

Copyright © 2013 Pearson Education, Inc.

42. If A = 18° and B = 36° we have

2 25sin18 sin 361.12257

sin(18 36)A R R

⎡ ⎤° °= ≈⎢ ⎥+ °⎣ ⎦

43. (a) 210

11.4 in. in. 8.77 in.13⋅ ≈

(b) ( )2 2sin18 sin 36

50 5 0.308 5.32 in.sin 18 36

A⎡ ⎤° °= ⋅ ≈⎢ ⎥° + °⎣ ⎦

44. red

Section 7.3 The Law of Cosines

1. a, b, and C

(a) SAS

(b) law of cosines

2. A, C, and c

(a) SAA

(b) law of sines

3. a, b, and A

(a) SSA

(b) law of sines

4. a, B, and C

(a) ASA

(b) law of sines

5. A, B, and c

(a) ASA

(b) law of sines

6. a, c, and A

(a) SSA

(b) law of sines

7. a, b, and c

(a) SSS

(b) law of cosines

8. b, c, and A

(a) SAS

(b) law of cosines

9. ( ) ( )( )22 21 4 2 2 1 4 2 cos 45

21 32 8 2 33 8 25

2

25 5

a

a

= + − °

⎛ ⎞= + − = − =⎜ ⎟⎝ ⎠= =

10. ( )( )2 2 23 8 2 3 8 cos 60

19 64 48 73 24 49

2

49 7

a

a

= + − °⎛ ⎞= + − = − =⎜ ⎟⎝ ⎠

= =

354 Chapter 7 Applications of Trigonometry and Vectors


11. ( )( )2 2 2 2 2 23 5 7

cos2 2 3 5

9 25 49 1120

30 2

b c a

bcθ

θ

+ − + −= =

+ −= = − ⇒ = °

12. ( )( )( )

22 22 2 2 1 3 1

cos2 2 1 3

1 3 1 3 3 330

22 3 2 3 3

b c a

bcθ

θ

+ −+ −= =

+ −= = ⋅ = ⇒ = °

13. A = 121°, b = 5, c = 3

Start by finding a with the law of cosines.

( )( )2 2 2

2 2 2

2 cos

5 3 2 5 3 cos121 49.5

7.04 7.0

a b c bc A

a

a

= + − ⇒= + − ° ≈ ⇒≈ ≈

Of the remaining angles B and C, C must be smaller since it is opposite the shorter of the two sides b and c. Therefore, C cannot be obtuse.

sin sin sin121 sin

7.04 33sin121

sin 0.36527016 21.47.04

A C C

a C

C C

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, B = 180° − 121° − 21.4° = 37.6°.

14. A = 61°, b = 4, c = 6 Start by finding a with the law of cosines.

( )( )2 2 2

2 2 2

2 cos

4 6 2 4 6 cos 61 28.729

5.36 5.4

a b c bc A

a

a

= + −= + − ° ≈≈ ≈

Of the remaining angles B and C, B must be smaller since it is opposite the shorter of the two sides b and c. Therefore, B cannot be obtuse.

sin sin sin 61 sin

5.36 44sin 61

sin 0.65270127 40.75.36

A B B

a b

B B

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, 180 61 40.7 78.3 .C = ° − ° − ° = °

15. a = 12, b = 10, c = 10

We can use the law of cosines to solve for any angle of the triangle. Since b and c have the same measure, so do B and C since this would be an isosceles triangle. If we solve for B, we obtain

( )( )

2 2 2

2 2 2

2 2 2

2 cos

cos2

12 10 10 144 3cos

2 12 10 240 5

53.1

b a c ac B

a c bB

ac

B

B

= + −+ −=

+ −= = = ⇒

≈ °

Therefore, 53.1C B= ≈ ° and

A = 180° − 53.1° − 53.1° = 73.8°. If we solve for A directly, however, we obtain

( )( )

2 2 2

2 2 2

2 2 2

2 cos

cos2

10 10 12 56 7cos

2 10 10 200 5

73.7

a b c bc A

b c aA

bc

A

A

= + −+ −=

+ −= = =

≈ °

The angles may not sum to 180° due to rounding.

16. a = 4, b = 10, c = 8

We can use the law of cosines to solve for any angle of the triangle. We solve for B, the largest angle. We will know that B is obtuse if

cos 0.B <

( )( )

2 2 2

2 2 2

2 2 2

2 cos

cos2

4 8 10 20 5cos

2 4 8 64 16

108.2

b a c ac B

a c bB

ac

B

B

= + −+ −=

+ − −= = = − ⇒

≈ °

We can now use the law of sines or the law of cosines to solve for either A or C. Using the law of cosines to solve for A, we have

( )( )

2 2 2

2 2 2

2 2 2

2 cos

cos2

10 8 4 148 37cos 22.3

2 10 8 160 40

a b c bc A

b c aA

bc

A A

= + −+ −=

+ −= = = ⇒ ≈ °

Thus, C = 180° − 108.2° − 22.3° = 49.5°.

17. B = 55°, a = 90, c = 100 Start by finding b with the law of cosines.

( )( )2 2 2

2 2 2

2 cos

90 100 2 90 100 cos 55 7775.6

88.18

b a c ac B

b

b

= + −= + − ° ≈≈

(will be rounded as 88.2) Of the remaining angles A and C, A must be smaller since it is opposite the shorter of the two sides a and c. Therefore, A cannot be obtuse.

sin sin sin sin 55

90 88.1890sin 55

sin 0.83605902 56.788.18

A B A

a b

A A

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, C = 180° − 55° − 56.7° = 68.3°.



18. a = 5, b = 7, c = 9

We can use the law of cosines to solve for any angle of the triangle. We solve for C, the largest angle. We will know that C is obtuse if

cos 0.C <

( )( )

2 2 2

2 2 2

2 2 2

2 cos

cos2

5 7 9 7 1cos 95.7

2 5 7 70 10

c a b ab C

a b cC

ab

C C

= + −+ −=

+ − −= = = − ⇒ ≈ °

We can now use the law of sines or the law of cosines to solve for either A or B. Using the law of sines to solve for A, we have

sin sin sin sin 95.7

5 95sin 95.7

sin 0.55280865 33.69

A C A

a C

A A

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, B = 180° − 95.7° − 33.6° = 50.7°.

19. A = 41.4º, b = 2.78 yd, c = 3.92 yd First find a.

( )( )2 2 2

2 2 2

2 cos

2.78 3.92 2 2.78 3.92 cos 41.4

6.7460 2.60 yd

a b c bc A

a

a

= + −= + − °≈ ⇒ ≈

Find B next, since angle B is smaller than angle C (because b < c), and thus angle B must be acute.


2.78 2.5972.78sin 41.4

sin 0.7070911822.597

45.1

B A B

b a

B

B

°= ⇒ =

°= ≈ ⇒

≈ °

Finally, 180 41.4 45.1 93.5 .C = ° − ° − ° = °

20. C = 28.3°, b = 5.71 in., a = 4.21 in. First find c.

( )( )2 2 2

2 2 2

2 cos

4.21 5.71 2 4.21 5.71 cos 28.3

7.9964 2.83 in.

c a b ab C

c

c

= + − ⇒= + − °≈ ⇒ ≈

Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must be acute.


4.21 2.8284.21sin 28.3

sin 0.705767812.828

44.9

A C A

a C

A

A

°= ⇒ =

°= ≈ ⇒

≈ °

Finally, 180 28.3 44.9 106.8 .B = ° − ° − ° = °

21. C = 45.6°, b = 8.94 m, a = 7.23 m First find c.

( )( )2 2 2

2 2 2

2 cos

7.23 8.94 2 7.23 8.94 cos 45.6

41.7493 6.46 m

c a b ab C

c

c

= + − ⇒= + − °≈ ⇒ ≈



7.23 6.4617.23sin 45.6

sin 0.799510526.461

53.1

A C A

a c

A

A

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

Finally, 180 53.1 45.6 81.3 .B = ° − ° − ° = °

22. A = 67.3°, b = 37.9 km, c = 40.8 km First find a.

( )( )2 2 2

2 2 2

2 cos

37.9 40.8 2 37.9 40.8 cos 67.3

1907.5815 43.7 km

a b c bc A

a

a

= + − ⇒= + − °≈ ⇒ ≈



37.9 43.6837.9sin 67.3

sin 0.8004623143.68

53.2

B A B

b a

B

B

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

Finally, 180 67.3 53.2 59.5 .C = ° − ° − ° = °

23. a = 9.3 cm, b = 5.7 cm, c = 8.2 cm We can use the law of cosines to solve for any of angle of the triangle. We solve for A, the largest angle. We will know that A is obtuse if

cos 0.A <

( )( )

2 2 2

2 2 2

2 cos

5.7 8.2 9.3cos 0.14163457

2 5.7 8.2

82

a b c bc A

A

A

= + − ⇒+ −= ≈ ⇒

≈ °


sin sin sin sin 82

5.7 9.35.7sin 82

sin 0.60693849 379.3

B A B

b a

B B

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, 180 82 37 61 .C = ° − ° − ° = °



24. a = 28 ft, b = 47 ft, c = 58 ft Angle C is the largest, so find it first.

2 2 2

2 2 2

2 cos

28 47 58cos 0.14095745

2(28)(47)

98

c a b ab C

C

C

= + − ⇒+ −= ≈ − ⇒

≈ °


sin sin sin sin 98

28 5828sin 98

sin 0.47806045 2958

A C A

a c

A A

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, 180 29 98 53 .B = ° − ° − ° = °

25. a = 42.9 m, b = 37.6 m, c = 62.7 m Angle C is the largest, so find it first.

( )( )

2 2 2

2 2 2

2 cos

42.9 37.6 62.7cos

2 42.9 37.6

0.20988940

102.1 102 10

c a b ab C

C

C

= + − ⇒+ −=

≈ − ⇒≈ ° ≈ ° ′

Find B next, since angle B is smaller than angle A (because b < a), and thus angle B must be acute.

sin sin sin sin102.1

37.6 62.737.6sin102.1

sin 0.5863580562.7

35.9 35 50

B C B

b c

B

B

°= ⇒ = ⇒

°= ≈ ⇒

≈ ° ≈ ° ′

Thus,

180 35 50 102 10

180 138 42 00

A = ° − ° − °′ ′= ° − ° = ° ′

26. a = 189 yd, b = 214 yd, c = 325 yd Angle C is the largest, so find it first.

( )( )

2 2 2

2 2 2

2 cos

189 214 325cos 0.29802700

2 189 214

107.3 107 20

c a b ab C

C

C

= + − ⇒+ −= = − ⇒

≈ ° ≈ ° ′


sin sin sin sin107 20

214 325214sin107.3

sin 0.62867326325

39.0 39 00

B C B

b c

B

B

° ′= ⇒ = ⇒

°= ≈ ⇒

≈ ° ≈ ° ′

Thus,

180 39 00 107 20

179 60 146 20 33 40

A = ° − ° − °′ ′= ° − ° = °′ ′ ′

27. a = 965 ft, b = 876 ft, c = 1240 ft Angle C is the largest, so find it first.

( )( )

2 2 2

2 2 2

2 cos

965 876 1240cos 0.09522855

2 965 876

84.5 or 84 30

c a b ab C

C

C

= + − ⇒+ −= ≈ ⇒

≈ ° ° ′


sin sin sin sin 84 30

876 1240876sin 84 30

sin 0.703199251240

44.7 or 44 40

B C B

b c

B

B

° ′= ⇒ = ⇒

° ′= ≈ ⇒

≈ ° ° ′

Thus,

180 44 40 84 30

179 60 129 10 50 50

A = ° − ° − °′ ′= ° − ° = °′ ′ ′

28. a = 324 m, b = 421 m, c= 298 m Angle B is the largest, so find it first.

( )( )

2 2 2

2 2 2

2 cos

324 298 421cos 0.08564815

2 324 298

85.1 85 10

b a c ac B

B

B

= + − ⇒+ −= ≈ ⇒

≈ ° ≈ ° ′

Find C next, since angle C is smaller than angle A (because c < a), and thus angle B must be acute.


298 421298sin 85.1

sin 0.70525154421

44.8 44 50

C B C

c b

C

C

°= ⇒ = ⇒

°= ≈ ⇒

≈ ° ≈ ° ′

Thus, 180 85 10 44 50

180 130 50 00

A = ° − ° − °′ ′= ° − ° = ° ′

29. 80 40A = ° ′ b = 143 cm, c = 89.6 cm

First find a.

( )( )2 2 2

2 2 2

2 cos

143 89.6 2 143 89.6 cos80 40

24,321.25 156 cm

a b c bc A

a

a

= + − ⇒= + − ° ′≈ ⇒ ≈

Find C next, since angle C is smaller than angle B (because c < a), and thus angle C must be acute.


89.6 156.089.6sin 80 40

sin 0.56675534156.0

34.5 34 30

C A C

c a

C

C

° ′= ⇒ = ⇒

° ′= ≈ ⇒

≈ ° = ° ′

Finally,

180 80 40 34 30

179 60 115 10 64 50

B = ° − ° − °′ ′= ° − ° = °′ ′ ′



30. 72 40 ,C = ° ′ a = 327 ft, b = 251 ft

First find c.

( )( )2 2 2

2 2 2

2 cos

327 251 2 327 251 cos 72 40

121,023.55 348 ft

c a b ab C

c

c

= + − ⇒= + − ° ′≈ ⇒ ≈



251 347.9251sin 72 40

sin 0.68870795347.9

43.5 43 30

B C B

b C

B

B

° ′= ⇒ = ⇒

° ′= ≈ ⇒

≈ ° = ° ′

Finally,

180 72 40 43 30

179 60 116 10 63 50

A = ° − ° − °′ ′= ° − ° = °′ ′ ′

31. B = 74.8°, a = 8.92 in., c = 6.43 in. First find b.

( )( )

2 2 2

2 2 2

2 cos

8.92 6.43

2 8.92 6.43 cos 74.8

90.8353 9.53 in.

b a c ac B

b

b

= + − ⇒= +

− °≈ ⇒ ≈

Find C next, since angle C is smaller than angle A (because c < a), and thus angle C must be acute.


6.43 9.536.43sin 74.8

sin 0.65405618119.53

40.6

C B C

c b

C

C

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

Thus, A = 180° − 74.8° − 40.6° = 64.6°.

32. C = 59.7°, a = 3.73 mi, b = 4.70 mi First find c.

( )( )

2 2 2

2 2 2

2 cos

3.73 4.70

2 3.73 4.7 cos 59.7

18.3132 4.28 mi

c a b ab C

c

c

= + − ⇒= +

− °≈ ⇒ ≈



3.73 4.283.73sin 59.7

sin 0.75244518784.28

48.8

A C A

a c

A

A

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

Thus, B = 180° − 48.8° − 59.7° = 71.5°.

33. A = 112.8°, b = 6.28 m, c = 12.2 m First find a.

( )( )2 2 2

2 2 2

2 cos

6.28 12.2 2 6.28 12.2 cos112.8

247.658 15.7 m

a b c bc A

a

a

= + − ⇒= + − °≈ ⇒ ≈



6.28 15.746.28sin112.8

sin 0.3678081715.74

21.6

B A B

b a

B

B

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

Finally, C = 180° − 112.8° − 21.6° = 45.6°.

34. B = 168.2°, a = 15.1 cm, c = 19.2 cm First find b.

( )( )2 2 2

2 2 2

2 cos

15.1 19.2 2 15.1 19.2 cos168.2

1164.236 34.1 cm

b a c ac B

b

b

= + − ⇒= + − °≈ ⇒ ≈

Find A next, since angle A is smaller than angle C (because a < c), and thus angle A must be acute.


15.1 34.1215.1sin168.2

sin 0.0905008934.12

5.2

A B A

a b

A

A

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

Thus, C = 180° − 5.2° − 168.2° = 6.6°.

35. a = 3.0 ft, b = 5.0 ft, c = 6.0 ft Angle C is the largest, so find it first.

( )( )

2 2 2

2 2 2

2 cos

3.0 5.0 6.0 2 1cos

2 3.0 5.0 30 15

0.06666667 94

c a b ab C

C

C

= + − ⇒+ −= = − = −

≈ − ⇒ ≈ °


sin sin sin sin 94

3 63sin 94

sin 0.49878203 306

A C A

a c

A A

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, B = 180° − 30° − 94° = 56°.



36. a = 4.0 ft, b = 5.0 ft, c = 8.0 ft Angle C is the largest, so find it first.

( )( )

2 2 2

2 2 2

2 cos

4.0 5.0 8.0 23cos

2 4.0 5.0 40

0.57500000 125

c a b ab C

C

C

= + − ⇒+ −= = −

≈ − ⇒ ≈ °


sin sin sin sin125

4 84sin125

sin 0.40957602 248

A C A

a c

A A

°= ⇒ = ⇒

°= ≈ ⇒ ≈ °

Thus, B = 180° − 24° − 125° = 31°.

37. There are three ways to apply the law of cosines when a = 3, b = 4, and c = 10. Solving for A:

( )( )

2 2 2

2 2 2

2 cos

4 10 3 107cos 1.3375

2 4 10 80

a b c bc A

A

= + − ⇒+ −= = =

Solving for B:

( )( )

2 2 2

2 2 2

2 cos

3 10 4 93 31cos 1.55

2 3 10 60 20

b a c ac B

B

= + − ⇒+ −= = = =

Solving for C:

( )( )

2 2 2

2 2 2

2 cos

3 4 10 75 25cos 3.125

2 3 4 24 8

c a b ab C

C

= + − ⇒+ − −= = = − = −

Since the cosine of any angle of a triangle must be between –1 and 1, a triangle cannot have sides 3, 4, and 10.

38. Answers will vary.

39. A and B are on opposite sides of False River. We must find AB, or c, in the following triangle.

( )( )2 2 2

2 2 2

2

2 cos

286 350 2 286 350 cos 46.3

65,981.3 257

c a b ab C

c

c c

= + −= + − °≈ ⇒ ≈

The length of AB is 257 m.

40. X and Y are on opposite sides of a ravine. We must find XY.

( )( )( )( )

2 2 2

2 2 2

2 cos

153 103 2 153 103 cos 37.7

9080.216688

95.3 m

XY XZ YZ XZ YZ Z

XY

XY

= + −= + − °=≈

41. Using the law of cosines we can solve for the measure of angle A.

( )( )2 2 2

25.9 32.5 57.8cos

2 25.9 32.5

0.95858628 163.5

A

A

+ −=

≈ − ⇒ ≈ °

42. Find the diagonals, BD and AC, of the following parallelogram.

( )( )( )( )

( )( )( )( )

2 2 2

2 2 2

2

2 2 2

2 2 2

2

2 cos

4 6 2 4 6 cos 58

26.563875 5.2 cm

2 cos

4 6 2 4 6 cos122

77.436125 8.8 cm

BD AB AD AB AD A

BD

BD BD

AC AB BC AB BC B

AC

AC AC

= + −= + − °≈ ⇒ ≈= + −= + − °≈ ⇒ ≈

The lengths of the diagonals are 5.2 cm and 8.8 cm.



43. Find AC, or b, in the following triangle.

1 180 128 40 51 20m∠ = ° − ° = °′ ′

Angles 1 and 2 are alternate interior angles formed when parallel lines (the north lines) are cut by a transversal, line BC, so

2 1 51 20 .m m∠ = ∠ = ° ′

90 2 90 51 20 38 40m ABC m∠ = ° − ∠ = ° − ° = °′ ′

( )( )2 2 2

2 2 2

2 cos

359 450 2 359 450 cos38 40

79,106 281 km

b a c ac B

b

b

= + − ⇒= + − ° ′≈ ⇒ ≈

C is about 281 km from A.

44. Sketch a triangle showing the situation as follows.

The angle marked 130° is the corresponding

angle that measures 360° − 230° = 130°. The angle marked 55° is the supplement of the 125° angle. Finally, the 75° angle is marked as such because 180° − 55° − 50° = 75°. We can use the law of cosines to solve for the side of the triangle marked d.

( )( )2 2 2180 100 2 180 100 cos 75

33,082 181.9

d

d

= + − °≈ ⇒ ≈

The distance is approximately 180 mi. (rounded to two significant digits)

45. Let x = the distance between the ends of the two equal sides.

Use the law of cosines to find x.

( )( )2 2 2246.75 246.75

2 246.75 246.75 cos125 12

191,963.937 438.14

x

x

= +− ° ′

≈ ⇒ ≈

The distance between the ends of the two equal sides is 438.14 feet.

46. Let B be the harbor, AB is the course of one ship, BC is the course of the other ship. Thus, b = the distance between the ships.

( )( )2 2 2

2

402 402 2 402 402 cos135 40

554394.25 745 mi

b

b b

= + − ° ′≈ ⇒ =

47. Sketch a triangle showing the situation as follows.

90 45 20 44 40m A∠ = ° − ° = °′ ′ 308 40 270 38 40m C∠ = ° − ° = °′ ′ 180°

180 44 40 38 40 96° 40

m B A C∠ = − −= ° − ° − ° =′ ′ ′

Since we have only one side of a triangle, use the law of sines to find BC = a.

15.2

sin sin sin 44 40 sin 96 40

15.2sin 44 4010.8

sin 96 40

a b a

A B

a

= ⇒ = ⇒° ′ ° ′

° ′= ≈° ′

The distance between the ship and the rock is about 10.8 miles.

48. Let d = the distance between the submarine and the battleship.

24 10 17 30 6 40α = ° ′ − ° ′ = ° ′

17 30β = ° ′ since the angle of depression to

the battleship equals the angle of elevation from the battleship (They are alternate interior angles.)

(continued on next page)



(continued)

180 6 40 17 30

179 60 14 10 155 50

θ = ° − ° ′ − ° ′= ° − ° ′ = ° ′′

Since we have only one side of a triangle, use the law of sines to find d.

5120

sin 6 40 sin155 50

5120sin 6 401451.9

sin155 50

d

d

= ⇒° ′ ° ′

° ′= ≈° ′

The distance between the submarine and the battleship is 1450 ft. (rounded to three significant digits)

49. Use the law of cosines to find the angle, .θ

( )( )2 2 2

20 16 13 487cos

2 20 16 640

0.76093750 40

θ

θ

+ −= =

≈ ⇒ ≈ °

50. ( )( )2 2 220 13 16 313

cos2 20 13 520

0.6019230769 53

β

β

+ −= =

≈ ⇒ ≈ °

51. Let A = the angle between the beam and the 45-ft cable. Let B = the angle between the beam and the 60-ft cable.

2 2 245 90 60 6525 29

cos2(45)(90) 8100 36

0.80555556 36

A

A

+ −= = =

≈ ⇒ ≈ °

( )( )2 2

90 60 45 9675 43cos

2 90 60 10,800 48

0.89583333 26

B

B

2+ −= = =

≈ ⇒ ≈ °

52. AB is the horizontal distance between points A and B.

Using the laws of cosines, we have

2 2 2

2

10 10 2(10)(10)cos128

323.1 18 ft

AB

AB AB

= + − °⇒≈ ⇒ ≈

53. Let A = home plate; B = first base; C = second base; D = third base; P = pitcher’s rubber. Draw AC through P, draw PB and PD.

In triangle ABC, 90 ,m B∠ = ° and

45m A m C∠ = ∠ = ° .

2 2 290 90 2 90 90 2 and

90 2 60.5 66.8 ft

AC

PC

= + = ⋅ == − ≈

In triangle APB, 45m A∠ = ° .

( )( )( )( )

2 2 2

2 2 2

2

2 cos

60.5 90 2 60.5 90 cos 45

4059.86 63.7 ft

PB AP AB AP AB A

PB

PB PB

= + −= + − °≈ ⇒ ≈

Since triangles APB and APD are congruent,

63.7 ft.PB PD= =

The distance to second base is 66.8 ft and the distance to both first and third base is 63.7 ft.

54. Let A = home plate; B = first base; C = second base; D = third base; P = pitcher’s rubber. Draw AC through P, draw PB and PD.

(continued on next page)



(continued)

In triangle ABC, 90 ,m B∠ = ° and

45m A m C∠ = ∠ = ° .

2 2 260 60 2 60 60 2 and

60 2 46.0 38.9 ft

AC

PC

= + = ⋅ == − ≈

In triangle APB, 45m A∠ = ° .

( )( )( )( )

2 2 2

2 2 2

2

2 cos

46.0 60.0 2 46.0 60.0 cos 45

1812.77 42.6 ft

PB AP AB AP AB A

PB

PB PB

= + −= + − °≈ ⇒ ≈

Since triangles APB and APD are congruent,

42.6 ft.PB PD= =

The distance to second base is 38.9 ft and the distance to both first and third base is 42.6 ft.

55. Find the distance of the ship from point A.

1 189 180 9

2 360 317 43

1 2 9 43 52

m

m

m m

∠ = ° − ° = °∠ = ° − ° = °∠ + ∠ = ° + ° = °

Use the law of cosines to find .v

( )( )2 2 247.8 18.5 2 47.8 18.5 cos 52

1538.23 39.2 km

v

v

= + − °≈ ⇒ ≈

56. Let A = the man’s location; B = the factory whistle heard at 3 sec after 5:00; C = the factory whistle heard at 6 sec after 5:00.

Since sound travels at 344 m per sec and the

man hears the whistles in 3 sec and 6 sec, the factories are

( ) ( )3 344 1032 m and 6 344 2064 mc b= = = =

from the man. Using the law of cosines we have

( )( )2 2 21032 2064 2 1032 2064 cos 42.2

2,169, 221.3 1472.8

a

a

= + − °≈ ⇒ ≈

The factories are about 1473 m apart. (rounded to four significant digits)

57. Let c = the length of the property line that cannot be directly measured.

Using the law of cosines, we have

( )( )2 2 214.0 13.0 2 14.0 13.0 cos 70

240.5 15.5 ft

c

c

= + − °≈ ⇒ ≈

(rounded to three significant digits) The length of the property line is approximately 18.0 + 15.5 + 14.0 = 47.5 feet

58. Let A = the point where the ship changes to a bearing of 62°; C = the point where it changes to a bearing of 115°.

90 62 28m CAB∠ = ° − ° = °

180

180 62 118

m FCA m DAC∠ = ° − ∠= ° − ° = °

360

360 115 118 127

m ACB m FCB m FCA∠ = ° − ∠ − ∠= ° − ° − ° = °

180

180 127 28 25

m CBA m ACB m CAB∠ = ° − ∠ − ∠= ° − ° − ° = °

Since we have only one side of a triangle, use the law of sines to find CB.

50 50sin 2829.4

sin 28 sin127 sin127

CBCB

°= ⇒ = ≈° ° °

and

50 50sin 2526.5

sin 25 sin127 sin127

ACAC

°= ⇒ = ≈° ° °

The ship traveled 26.5 + 29.4 = 55.9 mi. To avoid the iceberg, the ship had to travel 55.9 – 50 = 5.9 mi farther.



59. Let c = the length of the tunnel.

Use the law of cosines to find c.

( )( )2 2 23800 2900 2 3800 2900 cos110

30,388,124 5512.5

c

c

= + − °≈ ⇒ ≈

The tunnel is 5500 meters long. (rounded to two significant digits)

60. Let x = the distance from the plane to the mountain when the second bearing is taken.

180 32.7 147.3θ = ° − ° = °

Since we have only one side of a triangle, use

the law of sines to find x.

7.92

sin 24.1 sin147.37.92sin 24.1

5.99sin147.3

x

x

=° °

°= ≈°

The plane is about 5.99 km from the mountain. (rounded to three significant digits)

61. Let a be the length of the segment from (0, 0) to (6, 8). Use the distance formula.

( ) ( )2 2 2 26 0 8 0 6 8

36 64 100 10

a = − + − = += + = =

Let b be the length of the segment from (0, 0) to (4, 3).

( ) ( )2 2 2 24 0 3 0 4 3

16 9 25 5

b = − + − = += + = =

Let c be the length of the segment from (4, 3) to (6, 8).

( ) ( )2 2 2 26 4 8 3 2 5

4 25 29

c = − + − = += + =

( )( )( )

2 2 2

22 2

cos2

10 5 29 100 25 29cos

2 10 5 100

0.96 16.26

a b c

abθ

θ

θ

+ −= ⇒

+ − + −= =

= ⇒ ≈ °

62. Let a be the length of the segment from (0, 0) to (8, 6). Use the distance formula.

( ) ( )2 2 2 28 0 6 0 8 6

64 36 100 10

a = − + − = += + = =

Let b be the length of the segment from (0, 0) to (12, 5).

( ) ( )2 2 2 212 0 5 0 12 5

144 25 169 13

b = − + − = += + = =

Let c be the length of the segment from (8, 6) to (12, 5).

( ) ( ) ( )2 2 2212 8 5 6 4 1 17c = − + − = + − =

( )( )( )

2 2 2

22 2

cos2

10 13 17cos 0.96923077

2 10 13

14.25

a b c

abθ

θ

θ

+ −=

+ −= ≈ ⇒

≈ °

63. Using 1

2bh= ⇒A

( )( )116 3 3 24 3 41.57.

2= = ≈A

To use Heron’s Formula, first find the semiperimeter,

( ) ( )1 1 16 14 16 36 18.

2 2 2s a b c= + + = + + = ⋅ =

Now find the area of the triangle.

( )( )( )( )( )( )( )( )( )

18 18 6 18 14 18 16

18 12 4 2 1728 41.57

s s a s b s c= − − −

= − − −

= = ≈

A

Both formulas give the same area.

64. Using 1

2bh= ⇒A

( )( )110 3 3 15 3 25.98.

2= = ≈A

To use Heron’s Formula, first find the semiperimeter,

( ) ( )1 1 110 6 14 30 15.

2 2 2s a b c= + + = + + = ⋅ =

Now find the area of the triangle.

( )( )( )( )( )( )( )( )( )

15 15 10 15 6 15 14

15 5 9 1 675 15 3 25.98

s s a s b s c= − − −

= − − −

= = = ≈

A

Both formulas give the same result.



65. a = 12 m, b = 16 m, c = 25 m

( ) ( )1 112 16 25

2 21

53 26.52

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( ) 2

26.5 26.5 12 26.5 16 26.5 25

26.5 14.5 10.5 1.5 78 m

s s a s b s c= − − −

= − − −

= ≈

A

(rounded to two significant digits)

66. a = 22 in., b = 45 in., c = 31 in.

( ) ( )1 122 45 31

2 21

98 492

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( ) 2

49 49 22 49 45 49 31

49 27 4 18 310 in.

s s a s b s c= − − −

= − − −

= ≈

A

(rounded to two significant digits)

67. a = 154 cm, b = 179 cm, c = 183 cm

( ) ( )1 1154 179 183

2 21

516 2582

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( ) 2

258 258 154 258 179 258 183

258 104 79 75 12,600 cm

s s a s b s c= − − −

= − − −

= ≈

A

(rounded to three significant digits)

68. a = 25.4 yd, b = 38.2 yd, c = 19.8 yd

( ) ( )1 125.4 38.2 19.8

2 21

83.4 41.72

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )

( )( )( )( ) 2

41.7 41.7 25.4 41.7 38.2

41.7 19.8

41.7 16.3 3.5 21.9 228 yd

s s a s b s c= − − −

− − ⋅=

−

= ≈

A


69. a = 76.3 ft, b = 109 ft, c = 98.8 ft

( ) ( )1 176.3 109 98.8

2 21

284.1 142.052

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )

( )( )( )( ) 2

142.05 142.05 76.3 142.05 109

142.05 98.8

142.05 65.75 33.05 43.25 3650 ft

s s a s b s c= − − −

− − ⋅=

−

= ≈

A


70. a = 15.89 m, b = 21.74 m, c = 10.92 m

( ) ( )1 115.89 21.74 10.92

2 21

48.55 24.2752

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( )

2

24.275 24.275 15.89

24.275 21.74 24.275 10.92

24.275 8.385 2.535 13.355

83.01 m (rounded to four

significant digits)

s s a s b s c= − − −

− ⋅=

− −

=≈

71. Perimeter: 9 + 10 + 17 = 36 feet, so the semi-

perimeter is 1

36 18 feet.2⋅ =

Use Heron’s Formula to find the area.

( )( )( )( )( )( )( )( )( )

18 18 9 18 10 18 17

18 9 8 1 1296 36 ft

s s a s b s c= − − −

= − − −

= = =

A

Since the perimeter and area both equal 36, the triangle is a perfect triangle.

72. (a) ( ) ( )1 111 13 20

2 21

44 222

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( )

22 22 11 22 13 22 20

22 11 9 2

4356 66, which is an integer

s s a s b s c= − − −

= − − −

== =

A

(b) ( ) ( )1 113 14 15

2 21

42 212

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( )

21 21 13 21 14 21 15

21 8 7 6

7056 84, which is an integer

s s a s b s c= − − −

= − − −

== =

A



(c) ( ) ( )1 17 15 20 21

2 2s a b c= + + = + + =

( )( )( )( )( )( )( )( )( )

21 21 7 21 15 21 20

21 14 6 1 1764 42,

s s a s b s c= − − −

= − − −

= = =

A

which is an integer.

(d) ( ) ( )1 19 10 17 18

2 2s a b c= + + = + + =

( )( )( )( )( )( )( )( )( )

18 18 9 18 10 18 17

18 9 8 1 1296 36,

s s a s b s c= − − −

= − − −

= = =

A

which is an integer.

73. Find the area of the Bermuda Triangle using Heron’s Formula.

( ) ( )1 1850 925 1300

2 21

3075 1537.52

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )( )( )( )

1537.5 1537.5 850

1537.5 925 1537.5 1300

1537.5 687.5 612.5 237.5

392,128.82

s s a s b s c= − − −

− ⋅=

− −

=≈

A

The area of the Bermuda Triangle is about

390,000 2mi .

74. Find the area of the region using Heron’s Formula.

( ) ( )1 175 68 85

2 21

228 1142

s a b c= + + = + +

= ⋅ =

( )( )( )( )( )( )

( )( )( )( ) 2

114 114 75 114 68 114 85

114 39 46 29 2435.3571 m

s s a s b s c= − − −

= − − −

= ≈

A

Number of cans needed 2

2

(area in m )

(m per can)= = 2435.3571

32.471428 cans75

=

She will need to open 33 cans.

75. (a) Using the law of sines, we have

sin sin sin sin 60

15 13

15sin 60 15 3sin 0.99926008

13 13 2

C A C

c a

C

°= ⇒ = ⇒

°= = ⋅ ≈

There are two angles C between 0° and 180° that satisfy the condition. Since

sin 0.99926008,C ≈ to the nearest tenth

value of C is 1 87.8 .C = °

Supplementary angles have the same sine value, so another possible value of C is

2 180 87.8 92.2 .B = ° − ° = °

(b) By the law of cosines, we have

( )( )

2 2 2

2 2 2

cos2

13 7 15 7 1cos

2 13 7 182 26

0.03846154 92.2

a b cC

ab

C

C

+ −=

+ − −= = = −

≈ − ⇒ ≈ °

(c) With the law of cosines, we are required to find the inverse cosine of a negative number; therefore; we know angle C is greater than 90°.

76. Using the law of cosines, we have

( )( )

2 2 2

2 2 2

cos2

6 5 4 36 25 16 3cos

2 6 5 60 4

a c bB

ac

B

+ −= ⇒

+ − + −= = =

( )( )

2 2 2

2 2 2

cos2

4 5 6 16 25 36 1cos

2 4 5 40 8

b c aA

bc

A

+ −= ⇒

+ − + −= = =

Since

22 3 9 16

2cos 1 2 1 24 16 16

B⎛ ⎞ ⎛ ⎞− = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 1cos ,

16 8A= = = A is twice the size of B.

77.

( ) ( )2 21 4 3 0 25 9 34a = − − + − = + =

( ) ( )2 22 4 5 0 4 25 29b = − + − = + =

( ) ( )2 21 2 3 5 9 4 13c = − − + − = + =

Chapter 7 Quiz 365


78. Use the law of cosines to find the measure of A∠ .

( ) ( ) ( ) ( )( )2 2 2

2 2 2

2 cos

34 29 13 2 29 13 cos 34 42 2 377 cos

48 2 377 cos cos

377

a b c bc A

A A

A A

= + −

= + − ⇒ = − ⇒

− = − ⇒ =

( )( ) 11 1 377 4sin 29 13 sin sin cos 9.5 sq units

2 2 2 377ab C C −⎛ ⎞

= = = =⎜ ⎟⎝ ⎠A

79. First find the semiperimeter. ( ) ( )1 134 29 13

2 2s a b c= + + = + +

Using Heron’s formula, we have

( )( )( )

( ) ( ) ( ) ( )1 1 1 134 29 13 34 29 13 34 34 29 13 29 34 29 13 13

2 2 2 2

9.5 sq units (found using a calculator)

s s a s b s c

+ + + + − + + − + + −

= − − −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⎝ ⎠ ⎝ ⎠ ⎝ ⎠=

A

80. ( ) ( ) ( ) ( ) ( ) ( )1 1 2 2 3 3, 2,5 ; , 1,3 ; , 4,0x y x y x y= = − =

( ) ( ) ( )( ) ( ) ( ) ( )1 2 1 2 2 3 2 3 3 1 3 1

1 12 3 5 1 1 0 3 4 4 5 0 2 9.5 sq units

2 2x y y x x y y x x y y x= − + − + − = − − + − − + − =A

Chapter 7 Quiz (Sections 7.1−7.3)

1. Using the law of sines, we have

sin sin sin 30.6 sin

7.42 4.544.54sin 30.6

sin 0.3114627.42

18.1

B C C

b c

C

C

°= ⇒ = ⇒

°= ≈ ⇒

≈ °

180

180 30.6 18.1 131.3 131

A B C= ° − −= ° − ° − ° = ° ≈ °


2. Using the law of cosines, we have 2 2 2

2 2 2

2 cos

75.0 135 2 75.0 135cos144

40, 232.59

201 m (rounded to three significant digits)

a b c bc A

a

a

= + −= + − ⋅ ⋅ °≈ ⇒≈

3. Using the law of cosines, we have 2 2 2

2 2 2

2 cos

21.2 28.4 16.9 2 28.4 16.9cos

642.73 959.92cos

642.73cos

959.9248.0 (rounded to three

significant digits)

c a b ab C

C

C

C

C

= + −= + − ⋅ ⋅

− = −

= ⇒

≈ °

4. ( )( )1 1sin 7 9 sin150

2 263

15.75 sq units4

ab C= = °

= =

A

5. First find the semiperimeter:

( )119.5 21.0 22.5 31.5

2s = + + =

Using Heron’s formula, we have

( )( )( )( )( )( )( )( )( )

2

31.5 31.5 19.5 31.5 21.0 31.5 22.5

31.5 12 10.5 9

35,721 189 km

s s a s b s c

− − −

= − − −

=

== =

A

6. Using the law of sines,we have


534 345534sin 25.4

sin 0.663917345

41.6 or 180 41.6 138.4

A C A

a c

A

A A

°= ⇒ = ⇒

°= ≈ ⇒

≈ ° ≈ ° − ° = °



7. 180 111 41 28C∠ = ° − ° − ° = °

Using the law of sines, we have

326

sin sin sin111 sin 28326sin111

648sin 28

326

sin sin sin 41 sin 28326sin 41

456sin 28

a c a

A C

a

b c b

B C

b

= ⇒ = ⇒° °

°= ≈°

= ⇒ = ⇒° °

°= ≈°

Note that both a and b have been rounded to three significant digits.

8. The height of the balloon is the length of the

altitude of ,XYZJ AZ.

We will use the fact that sin 42 10AZ

XZ° =′ to

find the length of AZ. First, we must find the length of XZ using the law of sines.

180 42 10 23 30 114 20

12.2

sin sin sin 23 30 sin114 2012.2sin 23 30

5.3390sin114 20

Z

XZ XY XZ

Y Z

XZ

= ° − ° − ° = °′ ′ ′

= ⇒ = ⇒° °′ ′

° ′= ≈° ′

sin 42 105.33905.3390sin 42 10 3.6 mi

AZ

AZ

° =′

= ° ≈′

9. AB = 22.47928 mi, AC = 28.14276 mi, A = 58.56989°

This is SAS, so use the law of cosines.

( )( )

( )( )

2 2 2

2 2 2

2

2 cos

28.14276 22.47928

2 28.14276 22.47928 cos 58.56989°

637.55393

25.24983

BC AC AB AC AB A

BC

BC

BC

= + −= +−

≈≈

is approximately 25.24983 mi.

(rounded to seven significant digits)

BC

10. To find the distance between the towns, d, use the law of cosines.

( )( )2 2 23428 5631 2 3428 5631 cos 43.33

15,376,718 3921.3

d

d

= + − °≈ ⇒ ≈

The distance between the two towns is about 3921 m (rounded to four significant digits.)

Section 7.4 Vectors, Operations, and the

Dot Product

1. Equal vectors have the same magnitude and

direction. Equal vectors are m and p; n and r.

2. Opposite vectors have the same magnitude but

opposite direction. Opposite vectors are m and

q, p and q, n and s, r and s.

3. One vector is a positive scalar multiple of another if the two vectors point in the same direction; they may have different magnitudes.

1 ; 2 ; 1 ; 2= = = =m p m t n r p t or

1 11 ; ; 1

2 2= = = =p m t m r n t p;

4. One vector is a negative scalar multiple of another if the two vectors point in the opposite direction; they may have different magnitudes.

m = –1q; p = –1q; r = –1s; q = –2t; n = –1s

5.

6.

7.

8.

9.

section 7.3 the law of cosines sin sin121 sin 7.04 3 3sin121 sin 0.36527016 21.4 ... 7.23 8.94 2...

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