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```Chapter 8
Applications of Trigonometric Functions
11. opposite = 2; adjacent = 3; hypotenuse = ?
(hypotenuse) 2 = 22 + 32 = 13
Section 8.1
1. a = 52 − 32 = 25 − 9 = 16 = 4
hypotenuse = 13
sin θ =
1
, 0 < θ < 90°
2
1
θ = tan −1   ≈ 26.6o
2
2. tan θ =
opp
2
2
13 2 13
=
=
⋅
=
hyp
13
13
13 13
3
3
13 3 13
=
=
⋅
=
hyp
13
13
13 13
opp 2
tan θ =
=
cos θ =
1
, 0 < θ < 90°
2
1
θ = sin −1 = 30°
2
3. sin θ =
csc θ =
hyp
13
=
opp
2
hyp
13
=
3
cot θ =
=
opp 2
sec θ =
4. False; sin 52° = cos 38°
5. True
6. angle of elevation
12. opposite = 3; adjacent = 3; hypotenuse = ?
(hypotenuse) 2 = 32 + 32 = 18
7. True
hypotenuse = 18 = 3 2
8. False
9. opposite = 5; adjacent = 12; hypotenuse = ?
(hypotenuse)2 = 52 + 122 = 169
sin θ =
hypotenuse = 169 = 13
opp 5
hyp 13
sin θ =
csc θ =
=
=
hyp 13
opp 5
hyp 13
cos θ =
sec θ =
=
=
hyp 13
opp 5
tan θ =
cot θ =
=
=
opp 5
cos θ =
opp
3
3
2
2
=
=
⋅
=
hyp 3 2 3 2 2
2
3
3
2
2
=
=
⋅
=
hyp 3 2 3 2 2
2
opp 3
tan θ =
= =1
csc θ =
hyp 3 2
=
= 2
opp
3
hyp 3 2
=
= 2
3
= =1
cot θ =
opp 3
sec θ =
10. opposite = 3; adjacent = 4, hypotenuse = ?
(hypotenuse) 2 = 32 + 42 = 25
hypotenuse = 25 = 5
opp 3
hyp 5
sin θ =
csc θ =
=
=
hyp 5
opp 3
hyp 5
cos θ =
sec θ =
=
=
hyp 5
opp 3
tan θ =
cot θ =
=
=
opp 3
791
Chapter 8: Applications of Trigonometric Functions
13. adjacent = 2; hypotenuse = 4; opposite = ?
(opposite) 2 + 22 = 42
(opposite)2 = 16 − 4 = 12
opposite = 12 = 2 3
opp 2 3
3
=
=
hyp
4
2
= =
cos θ =
hyp 4 2
sin θ =
tan θ =
opp 2 3
=
= 3
2
opp
2
=
= 2
1
csc θ =
hyp
3
=
=
opp
2
sec θ =
hyp
3
=
= 3
1
cot θ =
1
1
2
2
=
=
⋅
=
opp
2
2
2 2
3 2
6
⋅
=
2
2 2
16. opposite = 2; adjacent =
(hypotenuse) = 2 +
2
hyp
4
4
3 2 3
=
=
⋅
=
opp 2 3 2 3 3
3
hyp 4
sec θ =
= =2
csc θ =
cot θ =
tan θ =
2
( 3)
3 ; hypotenuse = ?
2
=7
hypotenuse = 7
2
2
3
3
=
=
⋅
=
opp 2 3 2 3 3
3
14. opposite = 3; hypotenuse = 4; adjacent = ?
(adjacent)2 = 16 − 9 = 7
sin θ =
opp
2
2
7 2 7
=
=
⋅
=
hyp
7
7
7 7
cos θ =
3
3 7
21
=
=
⋅
=
hyp
7
7
7 7
tan θ =
opp
2
2
3 2 3
=
=
⋅
=
3
3
3 3
csc θ =
hyp
7
=
opp
2
sin θ =
opp 3
=
hyp 4
sec θ =
hyp
7
7 3
21
=
=
⋅
=
3
3
3 3
cos θ =
7
=
hyp
4
cot θ =
3
=
opp
2
opp
3
3
7 3 7
=
=
⋅
=
7
7
7 7
hyp 4
=
csc θ =
opp 3
tan θ =
17. opposite = 1; hypotenuse =
(adjacent)2 = 5 − 1 = 4
hyp
4
4
7 4 7
=
=
⋅
=
sec θ =
7
7
7 7
cot θ =
sin θ =
7
=
opp
3
15. opposite =
( 2)
2
2
2
5 2 5
=
=
⋅
=
hyp
5
5
5 5
opp 1
tan θ =
=
+ 12 = 3
hypotenuse = 3
opp
sin θ =
=
hyp
2
=
3
opp
1
1
5
5
=
=
⋅
=
hyp
5
5
5 5
cos θ =
2 ; adjacent = 1; hypotenuse = ?
(hypotenuse) 2 =
( 5)
2
csc θ =
2 3
6
⋅
=
3
3 3
hyp
5
=
= 5
opp
1
hyp
5
=
2
cot θ =
= =2
opp 1
sec θ =
1
1
3
3
cos θ =
=
=
⋅
=
hyp
3
3
3 3
792
Section 8.1: Right Triangle Trigonometry; Applications
18. adjacent = 2; hypotenuse =
(opposite) + 2 =
2
2
( 5)
5 ; opposite = ?
25. tan 20º −
2
(opposite) = 5 − 4 = 1
2
opposite = 1 = 1
sin θ =
=0
opp
1
1
5
5
=
=
⋅
=
hyp
5
5
5 5
26. cot 40º −
2
2
5 2 5
cos θ =
=
=
⋅
=
hyp
5
5
5 5
opp 1
tan θ =
=
csc θ =
27. cos 35º ⋅ sin 55º + cos 55º ⋅ sin 35º
= cos 35º ⋅ cos(90º −55º ) + sin(90º −55º ) ⋅ sin 35º
= cos 35º ⋅ cos 35º + sin 35º ⋅ sin 35º
= cos 2 35º + sin 2 35º
=1
19. sin 38º − cos 52º = sin 38º − sin(90º − 52º )
28. sec35º ⋅ csc 55º − tan 35º ⋅ cot 55º
= sin 38º − sin 38º
= sec 35º ⋅ sec(90º −55º ) − tan 35º ⋅ tan(90º −55º )
= sec 35º ⋅ sec35º − tan 35º ⋅ tan 35º
=0
20. tan12º − cot 78º = tan12º − tan(90º − 78º )
= sec 2 35º − tan 2 35º
= tan12º − tan12º
= (1 + tan 2 35º ) − tan 2 35º
=1
=0
22.
sin 50º
cos(90º − 50º )
= cot 40º −
sin 40º
sin 40º
cos 40º
= cot 40º −
sin 40º
= cot 40º − cot 40º
=0
hyp
5
=
= 5
opp
1
hyp
5
sec θ =
=
2
cot θ =
= =2
opp 1
21.
cos 70º
sin(90º − 70º )
= tan 20º −
cos 20º
cos 20º
sin 20º
= tan 20º −
cos 20º
= tan 20º − tan 20º
cos10º sin(90º − 10º ) sin 80º
=
=
=1
sin 80º
sin 80º
sin 80º
29. b = 5, B = 20º
b
a
5
tan ( 20º ) =
a
tan B =
cos 40º sin(90º − 40º ) sin 50º
=
=
=1
sin 50º
sin 50º
sin 50º
a=
23. 1 − cos 2 20º − cos 2 70º = 1 − cos 2 20º − sin 2 (90º −70º )
= 1 − cos 2 20º − sin 2 (20º )
(
= 1 − cos 2 20º + sin 2 (20º )
5
5
≈
≈ 13.74
tan ( 20º ) 0.3640
b
c
5
sin ( 20º ) =
c
)
sin B =
= 1−1
=0
c=
24. 1 + tan 2 5º − csc 2 85º = sec2 5º − csc 2 85º
= sec2 5º − sec2 (90º − 85º )
5
5
≈
≈ 14.62
sin ( 20º ) 0.3420
A = 90º − B = 90º − 20º = 70º
= sec2 5º − sec2 5º
=0
793
Chapter 8: Applications of Trigonometric Functions
30. b = 4, B = 10º
33. b = 4, A = 10º
b
a
4
tan (10º ) =
a
a
b
a
tan (10º ) =
4
a = 4 tan (10º ) ≈ 4 ⋅ (0.1763) ≈ 0.71
tan B =
a=
tan A =
4
4
≈
≈ 22.69
tan (10º ) 0.1763
b
c
4
cos (10º ) =
c
cos A =
b
sin B =
c
4
sin (10º ) =
c
c=
4
4
c=
≈
≈ 23.04
sin (10º ) 0.1736
4
4
≈
≈ 4.06
0.9848
cos (10º )
B = 90º − A = 90º − 10º = 80º
A = 90º − B = 90º − 10º = 80º
34. b = 6, A = 20º
31. a = 6, B = 40º
b
a
b
tan ( 40º ) =
6
b = 6 tan ( 40º ) ≈ 6 ⋅ (0.8391) ≈ 5.03
a
b
a
tan ( 20º ) =
6
a = 6 tan ( 20º ) ≈ 6 ⋅ (0.3640) ≈ 2.18
a
c
6
cos ( 40º ) =
c
b
c
6
cos ( 20º ) =
c
tan A =
tan B =
cos B =
c=
cos A =
6
6
≈
≈ 7.83
cos ( 40º ) 0.7660
c=
A = 90º − B = 90º − 40º = 50º
B = 90º − A = 90º − 20º = 70º
32. a = 7, B = 50º
b
tan B =
a
b
tan ( 50º ) =
7
b = 7 tan ( 50º ) ≈ 7 ⋅ (1.1918) ≈ 8.34
35. a = 5, A = 25º
b
cot A =
a
b
cot ( 25º ) =
5
b = 5cot ( 25º ) ≈ 5 ⋅ ( 2.1445) ≈ 10.72
a
c
7
cos ( 50º ) =
c
cos B =
c=
6
6
≈
≈ 6.39
cos ( 20º ) 0.9397
c
a
c
csc ( 25º ) =
5
c = 5csc ( 25º ) ≈ 5 ⋅ ( 2.3662 ) ≈ 11.83
csc A =
7
7
≈
≈ 10.89
cos ( 50º ) 0.6428
A = 90° − B = 90° − 50° = 40°
B = 90° − A = B = 90º − A = 90° − 25° = 65°
794
Section 8.1: Right Triangle Trigonometry; Applications
36. a = 6, A = 40º
a
cot A =
b
b
cot ( 40º ) =
6
b = 6 cot ( 40º ) ≈ 6 ⋅ (1.1918) ≈ 7.15
39. a = 5, b = 3
c 2 = a 2 + b 2 = 52 + 32 = 25 + 9 = 34
c = 34 ≈ 5.83
a 5
=
b 3
5
A = tan −1   ≈ 59.0°
3
tan A =
c
a
c
csc ( 45º ) =
6
c = 6 csc ( 40º ) ≈ 6 ⋅ (1.5557 ) ≈ 9.33
csc A =
B = 90° − A ≈ 90° − 59.0° = 31.0°
40. a = 2, b = 8
c 2 = a 2 + b 2 = 22 + 82 = 4 + 64 = 68
B = 90° − A = 90° − 40° = 50°
c = 68 ≈ 8.25
37. c = 9, B = 20º
b
sin B =
c
b
sin ( 20º ) =
9
b = 9sin ( 20º ) ≈ 9 ⋅ (0.3420) ≈ 3.08
a 2 1
= =
b 8 4
1
A = tan −1   ≈ 14.0°
4
tan A =
B = 90° − A ≈ 90° − 14.0° = 76.0°
41. a = 2, c = 5
a
c
a
cos ( 20º ) =
9
a = 9 cos ( 20º ) ≈ 9 ⋅ (0.9397) ≈ 8.46
cos B =
c2 = a2 + b2
b 2 = c 2 − a 2 = 52 − 22 = 25 − 4 = 21
b = 21 ≈ 4.58
a 2
=
c 5
2
A = tan −1   ≈ 23.6º
5
sin A =
A = 90° − A = 90° − 20° = 70°
38. c = 10, A = 40º
a
c
a
sin ( 40º ) =
10
a = 10sin ( 40º ) ≈ 10 ⋅ (0.6428) ≈ 6.43
sin A =
B = 90° − A ≈ 90° − 23.6° = 66.4°
42. b = 4, c = 6
c2 = a 2 + b2
a 2 = c 2 − b2 = 62 − 42 = 36 − 16 = 20
b
cos A =
c
b
cos ( 40º ) =
10
b = 10 cos ( 40º ) ≈ 10 ⋅ (0.7660) ≈ 7.66
a = 20 ≈ 4.47
b 4 2
= =
c 6 3
2
A = tan −1   ≈ 48.2º
3
cos A =
B = 90° − A = 90° − 40° = 50°
B = 90º − A ≈ 90º − 48.2º = 41.8º
795
Chapter 8: Applications of Trigonometric Functions
43. c = 5, a = 2
Case 2: θ = 25º , b = 5
2
5
sin A =
c
2
A = sin   ≈ 23.6°
5
B = 90º − A ≈ 90º − 23.6º = 66.4º
The two angles measure about 23.6° and 66.4º .
a
−1
25º
5
cos ( 25º ) =
c=
44. c = 3, a = 1
B = 90º − A ≈ 90º − 19.5º = 70.5º
The two angles measure about 19.5° and 70.5º .
b.
45. c = 8, θ = 35º
5
5
≈
≈ 5.52 in.
cos ( 25º ) 0.9063
There are two possible cases because the
given side could be adjacent or opposite the
given angle.
48. Case 1: θ =
8
5
c
a
π
, a=3
8
a.
35º
b
a
8
a = 8sin ( 35º )
b
8
b = 8cos ( 35º )
≈ 8(0.5736)
≈ 8(0.8192)
≈ 4.59 in.
≈ 6.55 in.
sin ( 35º ) =
cos ( 35º ) =
c
8
b
π 3
sin   =
8 c
3
3
≈
≈ 7.84 m.
c=
 π  0.3827
sin  
8
46. c = 10, θ = 40º
10
a
40º
Case 2: θ =
π
, b=3
8
c
a
b
a
10
a = 10sin ( 40º )
b
10
b = 10 cos ( 40º )
≈ 10(0.6428)
≈ 10(0.7660)
≈ 6.43 cm.
≈ 7.66 cm.
sin ( 40º ) =
cos ( 40º ) =
_π
8
a.
b. There are two possible cases because the given
side could be adjacent or opposite the given
angle.
5
25º
b
sin ( 25º ) =
c=
3
π 3
cos   =
8 c
3
3
≈
≈ 3.25 m.
c=
 π  0.9239
cos  
8
47. Case 1: θ = 25º , a = 5
c
3
_π
49. tan ( 35º ) =
5
c
AC
100
AC = 100 tan ( 35º ) ≈ 100(0.7002) ≈ 70.02 feet
5
5
≈
≈ 11.83 in.
sin ( 25º ) 0.4226
796
Section 8.1: Right Triangle Trigonometry; Applications
56. opposite side = 10 feet, adjacent side = 35 feet
10
tan θ =
35
10
θ = tan −1   ≈ 15.9º
 35 
AC
50. tan ( 40º ) =
100
AC = 100 tan ( 40º ) ≈ 100(0.8391) ≈ 83.91 feet
51. Let x = the height of the Eiffel Tower.
x
tan ( 85.361º ) =
80
x = 80 tan ( 85.361º ) ≈ 80(12.3239) ≈ 985.91 feet
57. a.
52. Let x = the distance to the shore.
100 feet
The truck is traveling at 111.96 ft/sec, or
111.96 ft 1 mile 3600 sec
⋅
⋅
≈ 76.3 mi/hr .
sec
5280 ft
hr
25º
x
tan ( 25º ) =
x=
100
x
b.
100
100
≈
≈ 214.45 feet
tan ( 25º ) 0.4663
c.
20º
x
50
x
50
50
≈
≈ 137.37 meters
tan ( 20º ) 0.3640
54. Let x = the distance up the building
22 feet
x
22
x = 22sin ( 70º ) ≈ 22(0.9397) ≈ 20.67 feet
sin ( 70º ) =
300
=6
50
A = tan −1 6 ≈ 80.5º
The angle of elevation of the
30
30
≈
≈ 82.42 feet
tan ( 20º ) 0.3640
A ticket is issued for traveling at a speed of
60 mi/hr or more.
60 mi 5280 ft
1hr
⋅
⋅
= 88 ft/sec.
hr
mi
3600 sec
30
, the trooper should issue a
If tan θ <
88
 30 
ticket. Now, tan −1   ≈ 18.8° , so a ticket
 88 
is issued if θ < 18.8º .
58. If the camera is to be directed to a spot 6 feet
above the floor 12 feet from the wall, then the
“side opposite” the angle of depression is 3 feet.
(see figure)
12
3 1
=
tan A =
A
12 4
3
−1 1
A = tan
≈ 14.0°
4
The angle of depression should be about 14.0° .
x
70º
55.
30
x
The truck is traveling at 82.42 ft/sec, or
82.42 ft 1 mile 3600 sec
⋅
⋅
≈ 56.2 mi/hr .
sec
5280 ft
hr
50 meters
x=
tan ( 20º ) =
x=
53. Let x = the distance to the base of the plateau.
tan ( 20º ) =
Let x represent the distance the truck
traveled in the 1 second time interval.
30
tan (15º ) =
x
30
30
x=
≈
≈ 111.96 feet
tan (15º ) 0.2679
tan A =
300 feet
A
50 feet
797
Chapter 8: Applications of Trigonometric Functions
59. a.
(
)
61. Let h = the height of the monument.
4.22 ⋅ 5.9 × 1012 = 24.898 ×1012
h
789
h = 789 tan ( 35.1º )
tan ( 35.1º ) =
= 2.4898 ×1013
Proxima Centauri is about 2.4898 ×1013
miles from Earth.
b.
≈ 789(0.7028)
≈ 554.52 ft
Construct a right triangle using the sun,
Earth, and Proxima Centauri as shown. The
hypotenuse is the distance between Earth
and Proxima Centauri.
b
Sun
a
35.1º
789 ft
62. The elevation change is 11200 − 9000 = 2200 ft .
Let x = the length of the trail.
Proxima
Centauri
θ
h
x
Parallax
2200 ft
17º
c
2200
x
2200
2200
x=
≈
≈ 7524.67 ft.
sin (17º ) 0.2924
sin17º =
Earth
a
9.3 × 107
=
c 2.4898 × 1013
sin θ ≈ 0.000003735
sin θ =
63. Begin by finding angle θ = ∠BAC : (see figure)
40º
The parallax of Proxima Centauri is
0.000214° .
60. a.
(
11.14 ⋅ 5.9 × 10
12
i
E
Α
) = 65.726 ×10
12
C
1
=2
0.5
θ = 63.4º
∠DAC = 40º + 63.4º = 103.4º
∠EAC = 103.4º − 90º = 13.4º
Now, 90º −13.4º = 76.6º
The control tower should use a bearing of S76.6˚E.
tan θ =
= 6.5726 ×10
61 Cygni is about 6.5726 ×1013 miles from
Earth.
Construct a right triangle using the sun,
Earth, and 61 Cygni as shown. The
hypotenuse is the distance between Earth
and 61 Cygni.
b
Sun
61 Cygni
θ
64. Find ∠AMB and subtract from 80˚ to obtain θ
(see figure).
80º
M
a
1m
θ
13
b.
B
i
D
0.5
m
θ ≈ sin −1 0.000003735
θ ≈ 0.000214°
Parallax
Α
15
c
θ
30
C
Earth
a
9.3 × 107
sin θ = =
c 6.5726 × 1013
sin θ ≈ 0.000001415
B
∠CMA = 80°
30
=2
15
∠AMB = tan −1 2 ≈ 63.4º
θ = 80º − 63.4º = 16.6º
The bearing of the ship from port is S16.6ºE .
tan ∠AMB =
θ ≈ sin −1 0.000001415
θ ≈ 0.00008°
The parallax of 61 Cygni is 0.00008° .
798
Section 8.1: Right Triangle Trigonometry; Applications
65. Let y = the height of the embankment.
.8
y
36o2’
x
θ = 36°2 ' ≈ 36.033°
140º 40º
sin ( 40º ) =
x=
z=
43
27
34
a=
o
x
2593
x = 2593 ⋅ cos 34° ≈ 2150
The surveyor is located approximately 2150
feet from the building.
cos 34° =
b.
c.
d.
b=
1
a
1
≈ 1.1918 mi
tan ( 40º )
tan ( 50º ) =
x
1
z
1
≈ 1.3054 mi
sin ( 50º )
tan ( 40º ) =
h
1
x
1
≈ 1.5557 mi
sin ( 40º )
sin ( 50º ) =
a
93
b
3 mi
The length of the highway = x + y + z
Let h = the height of the building, a = the
height of the antenna, and x = the distance
between the surveyor and the base of the
building.
25
z
50º 130º
a
y
sin ( 36.033° ) =
51.8
y = 51.8sin ( 36.033° ) ≈ 30.5 meters
The embankment is about 30.5 meters high.
66. a.
1 mi
y
1 mi
51
67. Let x, y, and z = the three segments of the
highway around the bay (see figure).
1
b
1
≈ 0.8391 mi
tan ( 50º )
a+ y+b = 3
y = 3− a −b
h
sin 34° =
2593
h = 2593 ⋅ sin 34° ≈ 1450
The building is about 1450 feet high.
≈ 3 − 1.1918 − 0.8391 = 0.9691 mi
The length of the highway is about:
1.5557 + 0.9691 + 1.3054 ≈ 3.83 miles .
68. Let x = the distance from George at which the
camera must be set in order to see his head and
feet.
Let θ = the angle of inclination from the
surveyor to the top of the antenna.
2150
cos θ =
2743
 2150 
θ = cos−1 
 ≈ 38.4°
 2743 
x
20º
4 ft
tan ( 20º ) =
h + a 1450 + a
=
2743
2743
2743 ⋅ sin θ = 1450 + a
a = 2743 ⋅ sin θ − 1450
sin θ =
4
x
4
≈ 10.99 feet
tan ( 20º )
If the camera is set at a distance of 10 feet from
George, his feet will not be seen by the lens.
The camera would need to be moved back about
1 additional foot (11 feet total).
x=

 2150  
a = 2743sin  cos −1 
  − 1450
 2743  

a ≈ 253
The antenna is about 253 feet tall.
799
Chapter 8: Applications of Trigonometric Functions
69. We construct the figure below:
32º
h
23º
500 ft
32º
71. Let h represent the height of Lincoln's face.
23º
y
x
tan ( 32º ) =
500
x
b
tan ( 23º ) =
500
x=
tan ( 32º )
32º
500
y
800 feet
b
800
b = 800 tan ( 32º ) ≈ 499.90
tan ( 32º ) =
500
y=
tan ( 23º )
b+h
800
b + h = 800 tan ( 35º ) ≈ 560.17
Thus, the height of Lincoln’s face is:
h = (b + h) − b = 560.17 − 499.90 ≈ 60.27 feet
tan ( 35º ) =
Distance = x + y
500
500
=
+
tan ( 32º ) tan ( 23º )
≈ 1978.09 feet
70. Let h = the height of the balloon.
54º
35º
72. Let h represent the height of tower above the
Sky Pod.
61º
h
h
h
b
61º
20.1º
54º
4000 feet
x − 100
100 ft
x
tan ( 54º ) =
x=
b
4000
b = 4000 tan ( 20.1º ) ≈ 1463.79
tan ( 20.1º ) =
h
x
b+h
4000
b + h = 4000 tan ( 24.4º ) ≈ 1814.48
Thus, the height of tower above the Sky Pod is:
h = (b + h) − b = 1814.48 − 1463.79 ≈ 350.69 feet
tan ( 24.4º ) =
h
tan ( 54º )
h
x − 100
h = ( x − 100) tan ( 61º )
tan ( 61º ) =
73. Let x = the distance between the buildings.
h


h=
− 100  tan ( 61º )
 tan ( 54º )

h−
24.4º
tan(61°)
tan(54°)
h = −100 tan ( 61° )
1451
o
10.3
 tan(61°) 
h 1 −
 = −100 tan ( 61° )
 tan(54°) 
h=
x
1451
x
1451
x=
≈ 7984
tan (10.3° )
tan θ =
−100 tan ( 61° )
≈ 580.61
 tan ( 61° ) 
 1 − tan ( 54° ) 


Thus, the height of the balloon is approximately
580.61 feet.
The two buildings are about 7984 feet apart.
800
Section 8.1: Right Triangle Trigonometry; Applications
74. tan (α + β ) =
x +2 x +8
=
2
4
4( x + 2) = 2( x + 8)
4 x + 8 = 2 x + 16
2x = 8
x=4
Thus, the hypotenuse of the smaller triangle is
θ  2 1
4 + 2 = 6 . Now, sin   = = , so we have
2 6 3
2070
630
 2070 

 630 
 2070 
α = tan −1 
−β
 630 
 2070 
−1  67 
= tan −1 
 − cot  
 630 
 55 
≈ 33.69°
Let x = the distance between the Arch and the
boat on the Missouri side.
x
tan α =
630
x = 630 tan α = 630 tan ( 33.69° )
α + β = tan −1 
that:
x = 420
Therefore, the Mississippi River is
approximately 2070 − 420 = 1650 feet wide at
the St. Louis riverfront.
78. a.
75. The height of the beam above the wall is
46 − 20 = 26 feet.
26
tan θ =
= 2.6
10
θ = tan −1 2.6 ≈ 69.0º
The pitch of the roof is about 69.0º .
c.
3960
 d 
cos 
 = 3960 + h
7920


e.
2
θ/2
x
2
22
4
4
3960
θ 
cos   =
 2  3960 + h
d = 3960 θ
76. tan A =
77. A line segment drawn from the vertex of the
angle through the centers of the circles will
bisect θ (see figure). Thus, the angle of the two
θ
right triangles formed is .
1
= sin −1  
3
1
θ = 2 ⋅ sin −1   ≈ 38.9°
3
2
b.
d.
10 − 6 4
=
15
15
−1  4 
A = tan   ≈ 14.9º
 15 
The angle of elevation from the player’s eyes to
the center of the rim is about 14.9º .
θ
3960
 2500 
cos 
 = 3960 + h
7920


3960
0.9506 ≈
3960 + h
0.9506(3960 + h) ≈ 3960
3764 + 0.9506h ≈ 3960
0.9506h ≈ 196
h ≈ 206 miles
3960
3960
 d 
cos 
=
=
 7920  3960 + 300 4260
d
 3960 
= cos −1 

7920
 4260 
 3960 
d = 7970 cos −1 

 4260 
≈ 2990 miles
Let x = the length of the segment from the vertex
of the angle to the smaller circle. Then the
hypotenuse of the smaller right triangle is x + 2 ,
and the hypotenuse of the larger right triangle is
x + 2 + 2 + 4 = x + 8 . Since the two right
triangles are similar, we have that:
801
Chapter 8: Applications of Trigonometric Functions
79. Adding some lines to the diagram and labeling
special points, we obtain the following:
b.
5
D
C
B
α
1.8
α
Let y = the difference in height between
Freedom Tower and the office building.
Together with the result from part (a), we
get the following diagram
y
20°
2633
E
opposite
tan θ =
y
tan 20° =
2633
y ≈ 958
The Freedom Tower is about 958 feet taller
than the office building. Therefore, the
office building is 1776 − 958 = 818 feet tall.
1.2
1.5
A
1
3
If we let x = length of side BC, we see that, in
3
ΔABC , tan α = . Also, in ΔEDC ,
x
1.8
tan α =
. Therefore, we have
5− x
3 1.8
=
x 5− x
15 − 3x = 1.8 x
15 = 4.8 x
15
x=
= 3.125 ft
4.8
1 + 3.125 = 4.125 ft
The player should hit the top cushion at a point
that is 4.125 feet from upper left corner.
80. a.
81 – 82. Answers will vary.
83. Let θ = the central angle formed by the top of
the lighthouse, the center of the Earth and the
point P on the Earth’s surface where the line of
sight from the top of the lighthouse is tangent to
362
miles.
the Earth. Note also that 362 feet =
5280
The distance between the buildings is the
length of the side adjacent to the angle of
elevation in a right triangle.
1776
θ = cos−1 
3960

o
 ≈ 0.33715
+
3960
362
/
5280


34°
x
opposite
and we know the
angle measure, we can use the tangent to
find the distance. Let x = the distance
between the buildings. This gives us
1776
tan 34° =
x
1776
x=
tan 34°
x ≈ 2633
The office building is about 2633 feet from
the base of the tower.
Since tan θ =
Verify the airplane information:
Let β = the central angle formed by the plane, the
center of the Earth and the point P.
3960

β = cos −1 
 ≈ 1.77169°
+
3960
10,000
/
5280


Note that
d
d
tan θ = 1
and tan β = 2
3690
3690
d1 = 3960 tan θ
d 2 = 3960 tan β
So,
d1 + d 2 = 3960 tan θ + 3960 tan β
≈ 3960 tan(0.33715°) + 3960 tan(1.77169°)
≈ 146 miles
802
Section 8.2: The Law of Sines
To express this distance in nautical miles, we
express the total angle θ + β in minutes. That is,
Section 8.2
θ + β ≈ ( 0.33715o + 1.77169o ) ⋅ 60 ≈ 126.5
1. sin A cos B − cos A sin B
nautical miles. Therefore, a plane flying at an
altitude of 10,000 feet can see the lighthouse 120
miles away.
2. cos θ =
3
2
 3 π
 = = 30°
 2  6
θ = cos −1 
Verify the ship information:
The solution set is
{π } .
6
= 30°
3. For similar triangles, corresponding pairs of
sides occur in the same ratio. Therefore, we can
write:
x 5
15
=
or x =
3 2
2
15
The missing length is
.
2
Let α = the central angle formed by 40 nautical
miles then. α =
40 2
=
60 3
4. oblique
o
5.
3960
3960 + x
3960
cos ( (2 / 3)° − 0.33715° ) =
3960 + x
( 3960 + x ) cos ( (2 / 3)° − 0.33715° ) = 3960
cos(α − θ ) =
3960 + x =
x=
sin A sin B sin C
=
=
a
b
c
6. False
7. False: You must have at least one angle opposite
one side.
3960
cos ( (2 / 3)° − 0.33715° )
8. ambiguous case
9. c = 5, B = 45º , C = 95º
A = 180º − B − γ = 180º − 45º − 95º = 40º
3960
− 3960
cos ( (2 / 3)° − 0.33715° )
sin A sin C
=
a
c
sin 40º sin 95º
=
5
a
5sin 40º
a=
≈ 3.23
sin 95º
≈ 0.06549 miles
ft 

≈ ( 0.06549 mi )  5280 
mi


≈ 346 feet
Therefore, a ship that is 346 feet above sea level
can see the lighthouse from a distance of 40
nautical miles.
sin B sin C
=
b
c
sin 45º sin 95º
=
5
b
5sin 45º
b=
≈ 3.55
sin 95º
803
Chapter 8: Applications of Trigonometric Functions
10. c = 4, A = 45º , B = 40º
C = 180º − A − B = 180º − 45º − 40º = 95º
13. b = 7, A = 40º , B = 45º
C = 180º − A − B = 180º − 40º − 45º = 95º
sin A sin C
=
a
c
sin 45º sin 95º
=
4
a
4sin 45º
≈ 2.84
a=
sin 95º
sin A sin B
=
a
b
sin 40º sin 45º
=
7
a
7 sin 40º
≈ 6.36
a=
sin 45º
sin B sin C
=
b
c
sin 40º sin 95º
=
4
b
4 sin 40º
b=
≈ 2.58
sin 95º
sin C sin B
=
c
b
sin 95º sin 45º
=
7
c
7 sin 95º
≈ 9.86
c=
sin 45º
11. b = 3, A = 50º , C = 85º
B = 180º − A − C = 180º − 50º − 85º = 45º
14. c = 5, A = 10º , B = 5º
C = 180º − A − B = 180º − 10º − 5º = 165º
sin A sin B
=
a
b
sin 50º sin 45º
=
3
a
3sin 50º
a=
≈ 3.25
sin 45º
sin A sin C
=
a
c
sin10º sin165º
=
5
a
5sin10º
a=
≈ 3.35
sin165º
sin C sin B
=
c
b
sin 85º sin 45º
=
3
c
3sin 85º
c=
≈ 4.23
sin 45º
sin B sin C
=
b
c
sin 5º sin165º
=
5
b
5sin 5º
b=
≈ 1.68
sin165º
12. b = 10, B = 30º , C = 125º
A = 180º − B − C = 180º − 30º − 125º = 25º
15. b = 2, B = 40º , C = 100º
A = 180º − B − C = 180º − 40º − 100º = 40º
sin A sin B
=
a
b
sin 25º sin 30º
=
10
a
10 sin 25º
≈ 8.45
a=
sin 30º
sin A sin B
=
a
b
sin 40º sin 40º
=
2
a
2 sin 40º
a=
=2
sin 40º
sin C sin B
=
c
b
sin125º sin 30º
=
10
c
10sin125º
c=
≈ 16.38
sin 30º
sin C sin B
=
c
b
sin100º sin 40º
=
2
c
2 sin100º
c=
≈ 3.06
sin 40º
804
Section 8.2: The Law of Sines
16. b = 6, A = 100º , B = 30º
C = 180º − A − B = 180º − 100º − 30º = 50º
19. B = 70º , C = 10º , b = 5
A = 180º − B − C = 180º − 70º − 10º = 100º
sin A sin B
=
a
b
sin100º sin 30º
=
6
a
6 sin100º
a=
≈ 11.82
sin 30º
sin A sin B
=
a
b
sin100º sin 70º
=
5
a
5sin100º
a=
≈ 5.24
sin 70º
sin C sin B
=
c
b
sin 50º sin 30º
=
6
c
6sin 50º
c=
≈ 9.19
sin 30º
sin C sin B
=
c
b
sin10º sin 70º
=
5
c
5sin10º
c=
≈ 0.92
sin 70º
17. A = 40º , B = 20º , a = 2
C = 180º − A − B = 180º − 40º − 20º = 120º
20. A = 70º , B = 60º , c = 4
C = 180º − A − B = 180º − 70º − 60º = 50º
sin A sin B
=
a
b
sin 40º sin 20º
=
2
b
2sin 20º
b=
≈ 1.06
sin 40º
sin A sin C
=
a
c
sin 70º sin 50º
=
4
a
4 sin 70º
a=
≈ 4.91
sin 50º
sin C sin A
=
c
a
sin120º sin 40º
=
2
c
2sin120º
c=
≈ 2.69
sin 40º
sin B sin C
=
b
c
sin 60º sin 50º
=
4
b
4sin 60º
b=
≈ 4.52
sin 50º
18. A = 50º , C = 20º , a = 3
B = 180º − A − C = 180º − 50º − 20º = 110º
21. A = 110º , C = 30º , c = 3
B = 180º − A − C = 180º − 110º − 30º = 40º
sin A sin B
=
a
b
sin 50º sin110º
=
3
b
3sin110º
b=
≈ 3.68
sin 50º
sin A sin C
=
a
c
sin110º sin 30º
=
3
a
3sin110º
a=
≈ 5.64
sin 30º
sin C sin A
=
c
a
sin 20º sin 50º
=
3
c
3sin 20º
c=
≈ 1.34
sin 50º
sin C sin B
=
c
b
sin 30º sin 40º
=
3
b
3sin 40º
b=
≈ 3.86
sin 30º
805
Chapter 8: Applications of Trigonometric Functions
22. B = 10º , C = 100º , b = 2
A = 180º − B − C = 180º − 10º − 100º = 70º
25. a = 3, b = 2, A = 50º
sin B sin A
=
b
a
sin B sin ( 50º )
=
2
3
2sin ( 50º )
≈ 0.5107
sin B =
3
sin A sin B
=
a
b
sin 70º sin10º
=
2
a
2sin 70º
a=
≈ 10.82
sin10º
B = sin −1 ( 0.5107 )
B = 30.7º or B = 149.3º
The second value is discarded because
A + B > 180º .
C = 180º − A − B = 180º − 50º − 30.7º = 99.3º
sin C sin B
=
c
b
sin100º sin10º
=
2
c
2sin100º
c=
≈ 11.34
sin10º
sin C sin A
=
c
a
sin 99.3º sin 50º
=
3
c
3sin 99.3º
c=
≈ 3.86
sin 50º
One triangle: B ≈ 30.7º , C ≈ 99.3º , c ≈ 3.86
23. A = 40º , B = 40º , c = 2
C = 180º − A − B = 180º − 40º − 40º = 100º
sin A sin C
=
a
c
sin 40º sin100º
=
2
a
2sin 40º
a=
≈ 1.31
sin100º
26. b = 4, c = 3, B = 40º
sin B sin C
=
b
c
sin 40º sin C
=
4
3
3sin 40º
sin C =
≈ 0.4821
4
sin B sin C
=
b
c
sin 40º sin100º
=
2
b
2sin 40º
b=
≈ 1.31
sin100º
C = sin −1 ( 0.4821)
C = 28.8º or C = 151.2º
The second value is discarded because
B + C > 180º .
A = 180º − B − C = 180º − 40º − 28.8º = 111.2º
24. B = 20º , C = 70º , a = 1
A = 180º − B − C = 180º − 20º −70º = 90º
sin A sin B
=
a
b
sin 90º sin 20º
=
1
b
1sin 20º
b=
≈ 0.34
sin 90º
sin B sin A
=
b
a
sin 40º sin111.2º
=
4
a
4 sin111.2º
a=
≈ 5.80
sin 40º
One triangle: A ≈ 111.2º , C ≈ 28.8º , a ≈ 5.80
sin C sin A
=
c
a
sin 70º sin 90º
=
1
c
1sin 70º
≈ 0.94
c=
sin 90º
806
Section 8.2: The Law of Sines
27. b = 5, c = 3, B = 100º
29. a = 4, b = 5, A = 60º
sin B sin C
=
b
c
sin100º sin C
=
5
3
3sin100º
sin C =
≈ 0.5909
5
sin B sin A
=
b
a
sin B sin 60º
=
5
4
5sin 60º
sin B =
≈ 1.0825
4
There is no angle B for which sin B > 1 .
Therefore, there is no triangle with the given
measurements.
C = sin −1 ( 0.5909 )
C = 36.2º or C =143.8º
The second value is discarded because
B + C > 180º .
A = 180º − B − C = 180º − 100º − 36.2º = 43.8º
30. b = 2, c = 3, B = 40º
sin B sin C
=
b
c
sin 40º sin C
=
2
3
3sin 40º
sin C =
≈ 0.9642
2
C = sin −1 ( 0.9642 )
sin B sin A
=
b
a
sin100º sin 43.8º
=
5
a
5sin 43.8º
a=
≈ 3.51
sin100º
One triangle: A ≈ 43.8º , C ≈ 36.2º , a ≈ 3.51
C1 = 74.6º or C2 = 105.4º
For both values, B + C < 180º . Therefore, there
are two triangles.
28. a = 2, c = 1, A = 120º
sin C sin A
=
c
a
sin C sin120º
=
1
2
1sin120º
sin C =
≈ 0.4330
2
A1 = 180º − B − C1 = 180º − 40º − 74.6º = 65.4º
sin B sin A1
=
b
a1
sin 40º sin 65.4º
=
2
a1
C = sin −1 ( 0.4330 )
a1 =
C = 25.7º or C =154.3º
The second value is discarded because
A + C > 180º .
B = 180º − A − C = 180º − 120º − 25.7º = 34.3º
2sin 65.4º
≈ 2.83
sin 40º
A2 = 180º − B − C2 = 180º − 40º − 105.4º = 34.6º
sin B sin A2
=
b
a2
sin B sin A
=
b
a
sin 34.3º sin120º
=
2
b
2sin 34.3º
b=
≈ 1.30
sin120º
sin 40º sin 34.6º
=
2
a2
a2 =
2 sin 34.6º
≈ 1.77
sin 40º
Two triangles:
A1 ≈ 65.4º , C1 ≈ 74.6º , a1 ≈ 2.83 or
One triangle: B ≈ 34.3º , C ≈ 25.7º , b ≈ 1.30
A2 ≈ 34.6º , C2 ≈ 105.4º , a2 ≈ 1.77
807
Chapter 8: Applications of Trigonometric Functions
31. b = 4, c = 6, B = 20º
34. b = 4, c = 5, B = 95º
sin B sin C
=
b
c
sin 20º sin C
=
4
6
6sin 20º
sin C =
≈ 0.5130
4
C = sin −1 ( 0.5130 )
sin C sin B
=
c
b
sin C sin 95º
=
5
4
5sin 95º
sin C =
≈ 1.2452
4
There is no angle C for which sin C > 1 . Thus,
there is no triangle with the given measurements.
C1 = 30.9º or C2 = 149.1º
For both values, B + C < 180º . Therefore, there
are two triangles.
A1 = 180º − B − C1 = 180º − 20º − 30.9º = 129.1º
35. a = 2, c = 1, C = 25º
sin A sin C
=
a
c
sin A sin 25º
=
2
1
2sin 25º
sin A =
≈ 0.8452
1
A = sin −1 ( 0.8452 )
sin B sin A1
=
b
a1
sin 20º sin129.1º
=
4
a1
a1 =
4sin129.1º
≈ 9.08
sin 20º
A1 = 57.7º or A2 = 122.3º
For both values, A + C < 180º . Therefore, there
are two triangles.
A2 = 180º − B − C2 = 180º − 20º − 149.1º = 10.9º
sin B sin A2
=
b
a2
B1 = 180º − A1 − C = 180º − 57.7º − 25º = 97.3º
sin 20º sin10.9º
=
4
a2
sin B1 sin C
=
b1
c
4 sin10.9º
≈ 2.21
sin 20º
sin 97.3º sin 25º
=
b1
1
Two triangles:
A1 ≈ 129.1º , C1 ≈ 30.9º , a1 ≈ 9.08 or
b1 =
a2 =
A2 ≈ 10.9º , C2 ≈ 149.1º , a2 ≈ 2.21
1sin 97.3º
≈ 2.35
sin 25º
B2 = 180º − A2 − γ = 180º − 122.3º − 25º = 32.7º
32. a = 3, b = 7, A = 70º
sin B2 sin C
=
b2
c
sin B sin 70º
=
7
3
7 sin 70º
sin B =
≈ 2.1926
3
There is no angle B for which sin B > 1 . Thus,
there is no triangle with the given measurements.
sin 32.7º sin 25º
=
b2
1
b2 =
1sin 32.7º
≈ 1.28
sin 25º
Two triangles:
A1 ≈ 57.7º , B1 ≈ 97.3º , b1 ≈ 2.35 or
33. a = 2, c = 1, C = 100º
A2 ≈ 122.3º , B2 ≈ 32.7º , b2 ≈ 1.28
sin C sin A
=
c
a
sin100º sin A
=
1
2
sin A = 2 sin100º ≈ 1.9696
There is no angle A for which sin A > 1 . Thus,
there is no triangle with the given measurements.
808
Section 8.2: The Law of Sines
36. b = 4, c = 5, B = 40º
sin 60º sin 65º
=
b
150
150 sin 60º
b=
≈ 143.33 miles
sin 65º
Station Able is about 143.33 miles from the
ship, and Station Baker is about 135.58
miles from the ship.
sin B sin C
=
b
c
sin 40º sin C
=
4
5
5sin 40º
sin C =
≈ 0.8035
4
C = sin −1 ( 0.8035 )
b.
C1 = 53.5º or C2 = 126.5º
For both values, B + C < 180º . Therefore, there
are two triangles.
A1 = 180º − B − C1 = 180º − 40º − 53.5º = 86.5º
a 135.6
=
≈ 0.68 hours
200
r
min
0.68 hr ⋅ 60
≈ 41 minutes
hr
t=
38. Consider the figure below.
C
sin B sin A1
=
b
a1
sin 40º sin 86.5º
=
4
a1
b
4 sin 86.5º
a1 =
≈ 6.21
sin 40º
a
A2 = 180º − B − C2 = 180º − 40º − 126.5º = 13.5º
c B
A
We are given that A = 49.8974° ,
B = 180° − 49.9312° = 130.0688° , and
c = 300 km .
Using angles A and B, we can find
C = 180° − 130.0688° − 49.8974°
sin B sin A2
=
b
a2
sin 40º sin13.5º
=
4
a2
a2 =
4 sin13.5º
≈ 1.45
sin 40º
= 0.0338°
Using the Law of Sines, we can determine b and
a.
sin B sin C
=
b
c
c ⋅ sin B
b=
sin C
300 ⋅ sin130.0688°
=
sin 0.0338°
≈ 389,173.319
sin A sin C
=
a
c
c ⋅ sin A
a=
sin C
300 ⋅ sin 49.8974°
=
sin 0.0338°
≈ 388,980.139
At the time of the measurements, the moon was
Two triangles:
A1 ≈ 86.5º , C1 ≈ 53.5º , a1 ≈ 6.21 or
A2 ≈ 13.5º , C2 ≈ 126.5º , a2 ≈ 1.45
37. a.
Find C ; then use the Law of Sines (see
figure):
B
60º
a
C
150 mi
55º
b
A
C = 180º − 60º − 55º = 65º
sin 55º sin 65º
=
a
150
150 sin 55º
a=
≈ 135.58 miles
sin 65º
809
Chapter 8: Applications of Trigonometric Functions
39. ∠QPR = 180º − 25º = 155º
h
h
=
x 1126.57
h = (1126.57 ) sin 69.2º ≈ 1053.15 feet
The bridge is about 1053.15 feet high.
sin 69.2º =
∠PQR = 180º − 155º − 15º = 10º
Let c represent the distance from P to Q.
sin15º sin10º
=
c
1000
1000 sin15º
c=
≈ 1490.48 feet
sin10º
43. Let h = the height of the tree, and let x = the
distance from the first position to the center of
the tree (see figure).
40. From Problem 39, we have that the distance
from P to Q is 1490.48 feet. Let h represent the
distance from Q to D.
h
sin 25º =
1490.48
h = 1490.48sin 25º ≈ 629.90 feet
10º 60º h
20º
30º
40 ft
Using the Law of Sines twice yields two
equations relating x and h.
41. Let h = the height of the plane and x = the
distance from Q to the plane (see figure).
Equation 1:
R
h
40º
P
x
Equation 2:
35º
1000 ft
Q
∠PRQ = 180º − 40º − 35º = 105º
h
h
=
x 665.46
h = ( 665.46 ) sin 35º ≈ 381 .69 feet
The plane is about 381.69 feet high.
42. Let h = the height of the bridge, x = the distance
from C to point A (see figure).
880 ft
x
sin 20º sin 70º
=
h
x + 40
h sin 70º
x=
− 40
sin 20º
h sin 60º h sin 70º
=
− 40
sin 30º
sin 20º
 sin 60º sin 70º 
h
−
 = − 40
 sin 30º sin 20º 
−40
h=
sin 60º sin 70º
−
sin 30º sin 20º
≈ 39.4 feet
sin 35º =
69.2º
sin 30º sin 60º
=
h
x
h sin 60º
x=
sin 30º
Set the two equations equal to each other and
solve for h.
sin 40º sin105º
=
x
1000
1000sin 40º
x=
≈ 665.46 feet
sin105º
A
x
The height of the tree is about 39.4 feet.
B
44. Let x = the length of the new ramp (see figure).
65.5º
x
h
10 ft
12º 168º
h
18º
Using the Law of Sines:
sin162º sin12º
=
x
10
10sin162º
x=
≈ 14.86 feet
sin12º
The new ramp is about 14.86 feet long.
C
∠ACB = 180º − 69.2º − 65.5º = 45.3º
sin 65.5º sin 45.3º
=
x
880
880sin 65.5º
x=
≈ 1126.57 feet
sin 45.3º
810
Section 8.2: The Law of Sines
45. Note that ∠KOS = 57.7° − 29.6° = 28.1°
(See figure)
79.4o
K
sin10º sin162.875º
=
70
PQ
70sin162.875º
PQ =
≈ 118.7
sin10º
S
29.6o
es
mil
1
.
461
57.7
118.67
= 0.478 hour
250
The trip should have taken 0.4748 hour but,
because of the incorrect course, took 0.48 hour.
Thus, the trip took 0.0052 hour, or about 18.7
seconds, longer.
t=
o
28.1o
O
47. Let h = the perpendicular distance from R to PQ,
(see figure).
R
From the diagram we find that
∠KSO = 180° − 79.4° − 57.7° = 42.9° and
∠OKS = 180° − 28.1° − 42.9° = 109.0° .
We can use the Law of Sines to find the distance
between Oklahoma City and Kansas City, as
well as the distance between Kansas City and St.
Louis.
sin 42.9° sin109.0°
=
OK
461.1
461.1sin 42.9°
≈ 332.0
OK =
sin109.0°
184.5 ft
60º
P 123 ft Q
sin R sin 60º
=
123
184.5
123sin 60º
≈ 0.5774
sin R =
184.5
R = sin −1 ( 0.5774 ) ≈ 35.3º
sin 28.1° sin109.0°
=
KS
461.1
461.1sin 28.1°
≈ 229.7
KS =
sin109.0°
∠RPQ = 180º − 60º − 35.3º ≈ 84.7º
h
184.5
h = 184.5sin 84.7º ≈ 183.71 feet
sin 84.7º =
Therefore, the total distance using the connecting
flight is 332.0 + 229.7 = 561.7 miles. Using the
561.7 − 461.1 = 100.6 more frequent flyer miles.
48. Let θ = ∠AOP
sin θ sin15º
=
9
3
9sin15º
≈ 0.7765
sin θ =
3
θ = sin −1 ( 0.7765 ) ≈ 50.94º
46. The time of the actual trip was:
50 + 70 120
=
= 0.48 hour
t=
250
250
Q
or θ ≈ 180º − 50.94º = 129.06º
A = 114.06º or A = 35.94º
sin114.06º sin15º
=
3
a
3sin114.06º
a=
≈ 10.58 inches
sin15º
sin 35.94º sin15º
or
=
3
a
3sin 35.94º
a=
≈ 6.80 inches
sin15º
The approximate distance from the piston to the
center of the crankshaft is either 6.80 inches or
10.58 inches.
70 mi
10º
P
h
50 mi
R
RQ = 70, PR = 50, P = 10º
Solve the triangle:
sin10º sin Q
=
70
50
50sin10º
sin Q =
≈ 0.1240
70
Q = sin −1 ( 0.1240 ) ≈ 7.125º
R = 180º − 10º − 7.125º = 162.875º
811
Chapter 8: Applications of Trigonometric Functions
49. A = 180º − 140º = 40º ; B = 180º − 135º = 45º ;
C = 180º − 40º − 45º = 95º
b.
Use the Law of Sines:
sin 50º sin 75º
=
3
QR
3sin 75º
≈ 3.78 miles
QR =
sin 50º
The ship is about 3.78 miles from lighthouse Q.
c.
Use the Law of Sines:
sin 90º sin 75º
=
3.2
d
3.2sin 75º
≈ 3.10 miles
d=
sin 90º
The ship is about 3.10 miles from the shore.
A 140º
D
0.125 mi
C
2 mi
0.125 mi
E
B 135º
sin 40º sin 95º
=
2
BC
2sin 40º
BC =
≈ 1.290 mi
sin 95º
sin 45º sin 95º
=
2
AC
2sin 45º
AC =
≈ 1.420 mi
sin 95º
BE = 1.290 − 0.125 = 1.165 mi
AD = 1.420 − 0.125 = 1.295 mi
For the isosceles triangle,
180º − 95º
∠CDE = ∠CED =
= 42.5º
2
sin 95º sin 42.5º
=
DE
0.125
0.125sin 95º
≈ 0.184 miles
DE =
sin 42.5º
The approximate length of the highway is
AD + DE + BE = 1.295 + 0.184 + 1.165 ≈ 2.64 mi.
51. Determine other angles in the figure:
50º
40º
65º
88 in
65º
65º
Using the Law of Sines:
sin(65 + 40)° sin 25°
=
L
88
88sin 25°
L=
≈ 38.5 inches
sin105°
The awning is about 38.5 inches long.
52. The tower forms an angle of 95˚ with the ground.
Let x be the distance from the ranger to the tower.
B
50. Let PR = the distance from lighthouse P to the
ship, QR = the distance from lighthouse Q to the
ship, and d = the distance from the ship to the
shore. From the diagram, ∠QPR = 75º and
∠PQR = 55º , where point R is the ship.
45º
100 ft
A
a.
x
40º
5º
C
∠ABC = 180º − 95º − 40º = 45º
P
3 mi
95º
15º
d 35º
sin 40º sin 45º
=
x
100
100sin 45º
≈ 110.01 feet
x=
sin 40º
The ranger is about 110.01 feet from the tower.
R
Q
Use the Law of Sines:
sin 50º sin 55º
=
PR
3
3sin 55º
≈ 3.21 miles
PR =
sin 50º
The ship is about 3.21 miles from lighthouse
P.
812
Section 8.2: The Law of Sines
53. Let h = height of the pyramid, and let x = distance from the edge of the pyramid to the point beneath the tip of the
pyramid (see figure).
h
40.3º
46.27º
100 ft
100 ft
x
Using the Law of Sines twice yields two equations relating x and y:
sin 46.27º sin(90º − 46.27º )
Equation 1:
=
h
x + 100
( x + 100)sin 46.27º = h sin 43.73º
x sin 46.27º +100sin 46.27º = h sin 43.73º
h sin 43.73º −100sin 46.27º
sin 46.27º
sin 40.3º sin(90º − 40.3º )
=
h
x + 200
( x + 200 ) sin 40.3º = h sin 49.7º
x=
Equation 2:
x sin 40.3º +200sin 40.3º = h sin 49.7º
h sin 49.7º − 200sin 40.3º
x=
sin 40.3º
Set the two equations equal to each other and solve for h.
h sin 43.73º −100sin 46.27º h sin 49.7º − 200sin 40.3º
=
sin 46.27º
sin 40.3º
h sin 43.73º ⋅ sin 40.3º − 100sin 46.27º ⋅ sin 40.3º = h sin 49.7º ⋅ sin 46.27º − 200sin 40.3º ⋅ sin 46.27º
h sin 43.73º ⋅ sin 40.3º −h sin 49.7º ⋅ sin 46.27º = 100sin 46.27º ⋅ sin 40.3º − 200sin 40.3º ⋅ sin 46.27º
100sin 46.27º ⋅ sin 40.3º − 200sin 40.3º ⋅ sin 46.27º
h=
sin 43.73º ⋅ sin 40.3º − sin 49.7º ⋅ sin 46.27º
≈ 449.36 feet
The current height of the pyramid is about 449.36 feet.
_________________________________________________________________________________________________
54. Let h = the height of the aircraft, and let x = the
distance from the first sensor to a point on the
ground beneath the airplane (see figure).
5º 70º
15º
700 ft
Equation 2:
Set the equations equal to each other and solve for h.
h
h sin 70º h sin 75º
=
− 700
sin 20º
sin15º
 sin 70º sin 75º 
−
h
 = − 700
 sin 20º sin15º 
− 700
h=
sin 70º sin 75º
−
sin 20º sin15º
≈ 710.97 feet
20º
x
Using the Law of Sines twice yields two
equations relating x and h.
Equation 1:
sin15º sin 75º
=
h
x + 700
h sin 75º
− 700
x=
sin15º
sin 20º sin 70º
=
h
x
h sin 70º
x=
sin 20º
The height of the aircraft is about 710.97 feet.
813
Chapter 8: Applications of Trigonometric Functions
56. Using the Law of Sines:
55. Using the Law of Sines:
V enus
βB
Mercury
βB
x
57,910,000
γC
x
γ
C
Sun
15º
Earth
149,600,000
Earth
sin15º
sin B
=
57,910, 000 149, 600, 000
149, 600, 000 ⋅ sin15º
sin B =
57,910,000
14,960 ⋅ sin15º
=
5791
 14,960 ⋅ sin15º 
o
B = sin −1 
 ≈ 41.96
5791


or
10 º
108,200,000
Sun
149,600,000
sin10º
sin B
=
108, 200, 000 149, 600, 000
149, 600, 000 ⋅ sin10º
sin B =
108, 200, 000
1496 ⋅ sin10º
=
1082
1496
⋅ sin10º 

o
B = sin −1 
 ≈ 13.892
1082


or
B ≈ 138.04o
B ≈ 166.108o
C ≈ 180° − 13.892° − 10° = 156.108° or
C ≈ 180° − 166.108° − 10° = 13.892°
C ≈ 180o − 41.96o − 15º = 123.04o or
C ≈ 180o − 138.04o − 15º = 26.96o
sin10º
sin C
=
108, 200, 000
x
108, 200, 000 ⋅ sin C
x=
sin10º
108, 200, 000 ⋅ sin156.108°
x=
sin10º
≈ 252,363, 760.4 km
or
108, 200, 000 ⋅ sin 3.892°
x=
sin10º
≈ 42, 293, 457.3 km
So the approximate possible distances between
Earth and Venus are 42,300,000 km and
252,400,000 km.
sin15º
sin C
=
57,910, 000
x
57,910, 000 ⋅ sin C
x=
sin15°
57,910, 000 ⋅ sin123.04o
=
sin15º
≈ 187,564,951.5 km
or
57,910, 000 ⋅ sin 26.96o
x=
sin15º
≈ 101, 439,834.5 km
So the possible distances between Earth and
Mercury are approximately 101,440,000 km and
187,600,000 km.
57. Since there are 36 equally spaced cars, each car
360°
is separated by
= 10° . The angle between
36
the radius and a line segment connecting
170°
consecutive cars is
= 85° (see figure). If
2
we let r = the radius of the wheel, we get
sin10° sin 85°
=
22
r
22sin 85°
≈ 126
r=
sin10°
The length of the diameter of the wheel is
approximately d = 2r = 2 (126 ) = 252 feet.
814
Section 8.2: The Law of Sines
58.
59.
a + b a b sin A sin B
+
= + =
c
c c sin C sin C
sin A + sin B
=
sin C
A+ B 
A− B 
2sin 
cos 
2 
2 


=
C
C
2sin   cos  
2
 
2
π C
A− B 
sin  −  cos 

2
2


 2 
=
C
sin ( C ) cos  
2
C
A− B 
cos   cos 
2
2 


=
C
C
sin   cos  
2
 
2
A− B 
1

cos 
 cos  2 ( A − B ) 
2




=
=
C
1 


sin  
sin  C 
2
2 
60. a =
b sin A b sin [180º − ( B + γ ) ]
=
sin B
sin B
b
=
sin( B + C )
sin B
b
=
( sin B cos C + cos B sin C )
sin B
b sin C
= b cos C +
cos B
sin B
= b cos C + c cos B
a−b
a−b
= c
61.
a+b a+b
c
1
sin  ( A − B) 
2

1 

cos  C 
2 
=
1
cos  ( A − B) 
2

1 

sin  C 
2 
1
1
sin  ( A − B ) 
sin  C 
2

⋅
2 
=
1
1
cos  C 
cos  ( A − B) 
2 
2

1

1 
= tan  ( A − B)  tan  C 
2

2 
1

1

= tan  ( A − B)  tan  ( π − ( A + B) ) 
2
2




a − b a b sin A sin B sin A − sin B
−
=
= − =
sin C
c
c c sin C sin C
A− B 
A+ B 
2sin 
cos 
2 
2 


=
C
sin  2 ⋅ 
 2
A
B
−
 cos  A + B 
2sin 

 2 
2




=
C
C


2sin   cos  
2
2
A− B 
π C
sin 
cos  − 
2 
2 2


=
C
C
sin   cos  
2
2
A− B  C 
sin 
sin
2   2 

=
C
C
sin   cos  
2
 
2
A− B 
1

sin 
 sin  2 ( A − B ) 
2

=


=
1 
C


cos  
cos  C 
2
2 
 π  A + B 
1

= tan  ( A − B)  tan  − 

2

 2  2 
1

 A+ B 
= tan  ( A − B)  cot 

2

 2 
1
tan  ( A − B) 
2


=
1
tan  ( A + B ) 
2

62. sin B = sin ( ∠PQR ) = sin ( ∠PP ' R ) =
b
2r
sin B 1
=
b
2r
The result follows from the Law of Sines.
63 – 65. Answers will vary.
815
Chapter 8: Applications of Trigonometric Functions
c 2 = a 2 + b 2 − 2ab cos C
Section 8.3
( x2 − x1 )
1. d =
2
2
θ = 45° or
+ ( y2 − y1 )
2
a 2 + b2 − c 2
2ab
2.052 + 32 − 42 −2.7975
=
cos C =
2(2.05)(3)
12.3
cos C =
2
2. cos θ =
 −2.7975 
C = cos −1 
 ≈ 103.1º
 12.3 
π
4
The solution set is {45°} or
{π4 }
B = 180º − A − C ≈ 180º −30º −103.1º = 46.9º
3. Cosines
11. a = 2, b = 3, C = 95º
4. Sines
c 2 = a 2 + b 2 − 2ab cos C
5. Cosines
c 2 = 22 + 32 − 2 ⋅ 2 ⋅ 3cos 95º = 13 − 12 cos 95º
c = 13 − 12 cos 95º ≈ 3.75
6. False: Use the Law of Cosines
7. False
a 2 = b 2 + c 2 − 2bc cos A
8. True
b2 + c2 − a 2
2bc
32 + 3.752 − 22 19.0625
=
cos A =
2(3)(3.75)
22.5
cos A =
9. a = 2, c = 4, A = 45º
b 2 = a 2 + c 2 − 2ac cos B
b 2 = 22 + 42 − 2 ⋅ 2 ⋅ 4 cos 45º
 19.0625 
A = cos −1 
 ≈ 32.1º
 22.5 
2
= 20 − 16 ⋅
2
= 20 − 8 2
B = 180º − A − C ≈ 180º −32.1º −95º = 52.9º
12. a = 2, c = 5, B = 20º
b = 20 − 8 2 ≈ 2.95
b 2 = a 2 + c 2 − 2ac cos B
a 2 = b 2 + c 2 − 2bc cos A
b 2 = 22 + 52 − 2 ⋅ 2 ⋅ 5cos 20º = 29 − 20cos 20º
2bc cos A = b 2 + c 2 − a 2
b = 29 − 20 cos 20º ≈ 3.19
b2 + c2 − a2
2bc
2.952 + 42 − 22 20.7025
=
cos A =
2(2.95)(4)
23.6
cos A =
a 2 = b 2 + c 2 − 2bc cos A
cos A =
 20.7025 
A = cos −1 
 ≈ 28.7º
 23.6 
b2 + c2 − a 2
2bc
(
cos A =
C = 180º − A − B ≈ 180º −28.7º −45º ≈ 106.3º
29 − 20 cos 20º
)
2
+ 52 − 2 2
2( 29 − 20 cos 20º )(5)
≈ 0.97681
A = cos −1 ( 0.97681) ≈ 12.4º
10. b = 3, c = 4, A = 30º
C = 180º − A − B ≈ 180º −12.4º −20º = 147.6º
a 2 = b 2 + c 2 − 2bc cos A
a 2 = 32 + 42 − 2 ⋅ 3 ⋅ 4 cos 30º
 3
= 25 − 24 

 2 
= 25 − 12 3
a = 25 − 12 3 ≈ 2.05
816
Section 8.3: The Law of Cosines
13. a = 6, b = 5, c = 8
16. a = 4, b = 3, c = 4
a 2 = b 2 + c 2 − 2bc cos A
cos A =
a 2 = b 2 + c 2 − 2bc cos A
b 2 + c 2 − a 2 52 + 82 − 62 53
=
=
2bc
2(5)(8)
80
cos A =
 53 
A = cos −1   ≈ 48.5º
 80 
 9 
A = cos −1   ≈ 68.0º
 24 
b 2 = a 2 + c 2 − 2ac cos B
b 2 = a 2 + c 2 − 2ac cos B
a 2 + c2 − b2
2ac
62 + 82 − 52 75
cos B =
=
2(6)(8)
96
cos B =
cos B =
C = 180º − A − B ≈ 180º −68.0º −44.0º = 68.0º
17. a = 3, b = 4, C = 40º
C = 180º − A − B ≈ 180º −48.5º −38.6º = 92.9º
c 2 = a 2 + b 2 − 2ab cos C
14. a = 8, b = 5, c = 4
c 2 = 32 + 42 − 2 ⋅ 3 ⋅ 4 cos 40º = 25 − 24 cos 40º
a = b + c − 2bc cos A
2
2
c = 25 − 24 cos 40º ≈ 2.57
23
b 2 + c 2 − a 2 52 + 4 2 − 82
cos A =
=
=−
2bc
2(5)(4)
80
a 2 = b 2 + c 2 − 2bc cos A
 23 
A = cos −1  −  ≈ 125.1º
 80 
cos A =
a 2 + c 2 − b2 82 + 42 − 52 55
=
=
2ac
2(8)(4)
64
B = 180º − A − C ≈ 180º −48.6º −40º = 91.4º
 55 
B = cos−1   ≈ 30.8º
 64 
18. a = 2, c = 1, B = 10º
b 2 = a 2 + c 2 − 2ac cos B
C = 180º − A − B ≈ 180º −125.1º −30.8º = 24.1º
b 2 = 22 + 12 − 2 ⋅ 2 ⋅1cos10º = 5 − 4 cos10º
b = 5 − 4 cos10º ≈ 1.03
15. a = 9, b = 6, c = 4
a 2 = b 2 + c 2 − 2bc cos A
cos A =
a 2 = b 2 + c 2 − 2bc cos A
29
b 2 + c 2 − a 2 62 + 4 2 − 9 2
=
=−
2bc
2(6)(4)
48
cos A =
 29 
A = cos −1  −  ≈ 127.2º
 48 
1.9391
b 2 + c 2 − a 2 1.032 + 12 − 22
=
=−
2bc
2(1.03)(1)
2.06
 1.9391 
A = cos −1  −
 ≈ 160.3º
 2.06 
C = 180º − A − B ≈ 180º −160.3º −10º = 9.7º
b 2 = a 2 + c 2 − 2ac cos B
cos B =
b 2 + c 2 − a 2 42 + 2.57 2 − 32 13.6049
=
=
2bc
2(4)(2.57)
20.56
 13.6049 
A = cos −1 
 ≈ 48.6º
 20.56 
b 2 = a 2 + c 2 − 2ac cos B
cos B =
a 2 + c 2 − b 2 42 + 42 − 32 23
=
=
2ac
2(4)(4)
32
 23 
B = cos−1   ≈ 44.0º
 32 
 75 
B = cos −1   ≈ 38.6º
 96 
2
9
b 2 + c 2 − a 2 32 + 42 − 42
=
=
2bc
2(3)(4)
24
a 2 + c 2 − b 2 92 + 42 − 62 61
=
=
2ac
2(9)(4)
72
19. b = 1, c = 3, A = 80º
a 2 = b 2 + c 2 − 2bc cos A
 61 
B = cos   ≈ 32.1º
 72 
−1
a 2 = 12 + 32 − 2 ⋅1 ⋅ 3cos 80º = 10 − 6 cos80º
a = 10 − 6 cos 80º ≈ 2.99
C = 180º − A − B ≈ 180º −127.2º −32.1º = 20.7º
817
Chapter 8: Applications of Trigonometric Functions
23. a = 2, b = 2, C = 50º
b2 = a 2 + c 2 − 2ac cos B
cos B =
c 2 = a 2 + b 2 − 2ab cos C
a 2 + c 2 − b 2 2.992 + 32 − 12 16.9401
=
=
2ac
2(2.99)(3)
17.94
c 2 = 22 + 22 − 2 ⋅ 2 ⋅ 2 cos 50º = 8 − 8 cos 50º
 16.9401 
B = cos −1 
 ≈ 19.2º
 17.94 
c = 8 − 8 cos 50º ≈ 1.69
a 2 = b 2 + c 2 − 2bc cos A
C = 180º − A − B ≈ 180º −80º −19.2º = 80.8º
cos A =
20. a = 6, b = 4, C = 60º
 2.8561 
A = cos−1 
 ≈ 65.0º
 6.76 
c 2 = a 2 + b 2 − 2ab cos C
c 2 = 62 + 42 − 2 ⋅ 6 ⋅ 4 cos 60º = 28
B = 180º − A − C ≈ 180º −65.0º −50º = 65.0º
c = 28 ≈ 5.29
24. a = 3, c = 2, B = 90º
a 2 = b 2 + c 2 − 2bc cos A
cos A =
b 2 = a 2 + c 2 − 2ac cos B
4 + 5.29 − 6
7.9841
b +c −a
=
=
2bc
2(4)(5.29)
42.32
2
2
2
2
2
2
b 2 = 32 + 22 − 2 ⋅ 3 ⋅ 2 cos 90º = 13
 7.9841 
A = cos −1 
 ≈ 79.1º
 42.32 
b = 13 ≈ 3.61
a 2 = b 2 + c 2 − 2bc cos A
B = 180º − A − C ≈ 180º −79.1º −60º = 40.9º
cos A =
21. a = 3, c = 2, B = 110º
b = 3 + 2 − 2 ⋅ 3 ⋅ 2 cos110º = 13 − 12 cos110º
2
2
C = 180º − A − B ≈ 10º −56.3º −90º = 33.7º
b = 13 − 12 cos110º ≈ 4.14
25. a = 12, b = 13, c = 5
c 2 = a 2 + b 2 − 2ab cos C
cos C =
a 2 = b 2 + c 2 − 2bc cos A
3 + 4.14 − 2
22.1396
a +b −c
=
=
2ab
2(3)(4.14)
24.84
2
2
2
2
b 2 + c 2 − a 2 ( 13)2 + 22 − 32
=
≈ 0.55470
2bc
2( 13)(2)
A = cos −1 ( 0.55470 ) ≈ 56.3º
b 2 = a 2 + c 2 − 2ac cos B
2
b 2 + c 2 − a 2 22 + 1.692 − 22 2.8561
=
=
2bc
2(2)(1.69)
6.76
2
2
cos A =
 22.1396 
C = cos −1 
 ≈ 27.0º
 24.84 
b 2 + c 2 − a 2 132 + 52 − 122 50
=
=
2bc
2(13)(5)
130
 50 
A = cos−1 
 ≈ 67.4º
 130 
A = 180º − B − C ≈ 180º −110º −27.0º = 43.0º
b 2 = a 2 + c 2 − 2ac cos B
22. b = 4, c = 1, A = 120º
cos B =
a 2 = b 2 + c 2 − 2bc cos A
a 2 + c 2 − b 2 122 + 52 − 132
=
=0
2ac
2(12)(5)
B = cos −1 0 = 90º
a 2 = 42 + 12 − 2 ⋅ 4 ⋅1cos120º = 21
C = 180º − A − B ≈ 180º −67.4º −90º = 22.6º
a = 21 ≈ 4.58
c 2 = a 2 + b 2 − 2ab cos C
cos C =
a 2 + b 2 − c 2 4.582 + 42 − 12 35.9764
=
=
2ab
2(4.58)(4)
36.64
 35.9764 
C = cos −1 
 ≈ 10.9º
 36.64 
B = 180º − A − C ≈ 180º −120º −10.9º = 49.1º
818
Section 8.3: The Law of Cosines
26. a = 4, b = 5, c = 3
29. a = 5, b = 8, c = 9
a 2 = b 2 + c 2 − 2bc cos A
cos A =
a 2 = b 2 + c 2 − 2bc cos A
b 2 + c 2 − a 2 52 + 32 − 42
=
= 0.6
2bc
2(5)(3)
cos A =
A = cos −1 0.6 ≈ 53.1º
 120 
A = cos −1 
 ≈ 33.6º
 144 
b 2 = a 2 + c 2 − 2ac cos B
cos B =
b 2 = a 2 + c 2 − 2ac cos B
a 2 + c 2 − b2 42 + 32 − 52
=
=0
2ac
2(4)(3)
cos B =
B = cos −1 0 = 90º
27. a = 2, b = 2, c = 2
C = 180o − A − B ≈ 180o − 33.6o − 62.2o = 84.2o
a 2 = b 2 + c 2 − 2bc cos A
b 2 + c 2 − a 2 2 2 + 22 − 2 2
=
= 0.5
2bc
2(2)(2)
30. a = 4, b = 3, c = 6
a 2 = b 2 + c 2 − 2bc cos A
−1
A = cos 0.5 = 60º
cos A =
b 2 = a 2 + c 2 − 2ac cos B
cos B =
b 2 = a 2 + c 2 − 2ac cos B
C = 180º − A − B ≈ 180º −60º −60º = 60º
cos B =
28. a = 3, b = 3, c = 2
b 2 + c 2 − a 2 32 + 22 − 32 1
=
=
2bc
2(3)(2)
3
C = 180o − A − B ≈ 180o − 36.3o − 26.4o = 117.3o
1
A = cos −1   ≈ 70.5º
3
31. a = 10, b = 8, c = 5
a 2 = b 2 + c 2 − 2bc cos A
b = a + c − 2ac cos B
cos B =
2
2
cos A =
3 + 2 −3
1
a +c −b
=
=
2ac
2(3)(2)
3
2
2
2
2
a 2 + c 2 − b 2 42 + 62 − 32 43
=
=
2ac
48
2 ( 4 )( 6 )
 43 
B = cos −1   ≈ 26.4º
 48 
a 2 = b 2 + c 2 − 2bc cos A
2
b 2 + c 2 − a 2 32 + 62 − 42 29
=
=
2bc
2(3)(6)
36
 29 
A = cos−1   ≈ 36.3º
 36 
a 2 + c 2 − b 2 22 + 22 − 2 2
=
= 0.5
2ac
2(2)(2)
B = cos −1 0.5 = 60º
cos A =
a 2 + c 2 − b2 52 + 92 − 82 42
=
=
2ac
90
2 ( 5 )( 9 )
 42 
B = cos −1   ≈ 62.2o
 90 
C = 180º − A − B ≈ 180º −53.1º −90º = 36.9º
cos A =
b 2 + c 2 − a 2 82 + 92 − 52 120
=
=
2bc
2(8)(9)
144
2
2
11
b 2 + c 2 − a 2 82 + 52 − 102
=
=−
2bc
2(8)(5)
80
 11 
A = cos −1  −  ≈ 97.9º
 80 
1
B = cos   ≈ 70.5º
3
−1
b 2 = a 2 + c 2 − 2ac cos B
C = 180º − A − B ≈ 180º −70.5º −70.5º = 39.0º
cos B =
61
a 2 + c 2 − b 2 102 + 52 − 82
=
=
2ac
2(10)(5)
100
 61 
B = cos −1 
 ≈ 52.4º
 100 
C = 180o − A − B ≈ 180o − 97.9o − 52.4o = 29.7 o
819
Chapter 8: Applications of Trigonometric Functions
32. a = 9, b = 7, c = 10
b 2 = a 2 + c 2 − 2ac cos B
a 2 = b 2 + c 2 − 2bc cos A
cos A =
a 2 + c 2 − b2
2ac
2
6 + 9 2 − 82
53
=
=
2 ( 6 )( 9 )
108
cos B =
68
b 2 + c 2 − a 2 72 + 102 − 92
=
=
2bc
2(7)(10)
140
 68 
A = cos−1 
 ≈ 60.9º
 140 
B = cos −1
b = a + c − 2ac cos B
2
2
cos B =
2
c 2 = a 2 + b 2 − 2ab cos C
a 2 + c 2 − b2 92 + 102 − 72 132
=
=
2ac
2(9)(10)
180
a 2 + b2 − c2
2ab
2
6 + 82 − 92 19
=
=
2 ( 6 )( 8)
96
cos C =
 132 
B = cos −1 
 ≈ 42.8º
 180 
C = 180o − A − B ≈ 180o − 60.9o − 42.8o = 76.3o
C = cos −1
33. B = 20° , C = 75° , b = 5
sin B sin C
=
b
c
b sin C 5sin 75°
=
≈ 14.12
c=
sin B
sin 20°
19
≈ 78.6°
96
36. a = 14 , b = 7 , A = 85°
sin 85° sin B
=
14
7
sin 85°
sin B =
≈ 0.49810
2
B = sin −1 ( 0.49810 ) ≈ 29.9° or 150.1°
The second value is discarded since
A + B > 180° . Therefore, B ≈ 29.9° .
C = 180° − 29.9° − 85° = 65.1°
sin 85° sin 65.1°
=
14
c
14 ⋅ sin 65.1°
c=
≈ 12.75
sin 85°
A = 180° − 20° − 75° = 85°
sin A sin B
=
a
b
b sin A 5sin 85°
=
≈ 14.56
a=
sin B
sin 20°
34. A = 50° , B = 55° , c = 9
C = 180° − 50° − 55° = 75°
sin C sin B
=
c
b
c sin B 9sin 55°
=
≈ 7.63
b=
sin C
sin 75°
37. B = 35° , C = 65° , a = 15
A = 180° − 35° − 65° = 80°
sin C sin A
=
c
a
c sin A 9 sin 50°
=
≈ 7.14
a=
sin C
sin 75°
sin A sin C
=
a
c
a sin C 15sin 65°
=
≈ 13.80
c=
sin A
sin 35°
35. a = 6 , b = 8 , c = 9
sin A sin B
=
a
b
a sin B 15sin 35°
=
≈ 8.74
b=
sin A
sin 80°
a 2 = b 2 + c 2 − 2bc cos A
b2 + c2 − a 2
2bc
2
8 + 92 − 62 109
=
=
2 ( 8)( 9 )
144
cos A =
A = cos −1
53
≈ 60.6°
108
109
≈ 40.8°
144
820
Section 8.3: The Law of Cosines
38. a = 4 , c = 5 , B = 55°
41. b = 5 , c = 12 , A = 60°
b 2 = a 2 + c 2 − 2ac cos B
a 2 = b 2 + c 2 − 2bc cos A
b 2 = 42 + 52 − 2 ( 4 )( 5) cos 55°
a 2 = 52 + 122 − 2 ⋅ 5 ⋅12 ⋅ cos 60°
= 109
b 2 = 41 − 40 cos 55°
a = 109 ≈ 10.44
b = 41 − 40 cos 55° ≈ 4.25
sin 60°
sin B
5
109
5sin 60°
sin B =
≈ 0.414751
109
sin 55°
sin A
=
4
41 − 40 cos 55°
4 sin 55°
sin A =
≈ 0.77109
41 − 40 cos 55°
=
B = sin −1 ( 0.414751) ≈ 24.5° or 155.5°
A = sin −1 ( 0.77109 ) ≈ 50.5° or 129.5°
We discard the second value since A + B > 180° .
Therefore, A ≈ 50.5° .
We discard the second value because it would
give A + B > 180° . Therefore, B ≈ 24.5° .
C = 180° − 60° − 24.5° = 95.5°
C = 180° − 55° − 50.5° = 74.5°
42. a = 10 , b = 10 , c = 15
39. a = 3 , b = 10 , A = 10°
sin10° sin B
=
3
10
10 sin10°
sin B =
≈ 0.578827
3
B = sin −1 ( 0.578827 ) ≈ 35.4° or 144.6°
a 2 = b 2 + c 2 − 2bc cos A
b2 + c 2 − a 2
2bc
102 + 152 − 102 225 3
=
=
=
2 (10 )(15)
300 4
cos A =
Since both values yield A + B < 180° , there are
two triangles.
B1 = 35.4° and B2 = 144.6°
A = cos −1
3
≈ 41.4°
4
b 2 = a 2 + c 2 − 2ac cos B
C1 = 180° − 10° − 35.4° = 134.6°
a2 + c2 − b2
2ac
102 + 152 − 102 225 3
=
=
=
2 (10 )(15)
300 4
cos B =
C2 = 180° − 10° − 144.6° = 25.4°
Using the Law of Cosines we get
c1 = 32 + 102 − 2 ⋅ 3 ⋅10 ⋅ cos134.6°
≈ 12.29
B = cos −1
c2 = 32 + 102 − 2 ⋅ 3 ⋅10 ⋅ cos 25.4°
≈ 7.40
3
≈ 41.4°
4
c 2 = a 2 + b 2 − 2ab cos C
a 2 + b2 − c 2
2ab
102 + 102 − 152 −25
1
=
=
=−
2 (10 )(10 )
200
8
40. A = 65° , B = 72° , b = 7
sin 72° sin 65°
=
7
a
7 sin 65°
a=
≈ 6.67
sin 72°
cos C =
 1
C = cos −1  −  ≈ 97.2°
 8
C = 180° − 65° − 72° = 43°
sin 43° sin 72°
=
c
7
7 sin 43°
c=
≈ 5.02
sin 72°
821
Chapter 8: Applications of Trigonometric Functions
 124,148.39 
A = cos −1  −
 ≈ 153.6º
 138,570 
The captain needs to turn the ship through an
angle of 180° − 153.6° = 26.4° .
43. Find the third side of the triangle using the Law
of Cosines: a = 150, b = 35, C = 110º
c 2 = a 2 + b 2 − 2ab cos C
= 1502 + 352 − 2 ⋅150 ⋅ 35 cos110º
= 23, 725 − 10,500 cos110º
b.
c = 23, 725 − 10,500 cos110º ≈ 165
The ball is approximately 165 yards from the
center of the green.
44. a.
The angle inside the triangle at Sarasota is
180° − 50° = 130° . Use the Law of Cosines to
find the third side:
a = 150, b = 100, C = 130º
46. a.
c 2 = a 2 + b 2 − 2ab cos C
After 15 minutes, the plane would have
flown 220(0.25) = 55 miles. Find the
third side of the triangle:
a = 55, b = 330, γ = 10º
c 2 = a 2 + b 2 − 2ab cos C
= 1502 + 1002 − 2 ⋅150 ⋅100 cos130º
= 32,500 − 30, 000 cos130º
= 552 + 3302 − 2 ⋅ 55 ⋅ 330cos10º
= 111,925 − 36,300 cos10º
c = 32,500 − 30, 000cos130º ≈ 227.56 mi
b.
461.9 nautical miles
≈ 30.8 hours are
15 knots
required for the second leg of the trip. (The
total time for the trip will be about 40.8
hours.)
t=
c = 111,925 − 36,300cos10º ≈ 276
Use the Law of Sines to find the angle inside
the triangle at Ft. Myers:
sin A sin130º
=
100
227.56
100sin130º
sin A =
227.56
 100sin130º 
A = sin −1 
 ≈ 19.7º
 227.56 
Since the angle of the triangle is 19.7˚, the
pilot should fly at a bearing of N 19.7° E .
Find the measure of the angle opposite the
330-mile side:
a 2 + c2 − b2
cos B =
2ac
552 + 2762 − 3302
29, 699
=
=−
2(55)(276)
30,360
 29, 699 
B = cos −1  −
 ≈ 168.0º
 30,360 
The pilot should turn through an angle of
180° − 168.0° = 12.0° .
b.
45. After 10 hours the ship will have traveled 150
nautical miles along its altered course. Use the
Law of Cosines to find the distance from
a = 600, b = 150, C = 20º
c 2 = a 2 + b 2 − 2ab cos C
= 6002 + 1502 − 2 ⋅ 600 ⋅150 cos 20º
= 382,500 − 180, 000 cos 20º
If the total trip is to be done in 90 minutes,
and 15 minutes were used already, then
there are 75 minutes or 1.25 hours to
complete the trip. The plane must travel
276 miles in 1.25 hours:
276
r=
= 220.8 miles/hour
1.25
The pilot must maintain a speed of 220.8
mi/hr to complete the trip in 90 minutes.
c = 382,500 − 180, 000cos 20º
≈ 461.9 nautical miles
a.
Use the Law of Cosines to find the angle
opposite the side of 600:
b2 + c2 − a 2
cos A =
2bc
2
150 + 461.92 − 6002
124,148.39
=−
cos A =
2(150)(461.9)
138,570
822
Section 8.3: The Law of Cosines
47. a.
b.
Find x in the figure:
3rd
2nd
60
.5
ft
y
Home
7200 = (46 + y )2
90 ft
β
x
45°
46 + y = 7200 ≈ 84.85
y ≈ 38.85 feet
It is about 38.85 feet from the pitching
rubber to second base.
1st
90 ft
Use the Pythagorean Theorem to find y in
the figure:
602 + 602 = (46 + y )2
x 2 = 60.52 + 902 − 2(60.5)(90) cos 45º
 2
= 11, 760.25 − 10,980 

 2 
c.
= 11, 760.25 − 5445 2
x = 11, 760.25 − 5445 2 ≈ 63.7 feet
It is about 63.7 feet from the pitching rubber
to first base.
b. Use the Pythagorean Theorem to find y in
the figure:
902 + 902 = (60.5 + y ) 2
Find B in the figure by using the Law of
Cosines:
462 + 42.582 − 602 329.0564
cos B =
=
2(46)(42.58)
3917.36
 329.0564 
B = cos −1 
 ≈ 85.2º
 3917.36 
The pitcher needs to turn through an angle
of 85.2˚ to face first base.
49. a.
Find x by using the Law of Cosines:
16, 200 = (60.5 + y ) 2
60.5 + y = 16, 200 ≈ 127.3
y ≈ 66.8 feet
It is about 66.8 feet from the pitching rubber
to second base.
c.
y
Find B in the figure by using the Law of
Cosines:
60.52 + 63.72 − 902
382.06
cos B =
=−
2(60.5)(63.7)
7707.7
x = 260, 000 − 100, 000cos 80º ≈ 492.6 ft
The guy wire needs to be about 492.6 feet
long.
b.
46
ft
y
Home
45°
60 ft
Use the Pythagorean Theorem to find the
value of y:
y 2 = 1002 + 2502 = 72,500
y = 269.3 feet
The guy wire needs to be about 269.3 feet
long.
2nd
β
10°
x 2 = 5002 + 1002 − 2(500)(100) cos80º
= 260, 000 − 100, 000 cos80º
Find x in the figure:
3rd
80°
100 ft
 382.06 
B = cos −1  −
 ≈ 92.8º
 7707.7 
The pitcher needs to turn through an angle
of about 92.8˚ to face first base.
48. a.
250 ft
500 ft x
60 ft
x
1st
x 2 = 462 + 602 − 2(46)(60) cos 45º
 2
= 5716 − 5520 
 = 5716 − 2760 2
 2 
x = 5716 − 2760 2 ≈ 42.58 feet
It is about 42.58 feet from the pitching
rubber to first base.
823
Chapter 8: Applications of Trigonometric Functions
50. Find x by using the Law of Cosines:
53. Use the Law of Cosines:
45o
x
y
10
10
500 ft
95° 85°
5°
c
c 2 = a 2 + b2 − 2ab cos C
x 2 = 5002 + 1002 − 2(500)(100) cos 85º
= 260, 000 − 100, 000 cos85º
= 102 + 102 − 2 (10 )(10 ) cos ( 45° )
x = 260, 000 − 100, 000 cos 85º ≈ 501.28 feet
The guy wire needs to be about 501.28 feet long.
 2
= 100 + 100 − 200 

 2 


Find y by using the Law of Cosines:
y 2 = 5002 + 1002 − 2(500)100 cos 95º
(
= 200 − 100 2 = 100 2 − 2
(
= 260, 000 − 100, 000 cos 95º
)
)
c = 100 2 − 2 = 10 2 − 2 ≈ 7.65
y = 260, 000 − 100, 000 cos 95º ≈ 518.38 feet
The guy wire needs to be about 518.38 feet long.
The footings should be approximately 7.65 feet
apart.
51. Find x by using the Law of Cosines:
54. Use the Law of Cosines:
A
x
L
r
400 ft
45°
O
90 ft
θ
B
x
L2 = x 2 + r 2 − 2 x r cos θ
x 2 = 4002 + 902 − 2(400)(90) cos ( 45º )
x 2 − 2 x r cos θ + r 2 − L2 = 0
= 168,100 − 36, 000 2
x = 168,100 − 36, 000 2 ≈ 342.33 feet
It is approximately 342.33 feet from dead center
to third base.
x=
52. Find x by using the Law of Cosines:
x=
x
x=
280 ft
45°
2r cos θ + (2r cos θ )2 − 4(1)(r 2 − L2 )
2(1)
(
2r cos θ + 4r 2 cos 2 θ − 4 r 2 − L2
2
(
2r cos θ + 4 r cos 2 θ − r 2 + L2
2
2
2r cos θ + 2 r cos 2 θ − r 2 + L2
2
2
60 ft
x=
x 2 = 2802 + 602 − 2(280)(60) cos ( 45º )
x = r cos θ + r 2 cos 2 θ + L2 − r 2
= 82, 000 − 16,800 2
x = 82, 000 − 16,800 2 ≈ 241.33 feet
It is approximately 241.33 feet from dead center
to third base.
824
)
)
Section 8.4: Area of a Triangle
55. Use the Law of Cosines to find the length of side
d:
d 2 = r 2 + r 2 − 2 ⋅ r ⋅ r ⋅ cos θ
= 2r 2 − 2r 2 cos θ = 2r 2 (1 − cos θ )
1 − cos θ
 1 − cos θ 
= 4r 2 
 = 2r
2
2


θ 
= 2r sin  
2
If s = rθ is the length of the arc subtended by
θ
θ , then d < s , and we have 2r sin   < rθ or
2
θ
θ
θ 
2sin   < θ . Given sin θ = 2sin cos and
2
2
2
 
θ
θ
<θ .
2
2
Therefore, sin θ < θ for any angle θ > 0 .
cos
56. cos
≤ 1 , we have sin θ ≤ 2sin
58.
C
1 + cos C
=
2
2
=
1+
a 2 + b2 − c2
2ab
2
=
2ab + a 2 + b 2 − c 2
4ab
=
( a + b) 2 − c 2
4ab
=
=
4s( s − c)
=
4ab
=
s( s − c)
ab
=
4( s − b)( s − a)
4ab
=
( s − a)( s − b)
ab
cos A cos B cos C
+
+
a
b
c
2
2
2
2
b +c −a
a + c2 − b2 a 2 + b2 − c2
=
+
+
2bca
2acb
2abc
2
2
2
2
2
2
2
b + c − a + a + c − b + a + b2 − c2
=
2abc
a 2 + b2 + c2
=
2abc
1
bh
2
2. K =
1
ab sin C
2
s ( s − a)( s − b)( s − c) ;
K=
=
=
−(a 2 − 2ab + b2 − c 2 )
4ab
=
(2s − 2b)(2s − 2a)
4ab
1
(a + b + c)
2
5. a = 2, c = 4, B = 45º
2ab − a 2 − b2 + c 2
4ab
(
=
4. True
a2 + b2 − c2
2ab
2
− ( a − b) 2 − c 2
(a − b + c)(b + c − a)
4ab
1. K =
3.
C
1 − cos C
57. sin =
2
2
1−
=
Section 8.4
4ab
2s (2 s − c − c)
4ab
−(a − b + c)(a − b − c)
4ab
59 – 63. Answers will vary.
( a + b + c )( a + b − c )
=
=
1
1
ac sin B = (2)(4)sin 45º ≈ 2.83
2
2
6. b = 3, c = 4, A = 30º
1
1
K = bc sin α = (3)(4) sin 30º = 3
2
2
7. a = 2, b = 3, C = 95º
)
K=
1
1
ab sin C = (2)(3) sin 95º ≈ 2.99
2
2
4ab
825
Chapter 8: Applications of Trigonometric Functions
8. a = 2, c = 5, B = 20º
K=
16. a = 6, b = 4, C = 60º
1
1
ac sin B = (2)(5) sin 20º ≈ 1.71
2
2
K=
9. a = 6, b = 5, c = 8
17. a = 3, c = 2, B = 110º
1
1
19
( a + b + c ) = 2 ( 6 + 5 + 8) = 2
2
K = s ( s − a)( s − b)( s − c)
s=
K=
1
1
ac sin B = (3)(2)sin110º ≈ 2.82
2
2
18. b = 4, c = 1, A = 120º
3591
 19  7   9   3 
=        =
≈ 14.98
2
2
2
2
16
      
1
1
K = bc sin A = (4)(1) sin120º ≈ 1.73
2
2
10. a = 8, b = 5, c = 4
19. a = 12, b = 13, c = 5
1
1
17
s = ( a + b + c ) = (8 + 5 + 4 ) =
2
2
2
K = s ( s − a)( s − b)( s − c)
1
1
( a + b + c ) = 2 (12 + 13 + 5 ) = 15
2
K = s ( s − a)( s − b)( s − c)
s=
1071
 17  1  7  9 
=      =
≈ 8.18
16
 2  2  2  2 
=
(15 )( 3)( 2 )(10 ) =
900 = 30
20. a = 4, b = 5, c = 3
11. a = 9, b = 6, c = 4
1
1
( a + b + c ) = 2 ( 4 + 5 + 3) = 6
2
K = s ( s − a)( s − b)( s − c)
s=
1
1
19
s = ( a + b + c ) = (9 + 6 + 4) =
2
2
2
K = s ( s − a)( s − b)( s − c )
=
1463
 19   1   7  11 
=        =
≈ 9.56
16
 2   2   2  2 
( 6 )( 2 )(1)( 3) =
36 = 6
21. a = 2, b = 2, c = 2
1
1
( a + b + c ) = 2 ( 2 + 2 + 2) = 3
2
K = s ( s − a)( s − b)( s − c)
s=
12. a = 4, b = 3, c = 4
1
1
11
( a + b + c ) = 2 ( 4 + 3 + 4) = 2
2
K = s ( s − a)( s − b)( s − c)
s=
=
495
 11  3  5  3 
=      =
≈ 5.56
2
2
2
2
16
    
( 3)(1)(1)(1) =
3 ≈ 1.73
22. a = 3, b = 3, c = 2
1
1
( a + b + c ) = 2 (3 + 3 + 2) = 4
2
K = s ( s − a)( s − b)( s − c)
s=
13. a = 3, b = 4, C = 40º
K=
1
1
ab sin C = (6)(4) sin 60º ≈ 10.39
2
2
1
1
ab sin C = (3)(4) sin 40º ≈ 3.86
2
2
=
14. a = 2, c = 1, B = 10º
( 4 )(1)(1)( 2 ) =
8 ≈ 2.83
23. a = 5, b = 8, c = 9
1
1
K = ac sin B = (2)(1)sin10º ≈ 0.17
2
2
1
1
a + b + c ) = ( 5 + 8 + 9 ) = 11
(
2
2
K = s ( s − a)( s − b)( s − c)
s=
15. b = 1, c = 3, A = 80º
=
1
1
K = bc sin A = (1)(3)sin 80º ≈ 1.48
2
2
(11)( 6 )( 3)( 2 ) =
396 ≈ 19.90
826
Section 8.4: Area of a Triangle
30. A = 70º , B = 60º , c = 4
C = 180º − A − B = 180º − 70º − 60º = 50º
24. a = 4, b = 3, c = 6
1
1
13
( a + b + c ) = 2 ( 4 + 3 + 6) = 2
2
K = s ( s − a)( s − b)( s − c)
s=
 13  5  7  1 
=      =
 2  2  2  2 
K=
455
≈ 5.33
16
31. A = 110º , C = 30º , c = 3
B = 180º − A − C = 180º − 110º − 30º = 40º
K=
sin A sin B
25. From the Law of Sines we know
.
=
a
b
a sin B
. Thus,
Solving for b, so we have that b =
sin A
1
1  a sin B 
K = ab sin C = a 
sin C
2
2  sin A 
a 2 sin B sin C
=
2sin A
K=
π
7π
=
180 18
1
7π 112π 2
ASector = ⋅ 82 ⋅
=
ft
2
18
9
1
ATriangle = ⋅ 8 ⋅ 8sin 70º = 32 sin 70º ft 2
2
112π
ASegment =
− 32sin 70º ≈ 9.03 ft 2
9
34. Area of a sector =
1 2
r θ where θ is in radians.
2
π
2π
=
180 9
1
2π 25π 2
ASector = ⋅ 52 ⋅
=
in
2
9
9
1
25
ATriangle = ⋅ 5 ⋅ 5sin 40º = sin 40º in 2
2
2
25π 25
ASegment =
− sin 40º ≈ 0.69 in 2
9
2
θ = 40 ⋅
a 2 sin B sin C 22 sin 20º ⋅ sin120º
=
≈ 0.92
2sin A
2 sin 40º
28. A = 50º , C = 20º , a = 3
B = 180º − A − C = 180º − 50º − 20º = 110º
35. Find the area of the lot using Heron's Formula:
a = 100, b = 50, c = 75
a sin B sin C 3 sin110º ⋅ sin 20º
=
≈ 1.89
2 sin A
2 sin 50º
2
s=
29. B = 70º , C = 10º , b = 5
A = 180º − B − C = 180º − 70º − 10º = 100º
K=
1 2
r θ where θ is in radians.
2
θ = 70 ⋅
27. A = 40º , B = 20º , a = 2
C = 180º − A − B = 180º − 40º − 20º = 120º
K=
b 2 sin A sin C 22 sin 70º ⋅ sin100º
=
≈ 10.66
2sin B
2sin10º
33. Area of a sector =
sin A sin B sin C
=
=
, we have that
a
b
c
b sin C
c sin A
c=
and a =
. Thus,
sin B
sin C
1
1  b sin C 
K = bc sin A = b 
sin A
2
2  sin B 
b 2 sin A sin C
=
2sin B
1
1  c sin A 
K = ac sin B = 
c sin B
2
2  sin C 
c 2 sin A sin B
=
2sin C
2
c 2 sin A sin B 32 sin110º ⋅ sin 40º
=
≈ 5.44
2sin C
2 sin 30º
32. B = 10º , C = 100º , b = 2
A = 180º − B − C = 180º − 10º − 100º = 70º
26. From
K=
c 2 sin A sin B 42 sin 70º ⋅ sin 60º
=
≈ 8.50
2sin C
2sin 50º
1
1
225
( a + b + c ) = 2 (100 + 50 + 75) = 2
2
b 2 sin A sin C 52 sin100º ⋅ sin10º
=
≈ 2.27
2sin B
2sin 70º
827
Chapter 8: Applications of Trigonometric Functions
The triangle is a right triangle. Find the other leg:
82 + b 2 = 102
K = s ( s − a)( s − b)( s − c)
 225  25  125  75 
= 
 
 
 2  2  2  2 
b 2 = 100 − 64 = 36
b = 36 = 6
1
ATriangle = ⋅ 8 ⋅ 6 = 24 in 2
2
AShaded region = 12.5π − 24 ≈ 15.27 in 2
52, 734,375
16
≈ 1815.46
Cost = ( \$3)(1815.46 ) = \$5446.38
=
39. The area is the sum of the area of a triangle and a
sector.
1
1
ATriangle = r ⋅ r sin(π − θ ) = r 2 sin(π − θ )
2
2
1 2
ASector = r θ
2
36. Diameter of canvas is 24 feet; radius of canvas is
12 feet; angle is 260˚.
1
Area of a sector = r 2θ where θ is in radians.
2
π
13π
=
θ = 260 ⋅
180
9
1
13
π 936π
ASector = ⋅122 ⋅
=
= 104π ≈ 326.73 ft 2
2
9
9
K=
=
37. Divide home plate into a rectangle and a triangle.
12”
12”
=
17”
8.5”
=
8.5”
=
17”
ARectangle = lw = (17)(8.5) = 144.5 in 2
1 2
1
r sin ( π − θ ) + r 2θ
2
2
1 2
r ( sin(π − θ ) + θ )
2
1 2
r ( sin π cos θ − cos π sin θ + θ )
2
1 2
r ( 0 ⋅ cos θ − (−1)sin θ + θ )
2
1 2
r (θ + sin θ )
2
40. Use the Law of Cosines to find the lengths of the
diagonals of the polygon.
x 2 = 352 + 802 − 2 ⋅ 35 ⋅ 80 cos15º
Using Heron’s formula we get
ATriangle = s ( s − a)( s − b)( s − c)
= 7625 − 5600 cos15º
1
1
( a + b + c ) = (12 + 12 + 17 ) = 20.5
2
2
Thus,
ATriangle = (20.5)(20.5 − 12)(20.5 − 12)(20.5 − 17)
s=
x = 7625 − 5600 cos15º ≈ 47.072 feet
The interior angle of the third triangle is:
180° − 100° = 80° .
y 2 = 452 + 202 − 2 ⋅ 45 ⋅ 20 cos 80º
= (20.5)(8.5)(8.5)(3.5)
= 2425 − 1800 cos80º
= 5183.9375
y = 2425 − 1800 cos80º ≈ 45.961 feet
≈ 72.0 sq. in.
So, ATotal = ARectangle + ATriangle
Find the area of the three triangles:
1
( 35 + 80 + 47.072 ) = 81.036
2
s1 − a1 = 81.036 − 35 = 46.036
= 144.5 + 72.0
s1 ≈
= 216.5 in 2
The area of home plate is about 216.5 in 2 .
s1 − b1 = 81.036 − 80 = 1.036
s1 − c1 = 81.036 − 47.072 = 33.964
38. Find the area of the shaded region by subtracting
the area of the triangle from the area of the
semicircle.
Area of the semicircle
1
1
25
ASemicircle = π r 2 = π(5)2 =
π in 2
2
2
2
K1 ≈ 81.036(46.036)(1.036)(33.964) ≈ 362.31 ft 2
828
Section 8.4: Area of a Triangle
1
( 40 + 45.961 + 47.072 ) = 66.5165
2
s2 − a2 = 66.5165 − 40 = 26.5165
s2 ≈
Area ΔOAC =
b.
Area ΔOCB =
s2 − b2 = 66.5165 − 45.961 = 20.5555
s2 − c2 = 66.5165 − 47.072 = 19.4445
K 2 ≈ 66.5165(26.5165)(20.5555)(19.4445)
≈ 839.62 ft 2
1
(45 + 20 + 45.961) = 55.4805
2
s3 − a3 = 55.4805 − 45 = 10.4805
s3 ≈
1
OC ⋅ AC
2
AC
1 OC
= ⋅
⋅
2 1
1
1
= cos α sin α
2
1
= sin α cos α
2
43. a.
s3 − b3 = 55.4805 − 20 = 35.4805
s3 − c3 = 55.4805 − 45.961 = 9.5195
1
OC ⋅ BC
2
OC BC
2
1
= ⋅ OB ⋅
⋅
2
OB OB
1
OB
2
1
= OB
2
=
K3 ≈ 55.4805(10.4805)(35.4805)(9.5195)
≈ 443.16 ft 2
The approximate area of the lake is
362.31 + 839.62 + 443.16 = 1645.09 ft 2
c.
41. Use Heron’s formula:
a = 87 , b = 190 , c = 173
1
1
s = ( a + b + c ) = ( 87 + 190 + 173) = 225
2
2
K = s ( s − a )( s − b )( s − c )
= 225 (138 )( 35)( 52 )
d.
= 56,511, 000 ≈ 7517.4
The building covers approximately 7517.4 square
feet of ground area.
e.
42. Use Heron’s formula:
a = 1028 , b = 1046 , c = 965
1
1
s = ( a + b + c ) = (1028 + 1046 + 965 ) = 1519.5
2
2
cos β sin β
2
sin β cos β
1
BD ⋅ OA
2
1
= BD ⋅1
2
BD
1
= ⋅ OB ⋅
2
OB
Area ΔOAB =
=
= 225 ( 225 − 87 )( 225 − 190 )( 225 − 173)
2
1
OB sin(α + β )
2
OC
OA
OC OB
cos A
=
=
⋅
= OB
cos B
1
OC
OC
OB
Area ΔOAB = Area ΔOAC + Area ΔOCB
1
1
1
OB sin(α + β ) = sin α cos α + OB
2
2
2
2
sin β cos β
cosα
cos 2 α
sin(α + β ) = sin α cosα +
sin β cos β
cos β
cos 2 β
cos β
cosα
sin(α + β ) =
sin α cos α +
sin β cos β
cos α
cos β
sin(α + β ) = sin α cos β + cos α sin β
K = s ( s − a )( s − b )( s − c )
= 1519.5 ( 491.5 )( 473.5 )( 554.5)
≈ 442,816
The area of the Bermuda Triangle is approximately
442,816 square miles.
44. a.
Area of ΔOBC =
1
sin θ
⋅1 ⋅1 ⋅ sin θ =
2
2
b.
Area of ΔOBD =
1
tan θ
sin θ
⋅1 ⋅ tan θ =
=
2
2
2 cos θ
829
Chapter 8: Applications of Trigonometric Functions
 < Area ΔOBD
Area ΔOBC < Area OBC
1
1
sin θ
sin θ < θ <
2
2
2 cos θ
sin θ
sin θ < θ <
cos θ
sin θ
sin θ
θ
<
<
sin θ sin θ sin θ cos θ
1
θ
1<
<
sin θ cos θ
c.
Thus, the total grazing area is:
23,561.94 + 2(3499.66) + 2(292.19)
≈ 31,146 ft 2
46. We begin by dividing the grazing area into five
regions: three sectors and two triangles (see
figure). Region A1 is a sector representing
three-fourths of a circle with radius 100 feet:
3
2
Thus, A1 = π (100 ) = 7500π ≈ 23,561.9 ft 2 .
4
To find the areas of regions A2 , A3 , A4 , and
45. The grazing area must be considered in sections.
Region A1 represents three-fourth of a circle with
3
2
A1 = π (100 ) = 7500π ≈ 23,561.94 ft 2
4
A5 , we first position the rectangular barn on a
rectangular coordinate system so that the lower
right corner is at the origin. The coordinates of
the corners of the barn must then be O(0, 0) ,
P(−20, 0) , Q(−20,10) , and R(0,10) .
Angles are needed to find regions A2 and A3 :
(see the figure)
T
y
C
A2
D
B
10 ft
R
90 ft
A
A4
A5
U
x
Q(−20,10) and radius 90 feet. The equation of
the circle then is ( x + 20)2 + ( y − 10) 2 = 902 .
Likewise, region A5 is a sector of a circle with
center O(0, 0) and radius 80 feet. The equation
∠DAC = 130.49º − 90º = 40.49º
of this circle then is x 2 + y 2 = 802 . We use a
graphing calculator to find the intersection point
1
(10)(90)sin 40.49º ≈ 292.19 ft 2
2
S of the two sectors. Let Y1 = 802 − x 2 and
The angle for the sector A2 is
90º − 40.49º = 49.51º .
A2 =
A3
Now, region A2 is a sector of a circle with center
∠BAC = 180º − 45º − 4.51º = 130.49º
A3 =
80 ft
10 ft Q
P
O
20
ft
A1
In ΔABC , ∠CBA = 45º , AB = 10, AC = 90 .
Find ∠BCA :
sin ∠CBA sin ∠BCA
=
90
10
sin 45º sin ∠BCA
=
90
10
10 sin 45º
sin ∠BCA =
≈ 0.0786
90
 10sin 45º 
∠BCA = sin −1 
 ≈ 4.51º
90


S
90 ft
Y2 = 902 − ( x + 20)2 + 10 .
112
1
2
π 
≈ 3499.66 ft 2
90 )  49.51 ⋅
(
2
180 

Since the cow can go in either direction around
the barn, both A2 and A3 must be doubled.
−64
124
−12
830
Section 8.4: Area of a Triangle
K = s ( s − a )( s − b )( s − c )
The approximate coordinates are S (57.7,55.4) .
Now, consider ΔQRS (i.e. region A3 ). The
“base” of this triangle is 20 feet and the “height”
is approximately 55.4 − 10 = 45.4 feet (the ycoordinate of the intersection point minus the
side of the barn). Thus, the area of region A3 is
= 18 (18 − 9 )(18 − 10 )(18 − 17 )
= 18 ( 9 )( 8 )(1)
= 1296 = 36
Since the perimeter and area are numerically
equal, the given triangle is a perfect triangle.
1
A3 ≈ ⋅ 20 ⋅ 45.4 = 454 ft 2 . Likewise, consider
2
ΔORS (i.e. region A4 ). The “base” of this
triangle is 10 feet and the “height” is about 57.7
feet (the x-coordinate of the intersection point).
1
Thus, A4 ≈ ⋅10 ⋅ 57.7 = 288.5 ft 2 .
2
b.
Area:
1
1
s = ( a + b + c ) = ( 6 + 25 + 29 ) = 30
2
2
To find the area of sectors A2 and A5 , we must
determine their angles: ∠TQS and ∠SOU ,
K = s ( s − a )( s − b )( s − c )
respectively. Now, we know A3 ≈ 454 ft 2 .
Also, we know A3 =
Thus,
= 30 ( 30 − 6 )( 30 − 25 )( 30 − 29 )
1
⋅ 90 ⋅ 20sin ( ∠SQR ) .
2
= 30 ( 24 )( 5)(1)
1
⋅ 90 ⋅ 20 sin ( ∠SQR ) ≈ 454
2
sin ( ∠SQR ) ≈ 0.5044
= 3600 = 60
Since the perimeter and area are numerically
equal, the given triangle is a perfect triangle.
Since ∠TQR is a right angle, we have
∠TQS ≈
π
2
1
2K
2K
h a , so h1 =
. Similarly, h 2 =
2 1
a
b
2K
. Thus,
and h 3 =
c
1
1
1
a
b
c
+
+
=
+
+
h1 h 2 h 3 2 K 2 K 2 K
48. K =
− .5287 ≈ 1.0421 rad. So,
1 2
1
r θ ≈ ⋅ 902 ⋅1.0421 ≈ 4220.5 ft 2 .
2
2
1
Similarly, ⋅ 80 ⋅10sin ( ∠SOR ) ≈ 288.5
2
sin ( ∠SOR ) ≈ 0.72125
A2 =
So, ∠SOU ≈
A5 ≈
π
2
a + b + c 2s
=
2K
2K
s
=
K
=
− .8056 ≈ 0.7652 rad. and
1
⋅ 802 ⋅ 0.7652 ≈ 2448.6 ft 2
2
1
1
ah and K = ab sin C , which
2
2
means h = b sin C . From the Law of Sines, we
sin A sin B
a sin B
=
, so b =
. Therefore,
know
sin A
a
b
a sin B sin C
 a sin B 
h=
.
 sin C =
A
sin
sin A


49. We know K =
Thus, the total grazing area is:
23,561.9 + 4220.5 + 454 + 288.5 + 2448.6
= 30,973 ft 2
47. a.
Perimeter:
P = a + b + c = 6 + 25 + 29 = 60
Perimeter:
P = a + b + c = 9 + 10 + 17 = 36
Area:
1
1
s = ( a + b + c ) = ( 9 + 10 + 17 ) = 18
2
2
831
Chapter 8: Applications of Trigonometric Functions
52. Use the result of Problem 51:
A
B
C s −a s −b s −c
+
+
cot + cot + cot =
2
2
2
r
r
r
s −a + s −b+ s −c
=
r
3s − ( a + b + c )
=
r
3s − 2 s
=
r
s
=
r
a sin B sin C
where h is the altitude to side a.
sin A
In ΔOPQ , c is opposite ∠POQ . The two
50. h =
A
B
and . Then
2
2
A
B
sin
2
2 . Now, ∠POQ = π −  A + B  ,
r=


sin ( ∠POQ )
2 2
so
  A B 
sin ( ∠POQ ) = sin  π −  +  
  2 2 
c sin
53. K = Area ΔPOQ + Area ΔPOR + Area ΔQOR
1
1
1
= rc + rb + ra
2
2
2
1
= r (a + b + c)
2
= rs
 A B
= sin  + 
2 2
 A+ B 
= sin 

 2 
 π  A + B 
= cos  − 

 2  2 
 π − ( A + B) 
= cos 

2


C
= cos
2
A
B
c sin sin
2
2 .
Thus, r =
C
cos
2
C
cos
C
2 =
51. cot =
2 sin C
2
Now, K = s ( s − a)( s − b)( s − c) , so
rs = s ( s − a)( s − b)( s − c)
s ( s − a )( s − b)( s − c)
s
( s − a)( s − b)( s − c)
=
s
r=
54 – 56. Answers will vary.
A
B
⋅ sin
2
2
r
( s − a)( s − b)
ab
c ⋅ sin
Section 8.5
1. 5 = 5 ;
( s − b)( s − c) ( s − a)( s − c)
c⋅
bc
ac
=
( s − a)( s − b)
r
ab
c ( s − b)( s − c) ( s − a )( s − c)
ab
=
( s − a )( s − b)
r
bc
ac
2π π
=
4
2
2. simple harmonic; amplitude
3. simple harmonic; damped
4. True
5. d = − 5cos ( πt )
c ab( s − a )( s − b)( s − c) 2
=
r
abc 2 ( s − a)( s − b)
 2π 
6. d = −10 cos  t 
 3 
c ( s − c) 2 c s − c
= ⋅
r
r c
c2
s−c
=
r
=
7. d = − 6 cos ( 2t )
8. d = − 4 cos ( 4 t )
832
Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves
9. d = − 5sin ( π t )
18. d = − 2 cos(2t )
a. Simple harmonic
 2π 
10. d = −10sin  t 
 3 
11. d = − 6sin ( 2t )
12. d = − 4sin ( 4 t )
5 meters
c.
2π
seconds
3
d.
3
oscillation/second
2π
c.
π seconds
d.
1
oscillation/second
π
b.
2 meters
c.
1 second
d.
1 oscillation/second
20. d = 4 + 3sin(πt )
a. Simple harmonic
14. d = 4sin(2t )
a. Simple harmonic
b.
2 meters
19. d = 6 + 2 cos(2πt )
a. Simple harmonic
13. d = 5sin(3t )
a. Simple harmonic
b.
b.
4 meters
π seconds
1
oscillation/second
d.
π
c.
b.
3 meters
c.
2 seconds
d.
1
oscillation/second
2
21. d ( t ) = e −t / π cos ( 2t )
15. d = 6 cos(πt )
a. Simple harmonic
b.
6 meters
c.
2 seconds
d.
1
oscillation/second
2
π 
16. d = 5cos  t 
2 
a. Simple harmonic
b.
5 meters
c.
4 seconds
d.
1
oscillation/second
4
22. d ( t ) = e −t / 2π cos ( 2t )
1 
17. d = − 3sin  t 
2 
a. Simple harmonic
b.
3 meters
c.
4π seconds
d.
1
oscillation/second
4π
833
Chapter 8: Applications of Trigonometric Functions
23. d ( t ) = e −t / 2π cos t
27.
f ( x ) = x − sin x
28.
f ( x ) = x − cos x
29.
f ( x ) = sin x + cos x
30.
f ( x ) = sin ( 2 x ) + cos x
24. d ( t ) = e −t / 4π cos t
25.
26.
f ( x ) = x + cos x
f ( x ) = x + cos ( 2 x )
834
Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves
31. f ( x ) = sin x + sin ( 2 x )
b.
35. a.
32.
G ( x ) = cos ( 4 x ) cos ( 2 x )
1
[cos(4 x − 2 x) + cos(4 x + 2 x)]
2
1
= [ cos(2 x) + cos(6 x) ]
2
f ( x ) = cos ( 2 x ) + cos x
=
b.
33. a.
f ( x ) = sin ( 2 x ) sin x
1
[ cos(2 x − x) − cos(2 x + x)]
2
1
= [ cos( x) − cos(3 x)]
2
=
36. a.
b.
h ( x ) = cos ( 2 x ) cos ( x )
1
[cos(2 x − x) + cos(2 x + x) ]
2
1
= [ cos( x) + cos(3x)]
2
=
b.
34. a.
F ( x ) = sin ( 3x ) sin x
1
[ cos(3x − x) − cos(3x + x)]
2
1
= [ cos(2 x) − cos(4 x) ]
2
=
835
Chapter 8: Applications of Trigonometric Functions
37. a.
H ( x ) = 2sin ( 3 x ) cos ( x )
40. a.
1
= ( 2 )   [sin(3 x + x) + sin(3x − x) ]
2
= sin(4 x) + sin(2 x)
b.
 2π 2 (0.75)2
 
d = −15e −0.75 t / 2(20) cos    −
  6 
4(20)2

 π2 0.5625 
d = −15e −0.75 t / 40 cos 
t
−
 9
1600 


15
b.
30
0
−15
41. a.
38. a.
g ( x ) = 2sin ( x ) cos ( 3 x )
1
= ( 2 )   [sin( x + 3 x) + sin( x − 3x) ]
2
= sin(4 x) + sin(−2 x)
 π 2 (0.6)2
 
d = −18e −0.6 t / 2(30) cos    −
  2  4(30) 2

 π2 0.36 
d = −18e −0.6 t / 60 cos 
t
−
 4 3600 



t


18
b.
0
= sin(4 x) − sin(2 x)

t


20
−18
42. a.
 2π 2 (0.65)2
 
d = −16e−0.65 t / 2(15) cos    −
  5 
4(15) 2

 4π2 0.4225 
d = −16e−0.65 t / 30 cos 
t
−
 25
900 


16
b.
39. a.
 2π 2 (0.7)2
 
d = −10e−0.7t / 2(25) cos    −
  5  4(25) 2

 4π2 0.49 
d = −10e−0.7t / 50 cos 
t
−
 25 2500 


0

t


−16
43. a.
10
b.
0
25
25
 2π 2 (0.8)2
 
d = −5e−0.8 t / 2(10) cos    −
  3  4(10)2

 4π2 0.64 
d = −5e−0.8 t / 20 cos 
t
−
 9
400 


−10
836

t



t


Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves
5
b.
0
15
e.
−5
44. a.
  2π 2 (0.7) 2
d = −5e−0.7 t / 2(10) cos    −
  3  4(10)2


4π2 0.49 
−
d = −5e−0.7 t / 20 cos 
t
 9
400 


t


47. a.
b.
15
35
−30
The displacement of the bob at the start of
the second oscillation is about 28.47 meters.
e.
The displacement of the bob approaches
zero, since e −0.6 t / 80 → 0 as t → ∞ .
Damped motion with a bob of mass 20 kg
and a damping factor of 0.7 kg/sec.
20 meters leftward
20
c.
0
25
48. a.
−20
d.
30
d.
−5
b.
30 meters leftward
0
0
45. a.
Damped motion with a bob of mass 40 kg
and a damping factor of 0.6 kg/sec.
c.
5
b.
The displacement of the bob approaches
zero, since e −0.8 t / 40 → 0 as t → ∞ .
b.
The displacement of the bob at the start of
the second oscillation is about 18.33 meters.
Damped motion with a bob of mass 35 kg
and a damping factor of 0.5 kg/sec.
30 meters leftward
30
c.
0
e.
46. a.
b.
−30
The displacement of the bob approaches
zero, since e −0.7 t / 40 → 0 as t → ∞ .
d.
The displacement of the bob at the start of
the second oscillation is about 29.15 meters.
e.
The displacement of the bob approaches
zero, since e −0.5 t / 70 → 0 as t → ∞ .
Damped motion with a bob of mass 20 kg
and a damping factor of 0.8 kg/sec.
20 meters leftward
20
c.
0
25
49. a.
−20
d.
20
b.
The displacement of the bob at the start of
the second oscillation is about 18.10 meters.
Damped motion with a bob of mass 15 kg
and a damping factor of 0.9 kg/sec.
15 meters leftward
837
Chapter 8: Applications of Trigonometric Functions
15
c.
position at time t = 0 . Since a cos ( ωt ) peaks at
0
t = 0 if a > 0 and is at its lowest if a < 0 , we
select a cosine model and need a = −82.5 . Using
the model d = a cos (ωt ) + b , we have
30
d = −82.5cos ( wt ) + b . When t = 0 , the rider
should be 15 feet above the ground. That is,
15 = −82.5cos ( 0 ) + b
−15
d.
The displacement of the bob at the start of
the second oscillation is about 12.53 meters.
15 = −82.5 + b
97.5 = b
Therefore, the equation that describes the rider’s
motion is d = 97.5 − 82.5cos ( 3.2π t ) .
e.
50. a.
b.
The displacement of the bob approaches
zero, since e −0.9 t / 30 → 0 as t → ∞ .
53. The maximum displacement is the amplitude so
we have a = 0.01 . The frequency is given by
Damped motion with a bob of mass 25 kg
and a damping factor of 0.8 kg/sec.
f =
movement of the tuning fork is described by the
equation d = 0.01sin ( 880π t ) .
10 meters leftward
10
c.
0
54. The maximum displacement is the amplitude so
we have a = 0.025 . The frequency is given by
15
ω
= 329.63 . Therefore, ω = 659.26π and
2π
the movement of the tuning fork is described by
the equation d = 0.025sin ( 659.26π t ) .
f =
−10
d.
ω
= 440 . Therefore, ω = 880π and the
2π
The displacement of the bob at the start of
the second oscillation is 9.53 meters.
55. a.
V
1
e.
1
The displacement of the bob approaches
zero, since e −0.8 t / 50 → 0 as t → ∞ .
2
3
t
−1
51. The maximum displacement is the amplitude so
we have a = 0.80 . The frequency is given by
b.
ω
f =
= 520 . Therefore, ω = 1040π and the
2π
On the interval 0 ≤ t ≤ 3 , the graph of V
touches the graph of y = e −t / 3 when
t = 0, 2 . The graph of V touches the graph
motion of the diaphragm is described by the
equation d = 0.80 cos (1040π t ) .
of y = − e −t / 3 when t = 1, 3 .
c.
52. If we consider a horizontal line through the
center of the wheel as the equilibrium line, then
165
= 82.5 . The wheel
the amplitude is a =
2
2π
1
completes 1.6 so the period is
=
and
ω 1.6
ω = 3.2π . We want the rider to be at the lowest
We need to solve the inequality
−0.4 < e −t / 3 cos (π t ) < 0.4 on the interval
0 ≤ t ≤ 3 . To do so, we consider the graphs
of y = −0.4, y =e −t / 3 cos (π t ) , and y = 0.4 .
On the interval 0 ≤ t ≤ 3 , we can use the
INTERSECT feature on a calculator to
determine that y = e −t / 3 cos (π t ) intersects
838
Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves
y = 0.4 when t ≈ 0.35, t ≈ 1.75 , and
t ≈ 2.19, y = e
−t / 3
Let Y1 =
c.
cos (π t ) intersects
y = −0.4 when t ≈ 0.67 and t ≈ 1.29 and
the graph shows that
−0.4 < e −t / 3 cos (π t ) < 0.4 when t = 3 .
1
Therefore, the voltage V is between –0.4 and
0.4 on the intervals 0.35 < t < 0.67 ,
1.29 < t < 1.75 , and 2.19 < t ≤ 3 .
1.25
0
1.25
0
3.2 0
−1.25
1.25
d. f ( x ) =
3.2
3.2 0
4
−1
−1.25
1.25
0
1
1
sin ( 2π ) + sin ( 4π x )
2
4
1
1
+ sin ( 8π x ) + sin (16π x )
8
16
1
1
1
sin ( 2π x ) + sin ( 4π x ) + sin ( 8π x )
2
4
8
1
1
+ sin (16π x ) + sin ( 32π x )
16
32
57. Let Y1 = sin ( 2π ( 852 ) x ) + sin ( 2π (1209 ) x ) .
2.5
3.2
0
−1.25
1.25
−1.25
1.25
0
3.2 0
−1.25
56. a.
−2.5
58. The sound emitted by touching * is
y = sin ( 2π ( 941) t ) + sin ( 2π (1209 ) t ) .
3.2
Let Y1 = sin ( 2π ( 941) x ) + sin ( 2π (1209 ) x ) .
−1.25
2.5
1
1
Let Y1 = sin ( 2π ) + sin ( 4π x ) .
2
4
0
1
2
−2.5
4
0
2
59 – 60. CBL Experiments
−1
61. Let Y1 =
1
1
1
b. Let Y1 = sin ( 2π ) + sin ( 4π x ) + sin ( 8π x ) .
2
4
8
1
1
0
4
sin x
.
x
0
5π
−0.3
−1
As x approaches 0,
sin x
approaches 1.
x
839
Chapter 8: Applications of Trigonometric Functions
62. y = x sin x
 1 
y =  3  sin x
x 
3
0.05
2π
0
0
−5
−0.015
y = x 2 sin x
Possible observation: As x gets larger, the graph
 1 
of y =  n  sin x gets closer to y = 0 .
x 
5
2π
0
3π
−25
y = x3 sin x
Chapter 8 Review Exercises
10
1. opposite = 4; adjacent = 3; hypotenuse = ?
(hypotenuse)2 = 42 + 32 = 25
0
hypotenuse = 25 = 5
2π
opp 4
=
hyp 5
cos θ =
=
hyp 5
opp 4
tan θ =
=
sin θ =
−125
Possible observations: The graph lies between
the bounding curves y = ± x, y = ± x 2 , y = ± x3 ,
respectively, touching them at odd multiples of
π
. The x-intercepts of each graph are the
2
multiples of π .
2. opposite = 3; adjacent = 5; hypotenuse = ?
(hypotenuse)2 = 32 + 52 = 34
1
63. y =   sin x
x
hypotenuse = 34
1
opp
3
=
=
hyp
34
5
=
=
cos θ =
hyp
34
opp 3
=
tan θ =
sin θ =
0
5π
−0.3
 1
y= 2
x
0.1
0
hyp 5
=
opp 4
hyp 5
sec θ =
=
cot θ =
=
opp 4
csc θ =

 sin x

csc θ =
3
34
5
34
⋅
⋅
34
34
34
34
hyp
34
=
opp
3
hyp
34
=
5
=
cot θ =
opp 3
sec θ =
7π
−0.06
840
=
3 34
34
=
5 34
34
Chapter 8 Review Exercises
3. adjacent = 2; hypotenuse = 4; opposite = ?
(opposite)2 + 22 = 42
7.
sec55º csc(90º −55º ) csc 35º
=
=
=1
csc35º
csc 35º
csc 35º
8.
tan 40º
tan 40º
tan 40º
=
=
=1
cot 50º tan(90º − 50º ) tan 40º
(opposite) 2 = 16 − 4 = 12
opposite = 12 = 2 3
opp 2 3
3
=
=
hyp
4
2
cos θ =
= =
hyp 4 2
sin θ =
tan θ =
9. cos 2 40º + cos 2 50º = sin 2 (90º −40º ) + cos 2 50º
= sin 2 50º + cos 2 50º
=1
opp 2 3
=
= 3
2
(
hyp
4
4
3 2 3
=
=
⋅
=
csc θ =
opp 2 3 2 3 3
3
hyp 4
= =2
sec θ =
cot θ =
( 2)
2
= −1 + sec 40º − sec (90º −50º )
2
11. c = 10, B = 20º
b
c
b
sin 20º =
10
b = 10sin 20º ≈ 3.42
sin B =
2 ; hypotenuse = 2; adjacent = ?
(adjacent)2 = 4 − 2 = 2
a
c
a
cos 20º =
10
a = 10 cos 20º ≈ 9.40
cos B =
opposite = 2
sin θ =
opp
2
=
hyp
2
cos θ =
2
=
hyp
2
tan θ =
opp
=
csc θ =
hyp
2
2
2 2 2
=
=
⋅
=
= 2
opp
2
2
2 2
sec θ =
hyp
2
2
2 2 2
=
=
⋅
=
= 2
2
2
2 2
cot θ =
=
opp
2
2
2
2
2
= −1 + sec 2 40º − sec2 40º
= −1
2
2
3
3
=
=
⋅
=
opp 2 3 2 3 3
3
4. opposite =
)
10. tan 2 40º − csc 2 50º = −1 + sec2 40º − csc2 50º
A = 90º − B = 90º − 20º = 70º
12. a = 5, A = 35º
=1
a
c
5
sin 35º =
c
sin A =
c=
=1
5
≈ 8.72
sin 35º
a
b
5
tan 35º =
b
tan A =
5. cos 62º − sin 28º = cos 62º − cos(90º −28º )
= cos 62º − cos 62º
=0
b=
5
≈ 7.14
tan 35º
B = 90º − A = 90º − 35º = 55º
6. tan15º − cot 75º = tan15º − tan(90º − 75º )
= tan15º − tan15º
=0
841
Chapter 8: Applications of Trigonometric Functions
13. b = 2, c = 5
sin C sin B
=
c
b
sin 40º sin130º
=
2
b
2sin130º
≈ 2.38
b=
sin 40º
c 2 = a 2 + b2
a 2 = c 2 − b 2 = 52 − 22 = 25 − 4 = 21
a = 21 ≈ 4.58
b 2
=
c 5
2
B = sin −1   ≈ 23.6º
5
sin B =
B = 180º − A − C = 180º − 10º − 40º = 130º
17. A = 100º , c = 2, a = 5
sin C sin A
=
c
a
sin C sin100º
=
2
5
2 sin100º
≈ 0.3939
sin C =
5
 2sin100º 
C = sin −1 

5


C ≈ 23.2º or C ≈ 156.8º
The value 156.8º is discarded because
A + C > 180º . Thus, α ≈ 23.2º .
A = 90º − B ≈ 90º − 23.6º = 66.4º
14. b = 1, a = 3
c2 = a 2 + b2
c 2 = 32 + 12 = 9 + 1 = 10
c = 10 ≈ 3.16
b 1
=
a 3
1
B = tan −1   ≈ 18.4º
3
tan B =
B = 180º − A − C ≈ 180º − 100º − 23.2º = 56.8º
A = 90º − B ≈ 90º −18.4º = 71.6º
sin B sin A
=
b
a
sin 56.8º sin100º
=
b
5
5sin 56.8º
≈ 4.25
b=
sin100º
15. A = 50º , B = 30º , a = 1
C = 180º − A − B = 180º − 50º − 30º = 100º
sin A sin B
=
a
b
sin 50º sin 30º
=
1
b
1sin 30º
≈ 0.65
b=
sin 50º
18. a = 2, c = 5, A = 60º
sin C sin A
=
c
a
sin C sin 60º
=
5
2
5sin 60º
sin C =
≈ 2.1651
2
No angle C exists for which sin C > 1 .
Therefore, there is no triangle with the given
measurements.
sin C sin A
=
c
a
sin100º sin 50º
=
c
1
1sin100º
≈ 1.29
c=
sin 50º
16. A = 10º , C = 40º , c = 2
19. a = 3, c = 1, C = 110º
sin C sin A
=
c
a
sin 40º sin10º
=
2
a
2sin10º
≈ 0.54
a=
sin 40º
sin C sin A
=
c
a
sin110º sin A
=
1
3
3sin110º
sin A =
≈ 2.8191
1
No angle A exists for which sin A > 1 . Thus,
there is no triangle with the given measurements.
842
Chapter 8 Review Exercises
20. a = 3, c = 1, C = 20º
23. a = 2, b = 3, c = 1
sin C sin A
=
c
a
sin 20º sin A
=
1
3
3sin 20º
sin A =
≈ 1.0260
1
There is no angle A for which sin A > 1 . Thus,
there is no triangle with the given measurements.
a 2 = b 2 + c 2 − 2bc cos A
cos A =
A = cos −1 1 = 0º
No triangle exists with an angle of 0˚.
24. a = 10, b = 7, c = 8
a 2 = b 2 + c 2 − 2bc cos A
21. a = 3, c = 1, B = 100º
cos A =
b = a + c − 2ac cos B
2
2
2
= 32 + 12 − 2 ⋅ 3 ⋅1cos100º
b 2 + c 2 − a 2 72 + 82 − 102 13
=
=
2bc
2(7)(8)
112
 13 
A = cos −1 
 ≈ 83.3º
 112 
= 10 − 6 cos100º
b = 10 − 6 cos100º ≈ 3.32
b 2 = a 2 + c 2 − 2ac cos B
a 2 = b 2 + c 2 − 2bc cos A
cos A =
b 2 + c 2 − a 2 32 + 12 − 22
=
=1
2bc
2(3)(1)
cos B =
b 2 + c 2 − a 2 3.322 + 12 − 32 3.0224
=
=
2bc
2(3.32)(1)
6.64
a 2 + c 2 − b2 102 + 82 − 7 2 115
=
=
2ac
2(10)(8)
160
 115 
B = cos −1 
 ≈ 44.0º
 160 
C = 180º − A − B ≈ 180º − 83.3º − 44.0º ≈ 52.7º
 3.0224 
A = cos −1 
 ≈ 62.9º
 6.64 
C = 180º − A − B ≈ 180º −62.9º −100º = 17.1º
25. a = 1, b = 3, C = 40º
c 2 = a 2 + b 2 − 2ab cos C
22. a = 3, b = 5, B = 80º
= 12 + 32 − 2 ⋅1⋅ 3cos 40º
= 10 − 6 cos 40º
sin A sin B
=
a
b
sin A sin 80º
=
3
5
3sin 80º
≈ 0.5909
sin A =
5
 3sin 80º 
A = sin −1 

5


A ≈ 36.2º or A ≈ 143.8º
The value 143.8º is discarded because
A + B > 180º . Thus, A ≈ 36.2º .
c = 10 − 6 cos 40º ≈ 2.32
a 2 = b 2 + c 2 − 2bc cos A
cos A =
b 2 + c 2 − a 2 32 + 2.322 − 12 13.3824
=
=
2bc
2(3)(2.32)
13.92
 13.3824 
A = cos−1 
 ≈ 16.0º
 13.92 
B = 180º − A − C ≈ 180º −16.0º −40º = 124.0º
26. a = 4, b = 1, C = 100º
C = 180º − A − B ≈ 180º − 36.2º − 80º = 63.8º
c 2 = a 2 + b 2 − 2ab cos C
sin C sin B
=
c
b
sin 63.8º sin 80º
=
5
c
5sin 63.8º
c=
≈ 4.56
sin 80º
c 2 = 42 + 12 − 2 ⋅ 4 ⋅1cos100º = 17 − 8cos100º
c = 17 − 8cos100º ≈ 4.29
a 2 = b 2 + c 2 − 2bc cos A
cos A =
b 2 + c 2 − a 2 12 + 4.292 − 42 3.4041
=
=
2bc
2(1)(4.29)
8.58
 3.4041 
A = cos −1 
 ≈ 66.6º
 8.58 
B = 180º − A − C ≈ 180º −66.6º −100º = 13.4º
843
Chapter 8: Applications of Trigonometric Functions
27. a = 5, b = 3, A = 80º
sin A sin C2
=
a
c2
sin B sin A
=
b
a
sin B sin 80º
=
3
5
3sin 80º
≈ 0.5909
sin B =
5
 3sin 80º 
B = sin −1 

5


B ≈ 36.2° or B ≈ 143.8°
The value 143.8° is discarded because
A + B > 180º . Thus, B ≈ 36.2° .
sin 20º sin10.9º
=
c2
2
c2 =
2sin10.9º
≈ 1.11
sin 20º
Two triangles: B1 ≈ 30.9º , C1 ≈ 129.1º , c1 ≈ 4.54
or B2 ≈ 149.1º , C2 ≈ 10.9º , c2 ≈ 1.11
1
4
, c=
2
3
a 2 = b 2 + c 2 − 2bc cos A
29. a = 1, b =
C = 180º − A − B ≈ 180º − 80º − 36.2º = 63.8º
2
2
 1  +  4  − 12
   
27
b2 + c2 − a 2  2   3 
=
=
cos A =
2bc
37
1
4

2  
 
 2  3 
 27 
A = cos −1   ≈ 39.6º
 37 
sin C sin A
=
c
a
sin 63.8º sin 80º
=
5
c
5sin 63.8º
c=
≈ 4.56
sin 80º
b 2 = a 2 + c 2 − 2ac cos B
28. a = 2, b = 3, A = 20º
2
2
4
1
12 +   −  
3 2
91
a 2 + c 2 − b2

=
=
cos B =
2ac
96
4

2(1)  
3
 91 
B = cos−1   ≈ 18.6º
 96 
sin B sin A
=
b
a
sin B sin 20º
=
3
2
3sin 20º
≈ 0.5130
sin B =
2
 3sin 20º 
B = sin −1 

2


B1 = 30.9º or B2 =149.1º
For both values, A + B < 180º . Therefore, there
are two triangles.
C = 180º − A − B ≈ 180º − 39.6º − 18.6º ≈ 121.8º
30. a = 3, b = 2, c = 2
a 2 = b 2 + c 2 − 2bc cos A
cos A =
C1 = 180º − A − B1 ≈ 180º − 20º − 30.9º ≈ 129.1º
1
b 2 + c 2 − a 2 22 + 22 − 32
=
=−
2bc
2(2)(2)
8
 1
A = cos −1  −  ≈ 97.2º
 8
sin A sin C1
=
a
c1
sin 20º sin129.1º
=
c1
2
b 2 = a 2 + c 2 − 2ac cos B
cos B =
2sin129.1º
≈ 4.54
c1 =
sin 20º
a 2 + c 2 − b2 32 + 22 − 22 9
=
=
2ac
2(3)(2)
12
9
B = cos −1   ≈ 41.4º
 12 
C = 180º − A − B ≈ 180º − 97.2º − 41.4º ≈ 41.4º
C2 = 180º − A − B2 ≈ 180º − 20º − 149.1º ≈ 10.9º
844
Chapter 8 Review Exercises
31. a = 3, A = 10º , b = 4
33. c = 5, b = 4 , A = 70º
a 2 = b 2 + c 2 − 2bc cos A
sin B sin A
=
b
a
sin B sin10º
=
4
3
4sin10º
sin B =
≈ 0.2315
3
 4sin10º 
B = sin −1 

3


B1 ≈ 13.4º or B2 ≈ 166.6º
For both values, A + B < 180º . Therefore, there
are two triangles.
a 2 = 42 + 52 − 2 ⋅ 4 ⋅ 5cos 70º = 41 − 40cos 70º
a = 41 − 40 cos 70º ≈ 5.23
c 2 = a 2 + b 2 − 2ab cos C
a 2 + b2 − c 2
2ab
5.232 + 42 − 52 18.3529
=
=
2(5.23)(4)
41.48
cos C =
 18.3529 
C = cos −1 
 ≈ 64.0º
 41.48 
B = 180º − A − C ≈ 180º − 70º − 64.0º ≈ 46.0º
C1 = 180º − A − B1 ≈ 180º − 10º − 13.4º ≈ 156.6º
sin A sin C1
=
a
c1
34. a = 1, b = 2, C = 60º
c 2 = a 2 + b 2 − 2ab cos C
sin10º sin156.6º
=
3
c1
c1 =
= 12 + 22 − 2 ⋅1⋅ 2 cos 60º = 3
3sin156.6º
≈ 6.86
sin10º
c = 3 ≈ 1.73
a 2 = b 2 + c 2 − 2bc cos A
C2 = 180º − A − B2 = 180º − 10º − 166.6º ≈ 3.4º
6
3
b 2 + c 2 − a 2 22 + ( 3 ) − 12
cos A =
=
=
=
2bc
2
4 3
2(2) ( 3 )
 3
A = cos −1 
 = 30º
 2 
B = 180º − A − C = 180º − 30º − 60º = 90º
2
sin A sin C2
=
a
c2
sin10º sin 3.4º
=
3
c2
c2 =
3sin 3.4º
≈ 1.02
sin10º
35. a = 2, b = 3, C = 40º
Two triangles: B1 ≈ 13.4º , C1 ≈ 156.6º , c1 ≈ 6.86
K=
or B2 ≈ 166.6º , C2 ≈ 3.4º , c2 ≈ 1.02
1
1
ab sin C = (2)(3) sin 40º ≈ 1.93
2
2
36. b = 5, c = 5, A = 20º
32. a = 4, A = 20º , B = 100º
C = 180º − A − B = 180º − 20º − 100º = 60º
1
1
K = bc sin A = (5)(5)sin 20º ≈ 4.28
2
2
sin A sin B
=
a
b
sin 20º sin100º
=
4
b
4sin100º
b=
≈ 11.52
sin 20º
37. b = 4, c = 10, A = 70º
1
1
K = bc sin A = (4)(10)sin 70º ≈ 18.79
2
2
38. a = 2, b = 1, C = 100º
K=
sin C sin A
=
c
a
sin 60º sin 20º
=
c
4
4sin 60º
c=
≈ 10.13
sin 20º
1
1
ab sin C = (2)(1)sin100º ≈ 0.98
2
2
39. a = 4, b = 3, c = 5
1
1
(a + b + c) = (4 + 3 + 5) = 6
2
2
K = s ( s − a)( s − b)( s − c)
s=
=
( 6 )( 2 )( 3)(1) =
36 = 6
845
Chapter 8: Applications of Trigonometric Functions
46. c = 12 feet, a = 8 feet. We need to find A and
B (see figure).
40. a = 10, b = 7, c = 8
1
1
(a + b + c) = (10 + 7 + 8) = 12.5
2
2
K = s ( s − a)( s − b)( s − c)
s=
=
B
12 ft
(12.5)( 2.5 )( 5.5)( 4.5 )
A
= 773.4375 ≈ 27.81
sin A =
41. a = 4, b = 2, c = 5
1
1
s = (a + b + c) = (4 + 2 + 5) = 5.5
2
2
K = s ( s − a)( s − b)( s − c)
=
8
8
 A = sin −1   ≈ 41.8°
12
 12 
B = 180° − 90° − A = 180° − 90° − 41.8° = 48.2°
47. Let x = the distance across the river.
x
tan(25°) =
50
x = 50 tan(25°) ≈ 23.32
Thus, the distance across the river is 23.32 feet.
( 5.5 )(1.5 )( 3.5)( 0.5 )
= 14.4375
≈ 3.80
48. Let h = the height of the building.
h
tan(25°) =
80
h = 80 tan(25°) ≈ 37.30
Thus, the height of the building is 37.30 feet.
42. a = 3, b = 2, c = 2
1
1
s = (a + b + c) = (3 + 2 + 2) = 3.5
2
2
K = s ( s − a)( s − b)( s − c)
=
8 ft
( 3.5 )( 0.5 )(1.5 )(1.5)
49. Let x = the distance the boat is from shore (see
figure). Note that 1 mile = 5280 feet.
= 3.9375
≈ 1.98
1454 ft
43. A = 50º , B = 30º , a = 1
5º
C = 180º − A − B = 180º − 50º − 30º = 100º
K=
5280 ft
a 2 sin B sin C 12 sin 30º ⋅ sin100º
=
≈ 0.32
2sin A
2sin 50º
1454
x + 5280
1454
x + 5280 =
tan(5°)
1454
− 5280
x=
tan(5°)
≈ 16, 619.30 − 5280 = 11,339.30
Thus, the boat is approximately 11,339.30 feet,
11,339.30
≈ 2.15 miles, from shore.
or
5280
tan(5°) =
44. A = 10º , C = 40º , c = 3
B = 180º − A − C = 180º − 10º − 40º = 130º
K=
x
c 2 sin A sin B 32 sin10º ⋅ sin130º
=
≈ 0.93
2sin C
2sin 40º
45. To find the area of the segment, we subtract the
area of the triangle from the area of the sector.
1
1
π 

≈ 15.708 in 2
ASector = r 2θ = ⋅ 62  50 ⋅
2
2
180 

1
1
A Triangle = ab sin θ = ⋅ 6 ⋅ 6sin 50º ≈ 13.789 in 2
2
2
ASegment = 15.708 − 13.789 ≈ 1.92 in 2
846
Chapter 8 Review Exercises
52. Let θ = the inclination (grade) of the trail.
The “rise” of the trail is 4100 − 5000 = −900
feet (see figure).
50. Let x = the length of the lake, and let y = the
distance from the edge of the lake to the point on
the ground beneath the balloon (see figure).
500 ft
25º
y
900
4100
900 
θ = sin −1 
 ≈ 12.7º
 4100 
The trail is inclined about 12.7º from the lake to
the hotel.
sin θ =
500
x
500
x=
tan ( 65º )
tan ( 65º ) =
500
x+ y
500
x+ y =
tan ( 25º )
tan ( 25º ) =
53. Let h = the height of the helicopter, x = the
distance from observer A to the helicopter, and
γ = ∠AHB (see figure).
H
500
−y
tan ( 25º )
h
40º
25º
A
B
100 ft
γ = 180º − 40º − 25º = 115º
≈ 1072.25 − 233.15 = 839.10
Thus, the length of the lake is approximately
839.10 feet.
sin 40º sin115º
=
x
100
100 sin 40º
x=
≈ 70.92 feet
sin115º
51. Let x = the distance traveled by the glider
between the two sightings, and let y = the
distance from the stationary object to a point on
the ground beneath the glider at the time of the
second sighting (see figure).
h
h
≈
x 70.92
h ≈ 70.92 sin 25º ≈ 29.97 feet
The helicopter is about 29.97 feet high.
sin 25º =
40º
10º
γ
x
500
500
=
−
tan ( 25º ) tan ( 65º )
54. ∠ACB = 12º + 30º = 42º
∠ABC = 90º − 30º = 60º
∠CAB = 90º − 12º = 78º
200 ft
y
−900 ft
65º
x
x=
4100 ft
θ
2 mi
A
x
y
200
y = 200 tan(10°)
x+ y
tan(40°) =
200
x + y = 200 tan(40°)
tan(10°) =
B
P
x
b
12º
a
30º
C
a.
x = 200 tan(40°) − y
= 200 tan(40°) − 200 tan(10°)
sin 60º sin 42º
=
b
2
2sin 60º
b=
≈ 2.59 miles
sin 42º
≈ 167.82 − 35.27 = 132.55
The glider traveled 132.55 feet in 1 minute, so
the speed of the glider is 132.55 ft/min.
847
Chapter 8: Applications of Trigonometric Functions
56. a.
sin 78º sin 42º
=
a
2
2sin 78º
a=
≈ 2.92 miles
sin 42º
b.
Find the third side of the triangle to
determine the distance from the island:
x
2.59
x = 2.59 cos12º ≈ 2.53 miles
15º
55. α = 180º − 120º = 60º ; β = 180º − 115º = 65º ;
γ = 180º − 60º − 65º = 55º
D
3 mi
β
BWI
c
c 2 = a 2 + b 2 − 2ab cos C
c 2 = 722 + 2002 − 2 ⋅ 72 ⋅ 200 cos15º
= 45,184 − 28,800 cos15º
c ≈ 131.8 miles
The sailboat is about 131.8 miles from the
island.
0.25 mi
γ C
0.25 mi
E
b.
B 115º
sin 60º sin 55º
=
BC
3
3sin 60º
BC =
≈ 3.17 mi
sin 55º
Find the measure of the angle opposite the
200 side:
a 2 + c 2 − b2
cos B =
2ac
2
72 + 131.82 − 2002
17, 444.76
cos B =
=−
2(72)(131.8)
18,979.2
 17, 444.76 
B = cos −1  −
 ≈ 156.8º
 18,979.2 
The sailboat should turn through an angle of
180° − 156.8° = 23.2° to correct its course.
sin 65º sin 55º
=
AC
3
3sin 65º
AC =
≈ 3.32 mi
sin 55º
c.
BE = 3.17 − 0.25 = 2.92 mi
AD = 3.32 − 0.25 = 3.07 mi
For the isosceles triangle,
∠CDE = ∠CED =
72 mi
B
a = 72, b = 200, C = 15º
120º
Aα
200 mi
ST
cos12º =
c.
After 4 hours, the sailboat would have sailed
18(4) = 72 miles.
180º − 55º
= 62.5º
2
sin 55º sin 62.5º
=
0.25
DE
0.25sin 55º
DE =
≈ 0.23 miles
sin 62.5º
The original trip would have taken:
200
t=
≈ 11.11 hours . The actual trip takes:
18
131.8
t = 4+
≈ 11.32 hours . The trip takes
18
57. Find the third side of the triangle using the Law
of Cosines:
a = 50, b = 60, C = 80º
c 2 = a 2 + b 2 − 2ab cos C
The length of the highway is
2.92 + 3.07 + 0.23 = 6.22 miles.
= 502 + 602 − 2 ⋅ 50 ⋅ 60 cos 80º
= 6100 − 6000 cos 80º
c = 6100 − 6000 cos80º ≈ 71.12
The houses are approximately 71.12 feet apart.
848
Chapter 8 Review Exercises
60. Find angle AMB and subtract from 80˚ to obtain θ .
58. Find the lengths of the two unknown sides of the
middle triangle:
x 2 = 1002 + 1252 − 2 (100 )(125 ) cos 50º
80º
M
Α
10
= 25, 625 − 25, 000cos 50º
θ
x = 25, 625 − 25, 000 cos 50º ≈ 97.75 feet
40
y 2 = 702 + 502 − 2 ( 70 )( 50 ) cos100º
B
= 7400 − 7000 cos100º
40
=4
10
∠AMB = tan −1 4 ≈ 76.0º
θ ≈ 80º − 76.0º ≈ 4.0º
The bearing is S4.0°E .
tan ∠AMB =
y = 7400 − 7000 cos100º ≈ 92.82 feet
Find the areas of the three triangles:
1
K1 = (100)(125) sin 50º ≈ 4787.78 ft 2
2
1
K 2 = (50)(70) sin100º ≈ 1723.41 ft 2
2
1
s = (50 + 97.75 + 92.82) = 120.285
2
K3 = (120.285 )( 70.285 )( 22.535)( 27.465 )
61. Extend the tangent line until it meets a line
extended through the centers of the pulleys.
Label these extensions x and y. The distance
between the points of tangency is z. Two similar
triangles are formed.
≈ 2287.47 ft 2
The approximate area of the lake is
4787.78 + 1723.41 + 2287.47 = 8798.67 ft 2 .
z
βα
6.5
59. Construct a diagonal. Find the area of the first
triangle and the length of the diagonal:
d
y
2.5
24 + y 6.5
=
where 24 + y is the
y
2.5
hypotenuse of the larger triangle and y is the
hypotenuse of the smaller triangle. Solve for y:
6.5 y = 2.5(24 + y )
6.5 y = 60 + 2.5 y
4 y = 60
y = 15
B
40º
100 ft
K1 =
x
Thus,
20 ft
A 100º
50 ft
15
α
1
⋅ 50 ⋅100 ⋅ sin 40º ≈ 1606.969 ft 2
2
d 2 = 502 + 1002 − 2 ⋅ 50 ⋅100 cos 40º
= 12,500 − 10, 000 cos 40º
Use the Pythagorean Theorem to find x:
x 2 + 2.52 = 152
x 2 = 225 − 6.25 = 218.75
d = 12,500 − 10, 000 cos 40º ≈ 69.566914 feet
x = 218.75 ≈ 14.79
Use the Law of Sines to find B :
sin B
sin100º
=
20
69.566914
20sin100º
sin B =
69.566914
 20sin100º 
B = sin −1 
 ≈ 16.447º
 69.566914 
A ≈ 180º − 100º − 16.447º = 63.553º
Use the Pythagorean Theorem to find z:
( z + 14.79)2 + 6.52 = ( 24 + 15 )
2
( z + 14.79)2 = 1521 − 42.25 = 1478.75
z + 14.79 = 1478.75 ≈ 38.45
z ≈ 23.66
2.5
≈ 0.1667
15
2.5 
α = cos −1 
 15 
β ≈ π − 1.4033 ≈ 1.7383 radians
Find α : cos α =
Find the area of the second triangle:
1
K 2 = ⋅ 20 ⋅ 69.566914 ⋅ sin 63.553º ≈ 622.865 ft 2
2
The cost of the parcel is approximately
100(1606.969 + 622.865) = \$222,983.40 .
849
Chapter 8: Applications of Trigonometric Functions
π 
63. d = − 3cos  t 
2 
The arc length on the top of the larger pulley is
The arc length on the top of the smaller pulley is
π 
64. d = − 5cos  t 
3 
The distance between the points of tangency is
65. d = 6sin(2 t )
a. Simple harmonic
b. 6 feet
c. π seconds
The length of the belt is about:
2(11.30 + 3.51 + 23.66) = 76.94 inches.
62. Let x = the hypotenuse of the larger right triangle
in the figure below. Then 24 − x is the
hypotenuse of the smaller triangle. The two
triangles are similar.
24 − x
6.5
βα
z
x
y
d.
1
oscillation/second
π
66. d = 2 cos(4 t )
a. Simple harmonic
b. 2 feet
α
β
2.5
6.5
x
=
2.5 24 − x
2.5 x = 6.5(24 − x)
2.5 x = 156 − 6.5 x
9 x = 156
52
x=
3
20
24 − x =
3
c.
π
seconds
2
d.
2
oscillation/second
π
67. d = − 2 cos(π t )
a. Simple harmonic
b. 2 feet
c. 2 seconds
d.
6.5 3
=
52 8
3
3
α = cos−1   ≈ 67.98º
8
 
cos α =
1
oscillation/second
2
π 
68. d = − 3sin  t 
2 
a. Simple harmonic
b. 3 feet
c. 4 seconds
β = 180º − 67.98º = 112.02º
z = 6.5 tan 67.98º ≈ 16.07 in
y = 2.5 tan 67.98º ≈ 6.18 in
d.
The arc length on the top of the larger pulley is:
π 

6.5 112.02 ⋅
≈ 12.71 inches.
180 

69. a.
The arc length on the top of the smaller pulley is
π 

2.5 112.02 ⋅
≈ 4.89 inches.
180 

1
oscillation/second
4
 2π 2 (0.75) 2
 
d = −15e−0.75 t / 2(40) cos    −
  5 
4(40) 2

 4π2 0.5625 
d = −15e−0.75 t / 80 cos 
t
−
 25
6400 


b.
The distance between the points of tangency is
z + y ≈ 16.07 + 6.18 = 22.25 inches.
15
0
25
The length of the belt is about:
2(12.71 + 4.89 + 22.25) = 79.7 inches.
–15
850

t


Chapter 8 Review Exercises
70. a.
 π 2 (0.65) 2
 
d = −13e−0.65 t / 2(25) cos    −
 2
4(25) 2

 π
0.4225
d = −13e−0.65 t / 50 cos 
−
 4
2500

2

t


0

t


15
–20
13
b.
20
c.
d.
The displacement of the bob at the start of
the second oscillation is about 19.51 meters.
e.
It approaches zero, since e −0.5 t / 60 → 0 as
t→∞.
20
0
–13
71. a.
b.
c.
Damped motion with a bob of mass 20 kg
and a damping factor of 0.6 kg/sec.
15 meters leftward
73. y = 2sin x + cos ( 2 x ) ,
0 ≤ x ≤ 2π
x
74. y = 2 cos ( 2 x ) + sin ,
2
0 ≤ x ≤ 2π
15
25
0
–15
d.
e.
72. a.
b.
The displacement of the bob at the start of
the second oscillation is about 13.92 meters.
It approaches zero, since e
t→∞.
−0.6 t / 40
→ 0 as
Damped motion with a bob of mass 30 kg
and a damping factor of 0.5 kg/sec.
20 meters leftward
851
Chapter 8: Applications of Trigonometric Functions
Chapter 8 Test
4. Use the Law of Sines to find b :
a
b
=
sin A sin B
b
12
=
sin 41° sin 22°
12 ⋅ sin 22°
b=
≈ 6.85
sin 41°
C = 180° − A − B = 180° − 41° − 22° = 117°
Use the Law of Sines to find c:
a
c
=
sin A sin C
c
12
=
sin 41° sin117°
12 ⋅ sin117°
c=
≈ 16.30
sin 41°
1. opposite = 3; adjacent = 6; hypotenuse = ?
(hypotenuse)2 = 32 + 62 = 45
hypotenuse = 45 = 3 5
opp
3
=
=
hyp 3 5
6
cos θ =
=
=
hyp 3 5
opp 3 1
tan θ =
= =
sin θ =
csc θ =
1
5
2
5
⋅
⋅
5
5
5
5
=
5
5
=
2 5
5
hyp 3 5
=
= 5
opp
3
hyp 3 5
5
=
=
6
2
cot θ =
= =2
opp 3
sec θ =
5. Use the law of cosines to find A .
a 2 = b2 + c 2 − 2bc cos A
82 = (5)2 + (10)2 − 2(5)(10) cos A
64 = 25 + 100 − 100 cos A
100cos α = 61
cos A = 0.61
2. sin 40° − cos 50° = sin 40° − sin ( 90° − 50° )
= sin 40° − sin 40°
=0
A = cos −1 (0.61) ≈ 52.41°
3. Use the law of cosines to find a:
a 2 = b 2 + c 2 − 2bc cos A
Use the law of sines to find B .
a
b
=
sin A sin B
8
5
=
sin 52.41° sin B
5
sin B = ( sin 52.41° )
8
sin B ≈ 0.495
Since b is not the longest side of the triangle, we
have that B < 90° . Therefore
= (17)2 + (19) 2 − 2(17)(19) cos 52°
≈ 289 + 361 − 646(0.616)
= 252.064
a = 252.064 ≈ 15.88
Use the law of sines to find B .
a
b
=
sin A sin B
15.88
17
=
sin 52° sin B
17
sin B =
(sin 52°) ≈ 0.8436
15.88
Since b is not the longest side of the triangle, we
know that B < 90 . Therefore,
B = sin −1 (0.8436) ≈ 57.5°
B = sin −1 (0.495) ≈ 29.67°
C = 180° − A − B
= 180° − 52.41° − 29.67°
= 97.92°
6. A = 55º , C = 20º , a = 4
B = 180º − A − C = 180º − 55º − 20º = 105º
Use the law of sines to find b.
C = 180° − A − B = 180° − 52° − 57.5° = 70.5°
852
Chapter 8 Test
10. a = 8, b = 5, c = 10
sin A sin B
=
a
b
sin 55º sin105º
=
4
b
4 sin105º
b=
≈ 4.72
sin 55º
1
1
a + b + c ) = ( 8 + 5 + 10 ) = 11.5
(
2
2
K = s ( s − a)( s − b)( s − c)
s=
= 11.5(11.5 − 8)(11.5 − 5)(11.5 − 10)
= 11.5(3.5)(6.5)(1.5)
Use the law of sines to find c.
sin C sin A
=
c
a
sin 20º sin 55º
=
4
c
4sin 20º
c=
≈ 1.67
sin 55º
= 392.4375
≈ 19.81 square units
11. Let α = the angle formed by the ground and the
7. a = 3, b = 7, A = 40º
Use the law of sines to find B
sin B sin A
=
b
a
sin B sin 40º
=
7
3
7 sin 40º
sin B =
≈ 1.4998
3
There is no angle B for which sin B > 1 .
Therefore, there is no triangle with the given
measurements.
12 ft
A
Then sin A =
10.5
12
 10.5 
A = sin −1 
 ≈ 61.0°
 12 
The angle formed by the ladder and ground is
12. Let A = the angle of depression from the
balloon to the airport.
8. a = 8, b = 4, C = 70º
5 mi
A
c 2 = a 2 + b 2 − 2ab cos C
c 2 = 82 + 42 − 2 ⋅ 8 ⋅ 4 cos 70º
= 80 − 64 cos 70º
MSFC
c = 80 − 64 cos 70º ≈ 7.62
600 ft
AP
Note that 5 miles = 26,400 feet.
600
1
tan A =
=
26400 44
 1 
A = tan −1   = 1.3°
 44 
The angle of depression from the balloon to the
a 2 = b 2 + c 2 − 2bc cos A
cos A =
10.5 ft
b 2 + c 2 − a 2 42 + 7.622 − 82 10.0644
=
=
2bc
2(4)(7.62)
60.96
 10.0644 
A = cos −1 
 ≈ 80.5º
 60.96 
B = 180º − A − C ≈ 180º −80.5º −70º = 29.5º
13. We can find the area of the shaded region by
subtracting the area of the triangle from the area
of the semicircle.
Since triangle ABC is a right triangle, we can use
the Pythagorean theorem to find the length of the
third side.
a2 + b2 = c2
9. a = 8, b = 4, C = 70º
1
ab sin C
2
1
= (8)(4)sin 70° ≈ 15.04 square units
2
K=
a 2 + 6 2 = 82
a 2 = 64 − 36 = 28
a = 28 = 2 7
853
Chapter 8: Applications of Trigonometric Functions
sin A sin 40°
=
OB
AB
sin 70° sin 40°
=
5
AB
5sin 40°
AB =
≈ 3.420
sin 70°
Now, AB is the diameter of the semicircle, so the
3.420
= 1.710 .
2
1
1
2
ASemicircle = π r 2 = π (1.710 ) ≈ 4.593 sq. units
2
2
The area of the triangle is
1
1
A = bh = 2 7 ( 6 ) = 6 7 square cm .
2
2
The area of the semicircle is
2
1
1
A = π r 2 = π ( 4 ) = 8π square cm .
2
2
Therefore, the area of the shaded region is
8π − 6 7 ≈ 9.26 square centimeters .
( )
14. Begin by adding a diagonal to the diagram.
5
72°
11
1
ab sin(O )
2
1
= (5)(5)(sin 40°) ≈ 8.035 sq. units
2
ATriangle =
d
7
8
1
(5)(11)(sin 72°) ≈ 26.15 sq. units
2
By the law of cosines,
d 2 = (5)2 + (11)2 − 2(5)(11)(cos 72°)
Aupper  =
ATotal = ASemicircle + ATriangle
≈ 4.593 + 8.035 ≈ 12.63 sq. units
17. Using Heron’s formula:
5 x + 6 x + 7 x 18 x
s=
=
= 9x
2
2
K = 9 x(9 x − 5 x )(9 x − 6 x)(9 x − 7 x)
= 25 + 121 − 110(0.309)
= 112.008
d = 112.008 ≈ 10.58
= 9 x ⋅ 4 x ⋅ 3x ⋅ 2 x
(
Using Heron’s formula for the lower triangle,
7 + 8 + 10.58
s=
= 12.79
2
Alower  = 12.79(5.79)(4.79)(2.21)
)
= 216 x 4 = 6 6 x 2
Thus,
(6 6) x 2 = 54 6
x2 = 9
x=3
The sides are 15, 18, and 21.
= 783.9293
≈ 28.00 sq. units
Total Area = 26.15 + 28.00 = 54.15 sq. units
18. Since we ignore all resistive forces, this is simple
harmonic motion. Since the rest position ( t = 0 )
15. Use the law of cosines to find c:
c 2 = a 2 + b 2 − 2ab cos C
is the vertical position ( d = 0 ) , the equation will
= (4.2) 2 + (3.5) 2 − 2(4.2)(3.5) cos 32°
have the form d = a sin(ωt ) .
≈ 17.64 + 12.25 − 29.4(0.848)
= 4.9588
42°
c = 4.9588 ≈ 2.23
5 feet
a
16. Since ΔOAB is isosceles, we know that
180° − 40°
∠A = ∠B =
= 70°
2
Then,
854
Chapter 8 Cumulative Review
g ( x ) = x 2 − 3x − 4 ≥ 0 .
Now, the period is 6 seconds, so
2π
6=
ω=
ω=
x 2 − 3x − 4 ≥ 0
( x − 4 )( x + 1) ≥ 0
ω
2π
x = 4, x = −1 are the zeros.
6
π
Interval
Test Number
−∞ < x < −1
−2
−1 < x < 4
0
4< x<∞
5
3
From the diagram we see
a
= sin 42°
5
a = 5 ( sin 42° )
The domain of f ( x ) = x 2 − 3x − 4 is
 πt 
Thus, d = 5 ( sin 42° )  sin  or
3 

πt 
d ≈ 3.346 ⋅ sin   .
 3 
{x
x ≤ −1 or x ≥ 4 } .
4. y = 3sin (π x )
Amplitude:
Chapter 8 Cumulative Review
A = 3 =3
2π
Period:
T=
Phase Shift:
φ 0
= =0
ω π
3x2 + 1 = 4 x
1.
g ( x) Pos./Neg.
6
Positive
−4 Negative
6
Positive
π
=2
3x2 − 4 x + 1 = 0
( 3x − 1)( x − 1) = 0
x=
1
or x = 1
3
1 
The solution set is  ,1 .
3 
2. Center (−5, 1) ; Radius 3
 
π 
5. y = −2cos ( 2 x − π ) = −2 cos  2  x −  
2 


Amplitude: A = − 2 = 2
( x − h )2 + ( y − k )2 = r 2
( x − (−5) )2 + ( y − 1)2 = 32
( x + 5 )2 + ( y − 1)2 = 9
3.
2π
=π
2
Period:
T=
Phase Shift:
φ π
=
ω 2
f ( x ) = x 2 − 3x − 4
f will be defined provided
855
Chapter 8: Applications of Trigonometric Functions
6. tan θ = −2,
3π
< θ < 2π , so θ lies in quadrant IV.
2
f.
1
θ
2
5
a.
b.
c.
3π
< θ < 2π , so sin θ < 0 .
2
2
5
2 5
⋅
=−
sin θ = −
5
5 5
1 + cos θ
1 
=−
cos  θ  = −
2
2 
=−
7. a.
sin(2θ ) = 2 sin θ cos θ
0
cos(2θ ) = cos θ − sin θ
2
2
 5  2 5
= 
 −  −

5 
 5  
5 20
=
−
25 25
15
3
=−
=−
25
5
e.
y = ex , 0 ≤ x ≤ 4
60
 2 5  5 
= 2  −


5 

 5 
20
4
=−
=−
25
5
d.
b.
2
4
y = sin x , 0 ≤ x ≤ 4
0
4
−1.5
c.
y = e x sin x , 0 ≤ x ≤ 4
10
4
0
−50
5
1−
1 − cos θ
1 
5
=
sin  θ  =
2
2
2 
=
0
1.5
3π
< θ < 2π
2
3π 1
< θ <π
4 2
1
1 
Since θ lies in Quadrant II, sin  θ  > 0 .
2
2 
=
5
5
1+
2
5+ 5
5
=−
2
3π
< θ < 2π , so cos θ > 0
2
1
5
5
⋅
=
cos θ =
5
5 5
2
3π
< θ < 2π
2
3π 1
< θ <π
4 2
1
1 
Since θ lies in Quadrant II, cos  θ  < 0 .
2
2 
d.
y = 2 x + sin x , 0 ≤ x ≤ 4
8
5− 5
5
2
0
5− 5
10
0
4
856
5+ 5
10
Chapter 8 Cumulative Review
y=x
e.
y = ex
b.
y = x2
f.
y = ln x
c.
y= x
g.
y = sin x
h.
y = cos x
8. a.
d.
y = x3
857
Chapter 8: Applications of Trigonometric Functions
i.
y = tan x
10. 3x5 − 10 x 4 + 21x3 − 42 x 2 + 36 x − 8 = 0
Let f ( x ) = 3x5 − 10 x 4 + 21x3 − 42 x 2 + 36 x − 8 = 0
f ( x) has at most 5 real zeros.
Possible rational zeros:
p = ±1, ±2, ±4, ±8; q = ±1, ± 2, ±3;
p
1 1
2
4
8
= ±1, ± , ± , ±2, ± , ±4, ± , ±8, ±
q
2 3
3
3
3
9. a = 20, c = 15, C = 40o
sin C sin A
=
c
a
sin 40o sin A
=
15
20
20sin 40o
sin A =
15
 20sin 40o
A = sin −1 
15

Using the Bounds on Zeros Theorem:
10
8

f ( x) = 3  x5 − x 4 + 7 x3 − 14 x 2 + 12 x − 
3
3

10
8
a4 = − , a3 = 7, a2 = −14, a1 = 12, a0 = −
3
3

8
Max 1, − + 12 +
3

= Max {1, 39} = 39
1,833,832
sin C sin B1
=
b1
c
−15
sin 40o sin 81.0o
=
b1
15
sin 40
o
15
−1
3
−10
From the graph it appears that there are x1
intercepts at ,1, and 2.
3
sin C sin B2
=
b2
c
Using synthetic division with 1:
1 3 − 10 21 − 42 36 − 8
3 −7
14 − 28 8
sin 40o sin19.0o
=
b2
15
b2 ≈
3
−2,865,248
≈ 23.05
B2 = 180o − A2 − C ≈ 180o − 40o − 121.0o = 19.0o
15sin19.0o
3



The smaller of the two numbers is 15. Thus,
every zero of f lies between –15 and 15.
Graphing using the bounds: (Second graph has a
better window.)
B1 = 180o − A1 − C ≈ 180o − 40o − 59.0o = 81.0o
15sin 81.0o
10
 8
10 
1 + Max  −
, 12 , − 14 , 7 , −

3 
 3
= 1 + 14 = 15



A1 ≈ 59.0o or A2 ≈ 121.0o
For both values, A + C < 180º . Therefore, there
are two triangles.
b1 ≈
− 14 + 7 + −
3
≈ 7.60
− 7 14 − 28
8
0
Two triangles: A1 ≈ 59.0o , B1 ≈ 81.0o , b1 ≈ 23.05
Since the remainder is 0, x − 1 is a factor. The
other factor is the quotient:
3x 4 − 7 x3 + 14 x 2 − 28 x + 8 .
A2 ≈ 121.0o , B2 ≈ 19.0o , b2 ≈ 7.60 .
Using synthetic division with 2 on the quotient:
sin 40o
or
858
Chapter 8 Cumulative Review
2 3 − 7 14 − 28
8
6 −2
24 − 8
3 −1
−4
12
11. R( x) =
Since the remainder is 0, x − 2 is a factor. The
other factor is the quotient: 3x3 − x 2 + 12 x − 4 .
Domain:
(
x ≠ − 5, x ≠ 3}
2x +1 = 0
x−4 = 0
or
1
2
The y-intercept is
x=−
R(0) =
)
3x + 12 = 3 x + 4 = 3 ( x + 2i )( x − 2i ) .
x=4
2 ⋅ 02 − 7 ⋅ 0 − 4
R(− x) =
Factoring,
0 + 2 ⋅ 0 − 15
2
=
−4
4
= .
− 15 15
2(− x)2 − 7(− x) − 4
=
2x2 + 7 x − 4
(− x) + 2(− x) − 15 x 2 − 2 x − 15
is neither R ( x ) nor − R ( x) , so there is no
symmetry.
1

f ( x) = 3( x − 1) ( x − 2 )  x −  ( x + 2i )( x − 2i )
3

The real zeros are 1,2, and
{x
2 x2 − 7 x − 4 = 0
( 2 x + 1)( x − 4 ) = 0
0 12 0
2
(2 x + 1)( x − 4)
( x − 3)( x + 5)
The x-intercepts are the zeros of p( x) :
1
Since the remainder is 0, x − is a factor. The
3
other factor is the quotient:
2
=
R is in lowest terms.
1
on the quotient:
3
1
3 − 1 12 4
3
1 0 4
3
x + 2 x − 15
2
p( x) = 2 x 2 − 7 x − 4; q ( x ) = x 2 + 2 x − 15;
n = 2; m = 2
0
Using synthetic division with
2 x2 − 7 x − 4
1
. The imaginary
3
2
which
The vertical asymptotes are the zeros of q ( x) :
x 2 + 2 x − 15 = 0
( x + 5)( x − 3) = 0
zeros are −2i and 2i.
Therefore, over the complex numbers, the equation
3x5 − 10 x 4 + 21x3 − 42 x 2 + 36 x − 8 = 0 has solution
1


set  −2i, 2i, , 1, 2  .
3


x+5 = 0
or x − 3 = 0
x = −5
x=3
Since n = m , the line y = 2 is the horizontal
asymptote.
 26 
R ( x ) intersects y = 2 at  , 2  , since:
 11 
2 x2 − 7 x − 4
=2
x 2 + 2 x − 15
2 x 2 − 7 x − 4 = 2 x 2 + 2 x − 15
(
)
2 x − 7 x − 4 = 2 x + 4 x − 30
−11x = −26
26
x=
11
2
2
Graphing utility:
8
−10
8
−4
859
Chapter 8: Applications of Trigonometric Functions
Graph by hand:
14.
Test
Value
number of f
Interval
−6
( −5, −0.5 )
−1
( −0.5,3)
0
≈ 0.27
( 3, 4 )
3.5
≈ −0.94
( 4, ∞ )
5
0.55
a.
Location Point
Above
x-axis
Below
−0.3125
x-axis
( −∞, −5)
≈ 12.22
Above
x-axis
Below
x-axis
Above
x-axis
f ( x) = 4 x + 5 ; g ( x) = x 2 + 5 x − 24
f ( x) = 0
4x + 5 = 0
4 x = −5
5
x=−
4
( −6,12.22 )
( −1, −0.3125)
 5
The solution set is  −  .
 4
( 0, 0.27 )
( 3.5, −0.94 )
b.
f ( x ) = 13
4 x + 5 = 13
4x = 8
x=2
The solution set is {2} .
( 5, 0.55 )
c.
f ( x) = g ( x)
4 x + 4 = x 2 + 5 x − 24
0 = x 2 + x − 28
x=
=
12.
The solution set is 
( ) = ln (12)
x
x ln ( 3) = ln (12 )
x=
ln (12 )
ln ( 3)
−1 ± 117 −1 ± 3 13
=
2
2
 −1 − 3 13 −1 + 3 13 
,
.
2
2


3x = 12
ln 3
−1 ± 12 − 4(1)(−29)
2(1)
d.
≈ 2.26
The solution set is {2.26}.
13. log3 ( x + 8 ) + log3 x = 2
f ( x) > 0
4x + 5 > 0
4 x > −5
5
x>−
4

5
 5 
The solution set is  x x > −  or  − , ∞  .
4
 4 


log3 ( x + 8 )( x )  = 2
( x + 8 )( x ) = 32
g ( x) ≤ 0
e.
x2 + 8x = 9
x + 5 x − 24 ≤ 0
( x + 8 )( x − 3) ≤ 0
2
x2 + 8x − 9 = 0
( x + 9 )( x − 1) = 0
x = −9 or x = 1
x = −9 is extraneous because it makes the
original logarithms undefined. The solution set
is {1} .
x = −8, x = 3 are the zeros.
Interval Test number
−9
( −∞, −8 )
0
( −8,3)
4
( 3, ∞ )
The solution set is
{x
g ( x) Pos./Neg.
12
Positive
−24 Negative
12
Positive
− 8 ≤ x ≤ 3 } or [ −8,3] .
860
Chapter 8 Projects
f.
y = f ( x) = 4 x + 5
The graph of f is a line with slope 4 and yintercept 5.
 b 
y = f  −  = f (−2.5)
 2a 
= (−2.5)2 + 5(−2.5) − 24 = −30.25
The vertex is ( −2.5, −30.25) .
g.
y = g ( x) = x 2 + 5 x − 24
The graph of g is a parabola with y-intercept
−24 and x-intercepts −8 and 3. The xcoordinate of the vertex is
b
5
5
x=−
=−
= − = −2.5 .
2a
2 (1)
2
The y-coordinate of the vertex is
Chapter 8 Projects
C
Project I
C
1.
a
a b
O
B
c
Q
O
a
Q
sin a =
b
c
2.
A
CQ
, OC 2 + CQ 2 = OQ 2
OQ
For ΔOPQ , we have that
( PQ )2 = (OP )2 + (OQ )2 − 2(OQ)(OP) cos c
For ΔCPQ , we have that
P
( PQ )2 = (CQ )2 + (CP )2 − 2(CQ)(CP) cos C
Triangles ΔOCP and ΔOCQ are plane
right triangles.
3.
C
0 = (OP )2 + (OQ )2 − 2(OQ)(OP) cos c
− (CQ) 2 + (CP) 2 − 2(CQ )(CP) cos C 
2(OQ )(OP ) cos c = (OQ )2 − (CQ) 2 + (OP) 2
− (CQ )2 + 2(CQ )(CP ) cos C
b
O
sin b =
P
4.
CP
, OC 2 + CP 2 = OP 2
OP
From part a: (OC ) 2 = (OQ )2 − (CQ) 2
(OC ) 2 = (OP) 2 − (CP)2
861
Chapter 8: Applications of Trigonometric Functions
Thus,
2(OQ )(OP ) cos c = OQ 2 − CQ 2 + OP 2
2.
N
Lewiston/Clarkton
43.5°
− CQ 2 + 2(CQ)(CP) cos C
2(OQ )(OP ) cos c = OC 2 + OC 2
+ 2(CQ)(CP) cos C
45°
46.5° N
45° N
c
Lemhi
43.5°
2OC 2
2(CQ)(CP) cos C
cos c =
+
2(OQ )(OP )
2(OQ)(OP)
45°
117.0°W
113.5°W
2
(CQ)(CP) cos C
OC
cos c =
+
(OQ)(OP)
(OQ )(OP )
O = 117.0° − 113.5° = 3.5°
cos c = cos 45° cos 43.5° + sin 45° sin 43.5° cos 3.5°
cos c = 0.99875
c = 2.87°
OC 2
(CQ)(CP) cos C
+
(OQ )(OP )
(OQ)(OP)
OC OC CQ CP
=
⋅
+
⋅
⋅ cos C
OQ OP OQ OP
= cos a cos b + sin a sin b cos C
cos c =
5.
π 

s = rc = 3960  2.87° ⋅
 ≈ 198.4
°
180

It is about 198.4 miles from Lemhi to
Lewiston and Clarkston.
Project II
1.
Putting this on a circle of radius 1 in order to
apply the Law of Cosines from A:
N
3.
Great Falls
42.5°
45°
4.
47.5° N
45° N
c
They traveled 202.5 + 198.4 miles just to go
from Great Falls to Lewiston and Clarkston.
Lewiston
and
Clarkston
282 miles
42°
48.5°
45°
42.5°
0°
Great
Falls
x
y
Lemhi
0°
α
α = 180° − 48.5° − 42° = 89.5°
111.5°W
113.5°W
O = 113.5° − 111.3° = 2.2°
cos c = cos 45° cos 42.5° + sin 45° sin 42.5° cos 2.2°
cos c = 0.99870
c = 2.93°
π 

s = rc = 3960  2.93° ⋅
 ≈ 202.5
180° 

x
282
=
sin 48.5° sin 89.5°
282sin 48.5°
x=
≈ 211.2
sin 89.5°
y
282
=
sin 42° sin 89.5°
282sin 42°
y=
≈ 188.8
sin 89.5°
Using a plane triangle, they traveled
211.2 + 188.7 = 399.9 miles.
It is 202.5 miles from Great Falls to Lemhi.
The mileage by using spherical triangles and
that by using a plane triangle are relatively
close. The total mileage is basically the
same in each case. This is because compared
to the surface of the Earth, these three towns
are very close to each other and the surface
can be approximated very closely by a
plane.
862
Chapter 8 Projects
Project III
2.
f1 = sin(π t )
1
f3 = sin(3π t )
3
1
f5 = sin(5π t )
5
1
f 7 = sin(7π t )
7
1
f9 = sin(9π t )
9
a.
184 .5
ft
6°
84°
13 ft
b. Let h = the height of the tower with a lean of 6° .
ft
6°
184.5
1.
Project IV
f1 = sin(π t )
h
84°
1
f1 + f3 = sin(π t ) + sin(3π t )
3
1
1
f1 + f3 + f 5 = sin(π t ) + sin(3π t ) + sin(5π t )
3
5
f1 + f3 + f5 + f 7
1
1
1
= sin(π t ) + sin(3π t ) + sin(5π t ) + sin(7π t )
3
5
7
f1 + f3 + f5 + f 7 + f9
1
1
= sin(π t ) + sin(3π t ) + sin(5π t )
3
5
1
1
+ sin(7π t ) + sin(9π t )
7
9
h
= sin 84°
184.5
h = 184.5sin 84°
= 183.5 feet
13 ft
2
ft
3
1 84.5
ft
c. 13 ft − 16 in = 11 ft 8 in = 11
11 _23 ft
11.67
184.5
 11.67 
α = sin −1 
 ≈ 4°
 184.5 
d. sin α =
3. If one graphs each of these functions, one
observes that with each iteration, the function
becomes more square.
e.
1
4. f1 + f13 + f3 = sin(π t ) + cos(2π t ) + sin(3π t )
2
By adding in the cosine term, the curve does not
become as flat. The waves at the “tops” and the
“bottoms” become deeper.
h
= sin 86°
184.5
h = 184.5sin 86° ≈ 184.1 ft
f. The angles are relatively small and for part (d),
where 4° was acquired, it was arrived at by
rounding.
863
Chapter 8: Applications of Trigonometric Functions
Project V
Project VI
highest
point
a.
Mountain
b.
b.
50 yd
d
x
d
410
=
sin147.5° sin 22.5°
410sin147.5°
d=
sin 22.5°
d = 575.7 feet
70 yd
40°
Palm Tree
sin α sin 40°
=
70
50
70sin 40°
sin α =
≈ 0.8999
50
 70sin 40° 
α = sin −1 
 ≈ 64°
50


nc e
ista
d
l
α
im a
β−
min
l
d=
α 180−β β
x
L
c.
y = height
147.5° 32.5°
410’
α
l
410
=
sin10° sin 22.5°
410sin10°
l=
sin 22.5°
l = 186.0 feet
y = height
d
L
=
sin(180° − β ) sin( β − α )
L sin(180° − β )
d=
sin( β − α )
c. α = 180° − 64° = 116°
Mountain
50 yd
nc e
ista
d
l
a 22.5°
in im
l
=m
10°
Rock
l
L
=
sin α sin( β − α )
L sin α
l=
sin( β − α )
Rock
β
116°
x
fence
home plate
α
70 yd
d.
40°
Palm Tree
The third angle (i.e., at the rock) is 24° .
x
70
=
sin 24° sin116°
70sin 24°
x=
≈ 32
sin116°
The treasure is about 32 yards away from the
palm tree.
d
L
=
sin(180° − β ) sin( β − α )
L sin(180° − β )
d=
sin( β − α )
L sin(180°) cos(− β ) + cos180° sin(− β )
d=
sin( β − α )
L (0) cos( β ) + (−1) ( − sin( β ) ) 
sin( β − α )
L sin( β )
d=
sin( β − α )
d=
y
l
y = l sin β
sin β =
y=
L sin α sin β
sin( β − α )
864