48 Chapter 2 Right Triangle Trigonometry 2.1 Trigonometric Functions of an Acute Angle 2.1 Practice Exercises 1. hypotenuse 5 csc opposite 3 hypotenuse 5 sec adjacent 4 adjacent 4 cot opposite 3 hypotenuse 5 csc opposite 4 hypotenuse 5 sec adjacent 3 adjacent 3 cot opposite 4 A A A B B B = = = = = = = = = = = = 2. a. ( ) cos 43 sin 90 43 sin 47 °= °− °= ° b. ( ) cot 70 tan 90 70 tan 20 °= °− °= ° c. ( ) sec 54 csc 90 54 csc 36 °= °− °= ° 3. Because the cosine and sine are cofunctions, cos A = sin B if A and B are acute angles and A + B = 90°. Thus, ( ) ( ) cos 40 sin 40 90 2 50 25 θ θ θ θ θ θ + °= ⇒ + °+ = °⇒ = °⇒ = ° 4. First use the Pythagorean theorem to find b. 2 2 2 2 2 2 2 1 5 24 24 2 6 a b c b b b + = ⇒ + = ⇒ = ⇒ = = Now use a = 1 = opposite, 2 6 b = = adjacent, and c = 13 = hypotenuse. 1 sin csc 5 5 2 6 5 5 6 cos sec 5 12 2 6 1 6 tan 12 2 6 cot 2 6 a c A A c a b a A A c c a A b b A a = = = = = = = = = = = = = = 5. The distance that the boat is from the dock is represented by the side opposite the 60° angle. We are given the adjacent side, so we can use the tangent function. Let x = the length of the opposite side. tan 60 5 tan 60 53 8.7 5 x x °= ⇒ = °= ≈ The boat is approximately 8.7 feet from the dock. 6. a. b. 7. a. b.
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48
Chapter 2 Right Triangle Trigonometry
2.1 Trigonometric Functions of an Acute Angle
2.1 Practice Exercises
1.
hypotenuse 5csc
opposite 3hypotenuse 5
secadjacent 4
adjacent 4cot
opposite 3hypotenuse 5
cscopposite 4
hypotenuse 5sec
adjacent 3adjacent 3
cotopposite 4
A
A
A
B
B
B
= =
= =
= =
= =
= =
= =
2. a. ( )cos 43 sin 90 43 sin 47° = ° − ° = °
b. ( )cot 70 tan 90 70 tan 20° = ° − ° = °
c. ( )sec54 csc 90 54 csc36° = ° − ° = °
3. Because the cosine and sine are cofunctions, cos A = sin B if A and B are acute angles and A + B = 90°. Thus,
( ) ( )cos 40 sin 40 902 50 25
θ θ θ θθ θ
+ ° = ⇒ + ° + = °⇒= °⇒ = °
4.
First use the Pythagorean theorem to find b. 2 2 2 2 2 2 21 5 24
24 2 6
a b c b b
b
+ = ⇒ + = ⇒ = ⇒= =
Now use a = 1 = opposite, 2 6b = = adjacent, and c = 13 = hypotenuse.
1sin csc 5
52 6 5 5 6
cos sec5 122 61 6
tan122 6
cot 2 6
a cA A
c ab a
A Ac ca
Abb
Aa
= = = =
= = = = =
= = =
= =
5.
The distance that the boat is from the dock is represented by the side opposite the 60° angle. We are given the adjacent side, so we can use the tangent function. Let x = the length of the opposite side.
Section 2.1 Trigonometric Functions of an Acute Angle 53
57.
sin 40 30sin 40 19.30
BCBC° = ⇒ = ° ≈
The wire is attached about 19 feet up the flagpole.
58.
100 100
tan 40 119.tan 40
BCBC
° = ⇒ = ≈°
The tree was about 119 feet tall.
59.
35
tan tan 1.75 60 .20
θ θ θ= ⇒ = ⇒ ≈ °
The angle formed by the light beam and a vertical line through the spotlight is approximately 60°.
60.
40
tan tan 2.2222 66 .18
θ θ θ= ⇒ ≈ ⇒ ≈ °
The angle formed by the camera’s line of sight and a vertical line is approximately 66°.
2.1 C Exercises: Beyond the Basics
61.
In a cube, the length of each edge is equal, so
AB = BC = CD. In triangle ABC, angle B is a
right angle. Therefore, 2.AC AB= In triangle
ACD, 1 2
tan .22 2
DC AB
AC ABθ = = = =
62.
ABEF is a square, so AB = BE = EF = FA. We
are told that the length of the rectangle is twice the length of a side of the square, so AC = 2AB. The width of the rectangle is the same as the length of a side of the square, so CD = AB.
1tan 26.6
2 2
CD AB
AC ABθ θ= = = ⇒ ≈ °
63.
Triangle ABD is a 30°-60°-90° triangle, so
a = 2 and 2 3.b = Angle DBA measures 60°, so angle DBC measures 30°, and triangle BCD is also a 30°-60°-90° triangle. Since a = 2,
Since AC = AC, ABC ADC≅△ △ by hypotenuse-leg. Therefore,
30DAC BAC∠ = ∠ = ° since corresponding parts of congruent triangles are equal. Therefore, triangle ABC is a 30°-60°-90° triangle, and AC = 2BC = 4 units.
66.
We want sin and cos , so 1 unit.a b cθ θ= = =
67.
In a triangle the length of side opposite the
smaller angle is shorter than the side opposite the larger angle. So, b > a. We have
sin sina
a cc
α α= ⇒ = and sinb
cβ = ⇒
sin .b c β= Therefore, b a> ⇒
sin sin sin sin .c cβ α β α> ⇒ > Similarly,
cos cosb
b cc
α α= ⇒ = and
cos cos .a
a cc
β β= ⇒ = Therefore, b a> ⇒
cos cos cos cos .c cα β α β> ⇒ >
68.
We want tan , so 1 unit.a bθ= =
69.
In a triangle the length of side opposite the
smaller angle is shorter than the side opposite the larger angle. So, b > a. We have
1. The longest golf drive in competition was made by Mike Austin in 1974, and it was 515 yards. The range of values that would be reported as 515 yards is between 514.5 yards and 515.5 yards.
2. When solving a right triangle with side lengths given to three significant digits, the angles found should be given to the nearest 10 minutes or tenth of a degree.
3. If one acute angle in a right triangle is 26.4°, the other acute angle measures 63.6°.
4. If the hypotenuse of a right triangle measures 10 feet and one side measures 8 feet, then the remaining side measures 6 feet.
5. False. There is no uncertainty, so we don’t have to be concerned with rounding. (See page 67 in the text.)
6. True
7. A = 34°, c = 15 in. 90 34 56
sin sin 3415
15sin 34 8.4 in.
cos cos 3415
15cos34 12 in.
Ba a
Ac
ab b
Ac
b
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= ° ≈
8. B = 63°, c = 18 in. 90 63 27
sin sin 2718
18sin 27 8.2 in.
cos cos 2718
18cos 27 16 in.
Aa a
Ac
ab b
Ac
b
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= ° ≈
9. B = 52.37°, b = 11.41 90 52.37 37.63
tan tan 37.6311.41
11.41tan 37.63 8.796
11.41cos cos37.63
11.4114.41 in.
cos 37.63
A
a aA
ba
bA
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= š
10. A = 47.38°, a = 29.14 90 47.38 42.62B = ° − ° = °
29.14tan tan 47.38
29.1426.81
tan 47.3829.14
sin sin 47.38
29.1439.60
sin 47.38
aA
b b
b
aA
c c
c
= ⇒ ° = ⇒
= š
= ⇒ ° = ⇒
= š
11. a = 15.47, b = 33.54 2 2 2 2 2
2 2
1
15.47 33.54
15.47 33.54 36.9415.47
tan33.54
15.47tan 24.76
33.5490 24.76 65.24
c a b
ca
Ab
A
B
−
= + = + ⇒
= + ≈
= = ⇒
⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °
12. a = 31.29, b = 21.16 2 2 2 2 2
2 2
1
31.29 21.16
31.29 21.16 37.7731.29
tan21.16
31.29tan 55.93
21.1690 55.93 34.07
c a b
ca
Ab
A
B
−
= + = + ⇒
= + ≈
= = ⇒
⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °
Use the triangle below to help you identify the opposite and adjacent legs.
In exercises 15–18, one of the given measurements has three significant digits, while the other has four. The answer should have three significant digits.
15. B = 32.6°, c = 64.21 ft 90 32.6 57.4
sin sin 32.664.21
64.21sin 32.6 34.6 ft
cos cos 32.664.21
64.21cos 32.6 54.1 ft
Ab b
Bc
ba a
Bc
a
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= ° ≈
16. B = 37.6°, c = 14.42 ft 90 37.6 52.4
sin sin 37.614.42
14.42sin 37.6 8.80 ft
cos cos 37.614.42
14.42cos 37.6 11.4 ft
Ab b
Bc
ba a
Bc
a
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= ° ≈
17. A = 62.92°, b = 14.7 ft 90 62.92 27.08
tan tan 62.9214.7
14.7 tan 62.92 28.8 ft
14.7cos cos 62.92
14.732.3 ft
cos 62.92
B
a aA
ba
bA
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= š
18. A = 41.12°, b = 27.4 ft 90 41.12 48.88
tan tan 41.1227.4
27.4 tan 41.12 23.9 ft
27.4cos cos 41.12
27.436.4 ft
cos 41.12
B
a aA
ba
bA
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= š
19. B = 28.47°, a = 5.243 m 90 28.47 61.53
tan tan 28.475.243
5.243 tan 28.47 2.843 m
5.243cos cos 28.47
5.2435.964 m
cos 28.47
A
b bB
ab
aB
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= š
20. B = 71.43°, a = 22.14 m 90 71.43 18.57
tan tan 71.4322.14
22.14 tan 71.43 65.90 m
22.14cos cos 71.43
22.1469.52 m
cos 71.43
A
b bB
ab
aB
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= ° ≈
= ⇒ ° = ⇒
= š
21. A = 71.37°, a = 42.66 cm 90 71.37 18.63
42.66tan tan 71.37
42.6614.38 cm
tan 71.3742.66
sin sin 71.37
42.6645.02 ft
sin 71.37
B
aA
b b
b
aA
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= š
= ⇒ ° = ⇒
= š
22. B = 61.42°, b = 27.41 cm 90 61.42 28.58
27.41tan tan 61.42
27.4114.93 cm
tan 61.4227.41
sin sin 61.42
27.4131.21 cm
sin 61.42
A
bB
a a
a
bB
c c
c
= ° − ° = °
= ⇒ ° = ⇒
= š
= ⇒ ° = ⇒
= š
23. a = 15.6 cm, b = 12.4 cm 2 2 2 2
1
15.6 12.4 19.9 cm15.6
tan tan12.4
15.6tan 51.5
12.490 51.5 38.5
c a ba
A Ab
A
B
−
= + = + ≈
= ⇒ = ⇒
⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °
24. a = 210 cm, b = 514 cm Note that there are only two significant digits in 210, so there should be two significant digits in the answers.
In exercises 25–27, one of the given measurements has three significant digits, while the other has four. The answer should have three significant digits.
25. a = 5.71 m, b = 18.39 m 2 2 2 2
1
5.71 18.39 19.3 m5.71
tan tan18.39
5.71tan 17.2
18.3990 17.2 72.8
c a ba
A Ab
A
B
−
= + = + ≈
= ⇒ = ⇒
⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °
26. b = 60.4 m, c = 192.8 m 2 2 2 2 2 2
2 2 2
2 2
1
192.8 60.4
192.8 60.4
192.8 60.4 183 m60.4
cos cos192.8
60.4cos 71.7
192.890 71.7 18.3
c a b a
a
ab
A Ac
A
B
−
= + ⇒ = + ⇒= − ⇒
= − ≈
= ⇒ = ⇒
⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °
27. 34 7 , 5.278 ft
90 34 7 89 60 34 7 55 53
sin sin 34 75.278
5.278sin 34 7 2.960 ft
cos cos 34 75.278
5.278cos 34 7 4.370 ft
A c
B
a aA
ca
b bA
cb
= ° =′= ° − ° = ° − ° = °′ ′ ′ ′
= ⇒ ° = ⇒′
= ° ≈′
= ⇒ ° = ⇒′
= ° ≈′
28. 75 26 , 12.38 ft
90 75 26 89 60 75 26 14 34
sin sin 75 2612.38
12.38sin 75 26 11.98 ft
cos cos 75 2612.38
12.38cos 75 26 3.114 ft
A c
B
a aA
ca
b bA
cb
= ° =′= ° − ° = ° − ° = °′ ′ ′ ′
= ⇒ ° = ⇒′
= ° ≈′
= ⇒ ° = ⇒′
= ° ≈′
Use the figure below for exercises 29–36.
29. 37 , 49 , 17A BDC a= ° ∠ = ° =
In right triangle BDC, tanh
BDCx
∠ = ⇒
tan 49 tan 49 .h
h xx
° = ⇒ = °
In right triangle ABC, tanh
Aa x
∠ = ⇒+
( )tan 37 17 tan 37 .17
hh x
x° = ⇒ = + °
+ Now
equate the two expressions for h, and solve for x. ( )tan 49 17 tan 37x x° = + °⇒
( )
tan 49 tan 37 17 tan 37
tan 49 tan 37 17 tan 37
tan 49 tan 37 17 tan 37
17 tan 3732
tan 49 tan 37tan 49 32 tan 49 37
x x
x x
x
x
h x
° = ° + °⇒° − ° = °⇒° − ° = °⇒
°= ≈° − °
= ° = ° ≈
30. 21 , 53 , 8A BDC a= ° ∠ = ° =
In right triangle BDC, tanh
BDCx
∠ = ⇒
tan 53 tan 53 .h
h xx
° = ⇒ = °
In right triangle ABC, tanh
Aa x
∠ = ⇒+
( )tan 21 8 tan 21 .8
hh x
x° = ⇒ = + °
+ Now
equate the two expressions for h, and solve for x. ( )tan 53 8 tan 21x x° = + °⇒
62. Find the length of the third side using the Pythagorean theorem. Then, use the appropriate trigonometric function (based on the given sides) to find the measure of one of the acute angles. Finally, subtract the measure of this angle from 90° to find the measure of the other acute angle.
63. A right triangle formed given the measures of the two acute angles is not unique. Therefore, the length of the sides cannot be determined. Only the ratios of the sides can be determined.
Section 2.3 Additional Applications of Right Triangles 65
5.
( )
sin 87.14.7
sin 87.1 4.7 sin 87.1
4.7 sin 87.1 sin 87.1
4.7 sin 87.1 1 sin 87.1
4.7 sin 87.13665 mi
1 sin 87.1
r
rr r
r r
r
r
° =+
= ° + °° = − °° = − °
°= ≈− °
2.3 A Exercises: Basic Skills and Concepts
1. A bearing is the measure of a(n) acute angle from due north or due south.
2. The bearing that indicates the direction 35° northwest of a given location is written N 35° W.
3. If you walk at a steady rate for 15 minutes at a bearing of N 45° E and turn and walk 15 more minutes (at the same rate) at a bearing of N 45° W, your final position will be due north of your starting position.
4. Janet rides her bike at a steady rate for ten minutes at a bearing of S 30° E, then turns and rides her bike due north for ten more minutes (at the same rate). To return to her starting point from the second location, she should ride at a bearing of S 75° W.
5. True
6. False. If you turn 120° clockwise from due north, then the bearing is S 60° E.
Section 2.3 Additional Applications of Right Triangles 67
(continued from page 66)
76.6 13.4 90CBF FBA∠ +∠ = ° + ° = ° Since ABC△ is a right triangle, we can use the Pythagorean theorem to find AC.
2 26 10 11.7AC = + ≈ The hiker is about 11.7 miles from his starting point.
23.
The first oil rig is located at O, the second oil
rig is located at B, and the boat is located at A. Since the axes are parallel,
20.6 .OAB CBA∠ = ∠ = ° 3 3
tan 20.6 7.98tan 20.6
OAOA
° = ⇒ = ≈°
The boat is about 7.98 miles from the first rig.
24.
The first radar station is located at O, the
second radar station is located at B, and the plane is located at A. We are seeking AB.
14 14cos 61.7 29.5
cos 61.7AB
AB° = ⇒ = ≈
°
The plane is about 29.5 miles from the northernmost station.
25.
After one hour, the first ship is located at A, 14 miles from O, and the second ship is located at B, 20 miles from O. We are seeking .ABC∠ 67.3 22.7 180 90AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,
114 14tan tan 35
20 20OBA OBA − ⎛ ⎞∠ = ⇒ ∠ = ≈ °⎜ ⎟⎝ ⎠
Since the axes are parallel, 67.3 ,OBC∠ = ° so 67.3 35 32.3ABC OBC OBA∠ = ∠ − ∠ = ° − ° = °
The bearing from the second ship to the first ship after one hour is S 32.3° E.
26.
After two hours, the first plane is located at A,
600 miles from O, and the second plane is located at B, 900 miles from O. We are seeking
.CBA∠ 53.2 36.8 180 90AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,
1
600tan
9002
tan 33.7 .3
ABO
ABO −
∠ = ⇒
⎛ ⎞∠ = ≈ °⎜ ⎟⎝ ⎠
Since the N-S axes are parallel, 36.8 .CBO∠ = °
36.8 33.7 3.1CBA CBO ABO∠ = ∠ − ∠
= ° − ° ≈ °
The bearing from the second plane to the first plane after two hours is N 3.1° W.
In exercises 27–48, we don’t evaluate intermediate results in order to minimize rounding errors.
Section 2.3 Additional Applications of Right Triangles 71
43.
From geometry, we know that the diagonals of
a rhombus are perpendicular to each other and bisect each other. They also bisect the angles of the rhombus. Thus, angle DAE measures 24° and ED = 0.8m. In triangle DAE,
0.8sin sin 24
0.82.0
sin 24
EDDAE
AD AD
AD
∠ = ⇒ ° = ⇒
= š
The length of each side is about 2.0 m.
44.
The area of the parallelogram is 20h.
sin 49.3 16sin 49.3 .16
hh° = ⇒ = ° Thus, the
area = 20 16sin 49.3 242.6 sq m.⋅ ° ≈
45.
The height of the plane = BC + CD. We don’t
have information about any of the sides in TCD . We know that TC = AB and
8.1∠ = ∠ = °m CTB m ABT . Then 150
tan 8.1 1053.955° = ⇒ ≈ABAB
. Now find
CD: tan 67.4 2531.9641053.955
° ≈ ⇒ ≈CDCD .
The height of the plane is 150 2532 2682 feet.BC CD+ ≈ + ≈
46.
The height of the building AE = 600 − BC.
tan 75.24° = BC
AC and
600tan 60.05
−° = BC
AC.
( )
( )
600So, and
tan 75.24 tan 60.05600
tan 75.24 tan 60.05tan 60.05 tan 75.24 600
tan 60.05 600 tan 75.24 tan 75.24
tan 60.05 tan 75.24 600 tan 75.24
600 tan 75.24
tan 60.05 tan 75
BC BCAC AC
BC BC
BC BC
BC BC
BC
BC
−= = ⇒° °−= ⇒
° °° = ° − ⇒° = ° − °⇒° + ° = °⇒
°=° +
411.7329.24
š
.
Then 600 411.7329 188.27≈ − ≈AE meters. The building is about 188.27 m tall.
For exercises 47 and 48, we use the result of exercise
2.2.53, .cot cot
dh
α β=
−
47.
The plane is located at D, and the tracking
stations are located at A and B. We are seeking h. 500
5402 mcot 36.9 cot 38.9
h = ≈° − °
48.
The building is located at B, the plane’s first
position at E, and the plane’s second position at D. We are seeking DE. In triangle BED,
a rhombus bisect each other and the vertex angles. We are given that the length of one diagonal is 1.8 m and that the top angle measures 50°. Therefore, angle ABD = 25° and AD = 0.9 m. The sides of a rhombus are equal, so we are seeking AB.
0.9 0.9sin 25 2
sin 25AB
AB° = ⇒ = ≈
°
The sides of the rhombus are about 2 m long.
47.
First, use the Pythagorean theorem to find the
width of the rectangle. 2 2 2 2285.6 218.5
183.9BC AC AB= − = −
≈
The area of the rectangle equals (AB)(BC). ( )( ) ( )( )218.5 183.9 40,185 sq mAB BC = ≈
48.
90 37.4 52.6
sin sin 52.6160
160sin 52.6 127
cos cos 52.6160
160cos 52.6 97
BOA
AB ABBOA
OBAB
AO AOBOA
OBAB
∠ = ° − ° = °
∠ = ⇒ ° = ⇒
= ° ≈
∠ = ⇒ ° = ⇒
= ° ≈
The plane is about 127 miles north and 97 miles west from the airport.
49.
74 in.
tan 35 106xx
° = ⇒ ≈
The shadow is about 106 inches or 8 feet 10 inches long.
50.
tan12 1885
xx° = ⇒ ≈
The tree is about 18 feet tall.
51.
850
tan 40 1013xx
° = ⇒ ≈
The rescue ship is about 1013 ft from the crippled boat.
52.
After one hour, the first ship is located at A, 12
miles from O, and the second ship is located at B, 18 miles from O. We are seeking .CBA∠ 21.3 68.7 180 90AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,
112 2tan tan 33.7 .
18 3ABO ABO − ⎛ ⎞∠ = ⇒ ∠ = ≈ °⎜ ⎟⎝ ⎠
Since the N-S axes are parallel, 68.7 .CBO∠ = °
68.7 33.7 35CBA CBO ABO∠ = ∠ − ∠
= ° − ° ≈ °
The bearing from the second ship to the first ship after two hours is N 35° W.
8. a = 17.68 cm, b = 22.19 cm, right angle C. 2 217.68 22.19 28.37 cmc = + ≈
1
17.68tan tan
22.1917.68
tan 38.5522.19
90 38.55 51.45
aA A
b
A
B
−
= ⇒ = ⇒
⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠= ° − ° = °
9.
In right triangle BCD,
tan 54 tan 54 .h
h xx
° = ⇒ = °
In right triangle ABC,
tan 32 20 tan 32 tan 32 .20
hh x
x° = ⇒ = ° + °
+
Equate the two expressions for h, and solve for x.
( )
tan 54 20 tan 32 tan 32tan 54 tan 32 20 tan 32tan 54 tan 32 20 tan 32
20 tan 3217 ft
tan 54 tan 32
x xx xx
x
° = ° + °⇒° − ° = °⇒° − ° = °⇒
°= ≈° − °
10.
A = 32°, r = 10
90 32 5810
sin sin 3210
10sin 32 sin 32 10
sin 32 10 10sin 3210 10sin 32
8.9sin 32
B
rA
r x xx
x
x
= ° − ° = °
= ⇒ ° = ⇒+ +° + ° = ⇒° = − °⇒− °= ≈
°
11.
The boat is located at A and the dock at B. We
are seeking AB.
4
sin sin14
417
sin14
BCA
AB AB
AB
= ⇒ ° = ⇒
= š
The rope is about 17 ft long.
12.
100 100
cos10 101.5 ftcos10
xx
° = ⇒ = ≈°
13.
In right triangle ABO,
90 30 60 ,AOB∠ = ° − ° = ° and 30 .OAB∠ = ° In a 30°-60°-90° triangle, the leg opposite the 30° angle is half the hypotenuse, and the leg opposite the 60° angle equals the shorter leg
times 3. Thus, OB = 70, and
70 3 121.AB = ≈ The plane flew about 70 miles west and about 121 miles north.
14.
After one hour, the first ship is located at A, 12
miles from O, and the second ship is located at B, 16 miles from O. We are seeking .CBA∠ 15 75 180 90 .AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,