4-2 Degrees and Radians
Write each decimal degree measure in DMS form and each DMS measure in decimal degree form to
the nearest thousandth.
1. 11.773°
SOLUTION:
First, convert 0. 773° into minutes and seconds.
Next, convert 0.38' into seconds.
Therefore, 11.773° can be written as 11° 46′ 23″.
2. 58.244°
SOLUTION:
First, convert 0. 244° into minutes and seconds.
Next, convert 0.64' into seconds.
Therefore, 58.244° can be written as 58° 14′ 38″.
3. 141.549°
SOLUTION:
First, convert 0. 549° into minutes and seconds.
Next, convert 0.94' into seconds.
Therefore, 141.549° can be written as 141° 32′ 56″.
4. 273.396°
SOLUTION:
First, convert 0. 396° into minutes and seconds.
eSolutions
Manual
- Powered
byinto
Cognero
Next,
convert
0.76'
seconds.
Page 1
4-2 Degrees and Radians
Therefore, 141.549° can be written as 141° 32′ 56″.
4. 273.396°
SOLUTION:
First, convert 0. 396° into minutes and seconds.
Next, convert 0.76' into seconds.
Therefore, 273.396° can be written as 273° 23′ 46″.
5. 87° 53′ 10″
SOLUTION:
Each minute is
of a degree and each second is
of a minute, so each second is
of a degree.
of a minute, so each second is
of a degree.
Therefore, 87° 53′ 10″ can be written as about 87.886°.
6. 126° 6′ 34″
SOLUTION:
Each minute is
of a degree and each second is
Therefore, 126° 6′ 34″ can be written as about 126.109°.
7. 45° 21′ 25″
SOLUTION:
Each minute is
of a degree and each second is
of a minute, so each second is
of a degree.
Therefore, 45° 21′ 25″ can be written as about 45.357°.
8. 301° 42′ 8″
eSolutions
Manual - Powered by Cognero
SOLUTION:
Each minute is
of a degree and each second is
Page 2
of a minute, so each second is
of a degree.
4-2 Degrees and Radians
Therefore, 45° 21′ 25″ can be written as about 45.357°.
8. 301° 42′ 8″
SOLUTION:
Each minute is
of a degree and each second is
of a minute, so each second is
of a degree.
Therefore, 301° 42′ 8″ can be written as about 301.702°.
9. NAVIGATION A sailing enthusiast uses a sextant, an instrument that can measure the angle between two
objects with a precision to the nearest 10 seconds, to measure the angle between his sailboat and a lighthouse. He
will be able to use this angle measure to calculate his distance from shore. If his reading is 17° 37′ 50″, what is the
measure in decimal degree form to the nearest hundredth?
SOLUTION:
Convert 17° 37′ 50″ to decimal degree form. Each minute is
so each second is
of a degree and each second is
of a minute,
of a degree.
Therefore, 17° 37′ 50″ can be written as about 17.63°.
Write each degree measure in radians as a multiple of π and each radian measure in degrees.
10. 30°
SOLUTION:
To convert a degree measure to radians, multiply by
11. 225°
SOLUTION:
To convert a degree measure to radians, multiply by
eSolutions Manual - Powered by Cognero
Page 3
4-2 Degrees and Radians
11. 225°
SOLUTION:
To convert a degree measure to radians, multiply by
12. –165°
SOLUTION:
To convert a degree measure to radians, multiply by
13. –45°
SOLUTION:
To convert a degree measure to radians, multiply by
14.
SOLUTION:
To convert a radian measure to degrees, multiply by
15.
SOLUTION:
To convert a radian measure to degrees, multiply by
eSolutions Manual - Powered by Cognero
Page 4
4-2 Degrees and Radians
15.
SOLUTION:
To convert a radian measure to degrees, multiply by
16.
SOLUTION:
To convert a radian measure to degrees, multiply by
17.
SOLUTION:
To convert a radian measure to degrees, multiply by
Identify all angles that are coterminal with the given angle. Then find and draw one positive and one
negative angle coterminal with the given angle.
18. 120°
SOLUTION:
All angles measuring
are coterminal with an
angle.
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 5
4-2 Degrees and Radians
Identify all angles that are coterminal with the given angle. Then find and draw one positive and one
negative angle coterminal with the given angle.
18. 120°
SOLUTION:
All angles measuring
are coterminal with an
angle.
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
19. –75°
SOLUTION:
All angles measuring
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 6
4-2 Degrees and Radians
19. –75°
SOLUTION:
All angles measuring
are coterminal with a
angle.
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
20. 225°
SOLUTION:
All angles measuring
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
21. –150°
Page 7
4-2 Degrees and Radians
20. 225°
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
21. –150°
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 8
4-2 Degrees and Radians
21. –150°
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
22.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 9
4-2 Degrees and Radians
22.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
23.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 10
4-2 Degrees and Radians
23.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
24.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 11
4-2 Degrees and Radians
24.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
25.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
eSolutions Manual - Powered by Cognero
Page 12
4-2 Degrees and Radians
25.
SOLUTION:
All angles measuring
are coterminal with a
angle.
Sample answer: Let n = 1 and −1.
26. GAME SHOW Sofia is spinning a wheel on a game show. There are 20 values in equal-sized spaces around the
circumference of the wheel. The value that Sofia needs to win is two spaces above the space where she starts her
spin, and the wheel must make at least one full rotation for the spin to count. Describe a spin rotation in degrees
that will give Sofia a winning result.
SOLUTION:
If the wheel is divided into 20 equal-sized spaces, then the central angle of each sector has a measure of 360° ÷ 20
or 18°. Because the value that Sofia needs to win is two spaces above the space where she starts her spin, she
needs a spin rotation of 360° + 2(18°) or 396°.
Find the length of the intercepted arc with the given central angle measure in a circle with the given
radius. Round to the nearest tenth.
eSolutions Manual - Powered by Cognero
27.
, r = 2.5 m
SOLUTION:
Page 13
SOLUTION:
4-2
If the wheel is divided into 20 equal-sized spaces, then the central angle of each sector has a measure of 360° ÷ 20
or 18°. Because
value that Sofia needs to win is two spaces above the space where she starts her spin, she
Degrees
and the
Radians
needs a spin rotation of 360° + 2(18°) or 396°.
Find the length of the intercepted arc with the given central angle measure in a circle with the given
radius. Round to the nearest tenth.
27.
, r = 2.5 m
SOLUTION:
28.
, r = 3 in.
SOLUTION:
29.
, r = 4 yd
SOLUTION:
30. 105°, r = 18.2 cm
SOLUTION:
Method 1
Convert 105° to radian measure, and then use s = rθ to find the arc length.
Substitute r = 18.2 and θ =
eSolutions Manual - Powered by Cognero
.
Page 14
4-2 Degrees and Radians
30. 105°, r = 18.2 cm
SOLUTION:
Method 1
Convert 105° to radian measure, and then use s = rθ to find the arc length.
Substitute r = 18.2 and θ =
.
Method 2
Use s =
to find the arc length.
31. 45°, r = 5 mi
SOLUTION:
Method 1
Convert 45° to radian measure, and then use s = rθ to find the arc length.
Substitute r = 5 and θ =
.
Method 2
Use s =
to find the arc length.
eSolutions Manual - Powered by Cognero
Page 15
4-2 Degrees and Radians
31. 45°, r = 5 mi
SOLUTION:
Method 1
Convert 45° to radian measure, and then use s = rθ to find the arc length.
Substitute r = 5 and θ =
.
Method 2
Use s =
to find the arc length.
32. 150°, r = 79 mm
SOLUTION:
Method 1
Convert 150° to radian measure, and then use s = rθ to find the arc length.
Substitute r = 79 and θ =
.
Method 2
Use s =
to find the arc length.
eSolutions Manual - Powered by Cognero
Page 16
4-2 Degrees and Radians
32. 150°, r = 79 mm
SOLUTION:
Method 1
Convert 150° to radian measure, and then use s = rθ to find the arc length.
Substitute r = 79 and θ =
.
Method 2
Use s =
to find the arc length.
33. AMUSEMENT PARK A carousel at an amusement park rotates 3024° per ride.
a. How far would a rider seated 13 feet from the center of the carousel travel during the ride?
b. How much farther would a second rider seated 18 feet from the center of the carousel travel during the ride
than the rider in part a?
SOLUTION:
a. In this problem, θ = 3024° and r = 13 feet. Convert 3024º to radian measure.
Substitute r = 13 and θ = 16.8π into the arc length formula s = rθ.
b. Let r1 represent the distance the first rider is from the center of the carousel and r2 represent the distance the
second rider is from the center. For the second rider, r = 18 feet.
Therefore,
the second
rider
eSolutions
Manual - Powered
by Cognero
would travel about 264 feet farther than the first rider.
Find the rotation in revolutions per minute given the angular speed and the radius given the linear
speed and the rate of rotation.
Page 17
4-2 Degrees and Radians
33. AMUSEMENT PARK A carousel at an amusement park rotates 3024° per ride.
a. How far would a rider seated 13 feet from the center of the carousel travel during the ride?
b. How much farther would a second rider seated 18 feet from the center of the carousel travel during the ride
than the rider in part a?
SOLUTION:
a. In this problem, θ = 3024° and r = 13 feet. Convert 3024º to radian measure.
Substitute r = 13 and θ = 16.8π into the arc length formula s = rθ.
b. Let r1 represent the distance the first rider is from the center of the carousel and r2 represent the distance the
second rider is from the center. For the second rider, r = 18 feet.
Therefore, the second rider would travel about 264 feet farther than the first rider.
Find the rotation in revolutions per minute given the angular speed and the radius given the linear
speed and the rate of rotation.
34.
=
SOLUTION:
The angular speed is
radians per second.
Because each revolution measures 2π radians, the angle of rotation is 40π ÷ 2π or 20 revolutions per minute.
35.
= 135π
SOLUTION:
The angular speed is 135π radians per hour.
eSolutions Manual - Powered by Cognero
Page 18
Because each revolution measures 2π radians, the angle of rotation is 2.25π ÷ 2π or 1.125 revolutions per minute.
4-2 Degrees
and
Radians
Because each
revolution
measures 2π radians, the angle of rotation is 40π ÷ 2π or 20 revolutions per minute.
35.
= 135π
SOLUTION:
The angular speed is 135π radians per hour.
Because each revolution measures 2π radians, the angle of rotation is 2.25π ÷ 2π or 1.125 revolutions per minute.
36.
= 104π
SOLUTION:
The angular speed is 104π radians per minute.
Because each revolution measures 2π radians, the angle of rotation is 104π ÷ 2π or 52 revolutions per minute.
37. v = 82.3
, 131
SOLUTION:
The linear speed is 82.3 meters per second with an angle of rotation of 131 × 2π or 262π radians per
minute.
So, the radius is about 6 m.
38. v = 144.2
, 10.9
SOLUTION:
The linear speed is 144.2 feet per minute with an angle of rotation of 10.9 × 2π or 21.8π radians per
minute.
eSolutions Manual - Powered by Cognero
Page 19
4-2 Degrees and Radians
So, the radius is about 6 m.
38. v = 144.2
, 10.9
SOLUTION:
The linear speed is 144.2 feet per minute with an angle of rotation of 10.9 × 2π or 21.8π radians per
minute.
So, the radius is about 2.1 ft.
39. v = 553
, 0.09
SOLUTION:
The linear speed is 553 inches per hour with an angle of rotation of 0.09 × 2π or 0.18π radians per
minute.
So, the radius is about 16.3 in.
40. MANUFACTURING A company manufactures several circular saws with the blade diameters and motors
speeds shown below.
a. Determine the angular and linear speeds of the blades in each saw. Round to the nearest tenth.
b. How much faster is the linear speed of the 6
-inch saw compared to the 3-inch saw?
SOLUTION:
a. Determine the angular and linear speeds of the blades in each saw. Round to the nearest tenth.
eSolutions Manual - Powered by Cognero
Page 20
a. Determine the angular and linear speeds of the blades in each saw. Round to the nearest tenth.
b. How much faster is the linear speed of the 6
-inch saw compared to the 3-inch saw?
4-2 Degrees
and Radians
SOLUTION:
a. Determine the angular and linear speeds of the blades in each saw. Round to the nearest tenth.
eSolutions
by Cognero
b. Manual
The 6 - Powered
-inch saw
is about
105832.4 – 26389.4 or 79,443 in./s faster than the 3-inch saw.
41. CARS On a stretch of interstate, a vehicle’s tires range between 646 and 840 revolutions per minute. The
Page 21
4-2 Degrees and Radians
b. The 6
-inch saw is about 105832.4 – 26389.4 or 79,443 in./s faster than the 3-inch saw.
41. CARS On a stretch of interstate, a vehicle’s tires range between 646 and 840 revolutions per minute. The
diameter of each tire is 26 inches.
a. Find the range of values for the angular speeds of the tires in radians per minute.
b. Find the range of values for the linear speeds of the tires in miles per hour.
SOLUTION:
a. Because each rotation measures 2π radians, 646 and 840 revolutions correspond to angles of rotation of 646 ×
2π or 1292π radians and 840 × 2π or 1680π radians.
Therefore, the angular speeds of the tires range from 1292π
to 1680 π
.
b. The diameter of each tire is 26 inches, so the radius of each tire is 26 ÷ 2 or 13 in.
Use dimensional analysis to convert each speed from inches per minute to miles per hour.
Therefore, the linear speeds range of the tires range from 50 mi/h to 65 mi/h.
42. TIME A wall clock has a face diameter of 8
inches. The length of the hour hand is 2.4 inches, the length of the
minute hand is 3.2 inches, and the length of the second hand is 3.4 inches.
a. Determine the angular speed in radians per hour and the linear speed in inches per hour for each hand.
b. If the linear speed of the second hand is 20 inches per minute, is the clock running fast or slow? How much time
would it gain or lose per day?
SOLUTION:
eSolutions
- Powered
by Cognero
a. Manual
The hour
hand on
a clock
rotates through an angle of 360º ÷ 12 or 30º each hour, as shown below.
Page 22
4-2
a. Determine the angular speed in radians per hour and the linear speed in inches per hour for each hand.
b. If the linear speed of the second hand is 20 inches per minute, is the clock running fast or slow? How much time
would it gain
or lose
per day?
Degrees
and
Radians
SOLUTION:
a. The hour hand on a clock rotates through an angle of 360º ÷ 12 or 30º each hour, as shown below.
Convert this measure to radians.
Find the angular speed.
Find the linear speed.
The minute hand on a clock rotates completes 1 full rotation each hour. So, the angle of rotation is 2π radians. Find
the angular speed.
Find the linear speed.
The second hand on a clock rotates completes 60 full rotations in 1 hour. So, the angle of rotation is 2π × 60 or
120π. Find the angular speed.
Find the linear speed.
eSolutions Manual - Powered by Cognero
Page 23
4-2 Degrees and Radians
Find the linear speed.
b. The linear speed of the second hand of the clock is 20 inches per minute. From part b, we know that the linear
speed should be 1281.8 inches per hour. Use dimensional analysis to convert this speed to inches per minute.
Therefore, the clock is running slow.
So, the slow second hand has a speed that is 20 ÷ 21.36 = 0.9363 or 93.63% of the speed of the normal second
hand. The slow second hand is running 60 – (60)(0.9363) or about 3.82 seconds slow each minute. Use
dimensional analysis to determine the amount of time lost in one day.
The amount of time lost in one day is 5500.8 seconds.
Use dimensional analysis to convert this time to hours.
Therefore, the amount of time lost in one day is about 1.53 hours.
Find the area of each sector.
43.
SOLUTION:
The measure of the sector’s central angle θ is 102° and the radius is 1.5 inches. Convert the central angle measure
to radians.
Use the central angle and the radius to find the area of the sector.
Therefore, the area of the sector is about 2.0 square inches.
eSolutions Manual - Powered by Cognero
Page 24
4-2 Degrees
Radians
Therefore, and
the area
of the sector is about 2.0 square inches.
44.
SOLUTION:
The measure of the sector’s central angle θ is
and the radius is 3.4 meters.
Therefore, the area of the sector is about 24.2 square meters.
45.
SOLUTION:
The measure of the sector’s central angle θ is
and the radius is 12 yards.
Therefore, the area of the sector is about 90.5 square yards.
46.
SOLUTION:
The measure of the sector’s central angle θ is 146° and the radius is 21.4 ÷ 2 or 10.7 kilometers. Convert the
central angle measure to radians.
Use the central angle and the radius to find the area of the sector.
eSolutions Manual - Powered by Cognero
Page 25
4-2 Degrees and Radians
Therefore, the area of the sector is about 90.5 square yards.
46.
SOLUTION:
The measure of the sector’s central angle θ is 146° and the radius is 21.4 ÷ 2 or 10.7 kilometers. Convert the
central angle measure to radians.
Use the central angle and the radius to find the area of the sector.
Therefore, the area of the sector is about 145.9 square kilometers.
47.
SOLUTION:
The measure of the sector’s central angle θ is 177° and the radius is 18 feet. Convert the central angle measure to
radians.
Use the central angle and the radius to find the area of the sector.
Therefore, the area of the sector is about 500.5 square feet.
48.
SOLUTION:
eSolutions
Manual - Powered by Cognero
The measure of the sector’s central angle θ is
Page 26
and the radius is 43.5 centimeters.
4-2 Degrees and Radians
Therefore, the area of the sector is about 500.5 square feet.
48.
SOLUTION:
The measure of the sector’s central angle θ is
and the radius is 43.5 centimeters.
Therefore, the area of the sector is about 247.7 square centimeters.
49. GAMES The dart board shown is divided into twenty equal sectors. If the diameter of the board is 18 inches,
what area of the board does each sector cover?
SOLUTION:
If the board is divided into 20 equal sectors, then the central angle of each sector has a measure of 360° ÷ 20 or
18°. So, the measure of a sector’s central angle is 18° and the radius is 18 ÷ 2 or 9 inches.
Convert the central angle measure to radians.
Use the central angle and the radius to find the area of a sector.
Therefore, each sector covers an area of about 12.7 square inches.
50. LAWN CARE A sprinkler waters an area that forms one third of a circle. If the stream from the sprinkler
extends 6 feet, what area of the grass does the sprinkler water?
SOLUTION:
Draw a diagram to model the situation. The radius of the circle is 6 feet and the central angle of the sector has a
measure of 360° ÷ 3 or 120°.
eSolutions Manual - Powered by Cognero
Page 27
4-2 Degrees and Radians
Therefore, each sector covers an area of about 12.7 square inches.
50. LAWN CARE A sprinkler waters an area that forms one third of a circle. If the stream from the sprinkler
extends 6 feet, what area of the grass does the sprinkler water?
SOLUTION:
Draw a diagram to model the situation. The radius of the circle is 6 feet and the central angle of the sector has a
measure of 360° ÷ 3 or 120°.
Convert the central angle measure to radians.
Use the central angle and the radius to find the area of a sector.
Therefore, the sprinkler waters about 37.7 square feet of grass.
The area of a sector of a circle and the measure of its central angle are given. Find the radius of the
circle.
51. A = 29 ft2, θ = 68°
SOLUTION:
Convert the central angle measure to radians.
Substitute the area and the central angle into the area formula for a sector to find the radius.
eSolutions Manual - Powered by Cognero
Therefore, the radius is 7 ft.
Page 28
4-2 Degrees and Radians
Therefore, the sprinkler waters about 37.7 square feet of grass.
The area of a sector of a circle and the measure of its central angle are given. Find the radius of the
circle.
51. A = 29 ft2, θ = 68°
SOLUTION:
Convert the central angle measure to radians.
Substitute the area and the central angle into the area formula for a sector to find the radius.
Therefore, the radius is 7 ft.
52. A = 808 cm2, θ = 210°
SOLUTION:
Convert the central angle measure to radians.
Substitute the area and the central angle into the area formula for a sector to find the radius.
eSolutions
Manual - Powered
by Cognero
Therefore,
the radius
is 21 cm.
53. A = 377 in2, θ =
Page 29
4-2 Degrees
Radians
Therefore, and
the radius
is 7 ft.
52. A = 808 cm2, θ = 210°
SOLUTION:
Convert the central angle measure to radians.
Substitute the area and the central angle into the area formula for a sector to find the radius.
Therefore, the radius is 21 cm.
53. A = 377 in2, θ =
SOLUTION:
Therefore, the radius is 12 in.
54. A = 75 m2, θ =
SOLUTION:
2
A = 75 m , θ =
eSolutions Manual - Powered by Cognero
Page 30
4-2 Degrees and Radians
Therefore, the radius is 12 in.
54. A = 75 m2, θ =
SOLUTION:
2
A = 75 m , θ =
Therefore, the radius is 8m.
55. Describe the radian measure between 0 and 2
of an angle θ that is in standard position with a terminal side that
lies in:
a. Quadrant I
b. Quadrant II
c. Quadrant III
d. Quadrant IV
SOLUTION:
a. An angle that is in standard position that lies in Quadrant I will have an angle measure between 0º and 90º, as
shown.
Convert 0º and 90º to radians.
Therefore, in Quadrant I, 0 < θ <
.
b. An angle that is in standard position that lies in Quadrant II will have an angle measure between 90º and 180º, as
shown.
eSolutions Manual - Powered by Cognero
Page 31
Therefore, in Quadrant I, 0 < θ <
.
b. An angleand
that is
in standard position that lies in Quadrant II will have an angle measure between 90º and 180º, as
4-2 Degrees
Radians
shown.
From part a, we know that 90º =
Therefore, in Quadrant II,
. Convert 180º to radians.
< θ < π.
c. An angle that is in standard position that lies in Quadrant III will have an angle measure between 180º and 270º,
as shown.
From part b, we know that 180º = π. Convert 270º to radians.
Therefore, in Quadrant III, π < θ <
.
d. An angle that is in standard position that lies in Quadrant IV will have an angle measure between 270º and 360º,
as shown.
From part c, we know that 270º =
eSolutions Manual - Powered by Cognero
Therefore, in Quadrant IV,
. Convert 360º to radians.
< θ < 2π.
Page 32
From part c, we know that 270º =
. Convert 360º to radians.
4-2 Degrees and Radians
Therefore, in Quadrant IV,
< θ < 2π.
56. If the terminal side of an angle that is in standard position lies on one of the axes, it is called a quadrantal angle.
Give the radian measures of four quadrantal angles.
SOLUTION:
Sample answer: Draw a diagram of four different angles in standard position with terminal sides that lie on one of
the axes. Then find each angle measure.
So, four possible quadrantal angle are 0,
, π, and
.
57. GEOGRAPHY Phoenix, Arizona, and Ogden, Utah, are located on the same line of longitude, which means that
Ogden is directly north of Phoenix. The latitude of Phoenix is 33° 26′ N, and the latitude of Ogden is 41° 12′ N. If
Earth’s radius is approximately 3963 miles, about how far apart are the two cities?
SOLUTION:
The distance between Ogden and Phoenix corresponds to the length of an intercepted arc of a circle.
Convert the latitude measures for Ogden and Phoenix to degrees.
eSolutions Manual - Powered by Cognero
Page 33
4-2 Degrees
and Radians
So, four possible
quadrantal angle are 0,
, π, and
.
57. GEOGRAPHY Phoenix, Arizona, and Ogden, Utah, are located on the same line of longitude, which means that
Ogden is directly north of Phoenix. The latitude of Phoenix is 33° 26′ N, and the latitude of Ogden is 41° 12′ N. If
Earth’s radius is approximately 3963 miles, about how far apart are the two cities?
SOLUTION:
The distance between Ogden and Phoenix corresponds to the length of an intercepted arc of a circle.
Convert the latitude measures for Ogden and Phoenix to degrees.
So, the central angle measure is 41.2º – 33.433º or 7.767º. Convert the central angle measure to radians.
Use the central angle and the radius to find the length of the intercepted arc.
Therefore, Ogden and Phoenix are about 537 miles apart.
Find the measure of angle θ in radians and degrees.
58.
SOLUTION:
So, θ = 1.8 radians, or
eSolutions Manual - Powered by Cognero
≈ 103.1º.
Page 34
Use the central angle and the radius to find the length of the intercepted arc.
4-2 Degrees and Radians
Therefore, Ogden and Phoenix are about 537 miles apart.
Find the measure of angle θ in radians and degrees.
58.
SOLUTION:
So, θ = 1.8 radians, or
≈ 103.1º.
59.
SOLUTION:
So, θ ≈ 3.3 radians, or
≈ 191º.
60.
SOLUTION:
So, θ = 5 radians, or
≈ 286.5º.
61.
SOLUTION:
eSolutions Manual - Powered by Cognero
Page 35
4-2 Degrees
and Radians
So, θ = 5 radians,
or
≈ 286.5º.
61.
SOLUTION:
So, θ = 1.5 radians, or
≈ 85.9º.
62. TRACK A standard 8-lane track has an inside width of 73 meters. Each lane is 1.22 meters wide. The curve of
the track is semicircular.
a. What is the length of the outside edge of Lane 4 in the curve?
b. How much longer is the inside edge of Lane 7 than the inside edge of Lane 3 in the curve?
SOLUTION:
a. The length of the outside edge of the curve of Lane 4 is equivalent to the arc length of a semi-circle, as shown
below. The width of each lane is 1.22 meters, so the distance from the inside edge of Lane 1 to the outside edge of
Lane 4 is 4(1.22) or 4.88 meters.
Use the arc length formula s = rθ to find the length of the outside edge of Lane 4.
eSolutions
- Powered
by Cognero
b. Manual
First, find
the length
of each
curve.
Lane 3
The distance from the inside edge of Lane 1 to the inside edge of Lane 3 is 2(1.22) or 2.44 meters.
Page 36
4-2 Degrees and Radians
Use the arc length formula s = rθ to find the length of the outside edge of Lane 4.
b. First, find the length of each curve.
Lane 3
The distance from the inside edge of Lane 1 to the inside edge of Lane 3 is 2(1.22) or 2.44 meters.
Lane 7
The distance from the inside edge of Lane 1 to the inside edge of Lane 7 is 6(1.22) or 7.32 meters.
So, the inside edge of Lane 7 is 137.66 – 122.33 or about 15.3 meters longer than the inside edge of Lane 3.
63. DRAMA A pulley with radius r is being used to remove part of the set of a play during intermission. The height
of the pulley is 12 feet.
a. If the radius of the pulley is 6 inches and it rotates 180°, how high will the object be lifted?
b. If the radius of the pulley is 4 inches and it rotates 900°, how high will the object be lifted?
SOLUTION:
a. Convert 180º to radians.
Use the arc length formula.
So, the object will be lifted about 18.8 inches.
b. Convert 900º to radians.
eSolutions Manual - Powered by Cognero
Use the arc length formula.
Page 37
4-2 Degrees and Radians
So, the inside edge of Lane 7 is 137.66 – 122.33 or about 15.3 meters longer than the inside edge of Lane 3.
63. DRAMA A pulley with radius r is being used to remove part of the set of a play during intermission. The height
of the pulley is 12 feet.
a. If the radius of the pulley is 6 inches and it rotates 180°, how high will the object be lifted?
b. If the radius of the pulley is 4 inches and it rotates 900°, how high will the object be lifted?
SOLUTION:
a. Convert 180º to radians.
Use the arc length formula.
So, the object will be lifted about 18.8 inches.
b. Convert 900º to radians.
Use the arc length formula.
So, the object will be lifted about 62.8 inches or about 5.2 feet.
64. ENGINEERING A pulley like the one in Exercise 63 is being used to lift a crate in a warehouse. Determine
which of the following scenarios could be used to lift the crate a distance of 15 feet the fastest. Explain how you
reached your conclusion.
I. The radius of the pulley is 5 inches rotating at 65 revolutions per minute.
II. The radius of the pulley is 4.5 inches rotating at 70 revolutions per minute.
III. The radius of the pulley is 6 inches rotating at 60 revolutions per minute.
SOLUTION:
Find the linear speed for each scenario.
Scenario I
Because each rotation measures 2π radians, 65 revolutions correspond to an angle of rotation of 65 × 2π or 130π
radians.
eSolutions Manual - Powered by Cognero
Page 38
Use the arc length formula.
4-2 Degrees and Radians
So, the object will be lifted about 62.8 inches or about 5.2 feet.
64. ENGINEERING A pulley like the one in Exercise 63 is being used to lift a crate in a warehouse. Determine
which of the following scenarios could be used to lift the crate a distance of 15 feet the fastest. Explain how you
reached your conclusion.
I. The radius of the pulley is 5 inches rotating at 65 revolutions per minute.
II. The radius of the pulley is 4.5 inches rotating at 70 revolutions per minute.
III. The radius of the pulley is 6 inches rotating at 60 revolutions per minute.
SOLUTION:
Find the linear speed for each scenario.
Scenario I
Because each rotation measures 2π radians, 65 revolutions correspond to an angle of rotation of 65 × 2π or 130π
radians.
Scenario II
Because each rotation measures 2π radians, 70 revolutions correspond to an angle of rotation of 70 × 2π or 140π
radians.
Scenario III
Because each rotation measures 2π radians, 60 revolutions correspond to an angle of rotation of 60 × 2π or 120π
radians.
Sample answer: The pulley has the greatest linear speed in Scenario III. Therefore, Scenario III would be the
fastest method to use to lift the crate a distance of 15 feet.
Find the area of each shaded region.
65.
SOLUTION:
First, convert each central angle measure to radians.
eSolutions Manual - Powered by Cognero
Page 39
Sample answer:
pulley has the greatest linear speed in Scenario III. Therefore, Scenario III would be the
4-2 Degrees
and The
Radians
fastest method to use to lift the crate a distance of 15 feet.
Find the area of each shaded region.
65.
SOLUTION:
First, convert each central angle measure to radians.
Find the area of each sector.
So, the total area of the shaded region is about 55.96 + 76.03 or 132 square inches.
66.
SOLUTION:
First, convert each angle measure to radians.
Find the area of a sector of the smaller circle with a radius of 8 cm and a central angle of 88º or
Find the area of a sector of the smaller circle with a radius of 8 cm and a central angle of 104º or
eSolutions Manual - Powered by Cognero
radians.
radians.
Page 40
4-2 Degrees
So, the totaland
area Radians
of the shaded region is about 55.96 + 76.03 or 132 square inches.
66.
SOLUTION:
First, convert each angle measure to radians.
Find the area of a sector of the smaller circle with a radius of 8 cm and a central angle of 88º or
Find the area of a sector of the smaller circle with a radius of 8 cm and a central angle of 104º or
radians.
radians.
Find the area of a sector of the larger circle with a radius of 12 + 8 or 20 cm and central angle of 104º or
radians.
The area of the shaded region between the edge of the smaller and larger circles is 363.03 – 58.08 or about 304.95
cm.
Therefore, the total area of the shaded regions is 304.95 + 49.15 or about 354 square centimeters.
67. CARS The speedometer shown measures the speed of a car in miles per hour.
a. If the angle between 25 mi/h and 60 mi/h is 81.1°, about how many miles per hour are represented by each
Page 41
degree?
b. If the angle of the speedometer changes by 95°, how much did the speed of the car increase?
eSolutions Manual - Powered by Cognero
The area of the shaded region between the edge of the smaller and larger circles is 363.03 – 58.08 or about 304.95
cm.
4-2 Degrees and Radians
Therefore, the total area of the shaded regions is 304.95 + 49.15 or about 354 square centimeters.
67. CARS The speedometer shown measures the speed of a car in miles per hour.
a. If the angle between 25 mi/h and 60 mi/h is 81.1°, about how many miles per hour are represented by each
degree?
b. If the angle of the speedometer changes by 95°, how much did the speed of the car increase?
SOLUTION:
a. 81.1º corresponds to 60 – 25 or 35 mi/h. Write a proportion to find the number of miles per hour that correspond
to 1º.
b. From part a, you know that 1º corresponds to about 0.43 mi/h.
Find the complement and supplement of each angle.
68.
SOLUTION:
Recall from Geometry that two angles are complementary if they have a sum of 90º and supplementary if they
have a sum of 180º.
So, the complement of
is
–
or
, and the supplement is π –
or
.
69.
SOLUTION:
Recall from Geometry that two angles are complementary if they have a sum of 90º and supplementary if they
have a sum of 180º.
eSolutions Manual - Powered by Cognero
Page 42
So, the complement of
4-2 Degrees
and Radiansis
–
or
, and the supplement is π –
or
.
69.
SOLUTION:
Recall from Geometry that two angles are complementary if they have a sum of 90º and supplementary if they
have a sum of 180º.
does not have a complement because it is greater than
or 90°. The supplement is π –
or
.
70.
SOLUTION:
Recall from Geometry that two angles are complementary if they have a sum of 90º and supplementary if they
have a sum of 180º.
So, the complement of
is
–
or
, and the supplement is π –
or
.
71.
SOLUTION:
does not have a complement or supplement because complements and supplements are not defined for
negative angles.
72. SKATEBOARDING A physics class conducted an experiment to test three different wheel sizes on a
skateboard with constant angular speed.
a. Write an equation for the linear speed of the skateboard in terms of the radius and angular speed. Explain your
reasoning.
b. Using the equation you wrote in part a, predict the linear speed in meters per second of a skateboard with an
angular speed of 3 revolutions per second for wheel diameters of 52, 56, and 60 millimeters.
c. Based on your results in part b, how do you think wheel size affects linear speed?
SOLUTION:
eSolutions
Manual - Powered by Cognero
a. v = r
; Sample answer: Linear speed is given by v =
Page 43
. Because s = rθ and
=
, the equation for linear
SOLUTION:
does not have a complement or supplement because complements and supplements are not defined for
4-2 Degrees and Radians
negative angles.
72. SKATEBOARDING A physics class conducted an experiment to test three different wheel sizes on a
skateboard with constant angular speed.
a. Write an equation for the linear speed of the skateboard in terms of the radius and angular speed. Explain your
reasoning.
b. Using the equation you wrote in part a, predict the linear speed in meters per second of a skateboard with an
angular speed of 3 revolutions per second for wheel diameters of 52, 56, and 60 millimeters.
c. Based on your results in part b, how do you think wheel size affects linear speed?
SOLUTION:
a. v = r
; Sample answer: Linear speed is given by v =
. Because s = rθ and
=
, the equation for linear
speed can also be written in terms of radius and angular speed as v = r .
b. Because each rotation measures 2π radians, 3 revolutions correspond to an angle of rotation θ of 3 × 2π or 6π
radians.
Find the angular speed for each wheel.
Use dimensional analysis to convert each speed to meters per second.
c. Sample answer: As the wheel diameter increases, the linear speed also increases.
73. ERROR ANALYSIS Sarah and Mateo are told that the perimeter of a sector of a circle is 10 times the length of
the circle’s radius. Sarah thinks that the radian measure of the sector’s central angle is 8 radians. Mateo thinks that
there is not enough information given to solve the problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Sample answer: The formula for the length of an intercepted arc is s = rθ. Therefore, the perimeter of the sector is
the sum of the length of the intercepted arc and twice the radius or P = rθ + 2r. Because P = 10r, using
substitution, 10r = rθ + 2r. Set P = rθ + 2r and P = 10r equal to each another and solve for θ.
Therefore, Sarah is correct.
74. CHALLENGE The two circles shown are concentric. If the length of the arc from A to B measures 8π inches
Page 44
eSolutions Manual - Powered by Cognero
and DB = 2 inches, find the arc length from C to D in terms of π.
4-2 Degrees and Radians
Therefore, Sarah is correct.
74. CHALLENGE The two circles shown are concentric. If the length of the arc from A to B measures 8π inches
and DB = 2 inches, find the arc length from C to D in terms of π.
SOLUTION:
First, convert 160º to radians.
Use the central angle measure and the length of the arc from A to B to find OB.
So, OB = 9 inches. Because OD = OB – DB, OD = 9 – 2 or 7 inches.
Use the central angle measure and the length of OD to find the arc length from C to D.
REASONING Describe how the linear speed would change for each parameter below. Explain.
75. A decrease in the radius
SOLUTION:
Sample answer: If the radius decreased, then the linear speed would also decrease because the linear speed is
directly proportional to the radius.
76. A decrease in the unit of time
SOLUTION:
Sample answer: If the time decreased, then the linear speed would increase because the linear speed is inversely
proportional to the unit of time.
77. An increase in the angular speed
SOLUTION:
Increase; sample answer: The equation for linear speed can also be written as v = rω. If the angular speed
Page 45
would also increase because the linear speed is directly proportional to the angular
speed.
eSolutions
Manual - Powered
Cognero
increased,
then thebylinear
speed
76. A decrease in the unit of time
SOLUTION:
Sample answer:
the time decreased, then the linear speed would increase because the linear speed is inversely
4-2 Degrees
and IfRadians
proportional to the unit of time.
77. An increase in the angular speed
SOLUTION:
Increase; sample answer: The equation for linear speed can also be written as v = rω. If the angular speed
increased, then the linear speed would also increase because the linear speed is directly proportional to the angular
speed.
78. PROOF If
=
, prove that θ1 = θ2.
SOLUTION:
Proof:
Given:
=
Prove: θ1 = θ2
1.
=
(Given)
2. s = rθ (Arc length formula)
3. s1 = r1θ 1, s2 = r2θ 2 (Substitution)
4.
= θ 1,
= θ 2 (Div. Prop. of Equality)
5. θ1 = θ2 (Trans. Prop. of Equality using 1 and 4 )
79. REASONING What effect does doubling the radius of a circle have on each of the following measures? Explain
your reasoning.
a. the perimeter of the sector of the circle with a central angle that measures θ radians
b. the area of a sector of the circle with a central angle that measures θ radians
SOLUTION:
a. Sample answer: The perimeter of a sector of a circle P is equal to the sum of the arc length s and two times the
radius r, so P = s + 2r. Because s = rθ, if the radius doubled, the arc length would become a new arc length s' =
(2r)θ, which is equal to s' = 2(rθ) or s' = 2s. So, the perimeter would be P = 2s + 2(2r) or P = 2(s + 2r). Therefore,
the perimeter would double.
b. Sample answer: Because A =
which is equal to A' =
2
2
r θ, if the radius doubled, the area would become a new area A' =
4r θ or A' =
2
(2r) θ,
. Therefore, the area would quadruple.
80. Writing in Math Compare and contrast degree and radian measures. Then create a diagram similar to the one on
page 231. Label the diagram using degree measures on the inside and radian measures on the outside of the circle.
SOLUTION:
Sample answer: Degree and radian measures are both used to describe the measures of angles. Degree measures
have units, while radian measures are considered unitless and can therefore be used in mathematical relationships
involving linear measures.
eSolutions Manual - Powered by Cognero
Page 46
b. Sample answer: Because A =
2
4-2 Degrees
andto Radians
which is equal
A' = 4r θ
2
r θ, if the radius doubled, the area would become a new area A' =
or A' =
2
(2r) θ,
. Therefore, the area would quadruple.
80. Writing in Math Compare and contrast degree and radian measures. Then create a diagram similar to the one on
page 231. Label the diagram using degree measures on the inside and radian measures on the outside of the circle.
SOLUTION:
Sample answer: Degree and radian measures are both used to describe the measures of angles. Degree measures
have units, while radian measures are considered unitless and can therefore be used in mathematical relationships
involving linear measures.
Use the given trigonometric function value of the acute angle θ to find the exact values of the five
remaining trigonometric function values of θ.
81. sin θ =
SOLUTION:
Draw a right triangle and label one acute angle θ. Because sin θ =
=
, label the opposite side 8 and the
hypotenuse 15.
By the Pythagorean Theorem, the length of the side adjacent to θ is
eSolutions Manual - Powered by Cognero
82. sec θ =
or
.
Page 47
4-2 Degrees and Radians
82. sec θ =
SOLUTION:
Draw a right triangle and label one acute angle θ. Because sec θ =
=
, label the hypotenuse 4
and
the side adjacent to θ 10.
By the Pythagorean Theorem, the length of the side opposite θ is
or
.
83. cot θ =
SOLUTION:
Draw a right triangle and label one acute angle θ. Because cot θ =
, label the side adjacent to θ 17
=
and the opposite side 19.
By the Pythagorean Theorem, the length of the hypotenuse is
eSolutions Manual - Powered by Cognero
or 5
.
Page 48
4-2 Degrees and Radians
83. cot θ =
SOLUTION:
Draw a right triangle and label one acute angle θ. Because cot θ =
, label the side adjacent to θ 17
=
and the opposite side 19.
By the Pythagorean Theorem, the length of the hypotenuse is
or 5
.
84. BANKING An account that Hally’s grandmother opened in 1955 earned continuously compounded interest. The
table below shows the balances of the account from 1955 to 1959.
a. Find a function that models the amount in the account. Use the number of years after Jan. 1, 1955, as the
independent variable.
b. Write the equation from part a in terms of base e.
c. What was the interest rate on the account if the interest were compounded continuously and no deposits or
withdrawals were made during the period in question?
SOLUTION:
a. Enter the data into your calculator, letting x = 0 represent 1955. Use the expReg feature to find the exponential
regression of the data.
eSolutions Manual - Powered by Cognero
Page 49
4-2 Degrees and Radians
84. BANKING An account that Hally’s grandmother opened in 1955 earned continuously compounded interest. The
table below shows the balances of the account from 1955 to 1959.
a. Find a function that models the amount in the account. Use the number of years after Jan. 1, 1955, as the
independent variable.
b. Write the equation from part a in terms of base e.
c. What was the interest rate on the account if the interest were compounded continuously and no deposits or
withdrawals were made during the period in question?
SOLUTION:
a. Enter the data into your calculator, letting x = 0 represent 1955. Use the expReg feature to find the exponential
regression of the data.
x
The exponential regression equation is approximately y = 2137.5192(1.0534) .
b. Find u such that eu = 1.0534.
0.052
Substitute e
for 1.0534.
0.052x
The new equation is y = 2137.5192e
.
rt
c. The continuous compounding interest formula is A = Pe where r is the interest rate. Therefore, if the interest
were earned continuously, the interest rate would be 5.2%.
Express each logarithm in terms of ln 2 and ln 5.
85. ln
SOLUTION:
eSolutions Manual - Powered by Cognero
Page 50
Substitute e
for 1.0534.
0.052x
The new equation is y = 2137.5192e
.
c. The continuous
compounding interest formula is A = Pe
4-2 Degrees
and Radians
rt
where r is the interest rate. Therefore, if the interest
were earned continuously, the interest rate would be 5.2%.
Express each logarithm in terms of ln 2 and ln 5.
85. ln
SOLUTION:
86. ln 250
SOLUTION:
87. ln
SOLUTION:
List all possible rational zeros of each function. Then determine which, if any, are zeros.
88. f (x) = x4 – x3 – 12x – 144
SOLUTION:
Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −144.
Therefore, the possible rational zeros of f
are
By using synthetic division, it can be determined that x = 4 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero.
Because (x – 4) and (x + 3) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f
2
2
(x) = (x − 4)(x + 3)(x + 12). Because the factor (x + 12) yields no real zeros, the rational zeros of f are –3 and 4.
eSolutions Manual3- Powered
2 by Cognero
89. g(x) = x – 5x – 4x + 20
SOLUTION:
Page 51
Because (x – 4) and (x + 3) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f
2
2
4-2 Degrees
and+ Radians
(x) = (x − 4)(x
3)(x + 12). Because the factor (x + 12) yields no real zeros, the rational zeros of f
are –3 and 4.
89. g(x) = x3 – 5x2 – 4x + 20
SOLUTION:
Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 20.
Therefore, the possible rational zeros of g are
By using synthetic division, it can be determined that x = 2 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero.
Because (x – 2) and (x – 5) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g
(x) = (x − 2)(x – 5)(x – 2). Therefore, the rational zeros of g are –2, 2, and 5.
90. g(x) = 6x4 + 35x3 – x2 – 7x – 1
SOLUTION:
The leading coefficient is 6 and the constant term is –1. The possible rational zeros are
By using synthetic division, it can be determined that x =
is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = –
Because
and
is a rational zero.
are factors of g(x), we can use the final quotient to write a factored form of g(x) as
2
Because the factor (6x + 36 + 6) yields no rational zeros, the rational zeros
g(x) =
eSolutions Manual - Powered by Cognero
of g are –
or
and
.
Page 52
Because (xand
(x – 5) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g
– 2) and
4-2 Degrees
Radians
(x) = (x − 2)(x – 5)(x – 2). Therefore, the rational zeros of g are –2, 2, and 5.
90. g(x) = 6x4 + 35x3 – x2 – 7x – 1
SOLUTION:
The leading coefficient is 6 and the constant term is –1. The possible rational zeros are
is a rational zero.
By using synthetic division, it can be determined that x =
By using synthetic division on the depressed polynomial, it can be determined that x = –
Because
and
is a rational zero.
are factors of g(x), we can use the final quotient to write a factored form of g(x) as
2
Because the factor (6x + 36 + 6) yields no rational zeros, the rational zeros
g(x) =
of g are –
or
and
.
Describe the end behavior of each function.
91. f (x) = 4x5 + 2x4 – 3x – 1
SOLUTION:
Graph the function.
From the graph of f (x), it appears that f (x)
–
as x
–
and f (x)
as x
.
92. g(x) = −x6 + x4 – 5x2 + 4
eSolutions Manual - Powered by Cognero
SOLUTION:
Graph the function.
Page 53
4-2 Degrees
andofRadians
From the graph
f (x), it appears that f (x)
–
as x
–
as x
and f (x)
.
92. g(x) = −x6 + x4 – 5x2 + 4
SOLUTION:
Graph the function.
From the graph of g(x), it appears that g(x)
–
–
as x
and g(x)
–
as x
.
93.
SOLUTION:
Graph the function.
From the graph of h(x), it appears that h(x)
2 as x
–
and h(x)
2 as x
.
Write each set of numbers in set-builder and interval notation, if possible.
94. n > −7
SOLUTION:
The set includes all real numbers greater than –7. In set-builder notation this set can be described as {n | n > –7, n
R}, and in interval notation it can be described as (–7, ).
95. −4 ≤ x < 10
SOLUTION:
The set includes all real numbers greater than –31 and less than or equal to 64. In set-builder notation this set can
be described as {x | −4 ≤ x < 10, x R}, and in interval notation it can be described as [–4, 10).
96. y < 1 or y ≥ 11
SOLUTION:
The set includes all real numbers less than 1 or greater than or equal to 11. In set-builder notation this set can be
described as {y | y < 1 or y ≥ 11, y R}, in interval notation, this set can be described as (–∞, 1) È [11, ∞).
97. SAT/ACT In the figure, C and D are the centers of the two circles with radii of 3 and 2, respectively. If the larger
shaded
hasbyanCognero
area of
eSolutions
Manualregion
- Powered
9, what is the area of the smaller shaded region?
Page 54
96. y < 1 or y ≥ 11
SOLUTION:
The set includes
real numbers less than 1 or greater than or equal to 11. In set-builder notation this set can be
4-2 Degrees
and all
Radians
described as {y | y < 1 or y ≥ 11, y
R}, in interval notation, this set can be described as (–∞, 1) È [11, ∞).
97. SAT/ACT In the figure, C and D are the centers of the two circles with radii of 3 and 2, respectively. If the larger
shaded region has an area of 9, what is the area of the smaller shaded region?
A3
B4
C5
D7
E8
SOLUTION:
The area of a sector of a circle is given by A =
2
r θ. The area of the larger shaded region is 9 and the radius of
the circle is 3. Find the measure of the central angle a.
So, the measure of the central angle is 2 radians. Use the value for a and radius of circle D to find the area of the
smaller shaded region.
Therefore, the correct answer is B.
98. REVIEW If cot θ = 1, then tan θ =
F −1
G0
H1
J3
SOLUTION:
Because cot θ = 1 and tan θ =
99. REVIEW If sec θ =
, tan θ =
or 1. Therefore, the correct answer is H.
, then sin θ =
A
eSolutions Manual - Powered by Cognero
B
Page 55
J3
SOLUTION:
4-2 Degrees
and
Because cot
1 and tan θ =
θ = Radians
99. REVIEW If sec θ =
, tan θ =
or 1. Therefore, the correct answer is H.
, then sin θ =
A
B
C
or
D
SOLUTION:
Draw a right triangle and label one acute angle θ. Because sec θ =
=
, label the hypotenuse 25 and the
adjacent side 7.
By the Pythagorean Theorem, the length of the side opposite θ is
or 24.
Therefore, the correct answer is C.
100. Which of the following radian measures is equal to 56°?
F
G
H
J
SOLUTION:
To convert a degree measure to radians, multiply by
Therefore, the correct answer is H.
eSolutions Manual - Powered by Cognero
Page 56
4-2 Degrees and Radians
Therefore, the correct answer is C.
100. Which of the following radian measures is equal to 56°?
F
G
H
J
SOLUTION:
To convert a degree measure to radians, multiply by
Therefore, the correct answer is H.
eSolutions Manual - Powered by Cognero
Page 57