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. 11.773 62/87,21 First, convert 0. 773

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:




3. 141.549°

SOLUTION:




4. 273.396°

SOLUTION:




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.

Therefore, 87° 53′ 10″ can be written as about 87.886°.

6. 126° 6′ 34″

SOLUTION:



7. 45° 21′ 25″

SOLUTION:



8. 301° 42′ 8″

SOLUTION:



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 of a degree and each second is of a minute,

so each second is of a degree.


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:


12. –165°

SOLUTION:


13. –45°

SOLUTION:


14.

SOLUTION:

To convert a radian measure to degrees, multiply by

15.

SOLUTION:


16.

SOLUTION:


17.

SOLUTION:


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.

19. –75°

SOLUTION:

All angles measuring are coterminal with a angle. Sample answer: Let n = 1 and −1.

20. 225°

SOLUTION:


21. –150°

SOLUTION:


22.

SOLUTION:

All angles measuring are coterminal with a angle.

Sample answer: Let n = 1 and −1.

23.

SOLUTION:



24.

SOLUTION:



25.

SOLUTION:



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 herspin, 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° ÷ 20or 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.

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 = .

Method 2

Use s = to find the arc length.

31. 45°, r = 5 mi


Substitute r = 5 and θ = .

Method 2


32. 150°, r = 79 mm



Method 2


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.

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.


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.

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.

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 timewould it gain or lose per day?

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.


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.


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 measureto 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.

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.


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 toradians.


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°.

Convert the central angle measure to 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°



Therefore, the radius is 21 cm.

53. A = 377 in2, θ =

SOLUTION:

Therefore, the radius is 12 in.

54. A = 75 m2, θ =

SOLUTION:

A = 75 m2, 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, asshown.

From part a, we know that 90º = . Convert 180º to radians.

Therefore, in Quadrant II, < 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 = . Convert 360º to 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.

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 ≈ 103.1º.

59.

SOLUTION:

So, θ ≈ 3.3 radians, or ≈ 191º.

60.

SOLUTION:

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 ofLane 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.

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.


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.


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 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 304.95cm. 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.


does not have a complement because it is greater than or 90°. The supplement is π – or .

70.



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:

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 ofthe circle’s radius. Sarah thinks that the radian measure of the sector’s central angle is 8 radians. Mateo thinks thatthere 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 isthe 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 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 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? Explainyour 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 = r2θ, if the radius doubled, the area would become a new area A ' = (2r)2θ,

which is equal to A ' = 4r2θ or A ' = . 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 or .

82. sec θ =

eSolutions Manual - Powered by Cognero Page 1

4-2 Degrees and Radians


1. 11.773°

SOLUTION:




2. 58.244°

SOLUTION:




3. 141.549°

SOLUTION:




4. 273.396°

SOLUTION:




5. 87° 53′ 10″

SOLUTION:



6. 126° 6′ 34″

SOLUTION:



7. 45° 21′ 25″

SOLUTION:



8. 301° 42′ 8″

SOLUTION:





SOLUTION:





SOLUTION:


11. 225°

SOLUTION:


12. –165°

SOLUTION:


13. –45°

SOLUTION:


14.

SOLUTION:


15.

SOLUTION:


16.

SOLUTION:


17.

SOLUTION:



18. 120°

SOLUTION:


19. –75°

SOLUTION:


20. 225°

SOLUTION:


21. –150°

SOLUTION:


22.

SOLUTION:



23.

SOLUTION:



24.

SOLUTION:



25.

SOLUTION:




SOLUTION:



27. , r = 2.5 m

SOLUTION:

28. , r = 3 in.

SOLUTION:

29. , r = 4 yd

SOLUTION:

30. 105°, r = 18.2 cm



Method 2


31. 45°, r = 5 mi



Method 2


32. 150°, r = 79 mm



Method 2









34. =

SOLUTION:



35. = 135π



36. = 104π



37. v = 82.3 , 131



38. v = 144.2 , 10.9



39. v = 553 , 0.09






























43.

SOLUTION:




44.

SOLUTION:



45.

SOLUTION:



46.

SOLUTION:




47.

SOLUTION:




48.

SOLUTION:




SOLUTION:






measure of 360° ÷ 3 or 120°.





51. A = 29 ft2, θ = 68°




52. A = 808 cm2, θ = 210°




53. A = 377 in2, θ =

SOLUTION:


54. A = 75 m2, θ =

SOLUTION:

A = 75 m2, θ =

























58.

SOLUTION:

So, θ = 1.8 radians, or ≈ 103.1º.

59.

SOLUTION:


60.

SOLUTION:


61.

SOLUTION:






















65.




66.





radians.







68.



69.



70.



71.

SOLUTION:


negative angles.


SOLUTION:




















SOLUTION: Proof:

Given: =

Prove: θ1 = θ2

1. = (Given)











81. sin θ =

SOLUTION:


hypotenuse 15.


82. sec θ =




1. 11.773°

SOLUTION:




2. 58.244°

SOLUTION:




3. 141.549°

SOLUTION:




4. 273.396°

SOLUTION:




5. 87° 53′ 10″

SOLUTION:

Each minute is of a degree and each second is of a minute, so each second is o

Documents

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. 11.773 62/87,21 First, convert 0. 773