4-2 Degrees and Radians - MOC-FV - MOC-Floyd Valley ... · Write each decimal degree measure in DMS...

Write each decimal degree measure in DMS form and each DMS measure in decimal degree form to the nearest thousandth.

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″.

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°.

9. NAVIGATION A sailing enthusiast uses a sextant, an instrument that can measure the angle between two objectswith 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.

11. 225°

SOLUTION:

To convert a degree measure to radians, multiply by

13. –45°

SOLUTION:

To convert a radian measure to degrees, multiply by

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.

19. –75°

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.

SOLUTION:

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:

29. , r = 4 yd

SOLUTION:

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.

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 speedand the rate of rotation.

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.

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.

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.

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.

Find the area of each sector.

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.

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.

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.

Therefore, the area of the sector is about 500.5 square feet.

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.

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.

53. A = 377 in2, θ =

SOLUTION:

Therefore, the radius is 12 in.

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.

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.

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.

SOLUTION:

So, θ ≈ 3.3 radians, or ≈ 191º.

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.

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.

eSolutions Manual - Powered by Cognero Page 1

4-2 Degrees and Radians

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

3. 141.549°

SOLUTION:

5. 87° 53′ 10″

SOLUTION:

11. 225°

SOLUTION:

13. –45°

SOLUTION:

19. –75°

SOLUTION:

21. –150°

SOLUTION:

27. , r = 2.5 m

SOLUTION:

29. , r = 4 yd

SOLUTION:

31. 45°, r = 5 mi

Method 2

35. = 135π

37. v = 82.3 , 131

39. v = 553 , 0.09

SOLUTION:

51. A = 29 ft2, θ = 68°

53. A = 377 in2, θ =

SOLUTION:

4-2 Degrees and Radians - MOC-FV - MOC-Floyd Valley ... · Write each decimal degree measure in DMS...

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Transcript of 4-2 Degrees and Radians - MOC-FV - MOC-Floyd Valley ... · Write each decimal degree measure in DMS...