Section. Applications of Radian Measure 07 (continued) 88. 80 radians = = 0 5 5 80 radians = = 00 7 5 = 5 radian s 0 = 0 radian s = radian 80 89. (a) In hours, the hour hand will rotate twice around the clock. One complete rotation is radians, the two rotations will measure = (b) In hours, the hour hand will rotate = of the way around the clock, which is = 90. (a) In each hour, the minute hand rotates radians, so it rotates = radians in hours. (b) In each hour, the minute hand rotates radians, so it rotates = radians in hours. 9. In each rotation around Earth, the space vehicle would rotate (a) In.5 orbits, the space vehicle travels.5 = 5 (b) In of an orbit, the space vehicle travels 8 = 9. In each rotation, the pulley would rotate (a) In 8 rotations, the pulley would turn 8 = (b) In 0 rotations, the pulley would turn 0 = 0 9. In each revolution, the horse revolves In revolutions, the horse revolves = 9. (a) 9 = 0 80 00 (b) For degrees, we have.5 0.9 =.5. For.5 0.055 radian. 00 Section. Applications of Radian Measure. r =, = = =. r =, = = =.. 5.. 5 r =, = 5 = = 0 s =, = s r = = = = 8 s =, = s r = = = = 7 s =, = s r = = = = 8 7 7 7. r =, s = s = = = r 8. s =, r = s = = = or.5 r 9. s = 0, r = 0 s 0 = = = r 0 Copyright 0 Pearson Education, Inc.
08 Chapter Radian Measure and the Unit Circle 0. To find the degree measure of a central angle in a circle if both the raidus and the length of the intercepted arc are known, first apply the formula to find the radian measure. Then multiply the radian measure by 80 to find the degree measure.. r =. cm, s =. = 8. 5.8 cm. r = 0.89 cm, s 0 = 0.89 0 = 0.98 cm.08 cm 5. r =.8 ft, s 5 =.8 =.5 ft. ft (rounded to three significant digits) 7. r =. mi, s 7 =. =.78 mi.9 mi 5. r =.8 m, = 0 Converting to = 0 = 0 radian = Thus, the arc is.8 =.8 = 5.05 m.. r = 7.9 cm, = 5 Converting to = 5 = 5 radian = Thus, the arc is 5.7 = 7.9 cm = cm 9 cm. 7. r = 5. in., = 0 Converting to 7 = 0 = 0 radian = Thus, the arc is 7 05.7 = 5. = 55. in. 8. r =. ft, = 0 Converting to = 0 = 0 radian = Thus, the arc is. =. = 7.ft 9. The formula for arc length is. Substituting r for r we obtain s = r = r. The length of the arc is doubled. 0. Recall that radians = 80. If is measured in degrees, then the formula becomes r s = r s = 80 80 For Exercises, note that since 00 has two significant digits and the angles are given to the nearest degree, we can have only two significant digits in the answers.. 9 N, 0 N = 0 9 = = radian 80 = 00 500 km. N, 9 N = 9 = = radian 80 = 00 500 km Copyright 0 Pearson Education, Inc.
Section. Applications of Radian Measure 09. N, S S = N = ( ) = 5 = 5 radian 5 80 5 = 00 5900 km. 5 N, S S = N = 5 ( ) = 79 = 79 radian 79 s 80 79 = 00 8800 km 5. r = 00 km, s = 00 km 00 00 = 00 = = 00 Converting radian to degrees, we have 80 =. The north-south distance between the two cities is. Let x = the latitude of Madison. x = x = N The latitude of Madison is N.. r = 00 km, s = 00 km 00 00 = 00 = = 00 Converting radian to degrees we have 80 = 0. The north-south distance between the two cities is 0. Let x = the latitude of Toronto. x = 0 x = N The latitude of Toronto is N. 7. The arc length on the smaller gear is 5 =.7 00 =.7 8.5 = cm An arc with this length on the larger gear corresponds to an angle measure where 8.5 8.5 = 7. =. 8.5 80 = 5. The larger gear will rotate through approximately 5º. 8. The arc length on the smaller gear is 7 =.8 5 =.8 = 8. in. An arc with this length on the larger gear corresponds to an angle measure where 8. 8. = 7. = 7. 8. 80 = 7. The larger gear will rotate through approximately º. 9. A rotation of = 0.0 radian = radians on the smaller wheel moves through an arc length of 5. = 5. = cm. (holding on to more digits for the intermediate steps) Since both wheels move together, the larger wheel moves 5. 5.8 cm, which rotates it through an angle, where 5. = 8. 5. 5. 80 = 8.5.8.8 The larger wheel rotates through 8.5. 0. A rotation of 5 = 50 radian = radians on the smaller wheel moves through an arc length of 5 =.8 = 5.7 cm. Since both wheels move together, the larger wheel moves 5.7 cm, which rotates it through an angle, where 5.7 =. 5.7 5.7 80 s = 8... The larger wheel rotates through 8.. Copyright 0 Pearson Education, Inc.
and 0 Chapter Radian Measure and the Unit Circle. The arc length s represents the distance traveled by a point on the rim of a wheel. Since the two wheels rotate together, s will be the same for both wheels. For the smaller wheel, = 80 = 80 s and 9 =.7 = 5. cm. 9 For the larger wheel, 5 = 50 = 50 radian. 8 Thus, we can solve 5 5. = r 8 8 r = 5. = 8.7 5 The radius of the larger wheel is 8.7 cm. (rounded to significant digits). The arc length s represents the distance traveled by a point on the rim of a wheel. Since the two wheels rotate together, s will be the same for both wheels. For the smaller wheel, = 0 = 0 s and 9. =. = in. For the larger wheel, = 0 = 0 radian. Thus, we can solve 9. = r 9. r = = 9. The radius of the larger wheel is 9. in. (rounded to significant digits). (a) The number of inches lifted is the arc length in a circle with r = 9.7 in. and = 7 50. 50 7 50 = ( 7+ 0 ) 80 50 s = 9.7 ( 7+ 0 ). 80 The weight will rise. in. (rounded to three significant digits) (b) When the weight is raised in., we have = 9.7 80 = 9.7 9.7 7.08 = 7 + 0.08 0 7 5 The pulley must be rotated through 7 5.. To find the radius of the pulley, first convert 5. to 5. = 5. = 5. s 80 Now substitute this value of and s =. cm into the equation solve for r. 5.. = r 80 r =..58 5. The radius of the pulley is.7 cm. (rounded to three significant digits) 5. A rotation of = 80 radian = The chain 80 moves a distance equal to half the arc length of the larger gear. So, for the large gear and pedal,.7. Thus, the chain moves.7 in. The small gear rotates through an angle as follows. s.7 = =. r.8 for the wheel and for the small gear are the same, or.. So, for the wheel, we have r =... The bicycle will move in. (rounded to three significant digits). (a) In one hour, the car travels 55 mi. The radius is given in inches, so convert 55 mi to inches: s = 55 mi = 55( 580 ) ft = 90, 00 ft = 90, 00 in. =,8,800 in. Solving for the radius, we have, 8,800 in. = ( in. ), 8,800 = 8,9.9 radians Each rotation is Thus, we 8,9.9 have = 9,5.9 Thus, the number of rotations is 9, (rounded to the nearest whole rotation). Copyright 0 Pearson Education, Inc.
Section. Applications of Radian Measure (b) Find s for the -in. wheel. s in. 8,9.9 =,98,8. in. ft mi (,98, 8. in. ) in. 580 ft.9 mi The car with the -in. tires has gone.9 mi in one hour, so its speed is.9 mph. Yes, the driver deserves a ticket. 7. Since 0 = rotation, we have 0 = ( ) =. Thus, s = in. 8. Since 0 = rotation, we have 0 = ( ) =. Thus, s = = in. =.5 = 9, we have 9. Since s r s = = 9 = 7 in. =.5 =, we have 0. Since s r s = = = 9 in.. Let t = the length of the train. t is approximately the arc length subtended by 0. First convert = 0 to 0 0 = 0 = + = 0 = radian 5 The length of the train is t = r t =.5 0.0 km long. 5 (rounded to two significant digits). Let r = the distance of the boat. The height of the mast, ft, is approximately the arc length subtended by 0. First convert = 0 to 0 0 = 0 = + = = radian radian = 080 We must now find the radius, r. s r = 080 r = = 8. 080 The boat is about 850 ft away. (rounded to two significant digits). r =, s = = = = = r = ( ) = ( ) =. r = 8, s = = 8 = = 8 = r = () 8 = ( ) = 5. r =, s = = = = = r = ( ) = 7. r = 0, s = 5 5 5 = 0 = = 0 A= r A = ( 0) = ( 00) = 75 7. = sq units, r = = r = ( ) = ( ) = 8 = s 8 80 radians = = 0 The measure of the central angle is 0º. Copyright 0 Pearson Education, Inc.
Chapter Radian Measure and the Unit Circle 8. A = 9 sq units, r = = r 9 = 9 = ( ) 9 = 7 9 = s 7 80 radian = = 0 The measure of the central angle is 0º. 9. A = sq units, r = A = r = = ( ) = = =.5 radians 50. A = 8 sq units, r = A = r 8= ( ) 8= ( ) 8= 8 = radian In Exercises 5 58, we will be rounding to the nearest tenth. 5 5. r = 9. m, s 0.57. The area of the sector is. m. (0 m 5. r = 59.8 km, s A = r A = ( 59.8) = ( 57.0) 7.80 The area of the sector is 7.8 km. (70 km 5. r = 0.0 ft, s A = r A = ( 0.0) = ( 900) = 5 70.858 The area of the sector is 70.9 ft. (707 ft 5. 5 r = 90.0 yd, s A = r 5 A = ( 90.0) = ( 800) = 75 0,0.875 The area of the sector is 0,0.9 yd. (0,00 yd 55. r =.7 cm, = 8 The formula A = r requires that be measured in Converting 8 to 9 = 8 radian = Since 0 9 9 A = (.7) = (.9).009, 0 0 the area of the sector is.0 cm. ( cm 5. r = 8. m, = 5 The formula A = r requires that be measured in Converting 5 to 5 = 5 radian = Since 5 5 A = 8. = (.89) 5.08, the area of the sector is 5.m. (5 m 57. r = 0.0 mi, = 5 The formula A = r requires that be measured in Converting 5 to = 5 radian = A = 0.0 = ( 00) = 00 88.955 The area of the sector is 885.0 mi. (880 mi Copyright 0 Pearson Education, Inc.
Section. Applications of Radian Measure 58. r = 90.0 km, = 70 59. 0. The formula A = r requires that be measured in Converting 70 to = 70 radian = = 90.0 = ( 800) = 075 9,085.75 The area of the sector is 9,085. km. (9,00 km A = in., r =.0 in. () 9 A = r = = = =. radians 9 9 (rounded to two significant digits) A = m, A = r = r = r 78 78 r = r = r = m (rounded to two significant digits). The formula A = r requires that be measured in Converting 0.0 to = 0.0 radian = 9 A = r = ( 5) 9,0 = (,0) = 9 9 800 yd. If = and r = r, then = r = ( r) = ( r ) = r = r.. If the measure of the angle is halved and the radius is doubled, then the new area is twice the original area. This result holds in general for any values of and r. A = 50 in., r = 5 in. First find. 5 = r 50 = 5 50 = = radians To find the arc length, apply the formula. s = 5 = 0 in. A = cm, s = cm Using the formula, we have = r r =. Substituting into A = r and solving for gives = r 8 = = = 8 9 9 80 05 = s = = 8 8 5. (a) The central angle in degrees measures 0 =. Converting to radians, we 7 have the following. = ( ) radian 0 = radian 7 (b) Since C = r, and r = 7 ft, we have C = 7 = 5 77.5. The circumference is about 78 ft. (c) Since r = 7 ft and =, we have 7 5 = 7 = 7.80. 7 7 Thus, the length of the arc is 7.7 ft. Note: If this measurement is approximated to be 0, then the 9 approximated value would be 7.8 ft, rounded to three significant digits. Copyright 0 Pearson Education, Inc.
Chapter Radian Measure and the Unit Circle (d) Area of sector with r = 7 ft and = 7 is as follows. = r 577 = ( 7 ) = ( 577) = 7 7 7 7.08 7 ft. The area cleaned is the area of the sector wiped by the total area and blade minus the area wiped by the arm only. We must first convert 95 to 7. (a) 9 95 = ( 95 ) radian s Since 0 7 =, the arm was in. long. Thus, we have 9 9 arm only = () = ( 9) 9 = 7. in. 8 9 9 arm and blade = ( 0) = ( 00) 75 = 8.90 in. 8 Since 8.90 7. = 75.8, the area of the region cleaned was about (b) 75. in.. The triangle formed by the central angle and the chord is isosceles. Therefore, the bisector of the central angle is also the perpendicular bisector of the chord. 50 50 sin = r = 0 ft r sin 50 r = ; = sin Converting to 7 radians = Solving 0 for the arc length, we have 50 7 5 s = = 0 ft sin 0 sin (c) The area of the portion of the circle can be found by subtracting the area of the triangle from the area of the sector. From the figure in part (a), we have 50 50 tan = so h =. h tan sector = r 50 7 sector = 75 ft sin 0 and triangle = bh 50 triangle = ( 00) 5 ft tan The area bounded by the arc and the chord is 75 5 = ft. 8. If the land area is circular, the area of a circle is = r, and we have 950,000 = r 950, 000 r = 950,000 r = 59.900 Thus, the radius is 550 m. (rounded to two significant digits) If the land area is a 5 º sector of a circle, find the radius by first converting = 5 to 7 radians, giving 5 radian. The area of a sector is = r, so 7 = r 950,000 = r 7 950,000 = r 7 7 8, 00, 000 r = 950, 000 = 7 7 8, 00,000 r = 7. 7 Thus, the radius is 800 m. (rounded to two significant digits) Copyright 0 Pearson Education, Inc.
Section. Applications of Radian Measure 5 9. Use the Pythagorean theorem to find the hypotenuse of the triangle, which is also the radius of the sector of the circle. r = 0 + 0 r = 900 + 00 r = 500 r = 50 The total area of the lot is the sum of the areas of the triangle and the sector. Converting = 0 to 0 radian = Atriangle = bh = ( 0 )( 0 ) = 00 yd Asector = r = ( 50 ) 50 = ( 500 ) = yd Total area 50 Atriangle + A sector = 00 + 908.999 or 900 yd, rounded to two significant digits. 70. Converting = = to 0 radian. Solving 0 0, 800 for the arc length, we have s = 9 =.59. 0,800 0 Thus, there are approximately.5 statute miles in nautical mile (rounded to two decimal places.) 7. Converting = 7 = 7+ = 7. to 7. 7. radian =. 80 5 Solving for the radius with the arc length formula, we have 9 = r 5 5, 00 r = 9 = 97.0 Thus, the radius is approximately 950 mi. Using this approximate radius, we can find the circumference of the Earth. Since C = r, we have C 0 950,800. Thus, the approximate circumference is,800 mi. 7. (a) The longitude at Greenwich is 0º. (b) Answers will vary. 7. If we let r =, r then A sector = ( r ) = ( r) Thus, the area, = ( r ) = r 7. r, is quadrupled. A sector = r, where is in r Thus, A sector = r =, 80 0 is in degrees. 75. The base area is A sector = r. Thus, the r h volume is V = r h or V =, where is in 7. To find the base area, we need to find the area of the outer sector and then subtract the are of the inner sector (the missing sector ) from it: A outer sector = r and A inner sector = r Aouter sector A inner sector = r r = ( r r ) Thus, the volume is V = ( r r ) h, where is in 77. L is the arc length, so L = r. Thus, 78. Since cos h =, h = r cos r 79. d = r h d = r rcos d = r cos 80. L r =, so L d = r cos d = cos L r =. Copyright 0 Pearson Education, Inc.