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Solution Manual for Analytic Trigonometry With Applications 11th Edition by Barnett

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   Chapter 1 Right Triangle Ratios  1 Chapter 1 Right Triangle Ratios  EXERCISE 1.1 Angles, Degrees, and Arcs 2. Since one complete revolution has measure 360°, 14  revolution has measure 14  (360°) = 90°. Similarly, 13 (360°) = 120°, 78 (360°) = 315°, 1112 (360°) = 330°. 4. Since one complete revolution has measure 360°, the fraction of a counterclockwise revolution that will form a 30° angle is 30360  = 112 . Similarly, 225360  = 58 , 240360  = 23 . 6. Since 88° angle is between 0° and 90°, this is an acute angle. 8. None of these. 10. Since 175° is between 90° and 180°, this is an obtuse angle. 12. It is a right angle. 14. A minute of angle measure is 160  of one degree; a second of angle measure is 160  of one minute. 16. Since 43' = 43 ° 60  and 30 = 30 ° 3,600 , then 71°43'30 = 71 + 4360 + 303,600   #   $   %   &   '   °  !  71.725° to three decimal places. 18. Since 3' = 3 ° 60  and 1 = 1 ° 3,600 , then 9°3'1 = 9 + 360 + 13,600   #   $   %   &   '   °  !  9.050° to three decimal places. 20. Since 11' = 11 ° 60  and 25 = 25 ° 3,600 , then 267°11'25 = 267 + 1160 + 253,600   #   $   %   &   '   °  !  267.190° to three decimal places. 22. 35.425° = 35°(0.425 !  60)' = 35°25.5' = 35°25'(0.5 !  60) = 35°25'30 24. 52.927° = 52°(0.927 !  60)' = 52°55.62' = 52°55'(0.62 !  60) !  52°55'37 26. 235.253° = 235°(0.253 !  60)' = 235°15.18' = 235°15'(0.18 !  60) !  235°15'11   Exercise 1.1 Angles, Degrees, and Arcs 2 28. There are two methods. a. Convert the first one to decimal hour form and compare with the second one. 3 + 4360  + 243,600   !  3.723 to three decimal places. Since 3.723 < 3.732, Runner  A  is faster. b. Convert the second one to HMS form and compare with the first one. 3.732 = 3 hr + (0.732 !  60) min = 3 hr + 43.92 min = 3 hr + 43 min + (0.92 !  60) sec !  3 hr, 43 min, 55 sec This is bigger than 3 hr, 43 min, 24 sec, and hence Runner  A  is faster. 30. To compare !   and   , we convert !   to decimal form. Since 51' = 51 ° 60  and 54 = 54 ° 3,600 , then 32°51'54 = 32 + 5160 + 543,600   #   $   %   &   '   ° = 32.865°. Thus !   =   . 32. To compare !   and   , we convert    to decimal form. Since 40' = 40 ° 60  and 20 = 20 ° 3,600 , then 80°40'20 = 80 + 4060 + 203,600   #   $   %   &   '   °  !  80.672°. Thus !   <   . 34. We convert    to decimal form. Since 18' = 18 ° 60  and 32 = 32 ° 3,600 , then 242°18'32 = 242 + 1860 + 323,600   #   $   %   &   '   °  !  242.309°. Thus !   >   . 36. 105°53'22 + 26°38'55  DMS 132°32'17 38. 180° – 121°51'22  DMS 58°8'38 40. The circumference of a circle with radius r   is 2 !  r  . If 2 !  r   = 6, then estimate of r   to the nearest whole number would be 1. The value of r   to two decimal places is r   = 62   !  0.95. 42. The arc length of a semicircle that has diameter 10 is 12 C   = 12  (2 !  r  ) = 12 (10 !  ) = 5 !  . An estimate of 5 !   to the nearest whole number is 16. The length of the semicircle of radius 5 (or diameter 10) to two decimal places is 15.71. 44. Since sC   =   360 ° , then 46. Since sC   =   360 ° , then 12108  =   360 °   s 740  = 72 ° 360 °   #   = 12108   #   $   %   &   '   360° = 40° s  = 72360 (740) = 148 mi   Chapter 1 Right Triangle Ratios  3 48. Since sC   =   360 °  and C   = 2   r  , then s 2 r   =   360 °   38,000 cm2 r   = 45.3 ° 360 °   r   = (38,000 cm)(360)(2 )(45.3)   !  48,000 cm (to the nearest 1,000 cm) 50. Since sC   =   360 °  and #   = 24°16'34 = 24 + 1660 + 343,600   #   $   %   &   '   °  !  24.276°, then 14.23m C    !   24.276 ° 360 °   C    !   (14.23m)(360)24.276   !  211.0 m (to one decimal place) 52. Since  A r  2  =   360 ° , then  A (7.38 ft) 2  = 24.6 ° 360 °    A  = 24.6360 (   )(7.38 ft) 2   !  11.7 ft 2  (to one decimal place) 54. Since  A r  2  =   360 ° , then 347 in 2 (32.4 in) 2  =   360 °   #   = 347 (32.4) 2  · 360° !  37.9° 58. Since sC   =   360 °  and C   = 2   r  , then s 2 r   =   360 °   #   = s 2 r   · 360° = 12.1 mm2( )(5.26 mm)  · 360° !  131.8° 60. Since sC   =   360 °  and C   = 2   r  , then s 2 r   =   360 °   r   = s 2  · 360 °    = 11.8 mm2( )  · 360117.9   !  5.73 mm 56. Note that the central angles corresponding to the arcs  AB  and CD are 72° and hence #  A  = 36°, #  D  = 36°. Since the sum of the angles in a triangle is 180°, the angle #  x  should be 180° - (36° + 36°) = 108°.  Exercise 1.1 Angles, Degrees, and Arcs 4 62. Since sC   =   360 °  and C   = 2   r  , then s 2 r   =   360 °   #   = 40°40' – 33°30' = 7°10' = 7 + 1060   #   $   %   &   '   °  s  = 2   r   ·   360 °  = 2(   )(3,960 mi) · 7 +  1060 360   !  495 mi 64. Since sC   =   360 °  and C   = 2   r  , then s 2 r   =   360 °   #   = 42°50' – 36°0' = 6°50' = 6 + 5060   #   $   %   &   '   °  s  = 2   r   ·   360 °  = 2(   )(3,960 mi) · 6 +  5060 360   !  472 mi 66. To find the length of s  in nautical miles, since 1 nautical mile is the length of 1' on the circle shown in the diagram, we need only to find how many minutes are in the angle #  . Since #   = 40°40' – 33°30' = 7°10' = (7 !  60 + 10)' = 430'. Therefore, s  = 430 nautical miles. 68. To find the length of s  in nautical miles, since 1 nautical mile is the length of 1' on the circle shown in the diagram, we need only to find how many minutes are in the angle #  . Since #   = 42°50' – 36°0' = 6°50' = (6 !  60 + 50)' = 410'. Therefore, s  = 410 nautical miles. 70.   The arc length of a circular sector is very close to the chord length if the central angle of the sector is small and the radius of the sector is large, which is the case in this problem. 1 minute of arc is 3  in. !  1.05 in. at 100 yd; the diameter of the quarter is slightly less than 1 in. 72. Since sC   =   360 °  and C   = 2 !  r  , then s 2 r   =   360 °   #   = s 2 r   · 360˚ = 864,0002( )(1,780,000,000)  · 360° !  0.028°. 74. Since sC   =   360 °  and C   = 2   r  , then s 2 r   =   360 °   r   = s 2  · 360 °    = 522( )  · 360 ° 0.075 °   !  40,000 miles.
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