Chapter 7 - Pearson

110 Chapter 7 Applications of Trigonometry and Vectors77. θ = 135 78. θ » 36.87 79. θ = 90 80. θ = 45 81. θ » 36.8782. θ » 78.93 83. - 684. - 685. - 2486. - 2487. the vectors are orthogonal.88. the vectors are orthogonal.89. the vectors are not orthogonal90. the vectors are not orthogonal.91. the vectors are not orthogonal.92. the vectors are not orthogonal93. Draw a line parallel to the x-axis and thevector u + v (shown as a dashed line)Since θ 1 = 110, its supplementary angle is70°. Further, since θ 2 = 260, the angle α is260-180= 80 . Then the angle CBAbecomes 180 – (80 + 70) = 180 – 150 = 30°.Magnitude: » 9.5208The direction angle is 119.064794. ab , » - 4.1042, 11.276395. cd , » -0.5209, - 2.954496. » - 4.6252,8.321997. Magnitude: » 9.5208; Angle: 119.0647(θ lies in quadrant II)98. They are the same. Preference of method is anindividual choice.Section 7.5Applications of Vectors1. 2640 lb at an angle of 167.2 with the 1480-lbforce2. 1800 lb at an angle of 167 (rounded to threesignificant digits) with the 840-lb force3. 93.94. 70.15. 190 lb and the magnitude of the resultant isabout 283 lb6. The magnitude of the resultant is 117 lb; thesecond force is 93.9 lb.7. 188. 800 lb.9. 2.4 tons10. 2.8 tons.11. 17.512. 22.013. 226 lb.14. The weight of the crate is 64.8 lb; the tensionis 61.9 lb.15. 13.5 mi; 50.416. 6.6 mi; 117.117. 39.2 km18. 14.5 km19. The speed of the current is 3.5 mph and theactual speed of the motorboat is 19.7 mph.20. (a) 6.3(b) 14 seconds(c) 2521. the bearing and ground speed of the planev » 470.1the bearing is 23722. the pilot turned at 1 hr 21 min after 2 P.M., orat 3:21 P.M.23. Let x = the airspeed and d = the groundspeed.x » 156 mph; d » 161 mph24. 173.125. The bearing is 7400¢ ; the ground speed is202 mph.26. The bearing is 6530¢181 mph.; the ground speed is27. The airspeed must be 170 mph.The bearing must be approximately 358°.Copyright © 2013 Pearson Education, Inc.