Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.5 Applications of Vectors - 7.5 Exercises - Page 337: 28

Answer

The ground speed is 198 mph The resulting bearing of the plane is $186.5^{\circ}$

Work Step by Step

Let $a = 192~mph$ Let $b = 23.0~mph$ Let angle $\theta$ be the angle between these two vectors. Then $\theta = 78.0^{\circ}$ Let $c$ be the resultant of these two vectors. Note that the magnitude of $c$ is the ground speed. We can use the parallelogram rule to find $c$: $c = \sqrt{a^2+b^2+2ab~cos~\theta}$ $c = \sqrt{(192~mph)^2+(23.0~mph)^2+(2)(192~mph)(23.0~mph)~cos~78.0^{\circ}}$ $c = \sqrt{39229.276~mph^2}$ $c = 198~mph$ The ground speed is 198 mph We can use the law of sines to find the angle B between the vector c and the vector a: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $B = arcsin(\frac{b~sin~C}{c})$ $B = arcsin(\frac{(23.0)~sin~102^{\circ}}{198})$ $B = arcsin(0.1136)$ $B = 6.5^{\circ}$ The resulting bearing of the plane is $180^{\circ}+6.5^{\circ} = 186.5^{\circ}$
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