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: 27

Answer

The bearing the pilot should fly is $358^{\circ}$ and the airspeed should be 169.4 mph

Work Step by Step

Let $a$ be the airspeed Let $b = 11~mph$ Let $c$ be the resultant of the two vectors $a$ and $b$. Note that the magnitude of $c$ is the ground speed. Therefore $c = \frac{400~mi}{2.5~hr} = 160~mph$ The vectors a, b, and c form a triangle. Let angle $A$ be the angle that subtends side $a$. Then $A$ is the angle between vector b and vector c. Then $A = 328^{\circ}-180^{\circ} = 148^{\circ}$ We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{(11~mph)^2+(160~mph)^2-(2)(11~mph)(160~mph)~cos~148^{\circ}}$ $a = \sqrt{28706.13~(mph)^2}$ $a = 169.4~mph$ The airspeed should be 169.4 mph We can use the law of sines to find angle $B$ which is the angle between vectors $a$ and $c$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $B = arcsin(\frac{b~sin~A}{a})$ $B = arcsin(\frac{(11)~sin~148^{\circ}}{169.4})$ $B = arcsin(0.03441)$ $B = 2.0^{\circ}$ The bearing the pilot should fly is $360^{\circ}-2.0^{\circ} = 358^{\circ}$
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