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: 26

Answer

The bearing the pilot should fly is $65^{\circ}30'$ The ground speed will be 180.9 mph

Work Step by Step

Let $a = 168~mph$ Let $b = 27.1~mph$ Let $c$ be the resultant of the two vectors $a$ and $b$. Note that the magnitude of $c$ is the ground speed. 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 $A = 57^{\circ}40'$ We can use the law of sines to find the angle B between the vectors a and c: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $B = arcsin(\frac{b~sin~A}{a})$ $B = arcsin(\frac{(27.1)~sin~57^{\circ}40'}{168})$ $B = arcsin(0.1363)$ $B = 7^{\circ}50'$ The bearing the pilot should fly is $57^{\circ}40'+7^{\circ}50' = 65^{\circ}30'$ We can find the angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-57^{\circ}40'-7^{\circ}50'$ $C = 114^{\circ}30'$ We can use the law of sines to find $c$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(168~mph)~sin~114^{\circ}30'}{sin~57^{\circ}40'}$ $c = 180.9~mph$ The ground speed will be 180.9 mph
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