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 335: 6

Answer

The magnitude of the second force is 93.9 lb The magnitude of the resultant is 116.7 lb

Work Step by Step

Let $a = 28.7~lb$ Let $b$ be the second force. Let $c$ be the resultant force. When we use $a$ and $b$ to complete the parallelogram, the angles $A, B,$ and $C$ form a triangle, where $a$ is opposite angle $A$, $b$ is opposite angle $B$, and $c$ is opposite angle $C$. Angle $B = 32^{\circ}40'$. Angle $A = 42^{\circ}10'-32^{\circ}40'$ which is $A = 9^{\circ}30'$ We can use the law of sines to find $b$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $b = \frac{a~sin~B}{sin~A}$ $b = \frac{(28.7~lb)~sin~32^{\circ}40'}{sin~9^{\circ}30'}$ $b = 93.9~lb$ The magnitude of the second force is 93.9 lb Angle $C = 180^{\circ}- 42^{\circ}10' = 137^{\circ}50'$ We can use the law of sines to find $c$, which is the resultant: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(28.7~lb)~sin~137^{\circ}50'}{sin~9^{\circ}30'}$ $c = 116.7~lb$ The magnitude of the resultant is 116.7 lb
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