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: 1

Answer

The magnitude of the equilibrant is 2638.7 lb and it is directed at an angle of $167.2^{\circ}$ from the force of 1480 lb

Work Step by Step

Let $a = 1240~lb$ Let $b = 1480~lb$ Let angle $C = 180^{\circ}-28.2^{\circ} = 151.8^{\circ}$ We can use the law of cosines to find $c$, the magnitude of the equilibrant: $c = \sqrt{a^2+b^2-2ab~cos~\theta}$ $c = \sqrt{(1240~lb)^2+(1480~lb)^2-(2)(1240~lb)(1480~lb)~cos~151.8^{\circ}}$ $c = \sqrt{6962736.19~lb^2}$ $c = 2638.7~lb$ The magnitude of the equilibrium is 2638.7 lb We can use the law of sines to find the angle $A$ between the $b$ and $c$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~C}{c}$ $A = arcsin(\frac{a~sin~C}{c})$ $A = arcsin(\frac{1240~lb~sin~151.8^{\circ}}{2638.7~lb})$ $A = arcsin(0.222)$ $A = 12.8^{\circ}$ The angle between the equilibrium and the force 1480 lb is $180^{\circ}-A = 180^{\circ}-12.8^{\circ} = 167.2^{\circ}$ The magnitude of the equilibrium is 2638.7 lb and it is directed at an angle of $167.2^{\circ}$ from the force of 1480 lb
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.