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 336: 13

Answer

The weight of the box is 226.2 lb

Work Step by Step

Let $a = 150~lb$ Let $b = 114~lb$ Let angle $\theta$ be the angle between these two vectors. Then $\theta = 180-62.4^{\circ}-54.9^{\circ} = 62.7^{\circ}$ We can use the parallelogram rule to find $c$, which is equal in magnitude to the weight of the box: $c = \sqrt{a^2+b^2+2ab~cos~\theta}$ $c = \sqrt{(150~lb)^2+(114~lb)^2+(2)(150~lb)(114~lb)~cos~62.7^{\circ}}$ $c = \sqrt{51181.8~lb^2}$ $c = 226.2~lb$ The weight of the box is 226.2 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.