Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Review Exercises - Page 344: 28

Answer

The distance between the ends of the wire on the ceiling is 15.8 feet

Work Step by Step

Let $a = 12.2~ft$ Let $b = 15.0~ft$ Let angle $C = 70.3^{\circ}$ We can use the law of cosines to find $c$, the distance between the ends of the wire on the ceiling: $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(12.2~ft)^2+(15.0~ft)^2-(2)(12.2~ft)(15.0~ft)~cos~70.3^{\circ}}$ $c = \sqrt{250.46~ft^2}$ $c = 15.8~ft$ The distance between the ends of the wire on the ceiling is 15.8 feet
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