Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 321: 45

Answer

The distance between the ends of the two sides is 438.14 feet.

Work Step by Step

Let $a = 246.75~ft$, let $b = 246.75~ft$, and let angle $C = 125^{\circ}12'$. We can use the law of cosines to find $c$, the length of the line opposite the angle $C$: $c^2 = a^2+b^2-2ab~cos~C$ $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(246.75~ft)^2+(246.75~ft)^2-(2)(246.75~ft)(246.75~ft)~cos~125^{\circ}12'}$ $c = \sqrt{191963.9~ft^2}$ $c = 438.14~ft$ The distance between the ends of the two sides is 438.14 feet.
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