Trigonometry (11th Edition) Clone

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

Chapter 6 - Review Exercises - Page 292: 56


$x = \frac{\sqrt{45}}{7}$

Work Step by Step

$\theta = arcsin~\frac{2}{7}$ $sin~\theta = \frac{2}{7} = \frac{opposite}{hypotenuse}$ Note that angle $\theta$ is in quadrant I. We can find the length of the adjacent side. $adjacent = \sqrt{7^2-2^2} = \sqrt{45}$ We can find the value of $cos~\theta$: $cos~\theta = \frac{adjacent}{hypotenuse}$ $cos~\theta = \frac{\sqrt{45}}{7}$ $\theta = arccos~\frac{\sqrt{45}}{7}$ Therefore, $arccos~\frac{\sqrt{45}}{7} = arcsin~\frac{2}{7}$ and thus $x = \frac{\sqrt{45}}{7}$
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