Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - 8.1 Right Triangle Trigonometry; Applications - 8.1 Assess Your Understanding - Page 520: 56

Answer

$\approx15.9^o.$

Work Step by Step

We know that $\tan{\theta}=\frac{\text{opposite}}{\text{adjacent}}$. We can denote that $\tan{(\theta)}=\frac{10}{35}$, hence $\theta=\tan^{-1}{(\frac{10}{35})}\approx15.9^o.$
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