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 519: 21



Work Step by Step

We know that $\sin.(\theta)=\cos(90^{o}-\theta)$. Hence, $\dfrac{\cos(10^{o})}{\sin(80^{o})} \\=\dfrac{\cos(10^{o})}{\cos(90^{o}-80^{o})} \\=\dfrac{\cos(10^{o})}{\cos(10^{o})} \\=1$
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