Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 7 - Applications of Trigonometric Functions - Section 7.1 Right Triangle Trigonometry ; Applications - 7.1 Assess Your Understanding - Page 540: 22

Answer

$1$

Work Step by Step

Recall the co-function identity: $ \sin (\theta)=\cos(90^{\circ}-\theta)$ Therefore, $\dfrac{\cos (40^{\circ})}{\sin (50^{\circ})} \\=\dfrac{\cos \ 40^{\circ}}{\cos(90^{\circ}- 50^{\circ})} \\=\dfrac{\cos \ 40^{\circ}}{\cos \ 40^{\circ}} \\=1$
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