Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set: 35

Answer

$20^o$

Work Step by Step

RECALL: The following are the means on how to find the reference angle of an angle $\theta$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $180^o-\theta$ (3) Quadrant III: $\theta - 180^o$ (4) Quadrant IV: $360^o-\theta$ $160^o$ is in Quadrant II. Thus, using forrmula (2) above gives: reference angle of $160^o$ = $180^o-160^o=20^o$
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