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: 46

Answer

$70^o$

Work Step by Step

RECALL: An angle $\theta$, where $-360^o \lt \theta \lt 0$, is coterminal with: $\theta+ 360^o$ Thus, the given angle is coterminal with: $=-250^o + 360^o \\=110^o$ $-250^o$ is coterminal with $110^o$. $110^o$ is in Quadrant II so $-250^o$ is also in Quadrant II. 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: $360^o-\theta$ Use formula (2) above to obtain: reference angle of $-250^o$ = reference angle of $110^o $, which is $180^o-110^o=70^o$
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