Precalculus (6th Edition) Blitzer

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

Chapter 4 - Review Exercises - Page 645: 55

Answer

$\dfrac{\pi}{6}$

Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta $ b) Quadrant -II: $(\pi-\theta)$ c) Quadrant- III: $(\theta - \pi)$ d) Quadrant - IV: $(2\pi - \theta)$ First we subtract $360^{\circ}=2\pi$: $\dfrac{17\pi}{6}-2\pi=\dfrac{5\pi}{6}$ We are in quadrant 2, so subtract from $\pi$: $\pi-\dfrac{5\pi}{6}=\dfrac{\pi}{6}$
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