Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Review Exercises - Page 94: 17


$120^{\mathrm{o}}, \qquad 240^{\mathrm{o}}$

Work Step by Step

Since $\cos\theta$ is negative, $\theta$ must lie in quadrants II or III. Since the absolute value of $\cos\theta$ is $\displaystyle \frac{1}{2}$, browsing through the table: Function Values of Special Angles, the reference angle, $\theta^{\prime}$ must be $60^{\mathrm{o}}$. Now, from $\left[\begin{array}{lllll} Quadr.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ In quadrant II $\theta'=180^{o}-\theta$ so $\theta=180^{0}-\theta^{\prime}=180^{0}-60^{0}=120^{\mathrm{o}}$, and In quadrant III , $\theta'=\theta-180^{o}$ so $\theta= 180^{0}+\theta^{\prime}=180^{0}+60^{0}=240^{\mathrm{o}}$.
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