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: 19


$150^{\mathrm{o}},\quad 210^{\mathrm{o}}$

Work Step by Step

Since $\sec\theta$ is negative, $\theta$ must lie in quadrants II or III. Since the absolute value of $\sec\theta$ is $\displaystyle \frac{2\sqrt{3}}{3}$, browsing through the table: Function Values of Special Angles, the reference angle, $\theta^{\prime}$ must be $30^{\mathrm{o}}$. 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 $\theta=180^{\mathrm{o}}-\theta^{\prime}=180^{\mathrm{o}}-30^{0}=150^{\mathrm{o}}$ In quadrant III $\theta'=\theta-180^{o}$ so $\theta=180^{0}+\theta^{\prime}=180^{\mathrm{o}}+30^{\mathrm{o}}=210^{\mathrm{o}}$.
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