## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Review Exercises - Page 94: 18

#### Answer

$210^{\mathrm{o}}, \quad 330^{\mathrm{o}}$

#### Work Step by Step

Since $\sin\theta$ is negative, $\theta$ must lie in quadrants III or IV. Since the absolute value of $\sin\theta$ is $\displaystyle \frac{1}{2}$, 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 III $\theta'=\theta-180^{o}$ so $\theta=180^{\mathrm{o}}+\theta^{\prime}=180^{\mathrm{o}}+30^{\mathrm{o}}=210^{\mathrm{o}}$. In quadrant IV $\theta'=360^{o}-\theta$ so $\theta=360^{\mathrm{o}}-\theta^{\prime}=360^{\mathrm{o}}-30^{\mathrm{o}}=330^{\mathrm{o}}$.

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