Trigonometry (11th Edition) Clone

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

Chapter 2 - Test - Page 97: 9

Answer

$240^{o},\quad 300^{o}$

Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quad.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ Cosecant is negative in quadrants III and IV. Browsing through: Function Values of Special Angles, we find that $\displaystyle \csc 60^{o}=\frac{2\sqrt{3}}{3}$, so the reference angle is $60^{o}$ In quadrant III, $\theta^{\prime}=\theta-180^{o}$ so $\theta=180^{o}+\theta^{\prime}=180^{o}+60^{o}=240^{o}$ In quadrant IV, $\theta^{\prime}=360^{o}-\theta$ so $\theta=360^{o}-\theta^{\prime}=360^{o}-60^{o}=300^{o}$
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