Trigonometry (11th Edition) Clone

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

Chapter 3 - Radian Measure and the Unit Circle - Section 3.1 Radian Measure - 3.1 Exercises - Page 105: 78



Work Step by Step

Convert the angle measure to degrees to obtain: $=\frac{5\pi}{6} \cdot \frac{180^o}{\pi} = 5(30^o)=150^0$ Thus, $\tan{(\frac{5\pi}{6})} \\= \tan{150^o}$ 150 degrees is in Quadrant II so : (i) its reference angle is $180^o-150^o=30^o$ (ii) its tangent value is negative From Section 2.1 (page 50) , we learned that: $\tan{30^o} = \frac{\sqrt3}{3}$ This means that: $\tan{150^o} = \tan{30^o} = \frac{\sqrt3}{3}$.
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