Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 644: 24

Answer

$\sqrt 3$

Work Step by Step

The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ Now, $\tan \theta= \dfrac{Opposite}{Adjacent}$ and $\dfrac{\sin \theta}{\cos \theta}= \dfrac{-\sqrt 3/2}{(-1/2)}$ So, our required answer is: $\tan \theta=\sqrt 3$
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