Answer
$60^o$
Work Step by Step
The value of $\theta$ must be in the interval $(-90^o, 90^o)$.
Note that $\tan{60^o} = \sqrt{3}$.
Thus, $\tan^{-1}{\sqrt3}=60^o$
Trigonometry (11th Edition) Clone
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
978-0-13-421743-7
ISBN 13:
978-0-13421-743-7
Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 38
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