## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Review Exercises - Page 94: 37

#### Answer

$47.1^{o},\qquad 132.9^{o}$

#### Work Step by Step

The calculator, set to DEGREES, for $\sin^{-1}$ (0.73254290) returns 47.1000001021 $\approx 47.1^{o}$ This is the reference angle, in quadrant I. Since sine is also positive in quadrant II, and for $\theta$ in quadrant II, we calculate the reference angle with $\theta^{\prime}=180^{\mathrm{o}}-\theta$, we find the angle $\theta$ with $\theta=180^{\mathrm{o}}-\theta^{\prime}$ $=180^{\mathrm{o}}-47.1^{o}=132.9^{o}$

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