Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.4 - Inverse Trigonometric Functions and Right Triangles - 6.4 Exercises - Page 507: 28

Answer

$53.1^\circ$ and $126.9^\circ$

Work Step by Step

Given equation is- $\sin\theta$ = $\frac{4}{5}$ To find one solution within domain, taking $\sin^{-1}$ of each side- $\theta$ = $\sin^{-1}\frac{4}{5}$ Using calculator in degree mode- $\sin^{-1}$ → $4\div5$ = 53.1301023542 i.e. $\theta$ = $\sin^{-1}(\frac{4}{5}) \approx 53.1^\circ$ (Rounded to one decimal place) Another solution with $\theta$ between $0^\circ$ and $180^\circ$ is obtained by taking the supplement of the angle as this will be the reference angle of desired angle. Therefore- Second solution with $\theta$ between $0^\circ$ and $180^\circ$ = $180^\circ - 53.1^\circ$ = $126.9^\circ$ So desired angles between $0^\circ$ and $180^\circ$ satisfying the given equation are $53.1^\circ$ and $126.9^\circ$
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