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: 23

Answer

$41.8^\circ$ and $138.2^\circ$

Work Step by Step

Given equation is- $\sin\theta$ = $\frac{2}{3}$ To find one solution within domain, taking $\sin^{-1}$ of each side- $\theta$ = $\sin^{-1}\frac{2}{3}$ Using calculator in degree mode- $\sin^{-1}$ → $2\div3$ = 41.8103148958 i.e. $\theta$ = $\sin^{-1}(\frac{2}{3}) \approx 41.8^\circ$ (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 $180^\circ$ = $180^\circ - 41.8^\circ$ = $138.2^\circ$ So desired angles between $0^\circ$ and $180^\circ$ satisfying the given equation are $41.8^\circ$ and $138.2^\circ$
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