Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-6 Trigonometric Ratios - Lesson Check - Page 637: 7

Answer

The student needed to take the inverse of sine since he is solving for an angle given the side lengths. $\angle$$A$$=64.16$$^{\circ}$

Work Step by Step

Since the student is solving for an unknown angle, he needs to solve $sin^{-1}$$(.9)$ to get 64.16$^{\circ}$.
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