Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 11 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 527: 18

Answer

$33^{\circ}$

Work Step by Step

$\frac{\sin\alpha}{6}=\frac{\sin65^{\circ}}{10}$ $\sin\alpha=\frac{6\sin65^{\circ}}{10}$ $\alpha=\sin^{-1}(\frac{6\sin65^{\circ}}{10})$ $\alpha\approx33^{\circ}$
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