Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.1 The Law of Sines - 8.1 Exercises - Page 754: 14

Answer

$10\sqrt {2}$

Work Step by Step

From general properties of triangle $\angle B=180-\left( \angle A+\angle C\right) =180-\left( 105+45\right) =30$ From the law of sines $\dfrac {a}{\sin 45}=\dfrac {10}{\sin B}\Rightarrow a=\dfrac {10\sin 45}{\sin 30}=\dfrac {10\times \dfrac {\sqrt {2}}{2}}{\dfrac {1}{2}}=10\sqrt {2}$
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