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: 13


$\sqrt 3$

Work Step by Step

Sum of Internal angles of triangle is $180^0$ Then $\angle C=180-\left( 60+75\right) =45$ From the laws of sines $\dfrac {\sqrt {2}}{\sin \angle C}=\dfrac {a}{\sin 60}\Rightarrow a=\dfrac {\sqrt {2}\times \sin 60}{\sin 45}=\dfrac {\sqrt {2}\times \dfrac {\sqrt {3}}{2}}{\dfrac {\sqrt {2}}{2}}=\sqrt {3} $
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