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: 15


$\angle C=95$ $| BC| =\dfrac {18\sin 37}{\sin 95}\approx 10.874;$ $\left| AC\right| =\dfrac {18\sin 48}{\sin 95}\approx 13.4$

Work Step by Step

Sum of internal angles of a triangle is $180^0$ then $\angle C=180-\left( 37+48\right) =95$ From the law of sines $\dfrac {18}{\sin \angle C}=\dfrac {\left| BC\right| }{\sin 37}=\dfrac {\left| AC\right| }{\sin 48}\Rightarrow \left| BC\right| =\dfrac {18\sin 37}{\sin 95}\approx 10.874;\left| AC\right| =\dfrac {18\sin 48}{\sin 95}\approx 13.4$
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