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: 16

Answer

$\angle A=180-\left( 29+52\right) =99$ $\left| AB\right| =\dfrac {43\sin 29}{\sin 99}\approx 21.1;$ $\left| AC\right| =\dfrac {43\sin 52}{\sin 99}\approx 34.3$

Work Step by Step

Sum of internal angles of triangle is $180^0$ then $\angle A=180-\left( 29+52\right) =99$ From the law of sines $\dfrac {43}{\sin \angle A}=\dfrac {\left| AB\right| }{\sin 29}\Rightarrow \left| AB\right| =\dfrac {43\sin 29}{\sin 99}\approx 21.1;$ $\left| AC\right| =\dfrac {43\sin 52}{\sin 99}\approx 34.3$
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