Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - 8.2 The Law of Sines - 8.2 Assess Your Understanding - Page 530: 16

Answer

$C= 50^\circ$ $a \approx11.82$ $c\approx9.19$

Work Step by Step

Step 1. Using the given figure and conditions, we have $C=180-100-30=50^\circ$ Step 2. Using the Law of Sines, we have $\frac{a}{sin100^\circ}=\frac{c}{sin50^\circ}=\frac{6}{sin30^\circ}$ Step 3. Thus $a=\frac{6sin100^\circ}{sin30^\circ} \approx11.82$ and $c=\frac{6sin50^\circ}{sin30^\circ}\approx9.19$
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