Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.2 - The Law of Cosines - Exercise Set - Page 730: 13


$b\approx4.7, B\approx75^\circ, C\approx55^\circ$

Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $b^2=a^2+c^2-2ac\ cosB$, or $b^2=(6)^2+(5)^2-2(6)(5)cos(50^\circ)\approx 22.43$, thus $b=\sqrt {22.43}\approx 4.7$ Step 2. Using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinC}{c}$, $SinC=\frac{5}{4.7}sin(50^\circ)\approx 0.8149$, thus $C=sin^{-1}(0.8149)\approx55^\circ$ Step 3. We can find the angle as $B=180^\circ-50^\circ-55^\circ=75^\circ$ Step 4. We solved the triangle with $b\approx4.7, B\approx75^\circ, C\approx55^\circ$
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