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: 5



Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $a^2=b^2+c^2-2bc\ cosA$, we have $(6)^2=(8)^2+(8)^2-2(8)(8)cosA$, thus $cosA\approx0.7188$ and $A=cos^{-1}(0.7188)\approx44^\circ$ Step 2. As $b=c$, the triangle is isosceles; thus $B=C=\frac{180-44}{2}=68^\circ$ Step 3. We solved the triangle with $A=44^\circ,B=C=68^\circ$
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