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: 6


$A\approx39^\circ, B\approx49^\circ, C\approx92^\circ$

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 $(10)^2=(12)^2+(16)^2-2(12)(16)cosA$, thus $cosA\approx0.7813$ and $A=cos^{-1}(0.7813)\approx39^\circ$ Step 2. Using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{12}{10}sin(39^\circ)\approx0.7552$, thus $B=sin^{-1}(0.7552)\approx49^\circ$ Step 3. We can find the angle as $C=180^\circ-39^\circ-49^\circ=92^\circ$ Step 4. We solved the triangle with $A\approx39^\circ, B\approx49^\circ, C\approx92^\circ,$
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