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: 24


$A\approx139^\circ, B\approx14^\circ, C\approx27^\circ$

Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $a^2=b^2+c^2-2bc\ cosA$, or $(66)^2=(25)^2+(45)^2-2(25)(45)cos(A)$, thus $cosA\approx-0.7582$ and $A=cos^{-1}(-0.7582)\approx139.3\approx139^\circ$ Step 2. Using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{25}{66}sin(139.3^\circ)\approx 0.247$, thus $B=sin^{-1}(0.247)\approx14^\circ$ Step 3. We can find the angle as $C=180^\circ-139^\circ-14^\circ=27^\circ$ Step 4. We solved the triangle with $A\approx139^\circ, B\approx14^\circ, C\approx27^\circ$
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