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: 17


$A\approx28^\circ, B\approx40^\circ, C\approx112^\circ$

Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $c^2=a^2+b^2-2ab\ cosC$, or $(10)^2=(5)^2+(7)^2-2(5)(7)cos(C)$, thus $cosC\approx-0.3714$ and $C=cos^{-1}(-0.3714)\approx112^\circ$ Step 2. Using the Law of Sines, we have $\frac{SinA}{a}=\frac{SinC}{c}$, $SinA=\frac{5}{10}sin(112^\circ)\approx 0.4636$, thus $A=sin^{-1}(0.4636)\approx28^\circ$ Step 3. We can find the angle as $B=180^\circ-112^\circ-28^\circ=40^\circ$ Step 4. We solved the triangle with $A\approx28^\circ, B\approx40^\circ, C\approx112^\circ$
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