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: 19


$A\approx19^\circ, B\approx100^\circ, C\approx61^\circ$

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 $(9)^2=(3)^2+(8)^2-2(3)(8)cos(B)$, thus $cosB\approx-0.1667$ and $B=cos^{-1}(-0.1667)\approx100^\circ$ Step 2. Using the Law of Sines, we have $\frac{SinA}{a}=\frac{SinB}{b}$, $SinA=\frac{3}{9}sin(100^\circ)\approx 0.3283$, thus $A=sin^{-1}(0.3283)\approx19^\circ$ Step 3. We can find the angle as $C=180^\circ-100^\circ-19^\circ=61^\circ$ Step 4. We solved the triangle with $A\approx19^\circ, B\approx100^\circ, C\approx61^\circ$
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