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 729: 4



Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $a^2=b^2+c^2-2bc\ cosA=(6)^2+(15)^2-2(6)(15)cos(22^\circ)\approx94.11$, thus $c\approx\sqrt {94.11}\approx9.7$ Step 2. To find the unknown angles, using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{6}{9.7}sin(22^\circ)\approx0.2317$, thus $B=sin^{-1}(0.2317)\approx13^\circ$ Step 3. We can find the angle as $C=180^\circ-22^\circ132^\circ=145^\circ$ Step 4. We solved the triangle with $a=9.7,B=13^\circ,C=145^\circ$
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