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: 2



Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $b^2=a^2+c^2-2ac\ cosB=(6)^2+(8)^2-2(6)(8)cos(32^\circ)\approx18.59$, thus $a\approx\sqrt {18.59}\approx4.3$ Step 2. To find the unknown angles, using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinA}{a}$, $SinA=\frac{6}{4.3}sin(32^\circ)\approx0.7394$, thus $A=sin^{-1}(0.7394)\approx48^\circ$ Step 3. We can find the angle as $C=180^\circ-32^\circ-48^\circ=100^\circ$ Step 4. We solved the triangle with $b=4.3,A=48^\circ,C=100^\circ$
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