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: 14


$b\approx5.7, A\approx35^\circ, C\approx90^\circ$

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 $b^2=(4)^2+(7)^2-2(4)(7)cos(55^\circ)\approx 32.88$, thus $b=\sqrt {32.88}\approx 5.7$ Step 2. Using the Law of Sines, we have $\frac{SinA}{a}=\frac{SinB}{b}$, $SinA=\frac{4}{\sqrt {32.88}}sin(55^\circ)\approx 0.5714$, thus $A=sin^{-1}(0.5714)\approx35^\circ$ Step 3. We can find the angle as $C=180^\circ-55^\circ-35^\circ=90^\circ$ Step 4. We solved the triangle with $b\approx5.7, A\approx35^\circ, C\approx90^\circ$
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