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: 10


$c\approx7.1, A\approx159^\circ, B\approx6^\circ$

Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $c^2=a^2+b^2-2ab\ cosC$, we have $c^2=(10)^2+(3)^2-2(10)(3)cos(15^\circ)\approx51.04$, thus $c=\sqrt {51.04}\approx7.1$ Step 2. Using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinC}{c}$, $SinB=\frac{3}{7.1}sin(15^\circ)\approx0.1094$, thus $B=sin^{-1}(0.1094)\approx6^\circ$ Step 3. We can find the angle as $A=180^\circ-15^\circ-6^\circ=159^\circ$ Step 4. We solved the triangle with $c\approx7.1, A\approx159^\circ, B\approx6^\circ$
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