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: 9


$c\approx4.7, A\approx45^\circ, B\approx93^\circ$

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 $c^2=(5)^2+(7)^2-2(5)(7)cos(42^\circ)\approx21.98$, thus $c=\sqrt {21.98}\approx4.7$ Step 2. Using the Law of Sines, we have $\frac{SinA}{a}=\frac{SinC}{c}$, $SinA=\frac{5}{4.7}sin(42^\circ)\approx0.7118$, thus $A=sin^{-1}(0.7118)\approx45^\circ$ Step 3. We can find the angle as $B=180^\circ-42^\circ-45^\circ=93^\circ$ Step 4. We solved the triangle with $c\approx4.7, A\approx45^\circ, B\approx93^\circ$
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