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: 23


$A\approx117^\circ, B\approx18^\circ, C\approx45^\circ$

Work Step by Step

Step 1. Based on the given conditions, using the Law of Cosines, we have $a^2=b^2+c^2-2bc\ cosA$, or $(63)^2=(22)^2+(50)^2-2(22)(50)cos(A)$, thus $cosA\approx-0.4477$ and $A=cos^{-1}(-0.4477)\approx116.6\approx117^\circ$ Step 2. Using the Law of Sines, we have $\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{22}{63}sin(116.6^\circ)\approx 0.3122$, thus $B=sin^{-1}(0.3122)\approx18^\circ$ Step 3. We can find the angle as $C=180^\circ-117^\circ-18^\circ=45^\circ$ Step 4. We solved the triangle with $A\approx117^\circ, B\approx18^\circ, C\approx45^\circ$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.