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: 16


$b\approx7.6, A\approx67^\circ, C\approx23^\circ$

Work Step by Step

Step 1. Based on the given conditions, using the Pythagorean Theorem, we have $b=\sqrt {a^2+c^2}=\sqrt {7^2+3^2}=\sqrt {58}\approx7.6$ Step 2. For angle $A$, in the right triangle, we have $tanA=\frac{a}{c}=\frac{7}{3}$, thus $A=tan^{-1}(\frac{7}{3})\approx67^\circ$ Step 3. We can find the angle as $C=90^\circ-67^\circ=23^\circ$ Step 4. We solved the triangle with $b\approx7.6, A\approx67^\circ, C\approx23^\circ$
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.