Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Cumulative Review Exercises - Page 1038: 25


$C= 44^\circ$ $b \approx14.4$ $c \approx10.5$

Work Step by Step

Step 1. Given angles $A, B$, we can find angle $C$ as: $C=180^\circ-64^\circ-72^\circ=44^\circ$ Step 2. Using the Law of Sines, we have $\frac{sinB}{b}=\frac{sinA}{a}$, thus $b=\frac{a\ sinB}{sinA}=\frac{13.6\ sin72^\circ}{sin64^\circ}\approx14.4$ Step 3. Using the Law of Sines, we have $\frac{sinC}{c}=\frac{sinA}{a}$, thus $c=\frac{a\ sinC}{sinA}=\frac{13.6\ sin44^\circ}{sin64^\circ}\approx10.5$
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.