## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 310: 19

#### Answer

The angles are $A=142.13^{\circ}, B=27.19^{\circ}$, and $C=10.68^{\circ}$

#### Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(5.432~ft)~sin~(142.13^{\circ})}{7.297~ft}$ $B = arcsin(0.457)$ $B = 27.19^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-142.13^{\circ}-27.19^{\circ}$ $C = 10.68^{\circ}$ The angles are $A=142.13^{\circ}, B=27.19^{\circ}$, and $C=10.68^{\circ}$ Note that we can also find another value for angle B. $B = 180-27.19^{\circ} = 152.81^{\circ}$ However, we can not form a triangle with this angle B and angle A since these two angles sum to more than $180^{\circ}$

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.