## 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: 6

#### Answer

We can draw one triangle with the given parts. $A = 40^{\circ}, B = 33^{\circ},$ and $C = 107^{\circ}$

#### Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(30)~sin~(40^{\circ})}{35}$ $B = arcsin(0.551)$ $B = 33^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-40^{\circ}-33^{\circ}$ $C = 107^{\circ}$ Note that we can also find another angle for B. $B = 180-33^{\circ} = 147^{\circ}$ 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.