Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 304: 14

Answer

The angles are $A=72.2^{\circ}, B=48.2^{\circ}$, and $C=59.6^{\circ}$ The angles are $A=107.8^{\circ}, B=48.2^{\circ}$, and $C=24.0^{\circ}$

Work Step by Step

We can use the law of sines to find the angle $A$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~A = \frac{a~sin~B}{b}$ $sin~A = \frac{(890~cm)~sin~(48.2^{\circ})}{697~cm}$ $A = arcsin(0.9519)$ $A = 72.2^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-72.2^{\circ}-48.2^{\circ}$ $C = 59.6^{\circ}$ The angles are $A=72.2^{\circ}, B=48.2^{\circ}$, and $C=59.6^{\circ}$ Note that we can also construct another triangle. $A = 180-72.2^{\circ} = 107.8^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-48.2^{\circ}-107.8^{\circ}$ $C = 24.0^{\circ}$ The angles are $A=107.8^{\circ}, B=48.2^{\circ}$, and $C=24.0^{\circ}$
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.