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: 13

Answer

The angles are $A=29.7^{\circ}, B=49.1^{\circ}$, and $C=101.2^{\circ}$ The angles are $A=29.7^{\circ}, B=130.9^{\circ}$, and $C=19.4^{\circ}$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(41.5~ft)~sin~(29.7^{\circ})}{27.2~ft}$ $B = arcsin(0.7559)$ $B = 49.1^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-29.7^{\circ}-49.1^{\circ}$ $C = 101.2^{\circ}$ The angles are $A=29.7^{\circ}, B=49.1^{\circ}$, and $C=101.2^{\circ}$ Note that we can also construct another triangle. $B = 180-49.1^{\circ} = 130.9^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-29.7^{\circ}-130.9^{\circ}$ $C = 19.4^{\circ}$ The angles are $A=29.7^{\circ}, B=130.9^{\circ}$, and $C=19.4^{\circ}$
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