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: 15

Answer

The angles are $A=112^{\circ}10', B=26^{\circ}30'$, and $C=41^{\circ}20'$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $sin~B = \frac{(25.9~m)~sin~(41.33^{\circ})}{38.4~m}$ $B = arcsin(0.4454)$ $B = 26^{\circ}30'$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-26^{\circ}30'-41^{\circ}20'$ $A = 112^{\circ}10'$ The angles are $A=112^{\circ}10', B=26^{\circ}30'$, and $C=41^{\circ}20'$
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