Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 310: 5


There is one possible triangle. $A = 95^{\circ}, B = 31^{\circ}$, and $C = 54^{\circ}$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(26)~sin~(95^{\circ})}{50}$ $B = arcsin(0.518)$ $B = 31^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-95^{\circ}-31^{\circ}$ $C = 54^{\circ}$ Note that we can also find another angle for B. $B = 180-31^{\circ} = 149^{\circ}$ However, we can not form a triangle with this angle B and angle A since these two angles sum to more than $180^{\circ}$
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