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: 7


There is one possible triangle with the given parts. $A = 48^{\circ}, B = 39^{\circ},$ and $C = 93^{\circ}$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(26)~sin~(48^{\circ})}{31}$ $B = arcsin(0.623)$ $B = 39^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-48^{\circ}-39^{\circ}$ $C = 93^{\circ}$ Note that we can also find another angle for B. $B = 180-39^{\circ} = 141^{\circ}$ 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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