## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 310: 8

#### Answer

There are two possible triangles with the given parts. $A = 46^{\circ}, B= 54^{\circ},$ and $C = 80^{\circ}$ $A = 26^{\circ}, B= 54^{\circ},$ and $C = 100^{\circ}$

#### Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $sin~C = \frac{c~sin~B}{b}$ $sin~C = \frac{(28)~sin~(54^{\circ})}{23}$ $C = arcsin(.9849)$ $C = 80^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-54^{\circ}-80^{\circ}$ $A = 46^{\circ}$ Note that we can also find another angle for C. $C = 180-80^{\circ} = 100^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-54^{\circ}-100^{\circ}$ $A = 26^{\circ}$

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