# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 310: 16

The angles are $A=37^{\circ}50', B=48^{\circ}50'$, and $C=93^{\circ}20'$

#### Work Step by Step

We can use the law of sines to find the angle $A$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~B}{b}$ $sin~A = \frac{(3850~in)~sin~(48^{\circ}50')}{4730~in}$ $A = arcsin(0.6127)$ $A = 37^{\circ}50'$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-37^{\circ}50'-48^{\circ}50'$ $C = 93^{\circ}20'$ The angles are $A=37^{\circ}50', B=48^{\circ}50'$, and $C=93^{\circ}20'$

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