# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 5

The required values are $C=95{}^\circ,b\approx 81.0,$ and $c\approx 134.1$.

#### Work Step by Step

At first we will find C. Property of a triangle: The sum of three angles is $A+B+C=180{}^\circ$ \begin{align} & A+B+C=180{}^\circ \\ & 48{}^\circ +37{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -85{}^\circ \\ & C=95{}^\circ \end{align} Now, to find the remaining sides, we will use the ratio $\frac{a}{\sin A}$ or $\frac{100}{\sin 48{}^\circ }$, Now, we will use the law of sines to find b. \begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{b}{\sin 37{}^\circ }=\frac{100}{\sin 48{}^\circ } \\ & b=\frac{100\sin 37{}^\circ }{\sin 48{}^\circ } \\ & b=81.0 \end{align} Again use the law of sines to find c \begin{align} & \frac{c}{\sin C}=\frac{a}{\sin A} \\ & \frac{c}{\sin 95{}^\circ }=\frac{100}{\sin 48{}^\circ } \\ & c=\frac{16\sin 95{}^\circ }{\sin 48{}^\circ } \\ & c=134.1 \end{align} The solution is $C=95{}^\circ,b\approx 81.0$, and $c\approx 134.1$.

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