## Precalculus (6th Edition) Blitzer

The two sets of possible values are $B=55{}^\circ,\text{ }C=86{}^\circ,\text{ and }c\approx 31.7\text{ or }B=125{}^\circ,\text{ }C=16{}^\circ,\text{ and }c\approx 8.8$.
For any triangle, The law of sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$ The law of cosines states that \begin{align} & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\ & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\ & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\ \end{align} Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{20}{\sin 39{}^\circ }=\frac{26}{\sin B} \\ & 20\cdot \sin B=26\cdot \sin 39{}^\circ \\ & \sin B=\frac{26\cdot \sin 39{}^\circ }{20} \end{align} We will simplify it further to get the value of B. \begin{align} & \sin B=0.81 \\ & B={{\sin }^{-1}}\left( 0.81 \right) \\ & B\approx 55{}^\circ,125{}^\circ \end{align} Using the angle sum property for both the possible values of angle B, we get: For $B=55{}^\circ$, \begin{align} & 39{}^\circ +55{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -94{}^\circ \\ & C=86{}^\circ \end{align} For $B=125{}^\circ$, \begin{align} & 39{}^\circ +125{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -164{}^\circ \\ & C=16{}^\circ \end{align} Now, obtain the value of side c by using both the possible values of angle C in the law of sines. For $C=16{}^\circ$, \begin{align} & \frac{20}{\sin 39{}^\circ }=\frac{c}{\sin 16{}^\circ } \\ & c\cdot \sin 39{}^\circ =20\cdot \sin 16{}^\circ \\ & c=\frac{20\cdot \sin 16{}^\circ }{\sin 39{}^\circ } \\ & c\approx 8.8 \end{align} For $C=86{}^\circ$, \begin{align} & \frac{20}{\sin 39{}^\circ }=\frac{c}{\sin 86{}^\circ } \\ & c\cdot \sin 39{}^\circ =20\cdot \sin 86{}^\circ \\ & c=\frac{20\cdot \sin 86{}^\circ }{\sin 39{}^\circ } \\ & c\approx 31.7 \end{align} So, the two possible sets of combinations are as follows: \begin{align} & B=55{}^\circ,\text{ }C=86{}^\circ,\text{ and }c\approx 31.7\text{ } \\ & B=125{}^\circ,\text{ }C=16{}^\circ,\text{ and }c\approx 8.8 \end{align}