## Precalculus (6th Edition) Blitzer

There are two triangles and the solutions are ${{C}_{1}}\approx 68{}^\circ,\ {{B}_{1}}\approx 54{}^\circ,\ {{b}_{1}}\approx 21.0$ and ${{C}_{2}}\approx 112{}^\circ,{{B}_{2}}\approx 10{}^\circ,\ {{b}_{2}}\approx 4.5$.
Using the ratio $\frac{a}{\sin A}$ or $\frac{22}{\sin 58{}^\circ }$. We will use the law of sines to find C. \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{22}{\sin 58{}^\circ }=\frac{24.1}{\sin C} \\ & \sin C=\frac{24.1\sin 58{}^\circ }{22} \\ & \sin C=0.9290 \\ & C={{\sin }^{-1}}(0.9290) \end{align} The two possible angles are: \begin{align} & {{C}_{1}}=68{}^\circ \\ & {{C}_{2}}=180{}^\circ -68{}^\circ \\ & =112{}^\circ \end{align} There are two triangles: First: \begin{align} & A=58{}^\circ \\ & {{B}_{1}}=180{}^\circ -{{C}_{1}}-A \\ & =180{}^\circ -68{}^\circ -58{}^\circ \\ & =54{}^\circ \end{align} Second: \begin{align} & A=58{}^\circ \\ & {{B}_{2}}=180{}^\circ -{{C}_{2}}-A \\ & =180{}^\circ -112{}^\circ -58{}^\circ \\ & =10{}^\circ \end{align} Using the law of sines we will find side ${{b}_{1}}$ and ${{b}_{2}}$: For ${{b}_{1}},$ \begin{align} & \frac{{{b}_{1}}}{\sin {{B}_{1}}}=\frac{a}{\sin A} \\ & \frac{{{b}_{1}}}{\sin 54{}^\circ }=\frac{22}{\sin 58{}^\circ } \\ & {{b}_{1}}=\frac{22\sin 54{}^\circ }{\sin 58{}^\circ } \\ & {{b}_{1}}\approx 21.0 \end{align} For ${{b}_{2}},$ \begin{align} & \frac{{{b}_{2}}}{\sin {{B}_{2}}}=\frac{a}{\sin A} \\ & \frac{{{b}_{2}}}{\sin 10{}^\circ }=\frac{22}{\sin 58{}^\circ } \\ & {{b}_{2}}=\frac{22\sin 10{}^\circ }{\sin 58{}^\circ } \\ & {{b}_{2}}\approx 4.5 \end{align} In one triangle, the solution is ${{C}_{1}}\approx 68{}^\circ,{{B}_{1}}\approx 54{}^\circ,{{b}_{1}}\approx 21.0$ In the other triangle, the solution is ${{C}_{2}}\approx 112{}^\circ,{{B}_{2}}\approx 10{}^\circ,{{b}_{2}}\approx 4.5$