Precalculus (6th Edition) Blitzer

There are two triangles possible and the solution are ${{B}_{1}}\approx 56{}^\circ,{{C}_{1}}\approx 112{}^\circ,{{c}_{1}}\approx 31.2$ and ${{B}_{2}}\approx 124{}^\circ,{{C}_{2}}\approx 44{}^\circ,{{c}_{2}}\approx 23.4$.
Using the ratio $\frac{a}{\sin A}$,or $\frac{7}{\sin 12{}^\circ }$, We will use the law of sines to find B. \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{7}{\sin 12{}^\circ }=\frac{28}{\sin B} \\ & \sin B=\frac{28\sin 12{}^\circ }{7} \\ & \sin B\approx 0.8316 \end{align} There are two angles possible for this value: \begin{align} & {{B}_{1}}=56{}^\circ \\ & {{B}_{2}}=180{}^\circ -56{}^\circ \\ & =124{}^\circ \end{align} This implies there are two triangles. In the first triangle, \begin{align} & A=12{}^\circ \\ & {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\ & =180{}^\circ -56{}^\circ -12{}^\circ \\ & =112{}^\circ \end{align} In the second triangle, \begin{align} & A=12{}^\circ \\ & {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\ & =180{}^\circ -124{}^\circ -12{}^\circ \\ & =44{}^\circ \end{align} Using the law of sines we will find the side ${{c}_{1}}$ and ${{c}_{2}}$. For ${{c}_{1}}$ consider, \begin{align} & \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{1}}}{\sin 112{}^\circ }=\frac{7}{\sin 12{}^\circ } \\ & {{c}_{1}}=\frac{7\sin 112{}^\circ }{\sin 12{}^\circ } \\ & {{c}_{1}}\approx 31.2 \end{align} For ${{c}_{2}}$ consider, \begin{align} & \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{2}}}{\sin 44{}^\circ }=\frac{7}{\sin 12{}^\circ } \\ & {{c}_{2}}=\frac{7\sin 44{}^\circ }{\sin 12{}^\circ } \\ & {{c}_{2}}\approx 23.4 \end{align} In case of the first triangle, ${{B}_{1}}\approx 56{}^\circ,{{C}_{1}}\approx 112{}^\circ,{{c}_{1}}\approx 31.2$ In case of the second triangle, ${{B}_{2}}\approx 124{}^\circ,{{C}_{2}}\approx 44{}^\circ,{{c}_{2}}\approx 23.4$ Hence, there are two triangles possible and the solutions are ${{B}_{1}}\approx 56{}^\circ,{{C}_{1}}\approx 112{}^\circ,{{c}_{1}}\approx 31.2$ and ${{B}_{2}}\approx 124{}^\circ,{{C}_{2}}\approx 44{}^\circ,{{c}_{2}}\approx 23.4$.