Answer
There are two triangles and the solutions are
${{C}_{1}}\approx 68{}^\circ,\ {{B}_{1}}\approx 54{}^\circ,\ {{b}_{1}}\approx 21.0$ and ${{C}_{2}}\approx 112{}^\circ,{{B}_{2}}\approx 10{}^\circ,\ {{b}_{2}}\approx 4.5$.
Work Step by Step
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
${{C}_{2}}\approx 112{}^\circ,{{B}_{2}}\approx 10{}^\circ,{{b}_{2}}\approx 4.5$
