Answer
The required solution corresponding to the two triangles is ${{A}_{1}}=55{}^\circ,{{B}_{1}}=83{}^\circ,{{b}_{1}}=19.3$ and ${{A}_{2}}=125{}^\circ,{{B}_{2}}=13{}^\circ,{{b}_{2}}=4.4$, respectively.
Work Step by Step
The provided angle and sides of the triangle are given below:
$C=42{}^\circ,a=16,c=13$
Now, by using the law of sines, we will find the angle A of the triangle. That is,
$\begin{align}
& \frac{\sin \,A}{a}=\frac{\sin \,C}{c} \\
& \frac{\sin \,A}{16}=\frac{\sin \,42{}^\circ }{13} \\
& \sin \,A=16\times \frac{\sin \,42{}^\circ }{13} \\
& \sin \,A=0.8235 \\
& A={{\sin }^{-1}}(0.8235)
\end{align}$
According to this measurement, two angles are possible. That is,
$\begin{align}
& {{A}_{1}}=55{}^\circ \\
& {{A}_{2}}=180{}^\circ -55{}^\circ \\
& =125{}^\circ
\end{align}$
Thus, there are two possible triangles.Therefore,
$\begin{align}
& {{B}_{1}}=180{}^\circ -C-{{A}_{1}} \\
& =180{}^\circ -42{}^\circ -55{}^\circ \\
& =83{}^\circ
\end{align}$
and
$\begin{align}
& {{B}_{2}}=180{}^\circ -C-{{A}_{2}} \\
& =180{}^\circ -42{}^\circ -125{}^\circ \\
& =13{}^\circ
\end{align}$
Using the law of sines we will find ${{b}_{1}}\ \text{ and }\ {{b}_{2}}$:
$\begin{align}
& \frac{\sin \,{{B}_{1}}}{{{b}_{1}}}=\frac{\sin \,C}{c} \\
& \frac{\sin \,83{}^\circ }{{{b}_{1}}}=\frac{\sin \,42{}^\circ }{13} \\
& {{b}_{1}}=13\times \frac{\sin \,83{}^\circ }{\sin \,42{}^\circ } \\
& {{b}_{1}}=19.3
\end{align}$
and
$\begin{align}
& \frac{\sin \,{{B}_{2}}}{{{b}_{2}}}=\frac{\sin \,C}{c} \\
& \frac{\sin \,13{}^\circ }{{{b}_{2}}}=\frac{\sin \,42{}^\circ }{13} \\
& {{b}_{2}}=13\times \frac{\sin \,13{}^\circ }{\sin \,42{}^\circ } \\
& {{b}_{2}}=4.4
\end{align}$
In one triangle, the solution is ${{A}_{1}}=55{}^\circ,{{B}_{1}}=83{}^\circ,{{b}_{1}}=19.3$.
In the second triangle, the solution is ${{A}_{2}}=125{}^\circ,{{B}_{2}}=13{}^\circ,{{b}_{2}}=4.4$.
