Answer
There are two triangles and the solutions are ${{B}_{1}}\approx 77{}^\circ,\ {{C}_{1}}\approx 43{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 12.6$, and
${{B}_{2}}\approx 103{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.4$.
Work Step by Step
We will use the law of sines to find B:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{16}{\sin 60{}^\circ }=\frac{18}{\sin B} \\
& \sin B=\frac{18\sin 60{}^\circ }{16} \\
& \sin B\approx 0.9743
\end{align}$
There are two angles between 0° and 180° for which $\sin B\approx 0.9743$ is possible:
${{B}_{1}}\approx 77{}^\circ $
As we know that the sine function is positive in the second quadrant, so the second angle is
$\begin{align}
& {{B}_{2}}\approx 180{}^\circ -77{}^\circ \\
& \approx 103{}^\circ
\end{align}$
So, there are two triangles.
In the first triangle, using the angle sum property we get
$\begin{align}
& {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\
& \approx 180{}^\circ -77{}^\circ -60{}^\circ \\
& \approx 43{}^\circ
\end{align}$
In the second triangle, using the angle sum property we get
$\begin{align}
& {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\
& \approx 180{}^\circ -103{}^\circ -60{}^\circ \\
& \approx 17{}^\circ
\end{align}$
Using the law of sines we will find the sides ${{c}_{1}}$ and ${{c}_{2}}$.
For ${{c}_{1}}$,
$\begin{align}
& \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{1}}}{\sin 43{}^\circ }=\frac{16}{\sin 60{}^\circ } \\
& {{c}_{1}}=\frac{16\sin 43{}^\circ }{\sin 60{}^\circ } \\
& \approx 12.6
\end{align}$
For ${{c}_{2}}$,
$\begin{align}
& \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{2}}}{\sin 17{}^\circ }=\frac{16}{\sin 60{}^\circ } \\
& {{c}_{2}}=\frac{16\sin 17{}^\circ }{\sin 60{}^\circ } \\
& \approx 5.4
\end{align}$
In case of the first triangle, the solution is
${{B}_{1}}\approx 77{}^\circ,\ {{C}_{1}}\approx 43{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 12.6$
In the other triangle, the solution is
${{B}_{2}}\approx 103{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.4$
