Answer
The distance (to the nearest tenth) after both ships are anchored is approximately $71.6\text{ miles}$.
Work Step by Step
Consider the origin at the harbor with the $y$ axis pointing north.
The position of the first ship is $\text{N}42{}^\circ \text{E}$ for $23\text{ miles}$, that is ${{x}_{1}}=r\cos \theta $.
Substitute, $r=23$ and $\theta =90{}^\circ -42{}^\circ $.
$\begin{align}
& {{x}_{1}}=r\cos \theta \\
& =23\cos \left( 90{}^\circ -42{}^\circ \right) \\
& =23\cos 48{}^\circ \\
& =15.4
\end{align}$
And
$\begin{align}
& {{y}_{1}}=r\sin \theta \\
& =23\sin 48{}^\circ \\
& =17.1
\end{align}$
The final position of the second ship is ${{x}_{2}}=r\cos \theta $.
Substitute, $r=72$ and $\theta =90{}^\circ +38{}^\circ $.
$\begin{align}
& {{x}_{2}}=r\cos \theta \\
& =72\cos \left( 90{}^\circ +38{}^\circ \right) \\
& =72\cos 128{}^\circ \\
& =-44.3
\end{align}$
And
$\begin{align}
& {{y}_{2}}=r\sin \theta \\
& =72\sin 128{}^\circ \\
& =56.7
\end{align}$
Now, the distance between the two points is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 15.4,17.1 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( -44.3,56.7 \right)$.
$\begin{align}
& d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} \\
& =\sqrt{{{\left( 15.4-\left( -44.3 \right) \right)}^{2}}+{{\left( 17.1-56.7 \right)}^{2}}} \\
& =71.6
\end{align}$
Hence, the two ships are approximately $71.6\text{ miles}$ apart.
