Two ships leave a port at the same time.

Solution:

After three hours, let the ships be at $P$ and $Q$ respectively.

Then,

$O P=24 \times 3=72 \mathrm{~km}$ and $O Q=32 \times 3=96 \mathrm{~km}$

From figure, we have

$\angle P O Q=180^{\circ}-\angle N O P-\angle S O Q$

$=180^{\circ}-38^{\circ}-52^{\circ}$

$=90^{\circ}$

Now,

Since $O P O$ is a right angled triangle

$\therefore P Q^{2}=O P^{2}+O Q^{2}$

$\Rightarrow P Q^{2}=72^{2}+96^{2}$

$\Rightarrow P Q^{2}=5184+9216$

$\Rightarrow P Q^{2}=14400$

$\Rightarrow P Q=\sqrt{14400}=120 \mathrm{~km}$

Hence, the distance between the ships after 3 hours is $120 \mathrm{~km}$.