Right Triangle Positions of Moving Ships

Let the port be $P$, slower ship be $S_1$, faster ship be $S_2$. Initially, $PS_1 = PS_2 = 24\text{ km}$, angle $\angle S_1 P S_2 = 60^\circ$.

When $S_1$ travels 8 km towards $P$, its distance from $P$ becomes $24 - 8 = 16\text{ km}$. Let $x$ be the distance traveled by $S_2$. Its distance from $P$ becomes $24 - x$.

In $\triangle S_1' P S_2'$, sides are $16$ and $24 - x$ with included angle $60^\circ$. For this triangle to be right-angled, either angle at $S_1'$ or at $S_2'$ is $90^\circ$.

Case 1: Angle at $S_1'$ is $90^\circ$. In right $\triangle S_1' P S_2'$ with $\angle P = 60^\circ$: $\cos 60^\circ = \frac{P S_1'}{P S_2'} = \frac{16}{24 - x} = \frac{1}{2} \implies 24 - x = 32 \implies x = -8 \text{ (invalid)}$

Case 2: Angle at $S_2'$ is $90^\circ$. $\cos 60^\circ = \frac{P S_2'}{P S_1'} = \frac{24 - x}{16} = \frac{1}{2} \implies 24 - x = 8 \implies x = 16$

So $S_2$ traveled 16 km while $S_1$ traveled 8 km. Speed ratio $v_2 / v_1 = 16 / 8 = 2$.

When the faster ship $S_2$ reaches the port (travels full 24 km), the slower ship $S_1$ travels $24 / 2 = 12\text{ km}$. Distance between $S_1$ and the port $= 24 - 12 = 12\text{ km}$.