Draw the necessary diagram for this question.

From figure,

Let the top of the lighthouse be T, the base of the lighthouse be L, and the two ships be A and B.

Given,

The height of the lighthouse, TL, is 80 m.

The angles of depression are 30° and 50°.

These are equal to the alternate interior angles of elevation at the ships.

∠TBL = 50° and ∠TAL = 30°

In triangle TLA,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 30° = \frac{TL}{LA} \\[1em] \Rightarrow LA = \frac{TL}{\tan 30°} \\[1em] \Rightarrow LA = \frac{80}{0.577} \\[1em] \Rightarrow LA = 138.65 \text{ m}.\end{aligned}$$

In triangle TLB,

$$\begin{aligned}\Rightarrow \tan 50° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 50° = \frac{TL}{LB} \\[1em] \Rightarrow LB = \frac{TL}{\tan 50°} \\[1em] \Rightarrow LB = \frac{80}{1.192} \\[1em] \Rightarrow LB = 67.11 \text{ m}.\end{aligned}$$

Distance between ships = LA + LB

= 138.65 m + 67.11 m = 205.76 m.

Hence, the distance between the two ships is 206 m.