Draw the necessary diagram for this question.

Let AD be the lighthouse and B and C be the two ships on the opposite sides of the lighthouse.

Given,

Height of the lighthouse (AD) = 120 m

The angles of depression of the two ships are 32° and 58°.

From figure,

⇒ ∠FAB = ∠ABD = 32° (alternate angles are equal)

⇒ ∠EAC = ∠ACD = 58° (alternate angles are equal)

In right angled triangle ABD,

$$\begin{aligned}\Rightarrow \tan 32° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 32° = \frac{AD}{BD} \\[1em] \Rightarrow BD = \frac{AD}{\tan 32°} \\[1em] \Rightarrow BD = \frac{120}{0.6249} \\[1em] \Rightarrow BD = 192.03 \text{ m}.\end{aligned}$$

In right angled triangle ACD,

$$\begin{aligned}\Rightarrow \tan 58° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 58° = \frac{AD}{CD} \\[1em] \Rightarrow CD = \frac{AD}{\tan 58°} \\[1em] \Rightarrow CD = \frac{120}{1.6003} \\[1em] \Rightarrow CD = 74.99 \text{ m}.\end{aligned}$$

From the figure,

Distance between the two ships (BC) = BD + CD

= 192.03 + 74.99

= 267.02 m ≈ 267 m.

Hence, the distance between the two ships = 267 m.