If from the top of a cliff, 100 m high, the angles of depression of two ships at sea are 60° and 30°, then the distance between the ships is approximately :

1. 57.6 m

2. 115.47 m

3. 173 m

4. 346 m

Let AB be the cliff and two ships be at point C and D.

In triangle ABC,

$\Rightarrow \tan 60^\circ = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \dfrac{100}{BC} \\[1em] \Rightarrow BC = \dfrac{100}{\sqrt3} \text{ m}$

In triangle ABD,

$\Rightarrow \tan 30^\circ = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{100}{BD} \\[1em] \Rightarrow BD = 100\sqrt3 \text{ m}.$

Distance between the ships CD is,

$\Rightarrow CD = BD - BC \\[1em] \Rightarrow CD = 100\sqrt{3} - \dfrac{100}{\sqrt{3}} \\[1em] \Rightarrow CD = 100 \Big( \sqrt{3} - \dfrac{1}{\sqrt{3}} \Big) \\[1em] \Rightarrow CD = 100 \Big(\dfrac{3 - 1}{\sqrt{3}}\Big) \\[1em] \Rightarrow CD = \dfrac{200}{\sqrt{3}} \\[1em] \Rightarrow CD = \dfrac{200}{1.732} = 115.47 \text{ m}.$

Hence, option 2 is the correct option.