The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.

Let CD be the light house, 60 m tall and ships at point A and B.

From figure,

∠DAC = ∠XDA = 60° (Alternate angles are equal)
∠DBC = ∠YDB = 45° (Alternate angles are equal)

Considering right angled △ACD, we get

$\Rightarrow \text{tan 60°} = \dfrac{CD}{AC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{60}{AC} \\[1em] \Rightarrow AC = \dfrac{60}{\sqrt{3}} = 34.64$

Considering right angled △BCD, we get

$\Rightarrow \text{tan 45°} = \dfrac{CD}{BC} \\[1em] \Rightarrow 1 = \dfrac{60}{BC} \\[1em] \Rightarrow BC = 60$

Distance between two ships (AB) = AC + BC = 34.64 + 60 = 94.64 meters.

Rounding off to nearest meter, AB = 95 meters.

Hence, the distance between two ships is 95 meters.