As observed from the top of a 80 m tall light house, the angles of depression of two ships on the same side of the light house in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to nearest meter.

Let AB be the tower of length 80 m and the ships be at point C and D.

From figure,

∠ADB = ∠EAD = 30° (Alternate angles are equal)
∠ACB = ∠EAC = 40° (Alternate angles are equal)

Considering right angled △ADB, we get

$\Rightarrow \text{tan 30°} = \dfrac{AB}{DB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{80}{DB} \\[1em] \Rightarrow DB = 80 \times \sqrt{3} = 138.56$

Considering right angled △ACB, we get

$\Rightarrow \text{tan 40°} = \dfrac{AB}{BC} \\[1em] \Rightarrow 0.8391 = \dfrac{80}{BC} \\[1em] \Rightarrow BC = \dfrac{80}{0.8391} \\[1em] \Rightarrow BC = 95.34$

Distance between two ships (DC) = DB - BC = 138.56 - 95.34 = 43.22 meters.

Rounding off to nearest meter DC = 43 meters.

Hence, the distance between two ships is 43 meters.