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Question
From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.
Solution
Let AB be the lighthouse and C be the position of the ship from the lighthouse.
Suppose the distance of the ship from the lighthouse be x m.
Here, AB = 100 m and ∠ACB = 30º.
In right ∆ABC,
\[\tan30^\circ = \frac{AB}{BC}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x}\]
\[ \Rightarrow x = 100\sqrt{3} m\]
Thus, the ship is \[100\sqrt{3}\] m away from the lighthouse.
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