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.

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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,

$\tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x} \Rightarrow x = 100 \sqrt{3} m$
Thus, the ship is $100 \sqrt{3}$ m away from the lighthouse.

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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.

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°=ABBC⇒13=100x⇒x=1003 m Thus, the ship is 1003 m away from the lighthouse.