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From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100m high, how far is P from the foot of the tower ?

Heights & Distances

ICSE

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Answer

Let MO be the tower with O being the foot of tower.

From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100m high, how far is P from the foot of the tower? Heights and Distances, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Given, angle of elevation of the top of a tower is 30° and height of tower is 100m.

∴ ∠MPO = 30° and OM = 100 m.

In △POM, ∠POM = 90°.

From △POM, we get

⇒tan 30°=OMOP⇒13=100OP⇒OP=100×3⇒OP=173.2\Rightarrow \text{tan 30°} = \dfrac{OM}{OP} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{100}{OP} \\[1em] \Rightarrow OP = 100 \times \sqrt{3} \\[1em] \Rightarrow OP = 173.2

Hence, the point P is at a distance of 173.2 metres from the foot of the tower.

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