Find the angle of elevation of the top of a 250√3 m high tower, from a point which is 250 m away from its foot.

Question

Question

This question was previously asked in

RRB NTPC Graduate Level CBT-I Official Paper (Held On: 06 Jun, 2025 Shift 2)

Answer 1

Given:

Height of the tower (h) = 250\(\sqrt{3}\) m

Distance of the point from the foot of the tower (b) = 250 m

Formula used:

In a right-angled triangle, tan\(\theta\) = \(\frac{Opposite\ side}{Adjacent\ side}\)

Here, Opposite side = Height of the tower

Adjacent side = Distance from the foot of the tower

Calculations:

Let \(\theta\) be the angle of elevation.

tan\(\theta\) = \(\frac{Height\ of\ the\ tower}{Distance\ from\ the\ foot\ of\ the\ tower}\)

tan\(\theta\) = \(\frac{250\sqrt{3}}{250}\)

⇒ tan\(\theta\) = \(\sqrt{3}\)

We know that tan(60°) = \(\sqrt{3}\)

⇒ \(\theta\) = 60°

∴ The angle of elevation is 60°.

Download Solution PDF