The angle of elevation of the top of a building from the foot of the tower is… - CUET UG 2025 General Test PYQ

CUET UG 2025 — General Test Geometry

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The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Official previous-year question

Held on 23 May 2025 · Verified 13 Jul 2026.

Options

A
10 m
B
20/3 m
C
50/3 m
D
40/3 m

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Solution

Let the height of the building be $h$ m and the distance between the building and tower be $d$ m.

The tower height is 50 m.

From the foot of the tower, the angle of elevation to the top of the building is 30°.

$\tan(30°) = \frac{h}{d}$

$\frac{1}{\sqrt{3}} = \frac{h}{d}$

$d = h\sqrt{3}$ ... (1)

From the foot of the building, the angle of elevation to the top of the tower is 60°.

$\tan(60°) = \frac{50}{d}$

$\sqrt{3} = \frac{50}{d}$

$d = \frac{50}{\sqrt{3}}$ ... (2)

The distance $d$ is the same in both cases. Equating (1) and (2):

$h\sqrt{3} = \frac{50}{\sqrt{3}}$

$h \times \sqrt{3} \times \sqrt{3} = 50$

$h \times 3 = 50$

$h = \frac{50}{3}$

Therefore, the height of the building is $\frac{50}{3}$ m.

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