Geometry

Trigonometry

Medium

An observer 2 m tall is 24 $\sqrt{3}$ m away from a building. The angle of elevation of the top of the building as seen by him is 30°. What is the height of the building?

24 m

20 m

30 m

26 m

✅ Correct Option: 4

The observer is 2 m tall and stands 24√3 m away from the building. The angle of elevation to the top of the building is 30°.

The angle of elevation is measured from the observer's eye level, not from the ground. Let the total height of the building be $H$ meters.

The height of the building above eye level is $(H - 2)$ meters.

Using the tangent ratio:

$\tan(30°) = \frac{\text{opposite}}{\text{adjacent}}$

$\tan(30°) = \frac{H - 2}{24\sqrt{3}}$

Since $\tan(30°) = \frac{1}{\sqrt{3}}$ :

$\frac{1}{\sqrt{3}} = \frac{H - 2}{24\sqrt{3}}$

Cross multiplying:

$1 \times 24\sqrt{3} = (H - 2) \times \sqrt{3}$

$24\sqrt{3} = (H - 2) \times \sqrt{3}$

Dividing both sides by $\sqrt{3}$ :

$24 = H - 2$

$H = 26$

Therefore, the height of the building is 26 m.

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CUET General Test Past Year Questions (Topic-Wise)

An observer 2 m tall is 24 $\sqrt{3}$ m away from a building. The angle of elevation of the top of the building as seen by him is 30°. What is the height of the building?