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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.

Solution

✅ Correct Option: 3

Let the height of the building be hh m and the distance between the building and tower be dd 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°)=hd\tan(30°) = \frac{h}{d}

13=hd\frac{1}{\sqrt{3}} = \frac{h}{d}

d=h3d = h\sqrt{3} ... (1)


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

tan⁡(60°)=50d\tan(60°) = \frac{50}{d}

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

d=503d = \frac{50}{\sqrt{3}} ... (2)


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

h3=503h\sqrt{3} = \frac{50}{\sqrt{3}}

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

h×3=50h \times 3 = 50

h=503h = \frac{50}{3}

Therefore, the height of the building is 503\frac{50}{3} m.

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