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Two men are on opposite side of tower. They measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 meters, find the distance between the two men.

Solution

✅ Correct Option: 1

A tower has two men standing on opposite sides. The height of the tower is 50 meters. Man 1 measures an angle of elevation of 30° to the top, and Man 2 measures an angle of elevation of 45°.

Let d1d_1 be the distance from Man 1 to the base of the tower, and d2d_2 be the distance from Man 2 to the base of the tower.


For Man 2 with angle of elevation 45°:

tan⁡(45°)=50d2\tan(45°) = \dfrac{50}{d_2}

Since tan⁡(45°)=1\tan(45°) = 1:

1=50d21 = \dfrac{50}{d_2}

d2=50d_2 = 50 m


For Man 1 with angle of elevation 30°:

tan⁡(30°)=50d1\tan(30°) = \dfrac{50}{d_1}

Since tan⁡(30°)=13\tan(30°) = \dfrac{1}{\sqrt{3}}:

13=50d1\dfrac{1}{\sqrt{3}} = \dfrac{50}{d_1}

d1=503d_1 = 50\sqrt{3}

d1=50×1.732d_1 = 50 \times 1.732

d1=86.6d_1 = 86.6 m


Since the men are on opposite sides of the tower, the total distance between them is:

Total Distance =d1+d2= d_1 + d_2

Total Distance =86.6+50= 86.6 + 50

Total Distance =136.6= 136.6 m

Therefore, the distance between the two men is 136.6 meters.

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