Two building are in front of each other on either side of a road of width 10 metres. From the top of the first building which is 40 metres high, the angle of elevation to the top of the second is 45°. What is the height of the second building?

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#### Solution

Let AB and CD represent two buildings. AB = 40 m, BC is the width of the road.

BC = 10 m

m∠MAD = 45° ---- (angle of elevation)

ABCM is a rectangle.

AM = BC = 10 m ---(1)

AB = MC = 40 m ---(2)

Let MD = x,

Then in right angled ΔAMC,

tan ∠MAD = tan45° = MD/MA

∴ 1 = x/10

∴ x = 10

Now,

CD = CM + MD = 34 + 10 = 50 m.

Thus the height of the second building is 50 m.

Concept: Heights and Distances

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