Question

Two buildings 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 20 metres high, the angle of elevation to the top of the second is 45°. What is the height of the second building?

Answer 1

Let AB and CD represent two buildings. AB = 20 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 = 20 m ---(2)

Let MD = x,

Then in right angled ΔAMC,

tan ∠MAD = tan 45° = `"MD"/"MA"`

∴ `1 = x/10`

∴ x = 10

Now,

CD = CM + MD = 20 + 10 = 30 m.

Thus the height of the second building is 30 m.