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Question
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?
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.
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