If the shadow of a building increases by 10 meters when the angle of elevation of the sun rays decreases from 60° to 45°, what is the height of the building?
Explanation:
Initially the shadow of CB is AB and later the shadow becomes DB.
Let BC = x
In ∆ABC, tan60° = √3 = BC/AB
⇒ AB = x/√3
In ∆DBC, tan45° = 1 = BC/DB
⇒ DB = x
We know, DB – AB = AD = 10 meters.
⇒ x – x/√3 = 10
⇒ (√3 - 1)x/√3 = 10
⇒ x = 10√3/(√3 - 1)
Hence, option (b).
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