(C) 50 m
Let the height of the building AB be h metres and the distance between AB and tower EC be x metres.
Given, ∠EAD = 30°, ∠EBC = 60°
In rt. Δ ADE,
tan 30° = \(\frac{ED}{AD}=\frac{75-h}{x}\)
⇒ \(\frac{1}{\sqrt3}= \frac{75-h}{x}\)
⇒ x = (75 - h)\(\sqrt3\) ...(i)
In rt. Δ EBC,
tan 60° = \(\frac{EC}{BC}\)
⇒ \(\sqrt3\) = \(\frac{75}{x}\)
⇒ \(x= \frac{75}{\sqrt3}\) ...(ii)
∴ From (i) and (ii),
\(\frac{75}{\sqrt3}\) = (75 – h)\(\sqrt3\)
⇒ 75 = (75 – h) × 3
⇒ 75 = 225 – 3h
⇒ 3h = 150
⇒ h = 50 m.