Answer :

Let hill be AB and tower be CD

Given: Height of the tower = 50 m

Hence, CD = 50 m

Angle of elevation of top of tower from the foot of a hill = 30°

Hence, ∠ CBD = 30°

Angle of elevation of top of the hill at the foot of tower = 60°

Hence, ∠ ADB = 60°

Now, In right ΔCBD, we have

⇒BD = 50√3

In right ΔADB, we have

⇒ h = (50√3) × (√3)

⇒ h = 150 m

Hence, the height of the hill is 150m

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