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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