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Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower. [CBSE 2011]

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Solution


Let PQ be the tower.

We have,

AB=10 m, MAP=30° and PBQ=60°Also, MQ=AB=10 mLet BQ=x and PQ=hSo, AM=BQ=x and PM=PQ-MQ=h-10In BPQ,tan60°=PQBQ3=hxx=h3 .....iNow, in AMP,tan30°=PMAM13=h-10xh3-103=xh3-103=h3 Using i3h-30=h3h-h=302h=30h=302 h=15 m

So, the height of the tower is 15 m.

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