Answer :
Let Height of the House CD, Height of the Hill is AB.
Given, CD = 12, AE = 12
∠BCE = 30o
∠BDA = 600
In ∆ABD,
AD√3 = BE + 12
………..(1)
In ∆BCE,
EC = BE…………..(2)
BE + 12 = BE√3
12 = BE(√3 – 1)
BE = 6(√3 + 1)
Height of the tower = BE + AE = 6√3 + 6 + 12
= 28.39
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