Answer :

Let Height of the House CD, Height of the Hill is AB.

Given, CD = 12, AE = 12

∠BCE = 30^{o}

∠BDA = 60^{0}

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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