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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