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A 1.5 m tall boy is standing at some ...

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from `30o` to `60o` as he walks towards the building. Find the distance he walked towards the build

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Let AB be the building and let CD and EF be the two positions of the boy. Draw `DFG || CEA`. Then,
`CD = EF = 1.5 m , angleGDB = 30^(2)` and `angleGFB = 60^(@)`.
`AB = 30 m , GB = 30 m - 1.5m = 28.5 m`
From right `DeltaDGB` , we have
`(DG)/(DB) = cot 30^(@) rArr (DG)/(28.5 m) = sqrt(3) rArr DG = (57sqrt(3))/(2) m`
From right `DeltaFGB`, we have
`(FG)/(GB)= cot 60^(@) rArr (FG)/(28.5 m) = 1/(sqrt(3)) rArr FG = (57)/(2sqrt(3)) m`.

`:. DF = DG - FG = ((57sqrt(3))/(2) - (57)/(2sqrt(3))) m = ((171 - 57)/(2sqrt(3))) m`
`= (114)/(2sqrt(3)) m = (57)/(sqrt(3)) m = (57)/(sqrt(3)) xx (sqrt(3))/(sqrt(3)) m = 19sqrt(3)m`.
Hence, the required distance is `19sqrt(3) m`.
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