Q22 of 107 Page 12

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 30° to 60° as he walks towards the building. Find the distance he walked towards the building.


Let the initial point of boy is Z and final point is x


Z x is the distance he walked.


OP = OY-PY


OP = 30-1.5 ⇒ 28.5 m.


In ∆OPA,


tan 30° =


=


AP = ------------(1)


In ∆OPB,


tan 60° =


√3 =


PB =


PB = m.


AB = AP-BP


AB = 28.5√3 -


AB = 28.5 ⇒ 28.5√3×


AB = ⇒ 19√3 m.


Therefore the walking distance of boy is 19√3 m.


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