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.