Question

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.

Answer 1

Let the boy was standing at point S initially. He walked towards the building and reached at point T. It can be observed that

PR = PQ − RQ = (30 − 1.5) m = 28.5 m= \(\frac{57}2 \)m

In ∆PAR, 

\(\frac{PR}{ AR} = tan 30°\)

\(\frac{57}{ 2AR} = \frac{1}{\sqrt3}\)

\(AR = (\frac{57}{2 \sqrt3})m\)

In ∆PRB,

\(\frac{PR}{ BR} = tan60°\)

\(\frac{57}{ 2BR} = \sqrt 3\)

\(BR = \frac{57}{2\sqrt3} = (\frac{19\sqrt3}{2})\,m\)

ST = AB = AR- BR = \((\frac{57/\sqrt3}{2} - \frac{19\sqrt3}{2} )\,m\)
=\( (\frac{38\sqrt3}2 ) \,m \) = 1\(19\sqrt3 \,m\)

Hence, he walked \(19\sqrt3 \,m\)  towards the building.