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.