$A$ $1.5 \, m$ 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 $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walked towards the building.

(N/A) Let the boy be standing at point $A$ initially. He walks towards the building and reaches point $B$.
Let $PQ$ be the building of height $30 \, m$. The boy's height is $1.5 \, m$.
Therefore,the height of the triangle $PR = PQ - RQ = 30 - 1.5 = 28.5 \, m = \frac{57}{2} \, m$.
In $\triangle PAR$,$\tan 30^{\circ} = \frac{PR}{AR} \implies \frac{1}{\sqrt{3}} = \frac{57/2}{AR} \implies AR = \frac{57\sqrt{3}}{2} \, m$.
In $\triangle PBR$,$\tan 60^{\circ} = \frac{PR}{BR} \implies \sqrt{3} = \frac{57/2}{BR} \implies BR = \frac{57}{2\sqrt{3}} = \frac{19\sqrt{3}}{2} \, m$.
The distance walked towards the building is $AB = AR - BR$.
$AB = \frac{57\sqrt{3}}{2} - \frac{19\sqrt{3}}{2} = \frac{38\sqrt{3}}{2} = 19\sqrt{3} \, m$.
Thus,the boy walked $19\sqrt{3} \, m$ towards the building.