A 2m tall boy is standing at some distance from a 29m 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 walks towards the building.
AB: Building of height 29m
CD: Initial position of boy of height 2m at a distance 'x' from the building
EF: Final position of boy after walking a distance 'y' towards the building
From the figure,
$AB=29m$; $CD=EF=PA=2m$; $AC=PD=x$
$AE=PF=x-y$; $BP=BA-PA=29-2=27m$
In triangle BPD, $\tan 30^{\circ} = \dfrac{BP}{PD}$
$\implies$ $PD=x=\dfrac{BP}{\tan 30^{\circ}} = \dfrac{27}{1/\sqrt{3}} = 27\sqrt{3}m$ $\cdots$ (1)
In triangle BPF, $\tan 60^{\circ} = \dfrac{BP}{PF}$
$\implies$ $PF = x-y = \dfrac{BP}{\tan 60^{\circ}} = \dfrac{27}{\sqrt{3}}= 9\sqrt{3}m$ $\cdots$ (2)
Substituting the value of x from equation (1) in equation (2) gives
$27\sqrt{3} - y = 9\sqrt{3}$
$\implies$ $y = 27\sqrt{3} - 9\sqrt{3} = 18\sqrt{3} = 31.18m$
Therefore, the boy walks 31.18m towards the building.