A 1.5m tall boy is standing at some distance from a 30m 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 AC be the tower, CFG be the ground, DA be the line of sight and GD be the original position of the boy from the building.
∴AC=30 m and DG=BC=1.5 m
Let FH be the new position of the boy.
Now, AC=AB+BC
∴AB=AC−BC
∴AB=30−1.5=28.5 m
In △ABD, tan30°=AB/BD=28.5/BD
1/√3=28.5/BD
∴BD=28.5√3 m
In △ABH, tan60°=AB/BH
BH=28.5√3/√3=9.5√3 m
Now, DH=BD−BH
∴DH=28.5√3−9.5√3
∴DH=19√3 m