Question

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 ${30}^o$ to ${60}^o$ as he walks towards the building. Find the distance he walked towards the building.

Answer 1

As we can see $BCFE$ and $CDGF$ are Rectangles

$BE=CF=DG$

Given the height of the Boy, $=DG=CF=BE=1.5m$

Also, we can see that $AB=AE-BE=30-1.5=28.5$

in $△ABC$

$\tan {60}^o=\frac{28.5}{BC}\Rightarrow BC=\frac{28.5}{\sqrt{3}}$

and in $△ABD$

$\tan {30}^o=\frac{28.5}{BD}\Rightarrow BD=28.5\sqrt{3}$

Distance the Boy Covered$=CD=BD-BC=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}$

$\Rightarrow \frac{85.5-28.5}{\sqrt{3}}$

$\Rightarrow \frac{57}{\sqrt{3}}$

$\Rightarrow 19\sqrt{3}m$