Question

A $1.5m$ tall boy is standing at some distance from a $30m$ tall building. The angles of elevation from his eyes to the top of the building increases from $30$ to ${60}^o$ as he walks towards the building. The distance he walked towards the building is:

• A. $19\sqrt{3}m$
• B. $57\sqrt{3}m$
• C. $38\sqrt{3}m$
• D. $18\sqrt{3}m$

Answer 1

The correct option is A $19\sqrt{3}m$

Consider the given figure:
$\tan {30}^0=\frac{28.5m}{AB+BC}$

Hence, $AB+BC=28.5\sqrt{3}m$ ...(i)

And, $\tan {60}^0$$=\frac{28.5m}{BC}$$=\sqrt{3}$

Hence, $BC=\frac{28.5}{\sqrt{3}}$

Hence, $AB=28.5\sqrt{3}-BC$
$=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}$
$=28.5\left[\frac{3-1}{\sqrt{3}}\right]$
$=28.5\left[\frac{2}{\sqrt{3}}\right]$
$=\frac{57}{\sqrt{3}}$
$=19\sqrt{3}m$