Question

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.

Answer 1

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.

$\therefore AC=30\text{m}$ and $DG=BC=1.5\text{m}$

Let $FH$ be the new position of the boy.

Now, $AC=AB+BC$

$\therefore AB=AC-BC$

$\therefore AB=30-1.5=28.5\text{m}$

In $△ABD$,

$\tan {30}^o=\frac{AB}{BD}=\frac{28.5}{BD}$

$\frac{1}{\sqrt{3}}=\frac{28.5}{BD}$

$\therefore BD=28.5\sqrt{3}$ m

In $△ABH$,

$\tan {60}^o=\frac{AB}{BH}$

$BH=\frac{28.5}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=9.5\sqrt{3}\text{m}$

Now, $DH=BD-BH$

$\therefore DH=28.5\sqrt{3}-9.5\sqrt{3}$

$\therefore DH=18\sqrt{3}\text{m}$