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$30°to60°$ as he walks toward the building. Find the distance he walked towards the building.

Answer 1

Step-1: Given Information

Height of the building from the ground $AZ=30m$

Height of the boy $BZ=CX=DY=1.5m$

Height of the building from the eye line of the boy, $AB=AZ-BZ=30-1.5=28.5m$

The angle of elevation from the eyes of the boy to the building from the position $D$is ${30}^{\circ }$

The angle of elevation from the eyes of the boy to the building from the position $C$is ${60}^{\circ }$

Step-2: Find the distance of the boy from the building from the position $D$

In $∆ABD$

$\tan \left({30}^{\circ }\right)$$=\frac{AB}{BD}$ ( As $\tan \left(\varnothing \right)$$=\frac{perpendicular}{base}$)

$\Rightarrow$$\frac{1}{\sqrt{3}}$$=$$\frac{28.5}{BC}$

$\Rightarrow$$BD=28.5\sqrt{3}$ (Cross multiplication)

Step-3: Find the distance of the boy from the building from the position $C$

In $∆ABC$

$\tan \left({60}^{\circ }\right)$$=\frac{AB}{BC}$

$\Rightarrow$$\sqrt{3}=\frac{28.5}{BC}$ ( Cross Multiplication)

$\Rightarrow$$BC=\frac{28.5}{\sqrt{3}}$

Step-4: Find the distance boy walked towards the building from position $D$to position $C$

Distance walked by the boy from position $D$to position $C$$=BD-BC$

$=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}$

$=\frac{28.5\sqrt{3}\times \sqrt{3}-28.5}{\sqrt{3}}$

$=\frac{28.5\times 3-28.5}{\sqrt{3}}$

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

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

$=\frac{19\times \sqrt{3}\times \sqrt{3}}{\sqrt{3}}$

$=19\sqrt{3}m$

Hence, the distance walked by the boy from position $D$to position $C$ $19\sqrt{3}m$