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 increase from ${30}^{\circ }$ to ${60}^{\circ }$ as he walks towards the building. Find the distance he walked towards the building.

Answer 1

REF.Image

AB= 30-1.5=28.5 cm

In $\Delta ABD:-tan{30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow BD=\left(28.5\right)\sqrt{3}$

In $\Delta ABC:-tan{60}^{\circ }=\frac{AB}{BC}$

$\Rightarrow BC=\frac{\left(28.5\right)}{\sqrt{3}}$

CD= BD-BC

$=28.5(\sqrt{3}-\frac{1}{\sqrt{3}})$

$=28.5\left(\frac{2}{\sqrt{3}}\right)=32.91m$