Question

Question 6
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 the boy initially stands at point Y with inclination ${30}^{\circ }$ and then he approaches the building to the point X with inclination ${60}^{\circ }$.
$\therefore$ XY is the distance he walked towards the building.
Also, XY = CD
Height of the building = AZ = 30 m
$AB=AZ-BZ=(30-1.5)=28.5m$
As per the question,
In right $\Delta$ABD,
tan ${30}^{\circ }=\frac{AB}{BD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{BD}$
$\Rightarrow BD=28.5\sqrt{3}m$
Also,
In right $\Delta ABC$,
tan ${60}^{\circ }=\frac{AB}{BC}$
$\Rightarrow \sqrt{3}=\frac{28.5}{BC}$
$\Rightarrow BC=\frac{28.5}{\sqrt{3}}=\frac{28.5\sqrt{3}}{3}$ m
$\therefore XY=CD=BD-BC=(28.5\sqrt{3}-\frac{28.5\sqrt{3}}{3})=28.5\sqrt{3}(1-\frac{1}{3})=28.5\sqrt{3}\times \frac{2}{3}=\frac{57}{\sqrt{3}}=19\sqrt{3}m$
Thus, the distance boy walked towards the building is $\frac{19}{\sqrt{3}}$ m.