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.

• A. $10\sqrt{3}$ m
• B. $11\sqrt{3}$ m
• C. $5\sqrt{3}$ m
• D. $19\sqrt{3}$ m

Answer 1

The correct option is D

$19\sqrt{3}$ m

Let AB be the building and let CD and EF be the two positions of the boy. Draw $DFG\parallel CEA$. Then,

CD = EF = 1.5 m, $\angle GDB={30}^{\circ }$ and $\angle GFB={60}^{\circ }$,

AB = 30 m, GB = 30 m - 1.5 m = 28.5 m

From right $\Delta DGB$, we have

$\frac{DG}{GB}=cot{30}^{\circ }\Rightarrow \frac{DG}{28.5m}=\sqrt{3}\Rightarrow DG=\frac{57\sqrt{3}}{2}m$

From right $\Delta FGB$, we have

$\frac{FG}{GB}=cot{60}^{\circ }\Rightarrow \frac{FG}{28.5m}=\frac{1}{\sqrt{3}}\Rightarrow FG=\frac{57}{2\sqrt{3}}m$



$\therefore DF=DG-FG=(\frac{57\sqrt{3}}{2}-\frac{57}{2\sqrt{3}})m=\left(\frac{171-57}{2\sqrt{3}}\right)m$

$=\frac{114}{2\sqrt{3}}m=\frac{57}{\sqrt{3}}m=\frac{57}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}m=19\sqrt{3}m$

Hence, the required distance is $19\sqrt{3}$ m