Question

A boy standing at some distance from a $30$m tall building and his eye level from the ground is $1.5$m. The angle of elevation from his eyes to the top of the building increases from ${30}^{\circ }$ and ${60}^{\circ }$ as he walks towards the building. Find the distance he walked towards the building.

• A. $19\sqrt{3}$ m
• B. $14\sqrt{3}$ m
• C. $12\sqrt{3}$ m
• D. None of these

Answer 1

The correct option is A $19\sqrt{3}$ m

Height of Building $30m$

Height of Eye from ground $1.5m$

Initial Angle ${30}^{\circ }$

Distance from Building is $\frac{EffectiveHeight}{\tan \theta }$

$=\frac{30-1.5}{\frac{1}{\sqrt{3}}}$

$=28.5\sqrt{3}$

Final Angle is ${60}^{\circ }$

Distance from Building is $\frac{EffectiveHeight}{\tan \theta }$

$=\frac{30-1.5}{\sqrt{3}}$

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

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

$=9.5\sqrt{3}$

Distance he walked $=28.5\sqrt{3}-9.5\sqrt{3}$

$=19\sqrt{3}m$