Question

A 1.8 m tall man is standing at some distance from a 31.8 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. 30 m
• B. $20\sqrt{3}$ m
• C. $20$ m
• D. $30\sqrt{3}$ m

Answer 1

The correct option is B

$20\sqrt{3}$ m

The situation can be represented by the figure above,
$tan\left(\angle ACB\right)=\frac{AB}{BC}\Rightarrow tan{60}^{\circ }=\frac{30}{BC}\Rightarrow BC=\frac{30}{tan{60}^{\circ }}=10\sqrt{3}m\Rightarrow tan\left(\angle ADB\right)=\frac{AB}{BD}\Rightarrow tan{30}^{\circ }=\frac{30}{BD}\Rightarrow BD=\frac{30}{\frac{1}{\sqrt{3}}}=30\sqrt{3}m\Rightarrow Distancewalked=DC=BD-BC=30\sqrt{3}-10\sqrt{3}=20\sqrt{3}m$