Question

A 1.8 m tall man is standing at some distance from a building. The angle of elevation from his eyes to the top of the building increases from ${30}^{\circ }$ to ${60}^{\circ }$ as he walks 20 m towards the building. Find the height of the building (in m). [Take $\sqrt{3}=1.7$]

• A. 18.8

Answer 1

The correct option is A 18.8

The situation can be represented by the figure above.
$tan\left(\angle ACB\right)=\frac{AB}{BC}\Rightarrow BC=\frac{AB}{tan{60}^{\circ }}.....\left(1\right)tan\left(\angle ADB\right)=\frac{AB}{BD}\Rightarrow tan{30}^{\circ }=\frac{AB}{BC+CD}\Rightarrow BC+CD=\frac{AB}{tan{30}^{\circ }}.....\left(2\right)\text{on subtracting equation(1) from (2)}\Rightarrow CD=\frac{AB}{tan{30}^{\circ }}-\frac{AB}{tan{60}^{\circ }}\Rightarrow AB=10\sqrt{3}$
Hence, the height of the building is
= AB + BG = $(10\sqrt{3}+1.8)m=17+1.8=18.8m$