Question

From the top of a building $100$ m high the angles of depression of the bottom and the top of another building just opposite to it are observed to be ${60}^{\circ }$ and ${45}^{\circ }$ respectively Find the height of the building

Answer 1

Let the height of the building be h metres. Let AC = BD = d metres
from $\Delta BDE\tan {45}^{\circ }=\frac{ED}{BD}$
$\Rightarrow 1=\frac{100-h}{d}$
$\Rightarrow d=100-h.....\left(1\right)$
From $\Delta ACE\tan {60}^{\circ }=CE/AC$
$\Rightarrow \sqrt{3}=\frac{100}{d}\Rightarrow \sqrt{3}d=100$
$\Rightarrow \sqrt{3}(100-h)=100$ (using (1))
Hence the height of the tower is $h=\frac{100(3-\sqrt{3})}{3}m$