Question

A tower is $\frac{100}{\sqrt{3}}$m high.Find the angle of elevation if the point of observation is $100$m away from its foot.

Answer 1

Let $AB$ be the tower of height $\frac{100}{\sqrt{3}}m$ and $C$ be the point at a distance of $100m$ from the foot of the tower.

Let $\theta$ be the angle of elevation of the top of the tower from point $C$

In right angled $△CAB,$

$\tan \theta =\frac{AB}{AC}=\frac{\frac{100}{\sqrt{3}}}{100}=\frac{1}{\sqrt{3}}$

$\therefore \theta ={60}^{\circ }=\frac{\pi }{6}$

Hence the angle of elevation of the top of the tower from a point $100m$ away from its foot is ${30}^{\circ }$.