Question

The height of a tower is $100\sqrt{3}\mathrm{m}$. The angle of elevation of its top from a point 100 m away from its foot is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

Answer 1

(c) 60°
Let $AB$ be the tower and $O$ be the point of observation.
We have:
$AB\hspace{0.17em}=100\sqrt{3}$ m and $OB=100$ m

In ∆$AOB$, we have:
$\frac{AB}{OB}=\tan \theta$

⇒ $\frac{100\sqrt{3}}{100}=\tan \theta$
⇒ $\sqrt{3}=\tan \theta$
⇒ $\tan 60°=\tan \theta$
⇒ $\theta ={60}^o$

Hence, the angle of elevation is ${60}^o$.