Question:
If the angle of elevation of a tower from a distance of 100 metres from its foot is 60°, then the height of the tower is
(a) $100 \sqrt{3} \mathrm{~m}$
(b) $\frac{100}{\sqrt{3}} m$
(c) $50 \sqrt{3}$
(d) $\frac{200}{\sqrt{3}} m$
Solution:
Let 
be the height of tower is 
meters
Given that: angle of elevation is 
from tower of foot and distance 
meters.
Here, we have to find the height of tower.
So we use trigonometric ratios.
In a triangle
,
$\Rightarrow \tan C=\frac{A B}{B C}$
$\Rightarrow \tan 60^{\circ}=\frac{A B}{B C}$
$\Rightarrow \sqrt{3}=\frac{h}{100}$
$\Rightarrow h=100 \sqrt{3}$
Hence correct option is a.