A pole $ 6 \mathrm{~m} $ high casts a shadow $ 2 \sqrt{3} \mathrm{~m} $ long on the ground, then the Sun's elevation is
(A) $ 60^{\circ} $
(B) $ 45^{\circ} $
(C) $ 30^{\circ} $
(D) $ 90^{\circ} $

Given: 

A pole $6\ m$ high casts a shadow $2\sqrt{3}\ m$ long on the ground.

To do:

We have to find the sun’s elevation.

Solution:

Let height $=6\ m$

length of shadow $=2\sqrt{3}\ m$

$\theta$ is angle of elevation

$tan\theta=\frac{height}{shadow-length}$

$=\frac{6}{2\sqrt{3}}$

$=\sqrt{3}$

$=tan60^o$

$\therefore \theta=60^o$

Thus, angle of elevation is $60^o$