If a pole 6 m high casts a shadow 2√3 m

(a) Let $B C=6 \mathrm{~m}$ be the height of the pole and $A B=2 \sqrt{3} \mathrm{~m}$ be

the length of the shadow on the ground. let the Sun's makes an angle $\theta$ on the ground.

Now, in $\triangle B A C$, $\tan \theta=\frac{B C}{A B}$

$\Rightarrow$ $\tan \theta=\frac{6}{2 \sqrt{3}}=\frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow$ $\tan \theta=\frac{3 \sqrt{3}}{3}=\sqrt{3}=\tan 60^{\circ}$ $\left[\because \tan 60^{\circ}=\sqrt{3}\right]$

$\therefore$ $\theta=60^{\circ}$

Hence, the Sun's elevation is $60^{\circ}$.