Let us assume PQ is the pillar and QR is the shadow of pillar.

Let us assume ‘h’ be the height of the pillar.

Then, length of the shadow of pillar is $\frac{1}{\sqrt{3}}h$

QR = \[\frac{1}{\sqrt{3}}h\]

In ΔPQR,

 $\tan \theta =\frac{PQ}{QR}$

$\tan \theta =\frac{h}{\frac{h}{\sqrt{3}}}$

$\tan \theta =\sqrt{3}$

As we know that,

\[\tan {{60}^{\circ }}=\sqrt{3}\]

Therefore on comparing, \[\theta ={{60}^{\circ }}\] Hence the angle of elevation of sun is ${{60}^{\circ }}$