If a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then the sun's elevation is

Pole height $= 6$ m

Shadow length $= 2\sqrt{3}$ m

The sun shining on a vertical pole creates a right triangle where the pole is vertical and the shadow is horizontal.

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Let $\theta =$ sun's elevation angle

Using the tangent ratio:

$\tan(\theta) = \dfrac{\text{height}}{\text{shadow}}$

$\tan(\theta) = \dfrac{6}{2\sqrt{3}}$

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$\tan(\theta) = \dfrac{6}{2\sqrt{3}}$

$\tan(\theta) = \dfrac{3}{\sqrt{3}}$

$\tan(\theta) = \dfrac{3}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\tan(\theta) = \dfrac{3\sqrt{3}}{3}$

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

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From standard trigonometric values, $\tan(60°) = \sqrt{3}$

Therefore $\theta = 60°$

The sun's elevation is $60°$.