(a) 60°
Here, pole height = 6 m, shadow length = $2\sqrt{3}$ m. $\tan\theta = \dfrac{6}{2\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, so $\theta = 60°$.
Set up a right triangle where the pole is the opposite side and the shadow is the adjacent side. Using $\tan\theta = \frac{\text{height}}{\text{shadow}}$ gives $\sqrt{3}$, which corresponds to 60°. Remember: $\tan 60° = \sqrt{3}$, $\tan 45° = 1$, $\tan 30° = \frac{1}{\sqrt{3}}$.