1. **State the problem:** Given that $\sin \theta = \frac{\sqrt{3}}{2}$ and $\theta$ is acute, find $\tan 2\theta$ in surd form. 2. **Identify $\theta$:** Since $\sin \theta = \frac{\sqrt{3}}{2}$ and $\theta$ is acute, $\theta = 60^\circ$ or $\frac{\pi}{3}$ radians. 3. **Recall the double angle formula for tangent:** $$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$ 4. **Find $\tan \theta$:** Since $\sin \theta = \frac{\sqrt{3}}{2}$ and $\cos \theta = \frac{1}{2}$ (from the unit circle for $60^\circ$), $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$ 5. **Calculate $\tan 2\theta$:** $$\tan 2\theta = \frac{2 \times \sqrt{3}}{1 - (\sqrt{3})^2} = \frac{2 \sqrt{3}}{1 - 3} = \frac{2 \sqrt{3}}{-2} = -\sqrt{3}$$ 6. **Final answer:** $\boxed{-\sqrt{3}}$ This corresponds to option (a).