1. The problem asks to find the value of $\sin 60^\circ$ as a fraction. 2. Recall the special angles in trigonometry: $\sin 60^\circ = \frac{\sqrt{3}}{2}$. 3. This value comes from the properties of an equilateral triangle split into two 30-60-90 right triangles. 4. In a 30-60-90 triangle, the sides are in the ratio $1 : \sqrt{3} : 2$, where the hypotenuse is 2, the side opposite 60 degrees is $\sqrt{3}$, and the side opposite 30 degrees is 1. 5. Therefore, $\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}$. 6. Among the given options, the correct fraction is $\frac{\sqrt{3}}{2}$. Final answer: $$\sin 60^\circ = \frac{\sqrt{3}}{2}$$