1. The problem is to find the value of $\sin 60^\circ$.\n\n2. We know from the properties of special right triangles that a 60-degree angle is present in an equilateral triangle split into two 30-60-90 right triangles.\n\n3. For a 30-60-90 triangle, the sides are in ratio $1 : \sqrt{3} : 2$, where the side opposite 60 degrees is $\sqrt{3}$ times the shorter leg.\n\n4. Therefore, $\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}$.\n\n5. Final answer: $$\sin 60^\circ = \frac{\sqrt{3}}{2}.$$