To solve \( 12 \sin 60^\circ \cdot \cos 60^\circ \), we first need to know the values of \( \sin 60^\circ \) and \( \cos 60^\circ \).
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 60^\circ = \frac{1}{2} \)
Now, substitute these values into the expression:
\[ 12 \sin 60^\circ \cdot \cos 60^\circ = 12 \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{1}{2} \right) \]
\[ = 12 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = 12 \cdot \frac{\sqrt{3}}{4} \]
\[ = 3 \sqrt{3} \]
Thus, \( 12 \sin 60^\circ \cdot \cos 60^\circ = 3 \sqrt{3} \).