Example 2. The radius of the base of a right circular cone is $\mathbf{R}$ (Figure 2), $\mathrm{AG}$ is the bisector of the base angle of the axial section $S A B$, and $B D$ is a chord of the base circle, forming a $30^{\circ}$ angle with $\mathrm{AB}$. Find the distance between $\mathrm{AC}$ and $\mathrm{BD}$.
Take any point M on AC, draw MN perpendicular to the base, with the foot N on the diameter AB of the base circle, and set MN = x. We have
$$
\begin{array}{c}
\Delta N = MN \operatorname{ctg} 30^{\circ} = \sqrt{3} x, \\
N B = 2 R - \sqrt{3} x .
\end{array}
$$

In the base plane, draw NP perpendicular to BD,
we have $\mathrm{NP} = \mathrm{NB} \sin 30^{\circ}$
$$
= -\frac{1}{2} (2 \mathbf{R} - \sqrt{3} x) = \mathbf{R} - \frac{\sqrt{3}}{2} \mathrm{x}.
$$

In $\mathrm{Rt} \triangle \mathrm{MNP}$, $\mathrm{MP}^{2} = \mathrm{MN}^{2} + \mathrm{NP}^{2}$
$$
\begin{array}{l}
= \mathrm{x}^{2} + \left(\mathrm{R} - \frac{\sqrt{3}}{2} \mathrm{x}\right)^{2} \\
= \frac{7}{4} \mathrm{x}^{2} - \left(\mathrm{x} - \frac{2 \sqrt{3}}{7} \mathrm{R}\right)^{2} + \frac{4}{7} \mathrm{R}^{2}.
\end{array}
$$
$\because$ MP $> 0, \therefore$ the minimum value of MP is $\frac{2 \sqrt{7}}{7} \mathrm{R}$, which is the distance between AC and BD.