Example 6. The radius of the base of a regular circular cone is $R$, the base angle $A$ of the axial section $\mathrm{SAB}$ is bisected by $\mathrm{AC}$, and $\mathrm{BD}$ is a chord of the base circle, forming a $30^{\circ}$ angle with $\mathrm{AB}$. Find the distance between the generatrix $\mathrm{AC}$ and the chord $\mathrm{BD}$.
Given Figure 7, construct
$\mathrm{SO} \perp$ plane $\mathrm{ABD}$, then $\mathrm{O}$ is the center of the base circle.
In the base $\mathrm{ABD}$, construct $\mathrm{OE} \perp \mathrm{BD}$ at $\mathrm{E}$, and connect $\mathrm{SE}$, then $\mathrm{SE} \perp \mathrm{BD}$.
$$
\begin{aligned}
\because O E & =\frac{1}{2} R, \\
\mathrm{SO}= & \sqrt{3 R}, \\
\therefore \mathrm{SE} & =\sqrt{(\sqrt{3} R)^{2}+\left(\frac{1}{2} R\right)^{2}} \\
& =\frac{\sqrt{13} R}{2} .
\end{aligned}
$$

Let $\angle \mathrm{SEO}=\theta$, then $\sin \theta=\frac{2 \sqrt{3}}{\sqrt{13}}$,
$$
\cos \theta=\frac{1}{\sqrt{13}}
$$

Let $\mathrm{F}$ be the midpoint of $\mathrm{BE}$; connect $\mathrm{CF}$, then $\mathrm{CF} \mathbb{\|} \frac{i}{2} S E$.
Thus, $\mathrm{CF} \perp \mathrm{BD}, \mathrm{CF}=\frac{\sqrt{13}}{4} \mathrm{R}$.
Connect $\mathrm{AD}$, we have $\mathrm{AD} \perp \mathrm{BD}, \mathrm{AD}=\mathrm{R}$, by the formula (1) for the distance between skew lines, the distance between $\mathrm{AC}$ and $\mathrm{BD}$ is
$$
\begin{aligned}
\mathrm{h} & =\frac{a b \sin \theta}{\sqrt{a^{2}+b^{2}-2 a b \cos \theta}} \\
& =\frac{2 \sqrt{7}}{7} R
\end{aligned}
$$