Example 7 In an acute triangle $\triangle ABC$, the circumcenter is $O$. The midpoints of segments $OA$ and $BC$ are $M$ and $N$ respectively, $\angle ABC = 4 \angle OMN$, and $\angle ACB = 6 \angle OMN$. Find $\angle OMN$.
(1995 Japan Olympiad Problem)
As shown in Figure 13-11, let $\angle O M N=\theta$, then $\angle A B C=4 \theta, \angle A C B=6 \theta$, $\angle B A C=180^{\circ}-(4 \theta+6 \theta)=180^{\circ}-10 \theta$.
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\text { Also, } \angle N O C=\frac{1}{2} \angle B O C=\angle B A C=180^{\circ}-10 \theta \text {, }
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$$
\angle M O C=\angle A O C=2 \angle A B C=8 \theta \text {. }
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Thus, $\angle M O N=8 \theta+\left(180^{\circ}-10 \theta\right)=180^{\circ}-2 \theta, \angle O N M=180^{\circ}-\left(180^{\circ}-2 \theta+\theta\right)=\theta=$ $\angle O M N$, which means $\triangle O M N$ is an isosceles triangle, hence $O N=O M=\frac{1}{2} O A=\frac{1}{2} O C$.
Also, $\angle O N C=90^{\circ}$, then $\angle N O C=60^{\circ}$.
Since $\angle N O C=180^{\circ}-10 \theta$, we have $\angle O M N=\theta=12^{\circ}$.