Five. (20 points) As shown in Figure 1, given that the circumcenter of acute triangle $\triangle ABC$ is $O$, the midpoints of line segments $OA$ and $BC$ are points $M$ and $N$ respectively, $\angle ABC = 4 \angle OMN$, and $\angle ACB = 6 \angle OMN$. Find the degree measure of $\angle OMN$.

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Translation:

Five. (20 points) As shown in Figure 1, given that the circumcenter of acute triangle $\triangle ABC$ is $O$, the midpoints of line segments $OA$ and $BC$ are points $M$ and $N$ respectively, $\angle ABC = 4 \angle OMN$, and $\angle ACB = 6 \angle OMN$. Find the degree measure of $\angle OMN$.
Let $\angle O M N=\theta$, then $\angle A B C=4 \theta, \angle A C B=6 \theta$.
Thus, $\angle B A C=180^{\circ}-10 \theta$.
Also, $\angle N O C=\frac{1}{2} \angle B O C=\angle B A C=180^{\circ}-10 \theta$.
$\angle M O C=\angle A O C=2 \angle A B C=8 \theta$,
Therefore, $\angle M O N=8 \theta+\left(180^{\circ}-10 \theta\right)=180^{\circ}-2 \theta$.
Since $\angle O N M=180^{\circ}-(\angle M O N+\angle O M N)$
$$
=180^{\circ}-\left(180^{\circ}-2 \theta+\theta\right)=\theta=\angle O M N \text {, }
$$

Therefore, $\triangle O M N$ is an isosceles triangle.
Hence $O N=O M=\frac{1}{2} O A=\frac{1}{2} O C$.
Since $\angle O N C=90^{\circ}$, thus, $\angle N O C=60^{\circ}$.
Also, $\angle N O C=180^{\circ}-10 \theta$, then $\angle O M N=\theta=12^{\circ}$.