10. (14 points) As shown in Figure 1, in $\triangle ABC$, $O$ is the midpoint of side $BC$, points $M$ and $N$ are on sides $AB$ and $AC$ respectively, and $AM=6$, $MB=4$, $AN=$ $4$, $NC=3$, $\angle MON=$ $90^{\circ}$. Find the size of $\angle A$.
10. Solution 1 As shown in Figure 3, extend $N O$ to point $P$, such that $O P=O N$. Connect $B P, M P$.
Since $B O=O C$, we know
$B P \parallel A C, B P=C N=3$.
Since point $M$ is on the perpendicular bisector of $N P$, therefore,
$M P=M N$.
Let $M N=a$. Then, in $\triangle A M N$ and $\triangle M B P$, by the cosine rule we get
$$
\begin{array}{l}
a^{2}=M N^{2}=6^{2}+4^{2}-2 \times 6 \times 4 \cos A \\
=52-48 \cos A, \\
a^{2}=M P^{2}=3^{2}+4^{2}+2 \times 3 \times 4 \cos A \\
=25+24 \cos A .
\end{array}
$$

Eliminating $a^{2}$ gives
$$
\begin{array}{l}
27-72 \cos A=0 \Rightarrow \cos A=\frac{3}{8} \\
\Rightarrow \angle A=\arccos \frac{3}{8} .
\end{array}
$$

Solution 2 Let $\overrightarrow{A M}=a, \overrightarrow{A N}=b$. Then
$$
|a|=6,|b|=4 \text {. }
$$

From the problem, we know $\overrightarrow{A B}=\frac{5}{3} a, \overrightarrow{A C}=\frac{7}{4} b$.
Since $O$ is the midpoint of side $B C$, therefore,
$$
\begin{array}{l}
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{5}{6} a+\frac{7}{8} b . \\
\text { Also } \overrightarrow{M O}=\overrightarrow{M A}+\overrightarrow{A O}=-\frac{1}{6} a+\frac{7}{8} b, \\
\overrightarrow{N O}=\overrightarrow{N A}+\overrightarrow{A O}=\frac{5}{6} a-\frac{1}{8} b .
\end{array}
$$

Since $M O \perp N O$, therefore,
$$
\begin{array}{l}
\overrightarrow{M O} \cdot \overrightarrow{N O}=-\frac{5}{36}|a|^{2}-\frac{7}{64}|b|^{2}+\frac{3}{4} a \cdot b=0 \\
\Rightarrow a \cdot b=9 \Rightarrow 6 \times 4 \cos A=9 \\
\Rightarrow \cos A=\frac{3}{8} \Rightarrow \angle A=\arccos \frac{3}{8}
\end{array}
$$