5. Given that $O$ is the circumcenter of acute triangle $\mathrm{V} A B C$, and let $\angle A=\theta$. If $\frac{\cos B}{\sin C} A B+\frac{\cos C}{\sin B} A C=2 m A O$, then $m=$

 untranslated part:
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Note: The last sentence is a note to the translator and should not be included in the translated text. Here is the final version:

5. Given that $O$ is the circumcenter of acute triangle $\mathrm{V} A B C$, and let $\angle A=\theta$. If $\frac{\cos B}{\sin C} A B+\frac{\cos C}{\sin B} A C=2 m A O$, then $m=$
5. $\sin \theta$

Analysis: As shown in the right figure, $O$ is the circumcenter of $\triangle A B C$, and $D, E$ are on $A C, A B$ respectively, with $O D / / A B, O E / / A C$. It is easy to know that $A O=A D+A E$, which means
$$
\begin{array}{l}
\left|\begin{array}{l}
A D \\
A C \\
A C
\end{array}\right|=\frac{\cos C}{2 m \sin B},\left|\begin{array}{c}
\text { un } \\
A E \\
A B
\end{array}\right|=\frac{\cos B}{2 m \sin C} \text { . Also, } \\
\left|\begin{array}{l}
\text { um } \\
A E \\
A B
\end{array}\right|=\left|\begin{array}{l}
\text { um } \\
A E \\
A O
\end{array}\right| \cdot\left|\begin{array}{l}
\text { um } \\
B O \\
\text { un } \\
A B
\end{array}\right|=\frac{\sin \angle A O E}{\sin \angle A E O} \cdot \frac{\sin \angle B A O}{\sin \angle A O B}=\frac{\sin \angle O A C}{\sin A} \cdot \frac{\sin \angle B A O}{\sin 2 C}=\frac{\cos B \cos C}{2 \sin A \sin C \cos C} \\
=\frac{\cos B}{2 \sin A \sin C} \text {, similarly, }\left|\begin{array}{l}
\operatorname{LWF} \\
A C
\end{array}\right|=\frac{\cos C}{2 \sin A \sin B} \text {, thus } m=\sin \theta \\
\end{array}
$$