Example 3 (2003 Hungarian Mathematical Competition) Given the circumradius of $\triangle A B C$ is $R$, if $\frac{a \cos \alpha+b \cos \beta+c \cos \gamma}{a \sin \beta+b \sin \gamma+c \sin \alpha}=$ $\frac{a+b+c}{9 R}$, where $a, b, c$, are the lengths of the three sides of $\triangle A B C$; $\alpha, \beta, \gamma$ are the degrees of $\angle A, \angle B, \angle C$ respectively. Find $\alpha, \beta, \gamma$.
Analyze the use of the sine theorem to unify the identity into a relationship of angles and then process it.
By the sine theorem: $a=2 R \sin \alpha, b=2 R \sin \beta, c=2 R \sin \gamma$, then
$$
\begin{array}{l} 
\frac{2 R(\sin \alpha+\sin \beta+\sin \gamma)}{9 R}=\frac{2}{9}(\sin \alpha+\sin \beta+\sin \gamma) \\
\frac{a \cos \alpha+b \cos \beta+c \cos \gamma}{a \sin \beta+b \sin \gamma+c \sin \alpha}=\frac{2 R(\sin \alpha \cos \alpha+\sin \beta \cos \beta+\sin \gamma \cos \gamma)}{2 R(\sin \alpha \sin \beta+\sin \beta \sin \gamma+\sin \gamma \sin \alpha)} \\
= \frac{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma}{2(\sin \alpha \sin \beta+\sin \beta \sin \gamma+\sin \gamma \sin \alpha)}
\end{array}
$$

Therefore, the equation is transformed into: $\frac{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma}{2(\sin \alpha \sin \beta+\sin \beta \sin \gamma+\sin \gamma \sin \alpha)}=\frac{2(\sin \alpha+\sin \beta+\sin \gamma)}{9}$
That is,
$$
\begin{aligned}
4 & (\sin \alpha+\sin \beta+\sin \gamma)(\sin \alpha \sin \beta+\sin \beta \sin \gamma+\sin \gamma \sin \alpha) \\
& =9(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) \\
& =9 \cdot 4 \sin \alpha \sin \beta \sin \gamma=36 \sin \alpha \sin \beta \sin \gamma
\end{aligned}
$$

The left side of equation (1) $\geqslant 4 \cdot 3 \sqrt[3]{\sin \alpha \sin \beta \sin \gamma} \cdot 3 \sqrt[3]{\sin ^{2} \alpha \sin ^{2} \beta \sin ^{2} \gamma}=36 \sin \alpha \sin \beta \sin \gamma$
Equality holds if and only if $\sin \alpha=\sin \beta=\sin \gamma$, i.e., $\alpha=\beta=\gamma=60^{\circ}$. Therefore, $\alpha=\beta=\gamma=60^{\circ}$