4. Given $\triangle A B C, \angle A, \angle B, \angle C$ have exterior angles in the ratio $\alpha: \beta: \gamma(\alpha, \beta, \gamma$ are all positive numbers $)$. Then $\angle A: \angle B: \angle C$ equals $\qquad$ (express the ratio using $\alpha, \beta, \gamma$).
4. $(\beta+\gamma-\alpha):(\gamma+\alpha-\beta):(\alpha+\beta-\gamma)$.

Let the exterior angles of $\angle A$, $\angle B$, and $\angle C$ be $\alpha x$, $\beta x$, and $\gamma x$ respectively. From the problem, we have
$$
\begin{array}{l} 
\alpha x+\beta x+\gamma x=360^{\circ} \\
\Rightarrow x=\frac{360^{\circ}}{\alpha+\beta+\gamma} \\
\Rightarrow \angle A=180^{\circ}-\frac{\alpha \cdot 360^{\circ}}{\alpha+\beta+\gamma} \\
= \frac{(\alpha+\beta+\gamma-2 \alpha) 180^{\circ}}{\alpha+\beta+\gamma} \\
= \frac{(\beta+\gamma-\alpha) 180^{\circ}}{\alpha+\beta+\gamma} .
\end{array}
$$

Similarly, $\angle B=\frac{(\gamma+\alpha-\beta) 180^{\circ}}{\alpha+\beta+\gamma}$,
$$
\angle C=\frac{(\alpha+\beta-\gamma) 180^{\circ}}{\alpha+\beta+\gamma} \text {. }
$$

Therefore, $\angle A: \angle B: \angle C$
$$
=(\beta+\gamma-\alpha):(\gamma+\alpha-\beta):(\alpha+\beta-\gamma) .
$$