4. Given an acute triangle $\triangle A B C$ with three interior angles satisfying $\angle A>\angle B>\angle C$, let $\alpha$ represent the minimum of $\angle A-\angle B$, $\angle B-\angle C$, and $90^{\circ}-\angle A$. Then the maximum value of $\alpha$ is $\qquad$
$$
\begin{array}{l}
\alpha .15^{\circ} . \\
\alpha=\frac{3 \alpha+2 \alpha+\alpha}{6} \\
\leqslant \frac{3\left(90^{\circ}-\angle A\right)+2(\angle A-\angle B)+(\angle B-\angle C)}{6} \\
=\frac{270^{\circ}-(\angle A+\angle B+\angle C)}{6}=\frac{90^{\circ}}{6}=15^{\circ} .
\end{array}
$$

On the other hand, when
$$
\angle A-\angle B=\angle B-\angle C=90^{\circ}-\angle A=15^{\circ}
$$

then, $\angle A=75^{\circ}, \angle B=60^{\circ}, \angle C=45^{\circ}$, that is, when the three interior angles of $\triangle A B C$ are $75^{\circ}, 60^{\circ}, 45^{\circ}$ respectively, $\alpha$ can reach its maximum value of $15^{\circ}$.