Example 1 Given an acute triangle $\triangle A B C$ with three interior angles satisfying $A>B>C$, let $\alpha$ represent the minimum of $A-B$, $B-C$, and $90^{\circ}-A$. Then the maximum value of $\alpha$ is $\qquad$ $(2005$, National Junior High School Mathematics League)
Given the conditions, we have
$$
\begin{aligned}
\alpha & =\frac{2 \alpha+\alpha+3 \alpha}{6} \\
& \leqslant \frac{2(A-B)+(B-C)+3\left(90^{\circ}-A\right)}{6} \\
& =\frac{270^{\circ}-(A+B+C)}{6} \\
& =15^{\circ} .
\end{aligned}
$$

When equality holds in equation (1), we have
$$
A-B=B-C=90^{\circ}-A=\alpha=15^{\circ},
$$

Thus, when $A=75^{\circ}, B=60^{\circ}, C=45^{\circ}$, $\alpha$ can achieve its maximum value of $15^{\circ}$.