7. Given an acute angle $\alpha$ satisfies the equation
$$
\begin{array}{l}
\sin \left(2 \alpha-20^{\circ}\right) \cdot \sin \left(2 \alpha-10^{\circ}\right) \\
=\cos \left(\alpha-10^{\circ}\right) \cdot \sin 10^{\circ} .
\end{array}
$$

Then $\alpha=$ . $\qquad$
7. $20^{\circ}$.

Given $\alpha \in\left(0^{\circ}, 90^{\circ}\right)$, we know $\cos \left(\alpha-10^{\circ}\right)>0$.
Dividing both sides of the given equation by $\cos \left(\alpha-10^{\circ}\right)$, we get
$2 \sin \left(\alpha-10^{\circ}\right) \cdot \sin \left(2 \alpha-10^{\circ}\right)=\sin 10^{\circ}$.
When $\alpha=20^{\circ}$, equation (1) clearly holds.
If $0^{\circ}<\alpha<20^{\circ}$, then $\sin \left(\alpha-10^{\circ}\right)<\sin 10^{\circ}$,
$2 \sin \left(2 \alpha-10^{\circ}\right)>2 \sin 30^{\circ}=1$.
Thus, $2 \sin \left(\alpha-10^{\circ}\right) \cdot \sin \left(2 \alpha-10^{\circ}\right)>\sin 10^{\circ}$.
If $80^{\circ} \leqslant \alpha<90^{\circ}$, then $\sin \left(\alpha-10^{\circ}\right)>\sin 70^{\circ}>\sin 30^{\circ}=1/2$,
$\sin \left(2 \alpha-10^{\circ}\right)>\sin 10^{\circ}$.
Thus, $2 \sin \left(\alpha-10^{\circ}\right) \cdot \sin \left(2 \alpha-10^{\circ}\right)>\sin 10^{\circ}$.
In summary, the value of $\alpha$ is $20^{\circ}$.