4. If the acute angle $\alpha$ satisfies
$$
\frac{1}{\sqrt{\tan \frac{\alpha}{2}}}=\sqrt{2 \sqrt{3}} \cdot \sqrt{\tan 10^{\circ}}+\sqrt{\tan \frac{\alpha}{2}},
$$

then $\alpha=$ $\qquad$
4. $50^{\circ}$.

The original expression can be transformed as follows:
$$
\begin{array}{l}
\frac{1}{\sqrt{\tan \frac{\alpha}{2}}}-\sqrt{\tan \frac{\alpha}{2}}=\sqrt{2 \sqrt{3}} \cdot \sqrt{\tan 10^{\circ}} \\
\Rightarrow \frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}+\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}-2=2 \sqrt{3} \tan 10^{\circ} \\
\Rightarrow \frac{2}{\sin \alpha}=2+2 \sqrt{3} \tan 10^{\circ} \\
\Rightarrow \sin \alpha=\frac{1}{1+\sqrt{3} \tan 10^{\circ}} \\
=\frac{\cos 10^{\circ}}{\cos 10^{\circ}+\sqrt{3} \sin 10^{\circ}} \\
=\frac{\cos 10^{\circ}}{2 \sin 40^{\circ}}=\frac{\cos 10^{\circ} \cdot \cos 40^{\circ}}{\sin 80^{\circ}} \\
=\cos 40^{\circ}=\sin 50^{\circ} .
\end{array}
$$