5. In a trapezoid with bases of lengths 1 and 3, an inscribed circle can be drawn. The maximum value of the angle between the two non-parallel sides is $\qquad$.

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5. The maximum value is $60^{\circ}$.

As shown in the figure, let trapezoid $A B C D$, from the given conditions we know the sum of the lengths of the two legs $A B+C D=A D+$ $B C=4$.
It is also easy to know $\frac{A B}{A E}=\frac{C D}{E D}=2 \Rightarrow \frac{A B+C D}{A E+E B}=2 \Rightarrow A E+E D=2$.
Let $A E=1-x$, then $E D=1+x, 0 \leqslant x \leqslant 1$, then by the cosine rule we get $\cos \alpha=\frac{(1-x)^{2}+(1+x)^{2}-1}{2\left(1-x^{2}\right)}=\frac{1+2 x^{2}}{2\left(1-x^{2}\right)} \geqslant \frac{1-x^{2}}{2\left(1-x^{2}\right)}=\frac{1}{2}$,
$\therefore \alpha \leqslant 60^{\circ}$, when $x=0$, that is, when the trapezoid is an isosceles trapezoid, we can get $\alpha=60^{\circ}$.