5. In tetrahedron $ABCD$,
$$
BD: CD: AB: AC: AD: BC = \sqrt{3}: \sqrt{2}: 1: 1: 1: 1 \text{. }
$$

Then the angle between $AD$ and $BC$ is $\qquad$
5. $60^{\circ}$.

As shown in Figure 3, let $AB = AC = AD = BC = 1$, and denote the midpoints of $BD$, $AB$, and $CD$ as $E$, $F$, and $G$ respectively.
Figure 3
From $AC^2 + AD^2 = CD^2$ and $G$ being the midpoint of $CD$, we get $AG = \sqrt{AD^2 - DG^2} = \frac{\sqrt{2}}{2}$.
Similarly, $BG = \frac{\sqrt{6}}{2}$.
Also, $AB = 1$, $AF = BF$, and using the median length theorem, we get $GF = \frac{\sqrt{2AG^2 + 2BG^2 - AB^2}}{2} = \frac{\sqrt{3}}{2}$.
Given $EF = \frac{1}{2} AD = \frac{1}{2}$ and $EG = \frac{1}{2} BC = \frac{1}{2}$, we know $\angle FEG = 120^{\circ}$, which means the angle between $EF$ and $EG$ is $60^{\circ}$. Therefore, the angle between $AD$ and $BC$ is $60^{\circ}$.