6. In the tetrahedron $S-ABC$, four edges are of length $\sqrt{2}$, and the other two edges are of lengths $\sqrt{3}$ and 2. Then the angle between the longer two edges is $\qquad$ .
6. $\frac{\pi}{2}$ or $\arccos \frac{5 \sqrt{3}}{12}$.
(1) When the two longer edges are opposite, let $S A=\sqrt{3}, R C$ $=2$. As shown in Figure 5, $D$ is the midpoint of $B C$, connect $S D$ and $A D$.
$\because A B=A C=S B=S C$,
$\therefore S D \perp B C, A D \perp B C$.
$\therefore B C \perp$ plane $S A D$.
$\therefore S A \perp B C$.
Thus, the angle between $S A$ and $B C$ is $\frac{\pi}{2}$.
(2) When the two longer edges are adjacent, let $S A=\sqrt{3}, A C$ $=\sqrt{2}, S C=2$. In $\triangle S A C$, by the cosine rule we get
$\cos \angle A S C=\frac{5 \sqrt{3}}{12}$.
$\therefore \angle A S C=\arccos \frac{5 \sqrt{3}}{12}$.
Thus, the angle between $S A$ and $S C$ is $\arccos \frac{5 \sqrt{3}}{12}$.
Therefore, the required angles are $\frac{\pi}{2}$ or $\arccos \frac{5 \sqrt{3}}{12}$.