In a tetrahedron $ABCD$, $AB=CD=a, BC=DA=b, CA=BD=c$, find the sizes of the angles opposite to the three pairs of opposite edges.

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5 As shown in Figure 20, let the midpoints of $AB, BC, CD, DA$ be $E, F, G, H$ respectively, then $E H \cong F G \cong \frac{1}{2} B D, E F \cong G H \geqq \frac{1}{2} A C$. Therefore, quadrilateral $E F G H$ is a rhombus with side length $\frac{1}{2} c$. In $\triangle C A D$, the square of the median $C H$ $=$ $\frac{1}{4}\left(2 a^{2}+2 c^{2}-b^{2}\right)$. Similarly, $B H=C H$, so $H F \perp B C$, and
$$
\begin{aligned}
F H^{2} & =C H^{2}-C F^{2}=\frac{1}{4}\left(2 a^{2}+2 c^{2}-b^{2}\right)-\frac{b^{2}}{4} \\
& =\frac{1}{2}\left(a^{2}+c^{2}-b^{2}\right) .
\end{aligned}
$$

In $\triangle F G H$,
$$
\cos \angle F G H=\frac{F G^{2}+G H^{2}-H F^{2}}{2 \cdot F G \cdot G H}=\frac{b^{2}-a^{2}}{c^{2}} .
$$

Therefore, the angle formed by $A C$ and $B D$ is
$$
\arccos \frac{\left|b^{2}-a^{2}\right|}{c^{2}} .
$$

Similarly, the angles formed by the other two pairs of opposite edges are
$$
\arccos \frac{\left|a^{2}-c^{2}\right|}{b^{2}}, \arccos \frac{\left|c^{2}-b^{2}\right|}{a^{2}} .
$$