79 In tetrahedron $ABCD$, it is known that $AB=BD=CD=3$, $AD=BC=4$, $AC=5$. Try to find the size of the dihedral angle $A-BD-C$.


79 As shown in Figure 31, it is easy to see that $\angle A B C=90^{\circ}$. Let $\angle A B D=\alpha, \angle C B D=\beta$. On $B D$, take $B E=$ 1, and draw a plane perpendicular to $B D$ through $E$ intersecting $A B$ at $F$ and $B C$ at $G$, then
$$
\begin{aligned}
B F= & \sec \alpha, E F=\tan \alpha, B G=\sec \beta, E G=\tan \beta . \\
\cos \angle F E G & =\frac{E G^{2}+E F^{2}-E G^{2}}{2 \times E G \times E F}=\frac{E G^{2}+E F^{2}-B G^{2}-B F^{2}}{2 \times E G \times E F} \\
& =-\frac{1}{\tan \alpha \tan \beta}=-\cot \alpha \cot \beta .
\end{aligned}
$$

In $\triangle A B D$,
$$
\cos \alpha=\frac{3^{2}+3^{2}-4^{2}}{2 \times 3 \times 3}=\frac{1}{9},
$$

so
$$
\cot \alpha=\frac{1}{\sqrt{80}} .
$$

Similarly,
$$
\begin{array}{c}
\cos \beta=\frac{3^{2}+4^{2}-3^{2}}{2 \times 3 \times 4}=\frac{2}{3} \\
\cot \beta=\frac{2}{\sqrt{5}} .
\end{array}
$$

Therefore,
$$
\cos \angle F E G=-\frac{1}{10} .
$$

Thus, the required dihedral angle is $\pi-\arccos \frac{1}{10}$.