Example 2. Given a rectangular cuboid $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with length, width, and height being $a, b, c(a>b)$ respectively. Find the angle formed by $A C^{\prime}$ and $B D$. (81 College Entrance Examination for Science Mathematics Supplementary Paper, Question 7)
Solve the dihedral angle $C^{\prime}-A C-D=90^{\circ}$.
Let the angle between $A C^{\prime}$ and $B D$ be $\theta$. By the corollary of the three-line theorem,
$$
\begin{array}{l}
\cos \theta=\cos \theta_{1} \cos \theta_{2} \\
=\cos \theta_{1} \cos 2 \theta_{3} \\
=\cos \theta_{1}\left(2 \cos ^{2} \theta_{3}-1\right) \\
=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+c^{2}}} \\
\text { - }\left(\frac{2 a^{2}}{a^{2}+b^{2}}-1\right) \\
=\frac{a^{2}-b^{2}}{\sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{a^{2}+b^{2}}} \text {. } \\
\text { Therefore, } \theta=\arccos \frac{a^{2}-b^{2}}{\sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{a^{2}+b^{2}}} \text {. } \\
\end{array}
$$