Example 10 Let point $A$ outside plane $M$ draw oblique lines $AB$ and $AC$, such that $\angle BAC = \alpha$. Also, the angles between $AB$ and $AC$ with plane $M$ are $\beta$ and $\gamma$, respectively. Find the angle formed by the projections of $AB$ and $AC$ on plane $M$.
(1962 Fujian Province Competition Problem)
Solve: As shown in Figure 24-8, draw $A D \perp$ plane $M$ at $D$, connect $B D, D C$, then $\angle A B D=$ $\beta, \angle A C D=\gamma, \angle B D C=\theta$ is the angle formed by the projections of $A B, A C$ on plane $M$, and also the plane angle of the dihedral angle $B-A D-C$.
In the trihedral angle $A-B C D$, by property 3, then
$$
\begin{aligned}
\cos \theta & =(\cos \alpha-\cos \angle B A D \cdot \cos \angle C A D) /(\sin \angle B A D \cdot \sin \angle C A D) \\
& =\sec \beta \cdot \sec \gamma \cdot \cos \alpha-\tan \beta \cdot \tan \gamma .
\end{aligned}
$$

Therefore, $\theta=\arccos (\sec \beta \cdot \sec \gamma \cdot \cos \alpha-\tan \beta \cdot \tan \gamma)$.