Example 9 In a tetrahedron, assuming the same as in Example 7, let $\omega=\frac{1}{2}(\alpha+\beta+\gamma)$, then the volume $V$ of the tetrahedron is $V=\frac{a b c}{3} \sqrt{\sin \omega \cdot \sin (\omega-\alpha) \cdot \sin (\omega-\beta) \cdot \sin (\omega-\gamma)}$.
Slightly prove as shown in Figure $24-6$, let the projection of point $A$ on plane $BCD$ be $E$, draw $AH \perp CD$ at $H$, connect $EH$, then $\angle AHE=\theta_{y}$.
From $S_{\triangle ACD}=\frac{1}{2} CD \cdot AH$, we have $AH=\frac{2}{c} S_{B}$. Also, $AE=AH \cdot \sin \theta_{\gamma}=\frac{2}{c} \cdot S_{B} \cdot \sin \theta_{\gamma}$.
Noticing property $3, V=\frac{1}{3} S_{A} \cdot AE=\frac{2}{3 c} S_{A} \cdot S_{B} \cdot \sin \theta_{\gamma}=\frac{a b c}{6} \sin \alpha \cdot \sin \beta \cdot \sqrt{1-\cos ^{2} \theta_{\gamma}}$
$$
\begin{array}{l}
=\frac{a b c}{6} \sqrt{1-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma+2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma} \\
=\frac{a b c}{3} \sqrt{\sin \omega \cdot \sin (\omega-\alpha) \cdot \sin (\omega-\beta) \cdot \sin (\omega-\gamma)},
\end{array}
$$

where $\omega=\frac{1}{2}(\alpha+\beta+\gamma)$.