Example 10 In the pyramid $P-ABC$, $PA \perp PB$, $PB \perp PC$, $PC \perp PA$. If $M$ is a point inside the base $\triangle ABC$, and it is known that $\sin \angle APM=\frac{4}{5}, \cos \angle BPM=\frac{\sqrt{3}}{3}$. Then the value of $\cos \angle CPM$ is
Analysis: This cone has one vertex where two edges are mutually perpendicular. By cutting out a rectangular parallelepiped with $PM$ as the diagonal, this can be achieved by drawing planes through point $M$ parallel to the faces $PBC$, $PCA$, and $PAB$. At this point, $\angle APM$, $\angle BPM$, and $\angle CPM$ become the angles at the vertex of the parallelepiped formed by the diagonals. Therefore, by the basic conclusion 13, the sum of the squares of the cosines of these angles is 1. Thus,
$$
\cos \angle CPM = \sqrt{1 - \left(\frac{3}{5}\right)^2 - \left(\frac{\sqrt{3}}{3}\right)^2} = \frac{\sqrt{69}}{15}.
$$