In the tetrahedron $P-ABC$, $\angle APB=\angle BPC=\angle CPA=90^{\circ}$, $M$ is a point within the base $ABC$, $\angle APM=\alpha, \angle BPM=\beta, \sin \alpha=\frac{\sqrt{6}}{3}, \cos \beta=\frac{\sqrt{6}}{6}$, then the value of $\angle CPM$ is
A. $30^{\circ}$
B. $45^{\circ}$
C. $60^{\circ}$
D. $75^{\circ}$
22 B. Extend it to form a rectangular parallelepiped with the segments $P A, P B, P C$ as edges, such that $P M$ lies exactly on one of its diagonals. Let $\angle C P M=\gamma$. From $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$, we get $\gamma=45^{\circ}$.