5. Let $P$ be the vertex of a cone, and $A, B, C$ be three points on the circumference of its base, satisfying $\angle A B C=90^{\circ}$, $M$ is the midpoint of $A P$. If $A B=1, A C=2, A P=\sqrt{2}$, then the size of the dihedral angle $M-B C-A$ is $\qquad$ .
Answer: $\arctan \frac{2}{3}$.
Solution: From $\angle A B C=90^{\circ}$, we know that $A C$ is the diameter of the base circle.
Let the center of the base be $O$, then $P O \perp$ plane $A B C$. It is easy to see that $A O=\frac{1}{2} A C=1$, thus $P O=\sqrt{A P^{2}-A O^{2}}=1$.
Let $H$ be the projection of $M$ on the base, then $H$ is the midpoint of $A O$.
In the base, draw $H K \perp B C$ at point $K$, then by the theorem of three perpendiculars,
we know $M K \perp B C$, thus $\angle M K H$ is the plane angle of the dihedral angle $M-B C-A$.
Since $M H=A H=\frac{1}{2}$, and combining $H K$ parallel to $A B$, we have $\frac{H K}{A B}=\frac{H C}{A C}=\frac{3}{4}$, i.e., $H K=\frac{3}{4}$, so $\tan \angle M K H=\frac{M H}{H K}=\frac{2}{3}$. Therefore, the size of the dihedral angle $M-B C-A$ is $\arctan \frac{2}{3}$.