5. Let $P$ be the vertex of a cone, and $A, B, C$ be three points on the circumference of the base circle, 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
5. $\arctan \frac{2}{3}$.

From $\angle A B C=90^{\circ}$, we know that $A C$ is the diameter of the base circle. As shown in Figure 3, let the center of the base be $O$.
Thus, $P O \perp$ plane $A B C$.
Therefore, $A O=\frac{1}{2} A C=1$
$\Rightarrow P O=\sqrt{A P^{2}-A O^{2}}=1$.
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 $K$.
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 $M-B C-A$.
Given $M H=A H=\frac{1}{2}$, and combining $H K / / A B$ we get
$$
\begin{array}{l}
\frac{H K}{A B}=\frac{H C}{A C}=\frac{3}{4} \Rightarrow H K=\frac{3}{4} \\
\Rightarrow \tan \angle M K H=\frac{M H}{H K}=\frac{2}{3} .
\end{array}
$$

Therefore, the size of the dihedral angle $M-B C-A$ is $\arctan \frac{2}{3}$.