3. Let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube, and point $M$ be the midpoint of edge $A A_{1}$. Then the plane angle $\theta$ (acute angle) of the dihedral angle $B_{1}-M C-A_{1}$ is $\qquad$ (express in radians or inverse trigonometric functions).
Figure 2
3. $\frac{\pi}{4}$.

As shown in Figure 2, let's assume $A B$ $=2$, then
$$
\begin{array}{l}
A M=M A_{1}=1, \\
M C^{2}=1^{2}+(2 \sqrt{2})^{2}=9, \\
B_{1} C^{2}=(2 \sqrt{2})^{2}=8, \\
M B_{1}^{2}=1^{2}+2^{2}=5 .
\end{array}
$$

Let $O_{1}$ be the midpoint of $A_{1} C_{1}$. Clearly, the projection of point $B_{1}$ on the plane $A_{1} C_{1} C A$ is exactly point $O_{1}$. Therefore, the projection of $\triangle B_{1} C M$ on the plane $A_{1} C_{1} C A$ is $\triangle O_{1} C M$. So,
$$
\cos \theta=\frac{S_{\triangle O_{1} C M}}{S_{\triangle B_{1} C M}} .
$$

It is easy to know that $M O_{1}^{2}=(\sqrt{2})^{2}+1^{2}=3, O_{1} C^{2}=(\sqrt{2})^{2}+2^{2}=6$.
Using Heron's formula, we get
$$
\begin{array}{l}
S_{\triangle B_{1} C M}=\frac{1}{4} \sqrt{2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-a^{4}-b^{4}-c^{4}} \\
=\frac{1}{4} \sqrt{2(9 \times 8+8 \times 5+5 \times 9)-81-64-25}=3, \\
S_{\triangle O_{1} C M}=\frac{1}{4} \sqrt{2(9 \times 3+3 \times 6+6 \times 9)-81-9-36}=\frac{3 \sqrt{2}}{2} .
\end{array}
$$

Thus, $\cos \theta=\frac{\sqrt{2}}{2}$, which means $\theta=\frac{\pi}{4}$.