2. In the acute $\triangle A B C$, take a point $P$ such that
$$
\angle P A C=60^{\circ}, \angle P C A=\angle P B A=30^{\circ} \text {, }
$$
$M 、 N$ are the midpoints of sides $A C 、 B C$ respectively. Then the degree of $\angle P N M$ is $\qquad$
2. $90^{\circ}$.

From the given conditions, it is easy to know that $\angle A P C=90^{\circ}$.
As shown in Figure 4, extend $C P$ to point $D$ such that $P D=C P$, and connect $A D$ and $D B$.
Then $\angle A P D=90^{\circ}, A D=A C$,
$\angle A D P=\angle A C P=30^{\circ}=\angle P B A$.
Thus, points $A, D, B, P$ are concyclic.
Therefore, $\angle D B A=\angle A P D=90^{\circ}$.
Since $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively, and $P$ is the midpoint of $C D$, we have $M N / / A B, P N / / D B$.
Therefore, $\angle P N M=\angle D B A=90^{\circ}$.