1. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$, $F$, and $G$ are the midpoints of edges $B C$, $C C_{1}$, and $C D$ respectively. Then the angle formed by line $A_{1} G$ and plane $D E F$ is $\qquad$
$=1.90^{\circ}$.
Since $E F \perp B_{1} C, E F \perp C D$, therefore, $E F \perp$ plane $B_{1} C D$. Thus, $E F \perp A_{1} G$.
In the square $A B C D$, $D G=C E$, so $A G \perp D E$. Since $A G$ is the projection of $A_{1} G$ in the plane $A B C D$, hence $D E \perp A_{1} G$.
Therefore, $A_{1} G \perp$ plane $D E F$.