1. As shown in Figure 2, pentagon $A B C D I$ is a regular pentagon, and hexagon DEFGHI is a regular hexagon. Then the size of $\angle I A H$ is $\qquad$
$$
=1.24^{\circ} \text {. }
$$

From the interior angles of a regular pentagon and a regular hexagon, we know that
$$
\angle A I D=108^{\circ}, \angle H I D=120^{\circ} \text {. }
$$

Therefore, $\angle A I H=132^{\circ}$.
Since $\triangle A I H$ is an isosceles triangle, we have
$$
\angle I A H=24^{\circ}
$$