1. As shown in Figure 2, the base of the regular triangular pyramid $A-BCD$ coincides with the side face $\triangle BCD$ of the regular tetrahedron $BCDE$. Connect $AE$. Then the angle between $AE$ and the plane $BCD$ is $\qquad$
$=, 1.90^{\circ}$. 
Draw $A H \perp$ plane $B C D$ intersecting $\triangle B C D$ at point $H$. From $A B=A C=A D$, we get $H B=H C=H D$, so $H$ is the circumcenter of $\triangle B C D$. Then draw $E H_{1} \perp$ plane $B C D$ intersecting $\triangle B C D$ at point $H_{1}$. From $E B=E C=E D$, we get $H_{1} B=H_{1} C=H_{1} D$, so $H_{1}$ is also the circumcenter of $\triangle B C D$. Therefore, point $H_{1}$ coincides with point $H$. Since $A H \perp$ plane $B C D$, we know $A E \perp$ plane $B C D$, meaning the angle between $A E$ and plane $B C D$ is $90^{\circ}$.