1. In tetrahedron $ABCD$, $AB=AC=AD=BC=1, BD=\sqrt{3}, CD=\sqrt{2}$, then the angle between $AD$ and $BC$ is $\qquad$.
1. $60^{\circ}$ Detailed Explanation: It can be proven that $\triangle B C D$ is a right triangle with $\angle B C D=90^{\circ}$. Also, $A B=A C=A D=1$, so the projection of $A$ onto the plane $B C D$ is the circumcenter of $\triangle A B C$. Since $\triangle B C D$ is a right triangle, its projection is the midpoint $O$ of $B D$. Construct $D D^{\prime}$ in the plane $B C D$ such that $D D^{\{\prime}} / / B C$ and $B D^{\prime} / / C D$, intersecting at $D^{\prime}$. Then $D D^{\prime}=B C=1$, and $A D^{\prime}=A C=1$, so $\triangle A D D^{\prime}$ is an equilateral triangle. Therefore, the angle between $A D$ and $B C$ is the same as the angle between $A D$ and $D D^{\prime}$, which is $60^{\circ}$.