All plane angles at vertex $D$ of the pyramid $A B C D$ are equal to $90^{\circ}, D A=1, D B=D C=\sqrt{2}$. Find the dihedral angles of this pyramid.
Since $A D \perp D B$ and $C D \perp B D$, then $A D C$ is the linear angle of the dihedral angle at the edge $B D$ of the given pyramid $A B C D$. According to the problem, $\angle A D C=90^{\circ}$. Therefore, the dihedral angle of the pyramid at the edge $B D$ is $90^{\circ}$. Similarly, the dihedral angles at the edges $C D$ and $A D$ are also $90^{\circ}$. The line $A D$ is perpendicular to two intersecting lines $C D$ and $B D$ in the plane $B C D$. Therefore, $A D$ is perpendicular to this plane. Let $A M$ be the height of the triangle $A B C$. Since $D M$ is the orthogonal projection of the slant $A M$ onto the plane $B C D$, by the theorem of three perpendiculars, $D M \perp B C$. Therefore, $D M$ is the height of the right triangle $B C D$, and $A M D$ is the linear angle of the dihedral angle at the edge $B C$. Further, we find:

$$
D M=\frac{1}{2} B C=\frac{1}{2} \cdot 2=1, \operatorname{tg} \angle A M D=\frac{A B}{D M}=1, \angle A M D=45^{\circ}
$$

Thus, the dihedral angle of the pyramid at the edge $B C$ is $45^{\circ}$. Using similar reasoning, we find that the dihedral angles at the edges $A C$ and $A B$ are each $60^{\circ}$.

## Answer

$90^{\circ} ; 90^{\circ} ; 90^{\circ} ; 45^{\circ} ; 60^{\circ} ; 60^{\circ}$.