Find the dihedral angles of a trihedral angle, the plane angles of which are $90^{\circ}, 90^{\circ}$ and $\alpha$.

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Consider the trihedral angle $PABC$ with vertex $P$. Let $\angle APC = \angle BPC = 90^{\circ}$, $\angle APB = \alpha$. Since $AP \perp CP$ and $BP \perp CP$, $APB$ is the linear angle of the dihedral angle with edge $CP$. According to the problem, $\angle APB = \alpha$. In the plane $APB$, draw a perpendicular $PM$ to the line $PB$. Then $CPM$ is the linear angle of the dihedral angle with edge $BP$.

The line $PC$ is perpendicular to two intersecting lines $AP$ and $BP$ in the plane $APB$, so $CP \perp MP$. Therefore, $\angle CPM = 90^{\circ}$. Similarly, the linear angle of the dihedral angle with edge $AP$ is also $90^{\circ}$.

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