[ Inequalities with trihedral angles ]

In space, points $A, B, C$ and $D$ are taken such that $A D=B D=C D, \angle A D B=90^{\circ}, \angle A D C=50^{\circ}, \angle B D C=$ $140^{\circ}$. Find the angles of triangle $A B C$.

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Suppose points $A, B, C$ and $D$ do not lie in the same plane. Then the dihedral angles of the trihedral angle $ABCD$ with vertex $D$ are $90^{\circ}, 50^{\circ}$, and $140^{\circ}$, which is impossible, since $90^{\circ} + 50^{\circ} = 140^{\circ}$, while the sum of any two dihedral angles of a trihedral angle must be greater than the third. Therefore, points $A, B, C$ and $D$ lie in the same plane, and ray $DA$ lies between rays $DB$ and $DC$. Since $AD = BD = CD$, points $A, B$, and $C$ lie on a circle with center $D$. Therefore, the angles of triangle $ABC$ are inscribed angles of this circle. Hence,

$$
\begin{aligned}
& \angle ABC = \frac{1}{2} \angle ADC = \frac{1}{2} \cdot 50^{\circ} = 25^{\circ} \\
& \angle ACB = \frac{1}{2} \angle ADB = \frac{1}{2} \cdot 90^{\circ} = 45^{\circ} \\
& \angle BAC = 180^{\circ} - 25^{\circ} - 45^{\circ} = 110^{\circ} .
\end{aligned}
$$

## Answer

$25^{\circ}, 45^{\circ}, 110^{\circ}$