12.414 One of the plane angles of a trihedral angle is equal to $\alpha$. The dihedral angles adjacent to this plane angle are equal to $\beta$ and $\gamma$. Find the other two plane angles.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-295.jpg?height=335&width=375&top_left_y=114&top_left_x=146)

Let the plane angle $\angle E_{2} F E_{3}$ of the given trihedral angle $F E_{1} E_{2} E_{3}$ be $\alpha$. From an arbitrary point $A$ on the ray $F E_{1}$, drop perpendiculars: $A O$ - to the plane $E_{2} F E_{3}$, $A B$ and $A C$ - to the rays $F E_{3}$ and $F E_{2}$.

Then $Q B \perp F E_{3}, O C \perp F E_{2}$ (Theorem 12$) \Rightarrow$ $\Rightarrow \angle A B O=\beta, \angle A C O=\gamma$ (by condition).

Extend $B O$ to intersect $F E_{2}$ at point $k$.

Then $F B=B K \cdot \operatorname{ctg} \alpha=(O B+O K) \operatorname{tg} \alpha=\left(O B+\frac{O C}{\cos \alpha}\right) \cdot \operatorname{ctg} \alpha=$

$$
\begin{gathered}
=\left(O A \cdot \operatorname{ctg} \beta+\frac{O A \cdot \operatorname{ctg} \gamma}{\cos \alpha}\right) \cdot \operatorname{ctg} \alpha=A B \cdot \sin \beta\left(\operatorname{ctg} \beta+\frac{\operatorname{ctg} \gamma}{\cos \alpha}\right) \cdot \operatorname{ctg} \alpha \Rightarrow \\
\Rightarrow \operatorname{ctg}(\angle A F B)=\frac{F B}{A B}=\operatorname{ctg} \alpha \cdot \sin \beta\left(\operatorname{ctg} \beta+\frac{\operatorname{ctg} \gamma}{\cos \alpha}\right) .
\end{gathered}
$$

Similarly, $\operatorname{ctg}(\angle A F C)=\operatorname{ctg} \alpha \cdot \sin \gamma\left(\operatorname{ctg} \gamma+\frac{\operatorname{ctg} \beta}{\cos \alpha}\right)$.

From this, we obtain the answer.

Answer: $\operatorname{arctg}\left(\operatorname{ctg} \alpha \cdot \sin \beta\left(\operatorname{ctg} \beta+\frac{\operatorname{ctg} \gamma}{\cos \alpha}\right)\right) ; \operatorname{arctg}\left(\operatorname{ctg} \alpha \cdot \sin \gamma\left(\operatorname{ctg} \gamma+\frac{\operatorname{ctg} \beta}{\cos \alpha}\right)\right)$.