12.2. A plane, containing an edge, divides a regular tetrahedron into two bodies, the volumes of which are in the ratio 3:5. Determine the measures of the angles in which the cutting plane divides the dihedral angle of the tetrahedron.
## Solution:

Let $a$ be the length of the edge of the tetrahedron, and $V K C$ be the cutting plane.

Consider that $\frac{v_{V K C B}}{v_{V K C A}}=\frac{3}{5}$. Then $\frac{B K}{A K}=\frac{3}{5}$ and $A K=\frac{5}{8} a, B K=\frac{3}{8} a$.

Consider the linear angle $A M B$ of the dihedral angle. Then $A M=B M=\frac{a \sqrt{3}}{2}$. Applying the cosine theorem in triangle $A C K$, we get $C K^{2}=A K^{2}+A C^{2}-A K \cdot A C$, which implies $C K=\frac{7}{8} a$. Since $C V \perp(A M B)$, we obtain that $K M \perp C V$, which implies $K M^{2}=C K^{2}-C M^{2}$ or $K M=\frac{a \sqrt{33}}{8}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_bc9e55b66943d01ac05ag-1.jpg?height=390&width=398&top_left_y=1301&top_left_x=1427)

Applying the cosine theorem in triangle $A M K$, we get

$A K^{2}=A M^{2}+K M^{2}-2 A M \cdot K M \cos m(\angle A M K)$, which implies $\cos m(\angle A M K)=\frac{7 \sqrt{11}}{33}$.

Applying the cosine theorem in triangle $B M K$, we get

$B K^{2}=B M^{2}+K M^{2}-2 B M \cdot K M \cos m(\angle B M K)$, which implies $\cos m(\angle B M K)=\frac{3 \sqrt{11}}{11}$.

Thus, we have obtained that the cutting plane divides the dihedral angle of the tetrahedron into angles of $\arccos \frac{7 \sqrt{11}}{33}$ and $\arccos \frac{3 \sqrt{11}}{11}$.