$\left[\begin{array}{l}{[\text { Right prism ] }} \\ {[\text { Dihedral angle }]\end{array}\right.$

In a regular triangular prism $B C D B 1 C 1 D 1$ ( $B B 1\|C C 1\| D D 1$ ), it is known that $B B 1: B C=5: 3$. Points $L, M$, and $N$ are taken on the lateral edges $B B 1, C C 1$, and $D D 1$ respectively, such that $B L: L B 1=3: 2, C M: M C 1=2: 3$, and $D N: N D 1=1: 4$. Find the dihedral angle between the planes $L M N$ and $B C D$.
Let $B B 1=5 a, B C=3 a$. Then

$$
\begin{gathered}
B L=\frac{3}{5} B B 1=\frac{3}{5} \cdot 5 a=3 a, C M=\frac{2}{5} C C 1=\frac{2}{5} B B 1=2 a \\
D N=\frac{1}{5} D D 1=\frac{1}{5} B B 1=a .
\end{gathered}
$$

Let the lines $L M$ and $B C$ intersect at point $P$, and the lines $L N$ and $B D$ intersect at point $Q$. From the similarity of triangles $P C M$ and $P B L$, it follows that

$$
\frac{P C}{P B B}=\frac{C M}{B L}, \text { or } \frac{P C}{P P C+3 a}=\frac{2}{3}
$$

From this, we find that $P C=6 a$. Similarly, we find that $D Q=\frac{3}{2} a$. Then

$$
B P=B C+C P=3 a+6 a=9 a, B Q=B D+D Q=3 a+\frac{3}{2} a=\frac{9}{2} a \text {. }
$$

Thus, in triangle $P B Q$, it is known that

$$
\angle P B Q=60^{\circ}, B P=9 a, B Q=\frac{9}{2} a
$$

Since $B P=2 B Q$ and $\angle P B Q=60^{\circ}$, triangle $P B Q$ is a right triangle, $\angle B Q P=90^{\circ}$. Therefore, $B Q \perp P Q$, and since $B Q$ is the orthogonal projection of the inclined line $L Q$ on the plane $B C D$, by the theorem of three perpendiculars, $L Q \perp P Q$. Thus, $L Q B$ is the linear angle of the dihedral angle between the planes $L M N$ and $B C D$. From the right triangle $L B Q$, we find that

$$
\operatorname{tg} \angle L Q B=\frac{B L}{B Q}=\frac{\frac{3 a}{\frac{3}{2}}}{\frac{1}{2}}=\frac{2}{3}
$$

## Answer

$\operatorname{arctg} \frac{2}{3}$