108801 topics: $[\quad$ Regular Tetrahedron $\quad]$ [ Angles between lines and planes ]

In a regular tetrahedron, find the angle between an edge and the plane of a face that does not contain this edge.

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Let $ABCD$ be a regular tetrahedron with edge $a$, and $M$ be the center of the face $ABC$. Since $DM$ is the height of the tetrahedron, the angle we are looking for between the edge $AD$ and the plane of the face $ABC$ is the angle $DAM$. Let's denote it as $\alpha$. In the right triangle $DAM$, it is known that $AD=a$, $AM=\frac{a \sqrt{3}}{3}$. Therefore,

$$
\cos \alpha = \cos \angle DAM = \frac{AM}{AD} = \frac{\frac{a \sqrt{3}}{3}}{a} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}
$$

## Answer

 $\arccos \frac{1}{\sqrt{3}}$ Find the angle between the faces of a regular tetrahedron.

## Solution

Let $ABCD$ be a regular tetrahedron with edge $a$, $M$ be the center of the face $ABC$, and $L$ be the midpoint of $BC$. Since $DL \perp BC$ and $LM \perp BC$, the linear angle of the dihedral angle between the planes $ABC$ and $DBC$ is the angle $DLM$.

Let's denote it as $\beta$. Since $DM$ is the height of the tetrahedron, the triangle $DLM$ is a right triangle. In it, it is known that $DL=\frac{a \sqrt{3}}{2}$, $LM=\frac{a \sqrt{3}}{6}$. Therefore,

$$
\cos \beta = \cos \angle DLM = \frac{LM}{DL} = \frac{\frac{a \sqrt{3}}{6}}{\frac{a \sqrt{3}}{2}} = \frac{1}{3}
$$

## Answer

 $\arccos \frac{1}{3}$