The plane angles at the vertex of a regular triangular pyramid are equal to $\alpha$. Find the dihedral angles at the base of the pyramid.

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Let $O$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, $M$ be the midpoint of $BC$, $\angle BDC = \angle ADB = \angle ADC = \alpha$. Denote $AB = BC = AC = a$, $\angle OMD = \beta$. Since $DM \perp BC$ and $OM \perp BC$, then $OMD$ is the linear angle of the dihedral angle of the pyramid at the edge $BC$, $\angle OMD = \beta$. Then

$$
\begin{gathered}
DM = BM \operatorname{ctg} \angle BDM = \frac{\frac{a}{2}}{2} \operatorname{ctg} \alpha \\
\cos \beta = \cos \angle OMD = \frac{OM}{DM} = \frac{\frac{a \sqrt{3}}{2}}{\frac{a}{2} \cdot \operatorname{ctg} \alpha} = \frac{\operatorname{ctg} \alpha}{\sqrt{3}}
\end{gathered}
$$

## Answer

 $\arccos \frac{\operatorname{ctg} \alpha}{\sqrt{3}}$