The radius of the inscribed sphere. At the base of a triangular pyramid $SABC$ is an isosceles triangle $ABC$, in which $AB = AC = a$ and the angle $BAC = \alpha$. All the sides are tilted to the plane of the base under the same angles and side $AC$ (or $AB$) forms with a lateral edge $SBC$ angle $\beta$. Determine the radius of the ball inscribed in this pyramid. I could not find the angle between the base of the pyramid and side faces.

For the lateral faces to form the same dihedral angle with the base it is necessary that the projection $H$ of vertex $S$ onto the base be equidistant from base sides: $HN=HM$ in the figure below, where $N$ is the midpoint of $BC$. If $K$ is the projection of $A$ onto the opposite face $BCS$, then $AK$ and $SH$ meet at a point $O$ which must be the center of the inscribed sphere. That entails that $OH=OK$, because both are radii of that sphere. Moreover, $\angle ABK=\beta$ because $BK$ is the projection of $AB$ onto $BCS$.
It is now only a matter of expressing $OK$ in terms of $a$, $\alpha$ and $\beta$. Notice that triangles $SKO$, $SHN$ are similar between them and equal to triangles $AHO$, $AKN$, so that
$$
SH=AK=a\sin\beta,\quad SN=AN=a\cos{\alpha\over2},\quad
HN=KN=a\sqrt{\cos^2{\alpha\over2}-\sin^2\beta}.
$$
From $OK:KN=SK:SH$ one finally obtains
$$
OK={HN\cdot (SN-KN)\over SH}={a\over\sin\beta}
\left(\cos{\alpha\over2}\sqrt{\cos^2{\alpha\over2}-\sin^2\beta}
-\cos^2{\alpha\over2}+\sin^2\beta\right).
$$

