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.
 A: 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
$$
OK={HN\cdot (SN-KN)\over SH}={a\over\sin\beta}
\left(\cos{\alpha\over2}\sqrt{\cos^2{\alpha\over2}-\sin^2\beta}
-\cos^2{\alpha\over2}+\sin^2\beta\right).
$$

