78. A sphere touches the base plane $ABCD$ of a regular quadrilateral pyramid $SABCD$ at point $A$ and, in addition, touches the inscribed sphere of the pyramid. A cutting plane is drawn through the center of the first sphere and the side of the base $BC$. Find the angle of inclination of this plane to the base plane, given that the diagonals of the section are perpendicular to the edges $SA$ and $SD$.
78. $\operatorname{arctg} \frac{1}{2 \sqrt{3}}$.

The arctangent (arctg) of $\frac{1}{2 \sqrt{3}}$ is the angle whose tangent is $\frac{1}{2 \sqrt{3}}$. This can be simplified to $\frac{\sqrt{3}}{6}$. The angle whose tangent is $\frac{\sqrt{3}}{6}$ is $\frac{\pi}{6}$ or 30 degrees. However, the exact value of $\operatorname{arctg} \frac{1}{2 \sqrt{3}}$ is not a standard angle, so it is typically left in its arctangent form.

Thus, the answer is:
\[
\operatorname{arctg} \frac{1}{2 \sqrt{3}}
\]