A regular three-sided pyramid (its base is an equilateral triangle) is inscribed in a sphere, the diameter of which is nine times the height of the pyramid. What angle do two side faces enclose?
Let the center of the circumscribed sphere of the pyramid $ABCD$ (base $ABC$) be $O$, the foot of the pyramid's height be $M$, and the height's length be 2 units. According to the condition, the radius of the sphere is 9 units, so $M$ is an internal point of $OD$.

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From the right triangle $OMB$,

$$
BM = \sqrt{OB^2 - OM^2} = \sqrt{9^2 - 7^2} = 4\sqrt{2}
$$

From the right triangle $MBD$, we get that the slant edge of the pyramid is

$$
BD = \sqrt{(4\sqrt{2})^2 + 2^2} = 6
$$

The pyramid is regular, so the base triangle $ABC$ is equilateral, and $M$ is also the orthocenter and centroid of this triangle. Therefore, $BM = \frac{2}{3} \cdot AB \frac{\sqrt{3}}{2}$, from which the base edge length $AB = 4\sqrt{6}$ is obtained.

Now, draw perpendiculars from $B$ and $C$ to the line $AD$. The triangles $ADC$ and $ADB$ are symmetric with respect to the plane $ADM$, so these two perpendiculars intersect the edge $AD$ at the same point $K$. It is known that the dihedral angle between two planes is the angle between the perpendiculars to their line of intersection, so the desired angle is $\angle BKC$.

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The angle $\angle ADB$ is obtuse, so $K$ lies on the extension of $AD$ beyond $D$. From the right triangles $BKD$ and $BKA$,

$$
DK^2 + KB^2 = BD^2 = 6^2
$$

and

$$
(6 + DK)^2 + KB^2 = AB^2 = (4\sqrt{6})^2
$$

From this, $DK = 2$ and $KB = 4\sqrt{2}$ are obtained, so $BK = BM$, which means the isosceles triangles $BKC$ and $BMC$ are congruent. Thus, $\angle BKC = \angle BMC = 120^\circ$.

Therefore, the dihedral angle between two lateral faces of the pyramid is $120^\circ$.