8,9

Through the lateral edge $P C$ of a regular triangular pyramid $A B C P$, a plane parallel to the side $A B$ of the base is drawn. The lateral edge $P A$ forms an angle $\arcsin \frac{\sqrt{2}}{3}$ with this plane. Find the angle between the lateral edge and the base plane of the pyramid.
Let $a$ and $b$ be the side of the base and the lateral edge of the given pyramid, and $\alpha$ be the required angle. Let $E$ be the orthogonal projection of point $A$ onto the given plane. According to the problem, $\sin \angle A P E = \frac{\sqrt{2}}{3}$, therefore

$$
A E = A P \sin \angle A P E = \frac{b \sqrt{2}}{3}
$$

Since the line $A B$ is parallel to the given plane, all its points are equidistant from this plane. Consider the section of the pyramid by the plane passing through points $C, P$, and the midpoint $K$ of edge $A B$. In this plane, the center $M$ of the base of the pyramid is located. The height $K D$ of triangle $C P K$ is perpendicular to the plane passing through line $C P$ parallel to $A B$, since $K D \perp P C$ and $K D \perp A B$ (line $A B$ is perpendicular to the plane $C P K$ containing $K D$). Therefore, $K D = A E = \frac{b \sqrt{2}}{3}$. Consider triangle $C P K$. In it,

$$
C K \cdot P M = C P \cdot K D, \quad \frac{a \sqrt{3}}{2} \cdot \sqrt{b^{2} - \left(\frac{a \sqrt{3}}{3}\right)^{2}} = b \cdot \frac{b \sqrt{2}}{3}
$$

After squaring and obvious simplifications, we get the equation

$$
8 b^4 - 27 a^2 b^2 + 9 a^4 = 0
$$

from which we find that $b = a \sqrt{3}$ or $b = \frac{a \sqrt{6}}{4}$. If $b = a \sqrt{3}$, then

$$
\cos \alpha = \cos \angle P A M = \frac{A M}{A P} = \frac{\frac{a \sqrt{3}}{3}}{b} = \frac{1}{3}, \quad \sin \alpha = \frac{2 \sqrt{2}}{3}
$$

If $b = \frac{a \sqrt{6}}{4}$, then

$$
\cos \alpha = \cos \angle P A M = \frac{A M}{A P} = \frac{\frac{a \sqrt{3}}{3}}{b} = \frac{2 \sqrt{2}}{3}, \quad \sin \alpha = \frac{1}{3}
$$

## Answer

$\arcsin \frac{2 \sqrt{2}}{3}$ or $\arcsin \frac{1}{3}$