6. As shown in the figure, in the tetrahedron $P-ABC$, $PA \perp$ plane $ABC, \angle ABC=120^{\circ}, PA=4$. If the radius of the circumscribed sphere of the tetrahedron $P-ABC$ is $2 \sqrt{2}$, then the tangent value of the angle formed by line $PC$ and plane $ABC$ is $\qquad$.
Solution:
As shown in the figure, let $AC = \sqrt{3} x$. Construct an isosceles triangle $O_1AC$ with $AC$ as the base and the vertex angle of $120^{\circ}$. Then $O_1$ is the circumcenter of the base $\triangle ABC$ with a radius of $x$. Draw $OO_1$ perpendicular to the base $ABC$. Since $OA = OP = 2\sqrt{2}$, it follows that $x = 2 \Rightarrow AC = 2\sqrt{3}$.
Therefore, the tangent value of the angle formed by the line $PC$ and the plane $ABC$ is $\frac{4}{2\sqrt{3}} = \frac{2\sqrt{3}}{3}$.