8. In a plane, given a convex decagon and all its diagonals, in such a graph, the number of triangles that have at least two vertices as vertices of the convex decagon is $\qquad$ (answer with a number).
8. 960 .

The number of triangles with all three vertices being vertices of a convex decagon is $\mathrm{C}_{10}^{3}$.

The number of triangles with only two vertices being vertices of the convex decagon, and the other vertex being the intersection of two diagonals, can be determined as follows: Two diagonals define four vertices of the convex decagon, and these four vertices and the intersection point of the two diagonals can form four triangles with only two vertices being vertices of the convex decagon. Therefore, the number of triangles with only two vertices being vertices of the convex decagon is $4 C_{10}^{4}$.
In summary, the total number of triangles is $C_{10}^{3}+4 C_{10}^{4}=960$