8. (for 9th grade. 2 points). In triangle $A B C$, angle $A$ is equal to $40^{\circ}$. The triangle is randomly thrown onto a table. Find the probability that vertex $A$ will be to the east of the other two vertices.
Solution. Draw a line through vertex $A$ running strictly from south to north. The event “Vertex $A$ is east of the other two” occurs if and only if vertices $B$ and $C$ are located in the western half-plane from the drawn line. The diagram shows the two extreme positions of the triangle where vertices $B$ and $C$ are not located in the eastern half-plane. It is clear that ray $AB$ must pass within the shaded angle, which is the exterior angle to angle $A$ of the triangle. Therefore, the probability of this event is $\frac{180^{\circ}-40^{\circ}}{360^{\circ}}=\frac{7}{18}$.

Answer: $\frac{7}{18}$.