Triangles have three angles, and the smallest angle is $20^{\circ}$. So the second smallest angle can be at least $20^{\circ}$ which leaves at most $180^{\circ}-(20^{\circ}+20^{\circ})=140^{\circ}$ for the largest angle (since the internal angles of a triangle sum to $180^{\circ}$).

*This problem is taken from the UKMT Mathematical Challenges.*