The exterior angles of a regular nonagon are $360^{\circ}\div 9 = 40^{\circ}$, whence the interior angles are $180^{\circ} - 40^{\circ}= 140 ^{\circ}$.

In the arrowhead quadrilateral whose rightmost vertex is X, three of the angles are $40^{\circ}$, $40^{\circ}$ and $360^{\circ} - 140^{\circ}=220^{\circ}$ and these add up to $300^{\circ}$.

So the agle at X is $60^{\circ}$.

[It is now posible to see that the entire nonagon can fit neatly inside an equilateral triangle and so the angle X $60^{\circ}$ ]

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