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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 angle 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 is $60^{\circ}$ ]
 
 

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.