Aregular 40-sided polygon is rotated with its center of rotation at its center.

what is the smallest degree of rotation needed to map the polygon back on to itself?

So one thing needed to solve this is the interior angle, or the angle between two sides. This can be solved with the equation 180(n-2)°, where n is the number of sides in a polygon

So if we substitute the sides for n, we get 180(40-2)° which gets us 180 * 38 which is 6840° Next we divide by 40. This will give us each angle of in the 40-sided polygon. That gets us 171°. Next we use 180 - 171 which gives us 9°

So the answer is 9°

Hopes this helps!