Question 782796
EITHER
there is a typo,
OR
there is no solution.
 
If all the {{{highlight(exterior)}}} angles of a polygon measure {{{18^o}}}, the polygon has {{{360^o/18^o=highlight(20)}}}{{{sides}}}.
 
The {{{highlight(interior)}}} angles of a polygon, are the angles between adjacent sides {{{highlight(inside)}}} the polygon.
 
The {{{highlight(exterior)}}} angles of a polygon, are the angles between adjacent sides {{{highlight(outside)}}} the polygon.
 
For example, in the polygon below,
all the exterior angles measure {{{72^o}}}, and
all the interior angles measure {{{108^o}}}.
(For any polygon, the measure of an interior angle and the measure of the adjacent exterior angle add up to {{{180^o}}}, because they are supplementary angles).
 
{{{drawing(300,300,-4,16,-2,18,
line(0,0,10,0),line(10,0,13.09,9.51),
line(13.09,9.51,5,15.39),line(5,15.39,-3.09,9.51),
line(0,0,-3.09,9.51),green(line(10,0,17,0)),
locate(10.5,1.5,72^o),locate(8,1.5,108^o)
)}}}
 
Each exterior angle is the angle that you deviate from your previous direction as you "turn the corner around a vertex. If you go one full turn around the whole polygon, the sum of all the (exterior) angles you turn is {{{360^o}}}.
 
IF all the exterior angles have the same measure, all the interior angles have the same measure (and vice versa).
IN THAT CASE, the measure of all those angles depends on the number of sides of the polygon. The measure of one exterior angle times the number of sides is the sum of all the exterior angles and equals {{{360^o}}}.
If {{{n}}}= number of sides and {{{E}}} is the measure each exterior angle, then
{{{n*E=360^o}}} <--> {{{n=360^o/E}}}.
{Also, {{{n*E=360^o}}} <--> {{{E=360^o/n}}}).
The polygon with the least sides, would have 3 sides, with exterior angles measuring {{{120^o}}} and interior angles measuring {{{60^o}}}. With more sides the exterior angles would be smaller and the interior angles would be larger. The exterior angles could measure {{{18^o}}} and less, but the inerior angles cannot measure less than {{{60^o}}}.