Question 785529
The exterior and interior angles of a polygon are supplementary *adding to 180^o}}}.
If {{{x}}} = measure of an interior angle in degrees.
{{{180-x}}} = measure of the corresponding exterior angle in degrees.
The problem says that
{{{180-x=x-100}}}, so
{{{180+100-x=x}}}
{{{280-x=x}}}
{{{280=x+x}}}
{{{280=2x}}}
{{{280/2=x}}}
{{{x=140}}}
and the exterior angles measure {{{180^o-140^o=40^o}}}
The exterior angles are the angles you have to turn (deviate from your original direction when going around the polygon, so they add to {{{360^o}}}, because going one full turn around you are doing a {{{360^o}}}.
This is a regular polygon, so it is very symmetrical:
all the exterior angles measure the same {{{40^o}}},
all the interior angles measure the same {{{140^o}}}, and
all the sides have the same length, which we do not know and do not care.
So if {{{n}}}= number of sides/exterior angles of the regular polygon, and all those exterior angles measure {{{40^o}}}
{{{n*(40^o)=360^o}}} so
{{{n=360^o/(40^o)}}}--> {{{highlight(n=9)}}}