Question 880955
{{{drawing(400,200,-10,10,-3,7,line(-10,0,0,0),line(0,0,8.426,7.071),
green(line(0,0,10,0)),arc(0,0,6,6,180,320),arc(0,0,7,7,320,360)
locate(3,2,exterior),locate(3.5,1,angle),
locate(-2.4,2,interior),locate(-2.8,1,angle))}}}
I like exterior angles.
Each exterior angle is the supplement of the adjacent interior angle.
If each interior angle measures {{{140^o}}} ,
then each exterior angle measures {{{180^o+140^o=40^o}}}
The exterior angle is the change in direction as you "turn the corner" at a vertex when going around the perimeter of the polygon.
Then, of course, the sum of all the exterior angles is {{{360^o}}} ,
because as you finish going around the polygon you have gone one whole turn and are headed in the same direction as when you started.
So if this polygon has {{{n}}} vertices and {{{n}}} sides,
{{{n(40^o)=360^o}}}--->{{{n=360^o/40^o}}}--->{{{highlight(n=9)}}}
 
THE ALTERNATE, LONGER WAY:
If you want to use the fact that in a polygon with {{{n}}} sides (and {{{n}}} interior angles) the sum of the measures of all those {{{n}}} interior angles is
{{{(n-2)*180^o}}} ,
knowing that each interior angle measures {{{140^o}}} ,
you will have to write the equation
{{{(n-2)*180^o=n*140^o}}} .
Solving:
{{{(n-2)*180^o=n*140^o}}}
{{{n*180^o-2*180^o=n*140^o}}}
{{{n*180^o-360^o=n*140^o}}}
{{{n*180^o-n*140^o=360^o}}}
{{{n*(180^o-140^o)=360^o}}}
{{{n*40^o=360^o}}}
{{{n=360^o/40^o}}}
{{{n=9}}}