Question 821048
The exterior and interior angles of a polygon come in exterior-interior supplementary pairs.
{{{drawing(300,300,-6,3,-3,4,line(0,-6,0,0),
line(0,0,-3.36,4),green(line(0,0,0,4)),
arc(0,0,5,5,90,230),locate(-2.4,0.5,interior),locate(-2.4,0,angle),
green(arc(0,0,7.5,7.5,230,270)),locate(-2.2,3,green(exterior)),
locate(-1.7,2.5,green(angle))
)}}}
So, if each exterior angle measures {{{x}}} degrees,
the measure of each interior angle, in degrees, is {{{180-x}}} .
 
"The measure of each interior angle of a regular polygon is 20 degrees more than three times the measure of each exterior angle" translates into the equation
{{{180-x=3x+20}}}
From that equation we find {{{x}}} .
{{{180-x=3x+20}}}
{{{180=x+3x+20}}}
{{{180=4x+20}}}
{{{180-20=4x}}}
{{{160=4x}}}
{{{160/4=x}}}
{{{x=40}}}
Each exterior angle measures {{{40^o}}} .
 
The measures of the exterior angles of a polygon add up to {{{360^o}}} because the exterior angle is the angle we turn at each corner (vertex) as we go around the polygon, and one turn around is going full circle, meaning {{{360^o}}} .
If the regular polygon has {{{n}}} sides, the sum of the equal {{{40^o}}} measures of the {{{n}}} angles is
{{{n*40^o=360^o}}} , so
{{{n=360^o/40^o}}}
{{{highlight(n=9)}}}
The polygon has {{{highlight(9)}}} sides.