Question 1043390
The sum of the measures of the exterior angles of a polygon is always {{{360^o}}} .
That is simpler and easier to see and remember than anything about interior angles.
Each exterior angle is supplementary to the adjacent interior angle,
so the measures (in degrees) of the exterior angles also form an arithmetic progression.
Starting from the smallest exterior angle, the terms of that progression are
{{{a[1]=180-172=8}}} ,
{{{a[2]=8+4=12}}} ,
{{{a[3]=8+2*4=16}}} ,
and so on, up to
{{{a[n]=8+4(n-1)}}}
The sum of the first {{{n}}} terms of an arithmetic progression is
{{{S[n]=n(a[1]+a[n])/2}}} .
In this case, {{{S[n]=n(8+8+4(n-1))/2}}} ,
and it equals {{{360}}} ,
because the sum of the measures of the exterior angles of any polygon is {{{360^o}}} .
So,
{{{n(8+8+4(n-1))/2=360}}}
{{{n(16+4n-4)/2=360}}}
{{{n(12+4n)=360*2}}}
{{{12n+4n^2=720}}}
{{{12n+4n^2-720=0}}}
{{{3n+n^2-180=0}}}
{{{n^2+3n-180=0}}}
{{{(n-12)(n+15)=0}}}
{{{highlight(n=12)}}} .
The number of sides of the polygon is {{{highlight(12)}}} .
 
NOTEs:
The problem does not ask for the measures of any other interior angles, but they range from {{{128^o}}} to {{{172^o}}} .
We cannot find what the lengths of the sides are,
because there are many different polygons with those same angle measures.
The possibilities are endless.
Here is what one of them looks like:
{{{drawing(300,300,-4.5,5.5,-1.3,8.7,
green(line(0,0,0.99,0.139)),line(2.056,0.527,0.99,0.139),
line(2.056,0.527,3.097,1.283),line(3.913,2.493,3.097,1.283),
line(3.913,2.493,4.2,4.123),line(3.62,5.908,4.2,4.123),
line(3.62,5.908,1.989,7.276),line(-0.419,7.444,1.989,7.276),
line(-0.419,7.444,-2.635,5.835),line(-3.174,2.776,-2.635,5.835),
line(-3.174,2.776,-1.005,0),red(line(0,0,-1.005,0)),
green(arrow(0,0,5.446,0.765)),green(arrow(0,0,5.5,0)),
green(arc(0,0,8,8,-8,0)),locate(4.1,0.7,green(8^o)),
red(arc(0,0,1,1,-180,-8)),locate(-0.2,1.3,red(172^o)),
arc(-1.005,0,0.9,0.9,-128,0),locate(-1.3,1.2,128^o)
)}}}