Question 450813
<pre>
The formula for the sum of the interior angles
of a polygon with n sides is

(n - 2)180°

We set that equal to 1800°

(n - 2)180° = 1800°

Divide both sides by 180°

n - 2 = 10
    n = 12

So the polygon has 12 sides, and therefore
12 interior angles.

Since one of the angles is 172°,

the other 11 interior angles must have sum
1800°-172°=1628°

Since they are all congruent, we divide 1628°/11 = 148°.
So each of the other 11 interior angles measures 148°.

So it's a 12-sided polygon.  I'll draw it with the 172°
angle at the top.  172° is close to 180°, so it looks almost
like the the two sides at the top form a straight line,
but they don't.  I'll color the sides of the 172° angle red
and green, so it will look like a 12-sided polygon, not
a 11-sided one. Here's what it looks like:
   {{{drawing(400,400,-1.5,1.5,-1.5,1.5,

line(cos(70*pi/180),sin(70*pi/180),cos(38*pi/180),sin(38*pi/180)),
line(cos(6*pi/180),sin(6*pi/180),cos(38*pi/180),sin(38*pi/180)),
line(cos(6*pi/180),sin(6*pi/180),cos(-26*pi/180),sin(-26*pi/180)),
line(cos(-58*pi/180),sin(-58*pi/180),cos(-26*pi/180),sin(-26*pi/180)),
line(cos(-58*pi/180),sin(-58*pi/180),cos(-90*pi/180),sin(-90*pi/180)),

line(-cos(70*pi/180),sin(70*pi/180),-cos(38*pi/180),sin(38*pi/180)),
line(-cos(6*pi/180),sin(6*pi/180),-cos(38*pi/180),sin(38*pi/180)),
line(-cos(6*pi/180),sin(6*pi/180),-cos(-26*pi/180),sin(-26*pi/180)),
line(-cos(-58*pi/180),sin(-58*pi/180),-cos(-26*pi/180),sin(-26*pi/180)),
line(-cos(-58*pi/180),sin(-58*pi/180),-cos(-90*pi/180),sin(-90*pi/180)),

green(line(0,sin(74*pi/180)/sin(88*pi/180),cos(70*pi/180),sin(70*pi/180))),
red(line(0,sin(74*pi/180)/sin(88*pi/180),-cos(70*pi/180),sin(70*pi/180)))


 )}}}

Edwin</pre>