Question 1154213
<pre>Any n-sided polygon has n sides, n vertices and n interior angles.
Pick any one of the n vertices.  The 2 vertices adjacent to that vertex 
make up 2 of the n sides.  If we join the chosen vertex to each of the 
other n-2 vertices we will have n-2 triangles and the sum of all their 
angles will give us (n-2)∙180° for the sum of all the n interior angles 
of the polygon. 

Now in a regular polygon all the angles are equal, so each angle will
measure 1/nth of the sum (n-2)∙180°, or

{{{((n-2)180^o)/n}}}

That is the answer AND WHY!

Edwin</pre>