Question 825165
ANOTHER WAY:
In a polygon with {{{n}}} sides,
the sum of the measures of the {{{n}}} exterior angles is {{{360^o}}} ,
and the sum of the {{{n}}} interior angles is {{{(n-2)*180^o}}} .
If the polygon is a regular polygon, all the exterior angles have the same measure, {{{E}}} , and all the interior angles have the same measure {{{I}}} .
So, for a regular polygon with {{{n}}} sides,
{{{n*E=360^o}}} , and {{{n*I=(n-2)*180^o}}} .
The ratio is
{{{n*E/(n*I)=360^o/((n-2)*180^o)}}} .
Simplifying, we get
{{{E/I=2/(n-2)}}} .
For the polygon of the problem, {{{E/I=2/13}}} ,
so {{{2/13=2/(n-2)}}}
{{{n-2=13}}}
{{{n=13+2}}}
{{{highlight(n=15)}}}
 
OR MAYBE:
{{{360^o/n}}}:{{{(n-2)*180^o/n)))=2:13
{{{360^o}}}:{{{(n-2)*180^o)))=2:13
{{{2}}}:{{{(n-2))))=2:13
{{{n-2=13}}}
{{{n=13+2}}}
{{{highlight(n=15)}}}