Question 1062720
{{{highlight(12)}}}
Relying on formulas, you would have memorized that,
for a polygon with {{{n}}} sides, that sum is
{{{(n-2)*180^o}}} ,
and in this case
{{{n-2=1800^o/180^o=10}}}---->{{{n=10+2=12}}} .
 
Relying on ideas, you would remember how, in a convex polygon,
all the diagonals drawn from one vertex divide it into triangles
(two less triangles than the number of sides);
the angles of those triangles are the interior angles of the polygon,
and that, with a sum of {{{180^o}}} per triangle, you would have
{{{1800^o/180^o=10}}} triangles.
 
Or maybe you would think of those incredibly useful exterior angles.
The exterior angles measure how much your direction changes at each vertex as you go around the polygon.
Of course, each exterior angle is attached to an exterior angle,
and they are supplementary, meaning that the pair of angles adds to {{{180^o}}} .
All the exterior angles add up to a whole turn around the polygon;
{{{360^o=2*180^o}}} .
For the polygon in this question, the sum of all exterior and interior angles is
{{{1800^o+2*180^o=10*180^o+2*180^o=12*180^o}}} .
That means {{{12}}} interior-exterior pairs,
meaning {{{12}}} angles in that polygon.