Question 949307
The measure of each (interior) angle in a convex polygon is less than {{{180^o}}} .
The sum of the measures of all the (interior) angles of a convex polygon with {{{n}}} sides is
{{{(n-2)*180^o}}}
Since {{{8*180^o=1440^o<1578^o}}} and {{{9*180^o=1620^o>1578^o}}} ,
we know that the polygon must have at least
{{{n-2=9}}}<--->{{{n=9+2=11}}} sides.
 
Could it have {{{n>=12}}}<--->{{{n-2>=10}}} ?
That would make the sum of the measures of all its angles at least
{{{10*180^8=1800^o}}} , and then, the measure of the remaining interior angle would be at least
{{{1800^o-1578^o=222^o>180^o}}} .
 
Then, {{{n-2=9}}}<--->{{{n=9+2=11}}} ,
and the sum of the measures of all the (interior) angles is
{{{9*180^o=1620^o>1578^o}}} .
The measure of the remaining interior angle is
{{{1620^o-1578^o=highlight(42^o)}}} .