Question 823339
I assume it is the sum of the measures of the interior angles, in degrees.
If you measure the angles in radians, there is no solution.
 
The number of diagonals in a polygon with {{{n}}} sides is
{{{n(n-3)/2}}} .
The sum of the interior angles, in degrees, is
{{{180(n-2)}}} .
The inequality to solve is
{{{n(n-3)/2>180(n-2)}}}
 
Solving:
{{{n(n-3)/2>180(n-2)}}}
{{{n(n-3)>360(n-2)}}}
{{{n^2-3n>360n-720}}}
{{{n^2-3n-360n+720>0}}}
{{{n^2-363n+720>0}}}
{{{P(n)=n^2-363n+720>0}}} is a quadratic function.
It graphs as a parabola with a minimum.
It may be negative for a certain interval of intermediate values,
but outside of that middle interval it will be positive, so we will have some solutions.
We know that {{{P(0)=720>0}}} ,
but we want to know the solutions for {{{n>=3}}} ,
because a polygon has to have at least 3 sides.
If we find the solutions to {{{P(n)=n^2-363n+720=0}}} , if any,
those solutions will be the ends of that middle interval where {{{P(n)<0}}} and 
we will have our answer.
Applying the quadratic formula to {{{n^2-363n+720=0}}} we find
{{{n = (363 +- sqrt(363^2-4*1*720 ))/(2*1)=(363 +- sqrt(131769-2880))/2=(363 +- sqrt(128889))/2 }}}
The approximate values are
{{{n[1]=1.995}}} and {{{n[2]=361.005}}}
For the number of diagonals to be greater than the sum of the measures of the interior angles in degrees, we need
either {{{n<n[1]}}} ,
or {{{n>n[2]}}} .
The inequality {{{n<n[1]}}} gives no polygon solution.
The inequality {{{n>n[2]}}} means that all polygons with {{{highlight(n>=362)}}} (362 or more sides) will have
a number of diagonals that is greater than the sum of the measures of the interior angles in degrees.