Originally Posted by

**Grandad** Hello magentaritaThe exterior angles of any polygon always add up to $\displaystyle 360^o$. So each exterior angle of a regulaer heptagon is $\displaystyle \frac{360^o}{7}$.

Now look at the quadrilateral $\displaystyle AEFG$. By the symmetry of the heptagon, it is a trapezium with $\displaystyle AE$ parallel to $\displaystyle GF$. So $\displaystyle \angle GAE =$ exterior angle at $\displaystyle G = \frac{360^o}{7}$. Similarly $\displaystyle \angle AGC = \frac{360^o}{7}$.

Therefore the obtuse angle between $\displaystyle AE$ and $\displaystyle CG = \frac{360^o}{7}+\frac{360^o}{7}$ (Exterior angle of a triangle = sum of interior opposite angles)

$\displaystyle = \frac{720^o}{7}$

$\displaystyle \Rightarrow m = 720, n = 7, m+n=727$.

Grandad