1. ## Regular Heptagon

In a given diagram, ABCDEFG is a regular heptagon. The degree measure of the obtuse angle formed by AE and CG is m/n where m and n are relatively prime positive integers. Find m + n.

I know that the sum of all the angles is 900 degrees. So, the measure of one interior angle is 128.57 degrees.

2. ## Angles in a heptagon

Hello magentarita
Originally Posted by magentarita
In a given diagram, ABCDEFG is a regular heptagon. The degree measure of the obtuse angle formed by AE and CG is m/n where m and n are relatively prime positive integers. Find m + n.

I know that the sum of all the angles is 900 degrees. So, the measure of one interior angle is 128.57 degrees.
The exterior angles of any polygon always add up to $360^o$. So each exterior angle of a regulaer heptagon is $\frac{360^o}{7}$.

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

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

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

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

3. ## great work

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

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

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

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

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