Finding the internal angles of a polygon inscribed in a heptagon

Question

Consider this heptagon whose side measure is $1$ and prove $\angle FAB = \displaystyle{4\pi\over 7}$.

I want to know the measure of the $\angle FAB$. My textbook states that it is given by $\frac{BF}{2}=\frac{4\cdot \frac{2\pi }{7}}{2}$, but doesn't explain much else.

Why can the angle be found this way. And why does $BF=4\cdot \frac{2\pi }{7}$?

Answer 1

If you imagine circumcircle of $ABCDEFG$ with center at $O$ then we see that perifer angle $\angle FAB $ is half of the central angle $\angle FOB$ which is 4 times $\color{red}{2\pi\over 7}$

So $$\angle FAB ={1\over 2}\angle FOB ={1\over 2}(4\cdot \color{red}{2\pi\over 7})$$

Answer 2

The angle at each vertex is $${(n-2)\pi\over n} = {5\pi\over 7}$$ Since $AGF$ is isosceles we have $$\angle GAF = {\pi -{5\pi\over 7}\over 2} = {\pi\over 7}$$

So the angle $$\angle FAB = {5\pi\over 7} - {\pi\over 7} = {4\pi\over 7}$$