[tex]\bf \textit{sum of interior angles in a regular polygon}
\\\\\\
n\theta=180(n-2)\qquad
\begin{cases}
n=\textit{number of sides}\\
\theta=\textit{angle in degrees}
\end{cases}
\\\\\\
\textit{notice, your two polygons, are just two regular octagons}\\
\textit{OCTAgons, thus OCTA = 8, 8 sides, thus}
\\\\\\
n\theta=180(n-2)\qquad n=8\implies 8\theta=180(8-2)
\\\\\\
\theta=\cfrac{180\cdot 6}{8}[/tex]
so... that's how much an internal angle is
now, subtract that from 180 and you get the angle outside, in the picture
subtract that outside angle twice from 180, and you get angle "2"
because angle 2 is in the same triangle as those two outside angles, and all internal angles in a triangle is 180, thus 180 - (those two angles) is angle "2"
