Sum of External Angles of Polygon equals Four Right Angles

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Let the external angles of a polygon be generated in the same direction going around the polygon.

Then the sum of all these external angles equals $4$ right angles, that is, $360 \degrees$.


Let $V$ be an arbitrary vertex of a polygon $P$.

Let $T$ be the sum of all the external angles of $P$

Let $A$ be the size of the internal angle of $V$.

Let $E$ be the size of the external angle of $V$.

By definition, $E = 180 \degrees - A$.

Let $S$ denote the sum of all internal angles of $P$.

Hence we have:

$T = n \times 180 \degrees - S$

From Sum of Internal Angles of Polygon:

$S = \paren {n - 2} 180 \degrees$

Hence we have:

\(\ds T\) \(=\) \(\ds n \times 180 \degrees - \paren {n - 2} 180 \degrees\)
\(\ds \) \(=\) \(\ds n \times 180 \degrees - n \times 180 \degrees + 2 \times 180 \degrees\) multiplying out
\(\ds \) \(=\) \(\ds 360 \degrees\)