# Angle Sum of a Polygon

by prasannapakkiam
Tags: angle, polygon
 P: 1,520 The sum of the interior angles is $$180n - 360$$. Let a set Q denote n angles whose sum is 180n - 360: $$[ {a_{1}, a_{2} , ... , a_{n} ]$$. For any of these angles, the external angle is equal to $$180 - a_{k}$$. Since there are n angles, the sum of all external angles is $$S = \sum_{k = 1}^{n} 180 - a_{k}$$ S is obviously equal to $$180n - (180n - 360) = 360$$