Angle Sum of a Polygon

prasannapakkiam

We all know that for the angle sum of the external angles of a non-concaved polygons is 360. How is/was this derived...

Werg22

The sum of the interior angles is [tex]180n - 360[/tex]. Let a set Q denote n angles whose sum is 180n - 360: [tex][ {a_{1}, a_{2} , ... , a_{n} ][/tex]. For any of these angles, the external angle is equal to [tex]180 - a_{k} [/tex]. Since there are n angles, the sum of all external angles is

[tex]S = \sum_{k = 1}^{n} 180 - a_{k} [/tex]

S is obviously equal to [tex] 180n - (180n - 360) = 360 [/tex]

prasannapakkiam

But... I see it as; the equation you started with was derived from the fact that the sum of exterior angles equalled 360... If not how was your starting equation derived?

AlephZero

Take any point inside the polygon. Join it to the vertices, to divide the polygon into n triangles.

The sum of the angles in a triangle is 180 (see any basic geometry textbook for a proof of that).

So the sum of the angles in all the triangles = 180n.

The sum of the angles round the interior point = 360.

So the sum of the interior angles if the polygon = 180n - 360.

prasannapakkiam

Brilliant; thanks for that.