Angle Sum of a Polygon

by prasannapakkiam
Tags: angle, polygon
May23-07, 04:09 PM
P: n/a
We all know that for the angle sum of the external angles of a non-concaved polygons is 360. How is/was this derived...
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Werg22 is offline
May23-07, 04:17 PM
P: 1,520
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]
May23-07, 05:05 PM
P: n/a
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 is offline
May23-07, 05:32 PM
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Angle Sum of a Polygon

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.
May23-07, 05:42 PM
P: n/a
Brilliant; thanks for that.

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