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Angle Sum of a Polygon 
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#1
May2307, 04:09 PM

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



#2
May2307, 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] 


#3
May2307, 05:05 PM

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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?



#4
May2307, 05:32 PM

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P: 6,959

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. 


#5
May2307, 05:42 PM

P: n/a

Brilliant; thanks for that.



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