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Angle Sum of a Polygon

  1. May 23, 2007 #1
    We all know that for the angle sum of the external angles of a non-concaved polygons is 360. How is/was this derived...
  2. jcsd
  3. May 23, 2007 #2
    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]
  4. May 23, 2007 #3
    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?
  5. May 23, 2007 #4


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    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.
  6. May 23, 2007 #5
    Brilliant; thanks for that.
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