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Angles of polygons

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    the measure of an interior angle of a regular polygon is given. need to find the number of sides in the polygon. i cannot find the formula to be able to do this.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 23, 2008 #2
    On any polygon, the measure of the exterior angles always adds up to 360 degrees and they are supplementary to the interior angles. Because the interior angles in a regular polygon are going to have the same measure, the exterior angles will as well, so the exterior angles will have the measure 360/n where n is the number of sides. See if you can use that to get started.
     
  4. Nov 23, 2008 #3

    HallsofIvy

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    Another way to do this is to draw a line from one vertex to every other vertex. The sides of polygon alredy connect that vertex to the vertex on either side so you draw n-3 "diagonals" and that divides the polygon into n-2 triangles. Since every triangle has angle sum 180 degrees, the n-1 triangles and so the total angles in the polygon have angle sum 180(n-2). Since there are n interior angles, what is the measure of each angle in a regular n-gon? Set that equal to the angle you are given and solve for n.
     
    Last edited: Nov 24, 2008
  5. Nov 23, 2008 #4
    I believe it is n-3 "diagonals" and n-2 triangles. A square (n=4) has 1 diagonal (n-3) and 2 triangles (n-2).
     
  6. Nov 24, 2008 #5

    HallsofIvy

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    Right. Thanks. I wrote too fast. I will edit what I wrote.
     
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