Question 163744
prove that one exterior angle of the polygon is 360/n.
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sum of the interior angles of a polygon is given by the equation sum of i = (n-2)*180.
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since a polygon of n sides has n interior angles, then each interior angle measures ((n-2)*180)/n so the formula for an interior angle is
i = ((n-2)*180)/n
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each exterior angle is a supplement of each interior angle, so each exterior angle measure 180 - ((n-2)*180)/n) so the formula for an exterior angle is
e = 180 - ((n-2)*180)/n)
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multiplying both sides of the equation by n we get
n*e = n*(180-((n-2)*180)/n)
which becomes
n*e = 180*n - (180*n - 360)
removing parentheses this equation becomes
n*e = 180*n - 180*n + 360)
combining like terms this becomes
n*e = 360
dividing both sides of the equation by n make it
e = 360/n
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