Question 799376
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The sum of the measures of the internal angles of an *[tex \Large n]-sided polygon is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (n\ -\ 2)180] degrees


Since there are *[tex \Large n] interior angles in an *[tex \Large n]-gon, and all of the interior angles in a regular *[tex \Large n]-gon are congruent, the size of one of the interior angles is the sum of the measures of the angles divided by the number of angles (or sides, same thing)


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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