Question 228839
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The measure of an exterior angle of a regular *[tex \Large n]-gon is:  *[tex \Large \frac{360}{n}]


The measure of an interior angle of a regular *[tex \Large n]-gon is:  *[tex \Large \frac{(n-2)180}{n}]


If the interior angle is *[tex \Large t] times the exterior angle, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(n-2)180}{n}\ =\ \left(\frac{360}{n}\right)t]


Just solve for *[tex \Large n] in terms of *[tex \Large t] 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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