Question 1160321
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The sum of the exterior angles of any convex polygon is *[tex \Large 360^\circ]


Hence the measure of one exterior angle of an *[tex \Large n]-sided regular polygon is *[tex \Large \frac{360}{n}^\circ]


Twice this value plus *[tex \Large 45^\circ] is then *[tex \Large 2\(\frac{360}{n}^\circ\)\ +\ 45^\circ]


The sum of the interior angles of a convex polygon is given by *[tex \Large (n\,-\,2)180^\circ], so the measure of one interior angle of a regular polygon is *[tex \Large \frac{(n\,-\,2)180^\circ}{n}]


So, for this polygon:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\(\frac{360}{n}^\circ\)\ +\ 45^\circ\ =\ \frac{(n\,-\,2)180^\circ}{n}]


Solve for *[tex \Large n]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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