Question 248425
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The sum of the measures of the interior angles of any polygon is given by:


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


That is because, by connecting one of the vertices to all of the other vertices, you create n minus 2 triangles and the sum of the interior angles of each triangle is 180.


If you have a regular polygon, then all of the angles are the same size, so you can calculate the measure of one of the interior angles by dividing the sum of the measures of all the angles by the number of angles:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{180(n\ -\ 2)}{n}]


So, set this equal to each of your interior angle measures and solve the resulting equation for *[tex \LARGE n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{180(n\ -\ 2)}{n}\ =\ 108]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{180(n\ -\ 2)}{n}\ =\ 120]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{180(n\ -\ 2)}{n}\ =\ 150]


I'll let you do all the arithmetic.


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