Question 617651
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If *[tex \LARGE A], *[tex \LARGE B], and *[tex \LARGE C] are the three interior angles of any triangle, and *[tex \LARGE C_e] is the exterior angle at the vertex *[tex \LARGE C], then you know two things.  First the sum of the three interior angles is 180 degrees and *[tex \LARGE C] and *[tex \LARGE C_e] are supplementary (by definition of exterior angle of a polygon).  So *[tex \LARGE m\angle{C}\ +\ m\angle{C_e}\ =\ 180].  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\angle{A}\ +\ m\angle{B}\ +\ m\angle{C}\ =\ m\angle{C_e}\ +\ m\angle{C}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\angle{A}\ +\ m\angle{B}\ =\ m\angle{C_e}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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