Question 608388
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Let *[tex \LARGE x] represent the measure of the smaller angle.  Then *[tex \LARGE x\ +\ 50] is the measure of the larger angle.  The sum of the measures of two complementary angles is *[tex \LARGE 90^\circ], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ x\ +\ 50\ =\ 90]


Solve for *[tex \LARGE x] and then calculate either *[tex \LARGE x\ +\ 50] or *[tex \LARGE 90\ -\ x] whichever tickles your fancy.


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