Question 523971
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The only way that the measure of a 3rd angle of a triangle can be equal to the sum of the measure of the other two angles is if the 3rd angle is a right angle.  That means the sum of the other two angles must be *[tex \Large 90^\circ].


Proof:


Let *[tex \Large x] represent the first angle and let *[tex \Large y] represent the second angle.  Then the third angle has to be *[tex \Large x\ +\ y]


The sum of the angles in any triangle is *[tex \Large 180^\circ], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ +\ (x\ +\ y)\ =\ 180]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 2y\ =\ 180]


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


Hence:


Since we have let *[tex \Large y] represent the second angle, we can say the first angle must be *[tex \Large 3y\ +\ 10], and then, by substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 3y\ +\ 10\ =\ 90]


Solve for *[tex \Large y] then calculate *[tex \Large 3y\ +\ 10]


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