Question 452178
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Sorry, we are not going to do your entire homework assignment for you.  One question per post, please.


Let *[tex \Large x] represent the long leg.  Then *[tex \Large x\ +\ 60] represents the hypotenuse.  And the short leg is 180.  Then write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ \left(x\ +\ 60\right)^2\ -\ \left(180\right)^2]


for which we owe a debt of gratitude to Mr. Pythagoras, late of Metapontum, Italy.


Solve for *[tex \Large x].



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