Question 461286
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The third side of the triangle is *[tex \Large \sqrt{39}] which is a little more than 6.  Hence the small angle is opposite the side that measures 5.


The small angle is then *[tex \Large \alpha\ =\ \sin^{-1}\left(\frac{5}{8}\right)]


The larger angle is then *[tex \Large \beta\ =\ 90^\circ\ -\ \alpha]


Make sure you set your calculator for degrees instead of radians.


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