Question 282047
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\phi\ +\ sin^2\phi\ =\ 1]


so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\phi\ =\ \sqrt{1\ -\ sin^2\phi}]


so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{A}\ =\ \sqrt{1\ -\ \left(\frac{3}{5}\right)^2}\ =\ \sqrt{\frac{16}{25}}\ =\ \frac{4}{5}\ =\ \frac{8}{10}]


Or you could just recognize that is is just a 3:4:5 right triangle, so the sides are 6 and 8 and the hypotenuse is 10, so *[tex \Large \cos{A}] has to be *[tex \Large \frac{8}{10}\ =\ \frac{4}{5}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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