# Suppose sin(x) = 9/11 and that #90 < x < 180, how do you find cos(x) and tan (x)?

May 24, 2015

$\sin x = \frac{9}{11} = 0.82$

${\cos}^{2} x = 1 - {\sin}^{2} x = 1 - 0.67 = 0.33 \to \cos x = - 0.57$
(90 < x < 180)

$\tan x = \frac{0.82}{-} 0.57 = - 1.44$