# If the point (3/5,4/5) corresponds to an angle in the unit circle, what is csc x?

Oct 27, 2015

Given point $M \left(\frac{3}{5} , \frac{4}{5}\right)$ the extremity of the arc x on the trig unit circle, $\tan x = \frac{y}{x} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$ --> $x = {53}^{\circ} 13$
$\csc x = \frac{1}{\sin} x = \frac{1}{0.8} = 1.25$