Answer
$\sin\theta=\frac{y}{r}=\frac{5}{\sqrt {34}}=\frac{5\sqrt {34}}{34}$
$\cos\theta=\frac{x}{r}=\frac{3}{\sqrt {34}}=\frac{3\sqrt {34}}{34}$
$\csc\theta=\frac{r}{y}=\frac{\sqrt {34}}{5}$
$\sec\theta=\frac{r}{x}=\frac{\sqrt {34}}{3}$
$\tan\theta=\frac{y}{x}=\frac{5}{3}$
$\cot\theta=\frac{x}{y}=\frac{3}{5}$
Work Step by Step
1. Terminal side of the angle defined by 5x-3y=0; a point this coterminal side can be (3,5), which satisfies the equation.
2. $x= 3; y=5$
$r=\sqrt {(x)^{2}+(y)^{2}}= \sqrt {(3)^{2}+(5)^{2}}=\sqrt {34}$