Answer
$\sin\theta=\frac{y}{r}=\frac{-\sqrt {59}}{8}$
$\cos\theta=\frac{x}{r}=\frac{\sqrt 5}{8}$
$\tan\theta=\frac{y}{x}=\frac{-\sqrt {59} }{\sqrt 5}=\frac{-\sqrt {295}}{5}$
$\cot\theta=\frac{x}{y}=\frac{\sqrt 5}{-\sqrt {59}}=\frac{-\sqrt {295}}{59}$
$\sec\theta=\frac{r}{x}=\frac{8}{\sqrt 5}=\frac{8\sqrt 5}{5}$
$\csc\theta=\frac{r}{y}=\frac{8}{-\sqrt {59}}=\frac{-8\sqrt {59}}{59}$
Work Step by Step
1. IV quadrant, x is positive and y is negative
$x=\sqrt 5$ and $r=8$
2. Calculate y using the distance formula
$8=\sqrt {(y)^{2}+(\sqrt 5)^{2}}$
$64=5+y^{2}$
$y=-\sqrt {59}$
3. Insert the values to find the trig functions
$\sin\theta=\frac{y}{r}=\frac{-\sqrt {59}}{8}$
$\cos\theta=\frac{x}{r}=\frac{\sqrt 5}{8}$
$\tan\theta=\frac{y}{x}=\frac{-\sqrt {59} }{\sqrt 5}=\frac{-\sqrt {295}}{5}$
$\cot\theta=\frac{x}{y}=\frac{\sqrt 5}{-\sqrt {59}}=\frac{-\sqrt {295}}{59}$
$\sec\theta=\frac{r}{x}=\frac{8}{\sqrt 5}=\frac{8\sqrt 5}{5}$
$\csc\theta=\frac{r}{y}=\frac{8}{-\sqrt {59}}=\frac{-8\sqrt {59}}{59}$