Answer
$\sin\theta=\frac{y}{r}=\frac{\sqrt {2}}{6}$
$\cos\theta=\frac{x}{r}=\frac{-\sqrt {34}}{6}$
$\tan\theta=\frac{y}{x}=\frac{\sqrt {2} }{-\sqrt {34}}=\frac{-\sqrt {17}}{17}$
$\cot\theta=\frac{x}{y}=\frac{-\sqrt {34}}{\sqrt {2}}=-\sqrt {17}$
$\sec\theta=\frac{r}{x}=\frac{6}{-\sqrt {34}}=\frac{-3\sqrt {34}}{17}$
$\csc\theta=\frac{r}{y}=\frac{6}{\sqrt {2}}=3\sqrt 2$
Work Step by Step
1. Quadrant II, x is negative and y is positive
2. $y=\sqrt 24$ and $r=6$
3. Find x using distance formula
$6=\sqrt {(x)^{2}+(\sqrt 2)^{2}}$
$36=2+x^{2}$
$x=-\sqrt {34}$
4. Plug values in to find the trig functions
$\sin\theta=\frac{y}{r}=\frac{\sqrt {2}}{6}$
$\cos\theta=\frac{x}{r}=\frac{-\sqrt {34}}{6}$
$\tan\theta=\frac{y}{x}=\frac{\sqrt {2} }{-\sqrt {34}}=\frac{-\sqrt {17}}{17}$
$\cot\theta=\frac{x}{y}=\frac{-\sqrt {34}}{\sqrt {2}}=-\sqrt {17}$
$\sec\theta=\frac{r}{x}=\frac{6}{-\sqrt {34}}=\frac{-3\sqrt {34}}{17}$
$\csc\theta=\frac{r}{y}=\frac{6}{\sqrt {2}}=3\sqrt 2$
