Answer
$\sin\theta=\frac{y}{r}=\frac{-\sqrt {39}}{8}$
$\cos\theta=\frac{x}{r}=\frac{-5}{8}=$
$\csc\theta=\frac{r}{y}=\frac{8}{-\sqrt {39}}=-\frac{8\sqrt {39}}{39}$
$\sec\theta=\frac{r}{x}=\frac{8}{-5}$
$\tan\theta=\frac{y}{x}=\frac{-\sqrt {39}}{-5}=\frac{\sqrt {39}}{5}$
$\cot\theta=\frac{x}{y}=\frac{-5}{-\sqrt {34}}=\frac{5\sqrt {39}}{39}$
Work Step by Step
We know that angle in III quadrant will have negative x and y values, therefore we get following values from the cosine $x=-5; r=8$, so lets find y:
$x^{2}+y^{2}=r^{2}$
$5^{2}+(y)^{2}=8^{2}$
$y^{2}+25=64$
$x^{2}=39$ (since we know that x is negative)
$x=\sqrt {39}$
