Answer
$\sin{\theta}= \dfrac{-20}{29}$
$\cos{\theta} =-\dfrac{21}{29}$
$\tan{\theta} =\dfrac{20}{21}$
$\csc{\theta} =-\dfrac{29}{20}$
$\sec{\theta} = -\dfrac{29}{21}$
$\cot{\theta} = \dfrac{21}{20}$
Work Step by Step
$\because \sin{\theta} = \dfrac{y}{r} = \dfrac{-20}{29}$
$\therefore y = -20 \hspace{10pt} \& \hspace{10pt} r = 29$
$$r^2 = x^2+y^2 \\ x^2 = r^2-y^2 \\ x = \pm \sqrt{r^2-y^2} =\pm \sqrt{29^2-(-20)^2} = \pm 21$$
As $\theta$ terminates in $QIII$, any point on its terminal side will have a negative $x$ coordinate.
$\therefore x = -21$
$\cos{\theta} = \dfrac{x}{r} = -\dfrac{21}{29}$
$\tan{\theta} = \dfrac{y}{x} = \dfrac{20}{21}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{29}{21}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = -\dfrac{29}{20}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = \dfrac{21}{20}$
