Answer
$\sin{\theta} =-\dfrac{3\sqrt{10}}{10}$
$\tan{\theta} =-3$
Work Step by Step
$y = -3x \\ \therefore \dfrac{y}{x} = -3 \\ \because \tan{\theta} = \dfrac{y}{x} \\ \therefore \tan{\theta} = -3$
$\because$ The terminal side of $\theta$ lies in $QIV%$, then $x$ is positive and $y$ is negative.
$\therefore $ Let $x=1 $ and $ y =-3$
$r = \sqrt{x^2+y^2} = \sqrt{(1)^2+(-3)^2} = \sqrt{10}$
$\sin{\theta} = \dfrac{y}{r} = \boxed{-\dfrac{3\sqrt{10}}{10}}$
$\tan{\theta} = \dfrac{y}{x} = \boxed{-3}$
