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 $QII%$, then $x$ is negative and $y$ is positive.
$\therefore$ 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 } $
