Answer
$\sin\theta=-\frac{2\sqrt{13}}{13}$
$\cos\theta=\frac{3\sqrt{13}}{13}$
$\tan\theta = -\frac{2}{3}$
Work Step by Step
By definition
$\sin\theta=\dfrac{y}{R}$
$\cos\theta=\dfrac{x}{R}$
$\tan\theta=\dfrac{y}{x}$
where $R^2=x^2+y^2.$
Our point has the coordinates $x=3$ and $y=-2$. then
$R=\sqrt{3^2+(-2)^2}=\sqrt{13}$.
So
$\sin\theta=\dfrac{-2}{\sqrt{13}}=-\dfrac{2\sqrt{13}}{13}$
$\cos\theta=\dfrac{3}{\sqrt{13}}=\dfrac{3\sqrt{13}}{13}$
$\tan\theta = -\dfrac{2}{3}$
