Answer
The point is $(0,-1)$
$r=1$
$\sin({-90^{\circ}}) =-1$
$\cos({-90}^{\circ}) = 0$
$\tan({-90}^{\circ})$ = undefined
Work Step by Step
The terminal side of $-90^{\circ}$ in standard position is represented by the blue line in the figure. It lies on the negative $y$ axix.
The coordinates of points on the terminal side of $-90^{\circ}$ can be given by $(0,-a)$, where $a$ is a positive number.
Choosing $a=1$ arbitrarily, the point is $(0,-1)$.
To find the distance from the origin to $(0,-1)$, we use the distance formula
$$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\r=\sqrt{(0-0)^2+(-1-0)^2}=1$$
$$\therefore r = \boxed{1}$$
$\sin({-90^{\circ}}) = \dfrac{y}{r} = \dfrac{-1}{1} = \boxed{-1}$
$\cos({-90}^{\circ}) = \dfrac{x}{r} = \dfrac{0}{1} = \boxed{0} $
$\tan({-90}^{\circ}) = \dfrac{y}{x} = \dfrac{-1}{0 } = \fbox{undefined}$
