Answer
The point is $(-1,-1)$
$r=\sqrt{2}$
$\sin({-135^{\circ}}) = -\dfrac{\sqrt{2}}{2}$
$\cos({-135}^{\circ}) = -\dfrac{\sqrt{2}}{2}$
$\tan({-135}^{\circ}) = 1$
Work Step by Step
The terminal side of $-135^{\circ}$ in standard position is represented by the blue line in the figure. It lies in the 3rd quadrant.
The coordinates of points on the terminal side of $-135^{\circ}$ can be given by $(-a,-a)$, where $a$ is a positive number.
Choosing $a=1$ arbitrarily, the point is $(-1,-1)$.
To find the distance from the origin to $(-1,-1)$, we use the distance formula
$$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\r=\sqrt{(-1-0)^2+(-1-0)^2}=\sqrt{2}$$
$$\therefore r = \boxed{\sqrt{2}}$$
$\sin({-135^{\circ}}) = \dfrac{y}{r} = \dfrac{-1}{\sqrt{2}} = \boxed{-\dfrac{\sqrt{2}}{2}}$
$\cos({-135}^{\circ}) = \dfrac{x}{r} = \dfrac{-1}{\sqrt{2}} = \boxed{-\dfrac{\sqrt{2}}{2}} $
$\tan({-135}^{\circ}) = \dfrac{y}{x} = \dfrac{-1}{-1 } = \boxed{1}$
