## Trigonometry (11th Edition) Clone

Two pairs of polar coordinates for this point are: $(2\sqrt{2}, 225^{\circ})$ $(-2\sqrt{2}, 45^{\circ})$
$(-2,-2)$ $r = \sqrt{(-2)^2+(-2)^2} = 2\sqrt{2}$ We can find the angle $\phi$ below the negative x-axis: $tan~\phi = \frac{2}{2}$ $\phi = arctan(\frac{2}{2})$ $\phi = 45^{\circ}$ Then $\theta = 180^{\circ} +45^{\circ} = 225^{\circ}$ Two pairs of polar coordinates for this point are: $(2\sqrt{2}, 225^{\circ})$ $(-2\sqrt{2}, 45^{\circ})$