# Convert the point (-sqrt3, -sqrt 3) in rectangular coordinates to polar coordinates that satisfies r > 0, -pi < theta leq pi Question
Parametric equations, polar coordinates, and vector-v Convert the point $$(-\sqrt3, -\sqrt 3)$$ in rectangular coordinates to polar coordinates that satisfies
$$r > 0, -\pi < \theta \leq \pi$$ 2020-11-27
We have to convert the rectangular coordinates to polar coordinates.$$(-\sqrt{3}, -\sqrt{3})$$ as we know that the rectangular coordinates are (x, y) and the polar coordinates are $$(r, \theta)$$ where $$r = \sqrt{x^2 + y^2}$$ here $$r = \sqrt{(-sqrt3)^2 + (-sqrt3)}^2$$
$$r = \sqrt{3 + 3}$$
$$r = \sqrt6$$ also $$\theta = tan^{-1} \frac{x}{y}$$ $$\theta = tan^{-1} \frac{(-sqrt3}{-sqrt3})$$
$$\theta = tan^{-1} 1$$
$$\theta = \frac{\pi}{4}$$ Therefore The polar coordinates are (\sqrt6, \frac{\pi}{4})

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