# Convert the point (-sqrt3, -sqrt 3) in rectangular coordinates to polar coordinates that satisfies r > 0, -pi < theta leq pi

Convert the point $\left(-\sqrt{3},-\sqrt{3}\right)$ in rectangular coordinates to polar coordinates that satisfies
$r>0,-\pi <\theta \le \pi$
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smallq9

We have to convert the rectangular coordinates to polar coordinates.$\left(-\sqrt{3},-\sqrt{3}\right)$ as we know that the rectangular coordinates are (x, y) and the polar coordinates are $\left(r,\theta \right)$ where $r=\sqrt{{x}^{2}+{y}^{2}}$ here $r={\sqrt{\left(-\sqrt{3}{\right)}^{2}+\left(-\sqrt{3}\right)}}^{2}$
$r=\sqrt{3+3}$
$r=\sqrt{6}$ also $\theta =ta{n}^{-1}\frac{x}{y}$ $\theta ={\mathrm{tan}}^{-1}\left(\frac{-\sqrt{3}}{-\sqrt{3}}\right)$
$\theta ={\mathrm{tan}}^{-1}1$
$\theta =\frac{\pi }{4}$ Therefore The polar coordinates are $\left(\sqrt{6},\frac{\pi }{4}\right)$