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

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

Answers (1)

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})
0

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