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Answer: \(\left(2,\dfrac{5\pi}{3}\right)\)

\(x=1\), \(y=-\sqrt3\). \(r=\sqrt{x^2+y^2}=\sqrt{1+3}=2\).

\(\cos\theta=\dfrac{x}{r}=\dfrac12\), \(\sin\theta=\dfrac yr=-\dfrac{\sqrt3}{2}\), so \(\theta\) is in the fourth quadrant: \(\theta=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}\).

Polar co-ordinates are \(\left(2,\dfrac{5\pi}{3}\right)\) (equivalently \(\left(2,-\dfrac{\pi}{3}\right)\)).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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