Question

The polar co-ordinates of (–3$\sqrt{3}$, –3) is-

• A. $\left( 6,\frac{\pi}{6} \right)$
• B. $\left( - 6,\frac{\pi}{6} \right)$
• C. $\left( - 6,\frac{- 5\pi}{6} \right)$
• D. None

Answer 1

Answer: (B)

$\left( - 6,\frac{\pi}{6} \right)$

Answer 2

(–3$\sqrt{3}$, – 3)

r = $\sqrt{x^{2} + y^{2}}$

r = $\sqrt{27 + 9}$= 6

a = tan^–1 $\left( \frac{3}{3\sqrt{3}} \right)$ = $\frac{\pi}{6}$

(–3$\sqrt{3}$, –3) lies in III quad.

\ q = – (p – a)

q = – $\left( \pi –\frac{\pi}{6} \right)$= – $\frac{5\pi}{6}$

$\left( 6,–\frac{5\pi}{6} \right)$ or $\left( –6,\frac{\pi}{6} \right)$.