Question

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

• A. $\left( 2\sqrt{3},\frac{\pi}{6} \right)$
• B. $\left( - 2\sqrt{3},\frac{5\pi}{6} \right)$
• C. $\left( 2\sqrt{3},\frac{5\pi}{6} \right)$
• D. None of these

Answer 1

Answer: (C)

$\left( 2\sqrt{3},\frac{5\pi}{6} \right)$

Answer 2

r = $\sqrt{x^{2} + y^{2}}$ = $\sqrt{9 + 3}$ = $2\sqrt{3}$

a = tan^–1 $\frac{|y|}{|x|}$

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

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

(–3, $\sqrt{3}$) lies in II quadrants.

\ q = p – a = p – $\frac{\pi}{6}$= $\frac{5\pi}{6}$

$\left( 2\sqrt{3},\frac{5\pi}{6} \right)$