The polar coordinates equivalent to `(-3,sqrt3)` are

Answer

D

Explanation

`r=sqrt((x^2+y^2))=sqrt(9+3)=2sqrt3`
`alpha=tan^(-1).|y|/|x|`
`alpha=tan^(-1)((sqrt3)/3)`
`alpha=tan^(-1)((1)/sqrt3)=(pi)/(6)`
Point `(-3,sqrt3)` lies in the second quadrant.
`therefore theta=pi-alpha=pi-(pi)/(6)=(pi)/(6)`
So, polar coordinates are `(2sqrt(3),(5pi)/(2))`.