Ans: C
Note that $\theta = 210^\circ – 180^\circ = 30^\circ$. Hence we have the $x$ coordinate of the point
$\begin{array}{cl}
= & – 6 \cos 30^\circ \\
= & – 6 \times \dfrac{\sqrt{3}}{2} \\
= & -3\sqrt{3}
\end{array}$
and the $y$ coordinate of the point
$\begin{array}{cl}
= & -6 \sin 30^\circ \\
= & -6 \times \dfrac{1}{2} \\
= & -3
\end{array}$
Therefore, the required coordinates are $(-3\sqrt{3},-3)$.
