If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm,

(c) $3 \sqrt{3} \mathrm{~cm}$

Given, $\mathrm{PA}$ and $\mathrm{PB}$ are tangents to circle with centre $\mathrm{O}$ and radius $3 \mathrm{~cm}$ and $\angle \mathrm{APB}=60^{\circ}$.

T angents drawn from an external point are equal; so, PA $=$ PB.

And $O P$ is the bisector of $\angle A P B$, which gives $\angle O P B=\angle O P A=30^{\circ}$.

OA $\perp$ PA. So, from right $-$ angled $\Delta$ OPA, we have :

$\frac{\mathrm{OA}}{\mathrm{AP}}=\tan 30^{\circ}$

$\Rightarrow \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{3}{\mathrm{AP}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \mathrm{AP}=3 \sqrt{3} \mathrm{~cm}$