Correct option is: (C) 5√3
\(\because\) \(\angle\) APB = 60°
= \(\angle\) APO = \(\frac {\angle APB}2 = \frac {60^\circ}{2}\) = 30°
Also OA \(\perp\) AP (Angle between radius and tangent at point of contact)
\(\therefore\) \(\angle\) OAP = 90°
Now, in right \(\triangle\) OAP
\(\frac {PA}{OP}\) = cos (\(\angle\)APO)
= \(\frac {PA}{10}\) = cos 30° (\(\because\) \(\angle\) OPA = 30° & OP = 10 cm).
= PA = 10 cos 30° = \(\frac {\sqrt3}2 \times 10 = 5\sqrt3\).