Two tangents PA and PB are …

A circle C(O, r). PA and PB are tangents to the circle from point P, outside the circle such that {tex}\angle{/tex}APB = 120°.
Construction: Join OA and OB. 
Proof. Consider {tex}\triangle{/tex}{tex}PAO \ and\ \triangle PBO {/tex}
{tex}PA = PB{/tex} [Tangents drawn from a point to circle are equal]
{tex} OP = OP{/tex}[Common]
{tex}\angle{/tex}OAP = {tex}\angle{/tex}OBP = 90°
{tex}\therefore{/tex}{tex}\triangle{/tex}OAP {tex}\cong{/tex} {tex}\triangle{/tex}OBP [by SAS cong.]
{tex}\therefore{/tex}{tex}\angle{/tex}OPA  = {tex}\angle{/tex}OPB = {tex}\frac { 1 } { 2 } \angle \mathrm { APB } = \frac { 1 } { 2 } \times 120 ^ { \circ } = 60 ^ { \circ }{/tex}
In right angled {tex}\triangle{/tex}OAP, {tex}\frac { \mathrm { AP } } { \mathrm { OP } }{/tex}= cos 60° = {tex}\frac {1}{2}{/tex} {tex}\Rightarrow{/tex} {tex}OP = 2AP.{/tex}