Saved Bookmarks
| 1. |
2tangent segment PA are drawn to a circlevwith center at O angle APB=120 prove that OP=2AP |
| Answer» 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.\xa0Proof. Consider {tex}\\triangle{/tex}{tex}PAO \\ and\\ \\triangle PBO\xa0{/tex}{tex}PA = PB{/tex}\xa0[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\xa0{tex}\\cong{/tex}{tex}\\triangle{/tex}OBP [by SAS cong.]{tex}\\therefore{/tex}{tex}\\angle{/tex}OPA =\xa0{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}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}OP = 2AP.{/tex} | |