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| 1. |
Op is equal to diameter of circle .Then prove that ABP is an equilateral triangle |
| Answer» Construction : Join A to Bwe have,OP = diameter{tex}\\Rightarrow {/tex}\xa0OQ + QP = diameter{tex}\\Rightarrow {/tex}\xa0Radius + QP = diameter{tex}\\Rightarrow {/tex}\xa0OQ - PQ = radiusThus OP is the hypotenuse of right angled\xa0{tex}\\triangle {/tex}AOPSo, In\xa0{tex}\\angle A O P , \\sin \\theta = \\frac { A O } { O P } = \\frac { 1 } { 2 }{/tex}\xa0{tex}\\theta = 30 ^ { \\circ }{/tex}Hence,\xa0{tex}\\angle A P B = 60 ^ { \\circ }{/tex}Now, In {tex}\\triangle A O P{/tex}AP = ABSo,\xa0{tex}\\angle P A B = \\angle P B A = 60 ^ { \\circ }{/tex}{tex}\\therefore \\triangle A P B{/tex} is an equilateral triangle. | |