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| 1. |
Two tangent PA and PB are drawn to circle such that angle APB=30Deegre .prove that OP=2AP |
| Answer» In {tex}\\triangle{/tex}AOP and {tex}\\triangle{/tex}BOP, we have,{tex}\\angle OAP = \\angle OBP = 90^\\circ{/tex}OP = OP [Common]PA = PB [{tex}\\because{/tex} Tangents from an external point are equal in length]So, by RHS congruence criterion, we have{tex}\\triangle AOP \\cong \\triangle BOP{/tex}{tex}\\therefore \\angle APO = \\angle BPO{/tex} [By C.P.C.T.]{tex}\\Rightarrow \\angle APO = \\angle BPO = \\frac{1}{2}\\angle APB{/tex}{tex} = \\frac{1}{2} \\times 120^\\circ{/tex}= 60°{tex} \\Rightarrow \\angle APO = \\angle BPO = 60^\\circ {/tex}In {tex}\\triangle{/tex}OAP, we have{tex}\\cos 60^\\circ = \\frac{{AP}}{{OP}}{/tex}{tex} \\Rightarrow \\frac{1}{2} = \\frac{{AP}}{{OP}}{/tex}{tex}\\Rightarrow{/tex} OP = 2APHence proved. | |