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| 1. |
PT and PS are 2 tangents are drawn to a circle and radius r OP=2r show angle ots=ost=30 |
| Answer» Given,In {tex}\\triangle{/tex}OTS,{tex}\u200b\u200b\u200bOT =OS {/tex}{tex}\\Rightarrow \\quad \\angle O T S = \\angle O S T{/tex}\xa0...(i)In right\xa0{tex}\\triangle OTP,{/tex}\xa0{tex}\\frac { \\mathrm { OT } } { \\mathrm { OP } } = \\sin \\angle \\mathrm { TPO }{/tex}{tex}\\Rightarrow \\frac { r } { 2 r } = \\sin \\angle \\mathrm { TPO }{/tex}{tex}\\sin \\angle \\mathrm { TPO } = \\frac { 1 } { 2 } \\Rightarrow \\angle \\mathrm { TPO } = 30 ^ { \\circ }{/tex}Similarly\xa0{tex}\\angle \\mathrm { OPS } = 30 ^ { \\circ }{/tex}{tex}\\Rightarrow \\angle T P S = 30 ^ { \\circ } + 30 ^ { \\circ } = 60 ^ { \\circ }{/tex}Also\xa0{tex}\\angle \\mathrm { TPS } + \\angle \\mathrm { SOT } = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad \\angle \\mathrm { SOT } = 120 ^ { \\circ }{/tex}In {tex}\\triangle{/tex}SOT,{tex}\\angle \\mathrm { SOT } + \\angle \\mathrm { OTS } + \\angle \\mathrm { OST } = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow 120 ^ { \\circ } + 2 \\angle \\mathrm { OTS } = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\angle \\mathrm { OTS } = 30 ^ { \\circ }{/tex}\xa0...(ii)From (i) and (ii)\xa0{tex}\\angle \\mathrm { OTS } = \\angle \\mathrm { OST } = 30 ^ { \\circ }{/tex} | |