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In any `Delta ABC` line joiningcircumcentre (O) and incentre (I) is parallel to AC, then OI is equal toA. `R|tan((A-C)/(2))|`B. `R|tan(A-C)|`C. `R|sec((A-C)/(2))|`D. `R|sec(A-C)|` |
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Answer» Correct Answer - A Distance of O from AC = Distance of I from AC `rArr R cos B = r` `rArr (r )/(R )=cos B` `rArr 4 sin.(A)/(2)sin.(B)/(2)sin.(C )/(2)-1=cos B` `rArr cos A + cos B + cos C -1 =cos B` `rArr cos A + cos C = 1` `OI = |AE - AD|` (where E nad D are feet of perpendiculars from O and I respectively on AC) `=|(b//2)-(s-a)|` `=(|a-c|)/(2)` `=R|sin A - sin C|` `=2 R|sin.(A-C)/(2)cos.(A+C)/(2)|` `=R|tan.(A-C)/(2)2cos.(A+C)/(2)cos.(A-C)/(2)|` `=R|tan.(A-C)/(2)(cos A + cos C)|` `=R|tan((A-C)/(2))|` |
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