In any `Delta ABC` line joiningcircumcentre (O) and incentre (I) is parallel to AC, then OI is equal to

Options

A. `R|tan((A-C)/(2))|`
B. `R|tan(A-C)|`
C. `R|sec((A-C)/(2))|`
D. `R|sec(A-C)|`

Answer

A

Explanation

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))|`