Given an isoceles triangle with equal side of length b and angle `alpha lt pi//4`, then
the inradius r is given by

Options

A. `(b sin 2 alpha)/(2(1 -cos alpha))`
B. `(b sin 2 alpha)/(2(1 + cos alpha))`
C. `(b sin alpha)/(2)`
D. `(b sin alpha)/(2(1+ sin alpha))`

Answer

B

Explanation

From sine rule, `(b)/(sin alpha) = 2R " or " R = (b)/(2) cosec alpha`
`r = (Delta)/(s) = ((1)/(2) b^(2) sin (pi - 2 alpha))/((1)/(2) (b + b+ 2b (cos alpha))) = (b sin 2 alpha)/(2(1 + cos alpha))`
Distance of incenter (I) from the side BC is r.
Distance of circumcenter (O) from the side BC is
`|R cos A | = |R cos (pi - 2 alpha)| = R cos 2 alpha`
Hence, distance between circumcenter and incenter is given by
`OI = |r+R cos 2 alpha|` ( `:'` circumcenter lies outside `Delta`)
`=|(b sin 2 alpha)/(2(1+cos alpha)) + (b " cosec " alpha cos 2 alpha)/(2)|`
`= b |(sin 2 alpha)/(2 cos^(2).(alpha)/(2)) + (cos 2 alpha)/(4 sin.(alpha)/(2) "cos" (alpha)/(2))|`
`b = |(cos 2 alpha "cos"(alpha)/(2) + sin 2 alpha "sin "(alpha)/(2))/(4 "sin"(alpha)/(2) "cos"^(2) (alpha)/(2))|`
`= |(b cos (3(alpha)/(2)))/(2 sin alpha "cos"(alpha)/(2))|`