Correct Answer - A::B::C::D
`sqrt2=(sin 45^@)/sinr`
`r=30^@`
`theta_C=sin^-1(1/mu)=45^@`
Applying sine law in `Delta`CPM,
`(CP)/(sin theta_C)=(CM)/(sin(90^@+r))`
`:. (CP)/((1//sqrt2))=R/cosr [R=radius]`
`:. CP=sqrt(2/3) R`
As we move away from C, angle PMC will
increse. Therefore, CP ? `sqrt(2/3) R.` Same is the case
on left side of C.