Correct Answer - A::D
`F_(1)=mg sin theta + mu mg cos theta`
`F_(2)=mg sin theta-mu mg cos theta`
But `mg=w mu= tan phi`
`F_(1)=w(sin theta+(sin phi)/(cosphi)costheta)`
`rArrF_(1)=w sin (theta+phi)secphi`
`F_(2)=w (sin theta-(sin phi)/(cos phi)cos theta)`
`rArr wsin theta (theta-phi)sec phi`
Now `F_(1)=2F_(2)`
`mg sin theta+mu mg cos theta=2 (mg sin theta -mu mg cos theta)`
`sin theta+mu cos theta=2sin theta-2mucos theta)`
`rArr 3mu cos theta=sintheta`
`=tan theta=3 mu tan theta=3tan phi`