Look at the figures is attachments. Here FF is the force applied and ff is force due to friction
F(cos α)=f cos β F (cos α)= f coβ (1)
Also,
f = μ N f = μ N (2)
where, μμ is the coefficient of friction and NN is the normal reaction by the floor on the block.
By force balance on the block along the vertical direction we get
N= F sin α + f sin β + mg N=F sin α +f sin β + mg (3)

subtracting 2 and 3 we get
N = F sin α + mg ( 1 − μ sin β ) N = F sin α + mg ( 1 − μ sin β) (4)

Now subtracting 2 and 4 we get
F cos α = μ F sin α + mg ( 1 − μ sin β )  cos β F cos α = μ F sin α + mg ( 1 − μ sin β ) cos β

F cos α ( 1− μ sin β ) =μ cos β ( F sin α + mg ) F cos α ( 1 – μ sin β ) = μ cos β ( F sin α + mg )

F cos α ( 1 − μ sin β ) = F μ cos β sin α + μ mg cos β F cos α ( 1−μ sin β) =F μ cos β  sin α+ μ mg cos β

F [cos α− μ cos α sin β− μ cos β sin α ]= μ mg cos β F [ cos α –μ cos α sin β− μ cos β sin α ] = μ mg cos β

F = μ mg cos β cos α – μ ( cos α sin β + cos β sin α ) F = μ mg cos β cos α – μ ( cos α sin β + cos β sin α )

F = μ mg cos β cos α – μ sin ( α + β ) F = μ mg cos β cos α – μ sin ( α + β )

since sin ( α + β ) = cos α sin β + cos β sin α sin ( α + β ) = cos α sin β + cos β sin α.

F = F min = μ mg cos β cos α – μ sin ( α + β )
F = F min = μ mg cos β cos α – μ sin ( α + β ).