Let the force F is applied at an angle θ with the horizontal. 
For horizontal equilibrium, Fcosθ=μR For vertical equilibrium, R+Fsinθ=mg or, R=mg−Fsinθ Substituting this value of R in eq. (i), we get Fcosθ=μ(mg−Fsinθ)=μmg−μFsinθ or, F(cosθ+μsinθ)=μmg or, F=cosθ+μsinθμmg For F to be minimum, the denominator (cosθ+μsinθ) should be maximum. ∴dθd(cosθ+μsinθ)=0 or, −sinθ+μcosθ=0 or, tanθ=μ or, θ=tan−1(μ) Then, sinθ=1+μ2μ and cosθ=1+μ21 Hence, Fmin =1+μ21+1+μ2μ2μw=1+μ2μw