Three identical rigid circular cylinders A, B and C are arranged on smooth inclined surfaces as shown in figure. The least value of θ that prevent the arrangement from collapse is

W = weight of each sphereN2 = normal reaction between,4 and inclined planeN1 = normal reaction between A and C = normal reaction between B and CFree body diagram of C:Resolving vertically 2N1cos 300 = Wor N1= W3-----(1)When the arrangement is on the point of collapsing, the reaction between A and B is zero.Free body diagram of A:Resolving horizontally and verticallyN2 sin θ = N1sin 300 or N2sin θ = W23--(2)N2 cos θ = W + N1 cos 300 = 3W2---(3)Dividing (2) by (3) we gettan θ = 133 or θ = tan-1(133)