To find the value of μ, we resolve forces parallel to the slopes for both particles.

For particle B on the 50° slope:

\(1.6g \sin 50^{\circ} - T - F_B = 0\)

For particle A on the 40° slope:

\(T - F_A - 1.2g \sin 40^{\circ} = 0\)

Normal reactions are:

\(R_A = 1.2g \cos 40^{\circ}\)

\(R_B = 1.6g \cos 50^{\circ}\)

Frictional forces are:

\(F_A = 1.2g \mu \cos 40^{\circ}\)

\(F_B = 1.6g \mu \cos 50^{\circ}\)

Substitute the frictional forces into the equations:

\(1.6g \sin 50^{\circ} - T - 1.6g \mu \cos 50^{\circ} = 0\)

\(T - 1.2g \mu \cos 40^{\circ} - 1.2g \sin 40^{\circ} = 0\)

Eliminate \(T\) by adding the equations:

\(1.6g \sin 50^{\circ} - 1.6g \mu \cos 50^{\circ} = 1.2g \sin 40^{\circ} + 1.2g \mu \cos 40^{\circ}\)

Solve for \(\mu\):

\(\mu = \frac{1.6g \sin 50^{\circ} - 1.2g \sin 40^{\circ}}{1.2g \cos 40^{\circ} + 1.6g \cos 50^{\circ}}\)

Calculating gives:

\(\mu = 0.233\)