Static equalibrium problem

An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and the boot together have a mass of 4.0 kg , and the doctor has decided to hang a 6 kg mass from the rope. The boot is suspended by the ropes and does not touch the bed.

The net traction force needs to pull straight out on the leg. What is the proper angle for the upper rope? (the bottom angle is 15)

If it helps you draw the picture there a 3 pulleys(frictionless). There's one attached near the top, one on the patients foot, and the other at the end of the bed. The rope is in the same x position at the beginning and at the end where the 6 kg is being hanged.

The tension in the rope T = Mw*g = 6.0*9.8 = 58.8 N

For the leg to be pulled straight out, the vertical components of forces on the leg must sum to zero.

The vertical component from the lower part of rope is -T*sin15º
The vertical component from the upper part of the rope is T*sinθ;
The vertical force from the weight of the foot and boot is -4.0*9.8 = -39.2 N

Then

-T*sin15º + T*sinθ - 39.2 = 0

sinθ = (39.2 + 58.8*sin15º)/58.8 = 0.925

θ = 68º

The net traction force is the sum of the horizontal components:

T*cosθ + T*cos15º = 79 N