The solution for this problem is:
For the torques to be equivalent the constituent of the
forces vertical to the rod must be equal and in the opposite direction
So 37.8 = 56.5 * sin (θ)
So θ = arc sin (37.8/56.5)
= arcsin 0.670212766 = sin⁻¹ 0.670212766 = 42° 5' 0.558"
= 42.08348831° + k×360° (k=..-1,0,1,..)
= -317.91651169°, 42.08348831°, 402.08348831°, ..
= 0.73449543rad + k×2π (k=..-1,0,1,..)
= -1.76620284π, 0.23379716π, 2.23379716π,..
= 42.1 degrees