If you look at the sketch I drew for the earlier part of this question, you'll see that, with respect to the positive horizontal (i.e. directly to the right), T₁ makes an angle of 180° - 54.9205° ≈ 125°, while T₂ makes an angle of 54.9205° ≈ 55°.
Split up the force acting on the block into vertical and horizontal components. We have
• net vertical force
∑ F = T₁ sin(125°) + T₂ sin(55°) - mg = 0
where m = 0.56 kg, and
• net horizontal force
∑ F = T₁ cos(125°) + T₂ cos(55°) = 0
Both net forces are 0 because the block is suspended in equilibrium.
Notice that cos(125°) = -cos(55°), so the second equation tells you that T₁ = T₂ and that the tensions in either string are the same. Also, sin(125°) = sin(55°).
Then in the first equation, we have
T₁ sin(125°) + T₂ sin(55°) - mg = 0
2 T₁ sin(55°) = mg
T₁ = mg/(2 sin(55°))
T₁ = (0.56 kg) (9.8 m/s²)/(2 sin(55°))
T₁ ≈ 3.35 N
