Forces that appear in the free body diagram
1) T2: horizontal tension to the left
2) T1: tension to the right, making angle 50° with the horizontal
Vertiical component of T1: T1y = T1sin(50°)
Horizontal component of T1 = T1x = T1. cos(50°)
3) Weight of the mass: 50 kg 9.8 m/s^2 = 490 N
4) Equilibrium analysis
Net vertical forces = 0 => T1y = weight => T1sin(50°) = 490 N => T1 = 490 N / sin (50°) = 639.7 N
T1 = 639.7 N
Net horizontal forces = 0 => T2 = T1x = T1cos (50°) = 639.7 N * cos(50°) = 411.2 N
Answer: 411.2 N