Answer: [tex] T_1 = 77.2 \text N \newline T_2= 91.87 \text N [/tex]
Assuming the figure as attached.
The sum of Horizontal components of the Tension in the rope is zero or equal as they balance out each other.[tex] T_1 sin40 \textdegree = T_2 sin50 \textdegree
\newline T_1 = 1.19 T_2 [/tex]
The sum of vertical components of the tension is equal to the weight of the block.[tex] T_1 cos 40 \textdegree +T_2cos50 \textdegree = 120 \text N
\newline 1.19T_2(0.766 )+T_2 0.64 = 120 \text N
T_2 = 77.2 \text N
T_1 = 1.19 T2 = 1.19 \times 77.2 \text N = 91.87 \text N [/tex]
