We can find the Tension using newton's second law:
[tex]\begin{gathered} \Sigma Fy=0 \\ mg-T1sin(40)-T2sin(20)=0 \\ so: \\ T2=\frac{mg-T1sin(40)}{sin(20)} \end{gathered}[/tex][tex]\begin{gathered} \Sigma Fx=0 \\ T1cos(40)-T2cos(20)=0 \\ \end{gathered}[/tex]
Replace T2 into the previous equation:
[tex]\begin{gathered} T1cos(40)-cos(20)(\frac{mg-T1sin(40)}{sin(20)})=0 \\ T1cos(40)-cot(20)mg+T1cot(20)sin(40)=0 \\ T1=\frac{cot(20)mg}{cos(40)+cot(20)sin(40)} \\ T1\approx9570.261N \end{gathered}[/tex]
Answer:
D