If ‘O’ is in equilibrium then tensions ${\text{T}}_{\text{1}}\text{and}{\text{T}}_{\text{2}}$ are $\left(\text{Tan}37°=\frac{3}{4}\right)$  

$\frac{{\text{T}}_{\text{1}}}{\sin \left(90°+53°\right)}=\frac{{\text{T}}_2}{\sin \left(90°+37°\right)}=100$                              

$\frac{{\text{T}}_1}{\frac{3}{5}}=100$

${\text{T}}_{\text{1}}=100\times \frac{3}{5}$

${\text{T}}_{\text{1}}=60\text{N}$

$\frac{{\text{T}}_2}{\cos 37°}=100$

$\frac{{\text{T}}_{\text{2}}}{\frac{\text{4}}{\text{5}}}=100$

${\text{T}}_{\text{2}}=100\times \frac{4}{5}$

${\text{T}}_{\text{2}}=80\text{N}$