A particle block of mass 'm' is in equilibrium being supported by light strings AB, BC and CD as shown. AB = 5a; CB = 3a; AC = 4a. A horizontal force F = mg in the plane of ABC acts parallel to AB as shown. Then,

${\text{T}}_2\sin 53°+{\text{T}}_{\text{1}}\sin 37°=\text{mg}$

${\text{T}}_2\cos 53-{\text{T}}_{\text{1}}\cos 37=\text{mg}$

${\text{T}}_{\text{2}}.\frac{4}{5}+{\text{T}}_{\text{1}}.\frac{3}{5}=\text{mg}$

${\text{4T}}_2+3T_1=5\text{mg}\times 3$

$3{\text{T}}_{\text{2}}-4{\text{T}}_{\text{1}}=5\text{mg}\times 4$

$12{\text{T}}_2+9{\text{T}}_{\text{1}}=15\text{mg}$

$\underset{\_}{12{\text{T}}_2-16{\text{T}}_{\text{1}}=20\text{mg}}$

$5{\text{T}}_{\text{1}}=-5\text{mg}$

${\text{T}}_1=\frac{-\text{mg}}{5}$

$1{\text{2T}}_2+9\frac{\left(-\text{mg}\right)}{5}=15\text{mg}$

$1{\text{2T}}_2=15\text{mg}+\frac{9}{5}\text{mg}$

$\text{=}\frac{75\text{mg}+9\text{mg}}{5}$

$12{\text{T}}_{\text{2}}=\frac{84}{5}\text{mg}$

${\text{60T}}_2=84\text{mg}$

${\text{T}}_2=\frac{84}{60}\text{mg}$

${\text{T}}_2=\frac{7}{5}\text{mg}$