Answer
$T_1=0.68KN$
$T_2=0.79KN$
Work Step by Step
We know that
$\Sigma F_x=-T_1cos18^{\circ}+T_2cos35^{\circ}=0$
$\implies T_2=T_1cos18^{\circ}/cos35^{\circ}$.......eq(1)
Similarly $F_y=T_1sin18^{\circ}+T_2sin35^{\circ}-mg=0$
$T_1sin18^{\circ}+(T_1\frac{cos18^{\circ}}{cos35^{\circ}})sin35^{\circ}-mg=0$
$\implies T_1(sin18^{\circ}+cos18^{\circ}tan35^{\circ})=mg$
$\implies T_1=\frac{(68)(9.81)}{sin18^{\circ}+cos18^{\circ}tan35^{\circ}}$
$T_1=684N=0.68KN$
From eq(1)
$T_2=T_1\frac{cos18^{\circ}}{cos35^{\circ}}$
$T_2=(0.68KN)\frac{cos18^{\circ}}{cos35^{\circ}}$
$T_2=0.79KN$
