Answer
The tension in the left cord is 260N and the tension in the left cord is 190N
Work Step by Step
Let $F_1$ be the tension in the right cord and $F_2$ be the tension in the left cord. We will sum the forces in each direction to solve for the tensions. So;
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$cos(37^{\circ})F_1-cos(53^{\circ})F_2=0$
$F_1=\frac{cos(53^{\circ})F_2}{cos(37^{\circ})}$
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_2sin53^{\circ}+F_1sin37^{\circ}-33kg*9.8m/s^2=0$
Now substitute in equation 1
$F_2sin53^{\circ}+(\frac{cos(53^{\circ})F_2}{cos(37^{\circ})})sin37^{\circ}-33kg*9.8m/s^2=0$
$F_2=(33kg*9.8m/s^2)/(sin53^{\circ}+(\frac{cos(53^{\circ})}{cos(37^{\circ})})sin37^{\circ})$
$F_2\approx260N$
$F_1=\frac{cos(53^{\circ})F_2}{cos(37^{\circ})}$
$F_1=\frac{cos(53^{\circ})260N}{cos(37^{\circ})}$
$F_1\approx190N$
