Answer
The force in the horizontal cord is 2900N and the force in the other cord is 3400N
Work Step by Step
Let $F_1$ be the tension in the horizontal cord and $F_2$ be the tension in the other cord. We will sum the forces in each direction to solve for the tensions. So;
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_2*sin\theta-m*g=0$
$F_2=\frac{m*g}{sin\theta}=\frac{190kg*9.8m/s^2}{sin(33^{\circ})}\approx3400N$
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$F_1-cos\theta*F_2=0$
$F_1=cos\theta*F_2=cos(33^{\circ})*3400N\approx2900N$
