Answer
$F_{DE}=392N$
$F_{CD}=340N$
$F_{CA}=243N$
$F_{CB}=275N$
Work Step by Step
We can find the required tension developed in each cord as follows:
$F_{FD}=mg=(20)(9.81)=196.2N$
Now we use the sum of forces at points D and C to find the required values
$\Sigma F_y=0$
$\implies F_{DE} sin 30-F_{FD}=0$
$\implies F_{DE} sin30-196.2$
$\implies F_{DE}=392N$
and $\Sigma F_x=0$
$\implies F_{DE} cos 30-F_{CD}=0$
$\implies 392 \space cos30-F_{CD}=0$
$\implies F_{CD}=340N$
Now we find $F_{CA}$ and $F_{CB}$ as
$\Sigma F_x=0$
$\implies -\frac{3}{5} F_{CA}+F_{CD}-F_{CB} cos 45=0$
We plug in the known values to obtain:
$F_{CA}=243N$
and $\Sigma F_y=0$
$\implies \frac{4}{5}F_{CA}-F_{CB} sin 45=0$
We plug in the known values to obtain:
$F_{CB}=275N$
