A system of connected flexible cables shown in diagram is supporting two vertical  forces 200 N and 250 N at points B and D. 
 

Of whole system is shown in figure.

Consider the equilibrium of point D :
$\begin{array}{c}{\sum }^{\text{}} F_x=0, or T_{1}sin{45}^{\circ }-T_{2}sin{60}^{\circ }=0\end{array}$  ….(i)
And  $\begin{array}{c}{\sum }^{\text{}} F_y=0, or T_{1}cos{45}^{\circ }+T_{2}cos{60}^{\circ }=250\end{array}$ …..(ii)
From (i), we have  $T_2=\frac{T_1\text{sin}{45}^{\circ }}{\text{sin}{60}^{\circ }}$
Now from (ii),  $T_1\text{cos}45+\frac{T_1\text{sin}{45}^{\circ }}{\text{sin}{60}^{\circ }}\times \text{cos}{60}^{\circ }=250$
 $or \frac{T_1}{\sqrt{2}}+\frac{T_1}{\sqrt{2}}\times \frac{1}{\sqrt{3}}=250 or T_1=224.4 \text{N}$
and  $\begin{array}{c}{T}_2=\frac{T_1\text{sin}{45}^{\circ }}{\text{sin}{60}^{\circ }}=\frac{224.4\times \left(\frac{1}{\sqrt{2}}\right)}{\left(\frac{\sqrt{3}}{2}\right)}=183.01 N\end{array}$
Consider the equilibrium of point  :
  $\begin{array}{c}{\sum }^{\text{}} F_x=0; or T_{2}sin{60}^{\circ }+T_{3}sin{30}^{\circ }-T_4=0\end{array}$ ….(i)
and   $\begin{array}{c}{\sum }^{\text{}} F_y=0; or T_2\cos {30}^{\circ }-T_2\cos {30}^{\circ }-200=0\end{array}$….(ii)

From (ii),  $T_3\text{cos}{30}^{\circ }-183.01\times \frac{1}{2}-200=0$
or  $T_3=336.60 \text{N}$
From equation (i),  $T_4=255.81 \text{N}$