A900 kg steel beam is supported by the two ropes shown in (figure 1) . calculate the tension in the rope.

Let T1 and T2 be tension in ropes1 and 2 respectively. 
since system is stationary (equilibrium), considering both ropes + beam as a system 

for horizontal equilibrium (no movement in that direction, so resultant force must be zero horizontally) 
T1sin(20) = T2sin(30) 
=> T1 = T2sin(30) / sin(20) 

for vertical equilibrium, (no movement in this direction, so resultant force must be zero vertically) 
T1cos(20) + T2cos(30) = mg 

m = 900kg, substituting for T1 
T2sin(30)*cos(20)/sin(20) + T2cos(30) = 900g 
2.328*T2 = 900*9.8 
T2 = 3788.65N 
so T1 from (1) 
T1 = 5535.21N