Direction of forces at E is known, i.e., 45⁰ to horizontal (along BE). Hence, resolved vertical and horizontal components of force F_E are equal i.e. F _EX=F_EY..

Now balancing horizontal forces:

F _EX = F_CX.

Taking moment about E,

1000 × 0.4 -(F_CY) × 1.6=0

F_CY =250N

Now balancing the vertical forces,

250 + F_EY - 1000=0

F_EY = 750N= F_EX

Thus the forces at C and E have been calculated. Likewise, other forces can be calculated.