∑F = 0

∑Horizontal forces = 0

F_{HH} + [-(T)] = 0

F_{HH} + [-(916N)] = 0

F_{HH} = 916N

∑Vertical forces = 0

F_{HV} + [-(W1)] + [-(W2)] = 0

F_{HV} + [-(100N)] + [-(500N)] = 0

F_{HV} = 600N

Now F_{HH} = F_{H} Cos θ

and F_{HV} = F_{H} Sin θ

So tan θ = [(F_{HV})/ (F_{HH})]

θ = 33.2°

F_{H} = F_{HV} / Sin θ

F_{H} = 600 _{} / Sin 33.2°

F_{H} = 1100N _{}

F_{H} = F_{H}_{H} / Cos θ

F_{H} = 916N _{} / Cos 33.2°

F_{H} = 1110N _{}