力學題目有人有詳細解題嗎？

固定A-B
AO = BO = 3ft
Cos∠A = 2/3
Sin∠A = 3/4 = Sin∠B
Ta = OA繩拉力
Ta Sin∠A + Ta Sin∠B = F = 200lb
1.5Ta = 200
Ta = 133lb

固定C-D
CO = DO = 3ft
Cos∠C = 1/3 = 0.333  
Sin∠C = √(3^2-1^2) / 3 = 2.828/3 = 0.94281 = Sin∠D
T = OA繩拉力
Tc Sin∠C + Tc Sin∠D= F = 200lb
Tc (2x0.94281) = 200
Tc = 106.07lb

因此固定在C-D吊的話，繩拉力會小一點=106lb(?)