The tension in the string marked A is determined as 331.35 N.
<h3>Angle made by A with respect to vertical</h3>
tanθ = Δy/Δx
tanθ = (6 - 1)/(7 - 1) = 0.8333
θ = arc tan(0.8333) = 39.81⁰
with respect to horizontal = 90 - 39.81 = 50.19⁰
<h3>Angle made by B with respect to horizontal</h3>
tanθ = Δy/Δx
tanθ = (9 - 6)/(7 - 5) = 1.5
θ = arc tan(1.5) = 56.31 ⁰
<h3>Angle made by C with respect to horizontal</h3>
tanθ = Δy/Δx
tanθ = (6 - 4)/(14 - 7) = 0.2857
θ = arc tan(0.2857) = 15.95 ⁰
Bcosθ + Aosθ = Ccosθ
Bcos(56.31) + A[cos(50.19)] = 56.3cos(15.95)
0.55B + 0.64A = 54.13 ----- (1)
Bsinθ + Asinθ = Csinθ
Bsin(56.31) + A[sin(50.19)] = 56.3sin(15.95)
0.832B + 0.77A = 15.47---- (2)
Solve (1) and (2)
0.2A = 66.27
A = 66.27/0.2
A = 331.35 N
Thus, the tension in the string marked A is determined as 331.35 N.
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