The magnitude of the tension in the string marked A is 52.5N
Generally, the equation for is mathematically given as
Let's take θ be an angle at A
So, tanθ = 3/8
Let's take α be an angle at B (Below X)
tanα = 5/4
Let's take β be an angle at C (Below x)
tanβ = 1/6
First we take the Horizontal Components
74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)
By solving the equation, we get
A = 78.9 - 0.668B … (1)
Now, we take the vertical components
74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)
By solving the equation, we get
40.07 = 1.015B
B = 39.5N
By substituting the value of B in equation (1)
A = 78.9 - 0.6668× 39.5
A = 52.5N
Hence, the magnitude of the tension in the string marked A is 52.5N
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