The magnitude of the tension in the string marked A is 52.5 N.
Given that the magnitude of the tension in the string labeled C is 56.3 N.
The angle at A is
tan θ = [tex]\frac{3}{8}[/tex]
below the negative x
B= tan Φ
tan Φ = [tex]\frac{5}{4}[/tex]
C = tan ρ
tan ρ = [tex]\frac{1}{6}[/tex]
Considering the Horizontal components only
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Considering the Vertical components only
74.9*Sin(9.46) + ASin(20.6) = BSin(51.3)
40.07 = 1.015B
B = 39.5 N
By substituting the value of B in the equation of A
Since, A = 78.9 - 0.668B
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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