The magnitude of the** tension** in the string marked A is **39.5 N.**

<h3>What is the tension in A?</h3>

The **tension** in A is determined thus:

The angle at A, θ = tan⁻¹(3/8) = 20.56

When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34

When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46

Taking the** horizontal components** of** tension;**

56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)

0.6247B= 55.53 - 0.936A

B = (55.53 - 0.936A)/0.6247 ----(1)

Taking the **vertical components** of** tension;**

56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)

9.25 + 0.35A = 0.78B ---- (2)

substitute the value (1) in (2)

9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}

(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A

0.22A + 0.73A = 43.31 - 5.78

0.93A = 37.53

A = 39.5 N

In conclusion, the tension in A is determined by solving for the **vertical **and **horizontal components** of **tension.**

Learn more about** tension force** at: brainly.com/question/24994188

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