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If point A is in equilibrium under the action of the applied force, then find the value of tension $T_{A C}$ .

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520 N

150 N

400 N

450 N

answer is C.

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Detailed Solution

$\frac{T_{A C}}{\sin 150^{0}} = \frac{800}{\sin 90^{0}} \Rightarrow T_{A C} = 400 N$

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