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Correct option is C)

or

$[(500)1+80(0.5)−(Tsin37_{0})(0.75)]=0$

(note that torque produced by $Tcos30_{0}$, $N_{x}$ and $N_{y}$ about the point A is equal to zero),Thus

$500+40−T(53 )(0.75)=0$

or

$T=3(0.75)540(5) =1200N$

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