For $W=490lb,$ determine the tension in each cable and the reaction at $C.($Include a minus sign if necessary.$)$

The tension in cable DAE is $$

$$ lb$.$

The tension in cable BF is $-$ lb$.$

The reaction at $C$ is $($ lb$)$ lb$)j+(-2486$

$$ x lb$)k.$

A $48-$in$.$ boom is held by a ball$-$and$-$socket joint at $C$ and by two cables BF and DAE; cable DAE passes around a frictionless pulley at $A.$