Determine the determine the required tension T in cable AB such that net effect of the two cable tensions is a downward force at a point A. Determine the magnitude R of this downward force.

The tension in Cables AC and AB act away from the point A and along the cables.

Therefore, their angles will be same as that of the cables.

Angle of AB $=α= \tan^{-1}$ (opposite / adjacent) $= \tan^{-1} (\text {height of AB /base of Ab }) = \tan^{-1} (40 / 50) = 38.66 \\ \text {Similarly, Angle of } AC = β= \tan^{-1} (60/40) = 56.31$

For resultant to be downward (vertical), its horizontal component should be zero. So, $∑Fx = 0$

$T_{AB} \cos α – T_{AC} \cos β = 0 \\ T_{AB} \cos (38.66) – 8 \cos (56.31) = 0 \\ T_{Ab}= 5.68 kN \\ Also, R= ∑Fy = - T_{AB} \sin α – T_{AC} \sin β = -5.68 \sin 38.66 - 8 \cos56.31 = -10.2 kN = 10.2 kN ↓ $