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) $= \tan^{-1}$ (height of AB /base of Ab) $= \tan^{-1} (40 / 50) = 38.66$

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, $R= ∑Fy = - T_{AB} \sin α – T_{AC} \sin β = -5.68 \sin38.66 – 8\cos56.31 = -10.2 kN = 10.2 kN ↓$