#### Answer

$F_{AD}=1.125KN$
$F_{AC}=844N$
$F_{AB}=506N$

#### Work Step by Step

We can find the required tension in the cables as follows:
The sum of the forces in the z-direction is
$\Sigma F_z=0$
$\implies F_{AD}(\frac{4}{5})-900=0$
$\implies F_{AD}=1125N=1.125KN$
The sum of the forces in the y-direction is
$\Sigma F_y=0$
$\implies F_{AC}(\frac{4}{5})-F_{AD}(\frac{3}{5})=0$
$\implies F_{AC}(\frac{4}{5})-(1125)(\frac{3}{5})=0$
This simplifies to:
$F_{AC}=844N$
The sum of the forces in the x-direction is
$\Sigma F_x=0$
$\implies F_{AB}-F_{AC}(\frac{3}{5})=0$
$\implies F_{AB}-(844)(\frac{3}{5})=0$
This simplifies to:
$F_{AB}=506N$