From the FBD of the Whole Section
$\Sigma M_G = 0$
$24A_V = 18(36) + 6(60)$
$A_V = 42 ~ \text{kN}$ answer
From the FBD to the section to the left of D
$\Sigma M_D = 0$
$9A_H + 6(36) =12(42)$
$A_H = 32 ~ \text{kN}$ answer
From the FBD of joint A
$\Sigma F_V = 0$
$F_{AB}(\frac{3}{5}) = 42$
$F_{AB} = 70 ~ \text{kN}$ answer
$\Sigma F_H = 0$
$F_{AC} + 32 = \frac{4}{5}F_{AB}$
$F_{AC} + 32 = \frac{4}{5}(70)$
$F_{AC} = 24 ~ \text{kN}$ answer