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*