From the FBD of the whole system
$\Sigma M_A = 0$
$5R_E = 5.5(24)$
$R_E = 26.4 ~ \text{kN}$
$\Sigma M_E = 0$
$5A_V = 0.5(24)$
$A_V = 2.4 ~ \text{kN}$
$\Sigma F_H = 0$
$A_H = 0$
From the FBD of the horizontal member
$\Sigma M_D = 0$
$3B_V = 1.5(24)$
$B_V = 12 ~ \text{kN}$
From the FBD of member BC
$\Sigma F_V = 0$
$C_V + 2.4 = 12$
$C_V = 9.6 ~ \text{kN}$
$\Sigma M_B = 0$
$3C_H + 1(2.4) = 1.5C_V$
$3C_H + 2.4 = 1.5(9.6)$
$C_H = 4 ~ \text{kN}$
$\Sigma F_H = 0$
$B_H = C_H$
$B_H = 4 ~ \text{kN}$
Answer:
$B_H = 4 ~ \text{kN to the right}$
$B_V = 12 ~ \text{kN upward}$
$C_H = 4 ~ \text{kN to the left}$
$C_V = 9.6 ~ \text{kN downward}$
