By symmetry, the vertical reactions at
A and
B are equal.
$A_V = B_V = 90 + 240$
$A_V = B_V = 330 ~ \text{kN}$
From the FBD of member AC
$\Sigma M_C = 0$
$2S \, A_H + 2S(90) + S(240) = 3S(330)$
$A_H = 285 ~ \text{kN}$ ← Answer for Part (1)
$\Sigma F_V = 0$ already, hence
$C_V = 0$ (not shown in the figure) ← Answer for Part (3)
Note:
$\Sigma F_H = 0$
$C_H = A_H = 285 ~ \text{kN}$
From the FBD of the whole system
$\Sigma F_H = 0$
$B_H = A_H$
$B_H = 285 ~ \text{kN}$
$R_B = \sqrt{{B_H}^2 + {B_V}^2}$
$R_B = \sqrt{285^2 + 330^2}$
$R_B = 436 ~ \text{kN}$ ← Answer for Part (2)
