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)