Solution :

Let , $X_A$ and $Y_A$ be the reactions at the hinge A and $R_B$ at the roller support B

F.B.D of given structure

Applying conditions of equilibrium

$1) \Sigma F_x =0$

$\therefore H_A-30cos40°=0 $

$\therefore H_A=22.98 KN$

$\space $

$2) \Sigma M_A^F =0$

$\therefore 10R_B-(5\times4\times2)-(25\times4)-(8\times 30sin40°)=0 $

$\therefore 10R_B-294.27=0$

$\therefore R_B=29.43KN$

$\space $

$3) \Sigma F_y =0$

$\therefore V_A-(5\times 4)-25-30sin40°+R_B=0$

$\therefore V_A+R_B=64.28 $

$\therefore V_A=43.86 KN $

$\space $

Resultant

$R_A=\sqrt {V_A^2+H_A^2}$

$R_A=\sqrt {34.86^2+22.98^2}$

$R_A=41.75KN$

Angle of resultant

$A=tan^-1(\frac{V_A}{H_A})$

$A=tan^-1(\frac{34.86}{22.98})$

$A=56.61°$