Answer
The tension in the wire and the horizontal reaction at A are 410N. The vertical reaction at A is 328N.
Work Step by Step
1) Sum the torques about point A
$(+ \circlearrowleft) \sum \tau=0$
$T*3.8m-12kg*9.8m/s^2*cos37^{\circ}*3.6m-21.5kg*9.8m/s^2*7.2m*cos37^{\circ}=0$
$T=(12kg*9.8m/s^2*cos37^{\circ}*3.6m+21.5kg*9.8m/s^2*7.2m*cos37^{\circ})/3.8m$
$T\approx410N$
2) Sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$R_A-12kg*g-21.5kg*g=0$
$R_A=12kg*g+21.5kg*g$
$R_A=12kg*9.8m/s^2+21.5kg*9.8m/s^2$
$R_A\approx328N$
3) Sum the forces horizontally.
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$R_X-T=0$
$R_X=T$
$R_X\approx410N$
