016 Components of a force parallel to supporting bars

By Sine Law:
$\dfrac{F_{AB}}{\sin 20^\circ} = \dfrac{8000}{\sin 40^\circ}$

$F_{AB} = 4256.71 \, \text{ N}$           answer
 

 

$\dfrac{F_{AC}}{\sin 120^\circ} = \dfrac{8000}{\sin 40^\circ}$

$F_{AC} = 10\,778.37 \, \text{ N}$           answer