Free body diagram of Joint A
For wire AB: By sine law (from the force polygon):
$\dfrac{T_{AB}}{\sin 40^{\circ}} = \dfrac{W}{\sin 80^{\circ}}$
$T_{AB} = 0.6527W$
$\sigma_{AB} \, A_{AB} = 0.6527W$
$30(0.4) = 0.6527W$
$W = 18.4 \, \text{kips}$
For wire AC:
$\dfrac{T_{AC}}{\sin 60^{\circ}} = \dfrac{W}{\sin 80^{\circ}}$
$T_{AC} = 0.8794W$
$T_{AC} = \sigma_{AC} \, A_{AC}$
$0.8794W = 30(0.5)$
$W = 17.1 \, \text{kips}$
For safe load W,
$W = 17.1 \, \text{ kips}$ answer