Answer
$N_C=163N$
$N_B=105N$
Work Step by Step
First we will sum the forces horizontally
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$125-N_C\cos40^{\circ}=0$
$N_C=125/\cos40^{\circ}=163.18N=163N$
Now we will sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$N_B-N_C\sin40^{\circ}=0$
$N_B=N_C\sin40^{\circ}=163.18\sin40^{\circ}=105N$
