[tex]\displaystyle\bf\\\sphericalangle BAC=x\\\sphericalangle CAN=x+90^o\\AN=AC\\\implies~\Delta AANC~este~triunghi~isoscel\\\\\\\sphericalangle ACN=\sphericalangle ANC=\frac{180-(x+90)}{2}=\frac{180-x-90}{2}=\frac{90-x}{2}\\\\\\\sphericalangle ACB=\frac{180-x}{2}\\\\\\\sphericalangle NCB=\sphericalangle ACB-\sphericalangle ACN=\frac{180-x}{2}-\frac{90-x}{2}=\\\\\\=\frac{180-x-(90-x)}{2}=\frac{180-x-90+x}{2}=\frac{180-90}{2}=\frac{90}{2}=\boxed{\bf45^o}[/tex]
