[tex]\displaystyle\\
\sphericalangle BAC = x\\
\sphericalangle CAN = x + 90\\
AN = AC\\
\Longrightarrow~~\Delta ANC~\text{ este triunghi isoscel}\\\\
\sphericalangle ACN = \frac{180 - (x+90)}{2} =\frac{180 - x-90}{2} = \frac{90 - x }{2} \\\\
\sphericalangle ACB = \frac{180 - x}{2}\\\\
\ \textless \ 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{45^o}
\\[/tex]
