Trong tam giác ABC ta có:

$\widehat{A}+\widehat{B}+\widehat{ACB}={{180}^{0}}\Rightarrow \widehat{B}+\widehat{ACB}={{180}^{0}}-{{100}^{0}}={{80}^{0}}$

$\widehat{ACB}+\widehat{ACx}={{180}^{0}}$ mà CN là tia phân giác góc ACx nên

$\widehat{BCA}+2\widehat{xCN}={{180}^{0}}\Rightarrow \widehat{xCN}=\widehat{ACN}=\frac{{{180}^{0}}-\widehat{BCA}}{2}$

Trong tam giác BNC ta có:

$\widehat{CBN}+\widehat{BNC}+\widehat{BCN}={{180}^{0}}$

$\frac{1}{2}\widehat{CBA}+\widehat{BNC}+{{180}^{0}}-\widehat{NCx}={{180}^{0}}$

$\frac{1}{2}\widehat{CBA}+\widehat{BNC}-\frac{{{180}^{0}}-\widehat{BCA}}{2}=0$

$\widehat{BNC}=\frac{{{180}^{0}}-\widehat{BCA}}{2}-\frac{1}{2}\widehat{CBA}=\frac{{{180}^{0}}-{{80}^{0}}}{2}={{50}^{0}}$