69. In triangle $ABC$, given are $\widehat{ABC}=70^{\circ}, \widehat{ACB}=$ $=50^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=40^{\circ}$, and on $AC$-point $N$ such that $\widehat{NBC}=50^{\circ}$. Find $\widehat{NMC}$.
69. Let's take point $K$ (Fig. 15) such that $\widehat{K B C}=\widehat{K C B}=30^{\circ}$, and denote by $L$ the intersection point of the lines $M C$ and $B K$. 

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-081.jpg?height=394&width=435&top_left_y=415&top_left_x=165)

Fig. 15. Since $\triangle B N C$ is isosceles $\left(\widehat{N B C}=\widehat{N C B}=50^{\circ}\right)$, then $\widehat{K N C}=$ $=40^{\circ}$. Point $L$ is the intersection of the angle bisectors of triangle $N K C$ ( $L K$ and $L C$ are angle bisectors). Therefore, $N L$ is also the angle bisector of $\angle K N C$ and $\widehat{L N B}=60^{\circ} ; B N$, in turn, is the angle bisector of $\angle M B L$; moreover, $B N \perp M L$; thus, $B N$ bisects $M L$ and $\widehat{M N B}=\widehat{B N L}=60^{\circ}$, and $\widehat{N M C}=$ $=30^{\circ}$.

Answer: $30^{\circ}$.