68. In triangle $ABC$, it is given that $|AB|=|BC|, \widehat{ABC}=$ $=20^{\circ}$; on $AB$ a point $M$ is taken such that $\widehat{MCA}=60^{\circ}$; on side $CB$ - a point $N$ such that $\widehat{NAC}=50^{\circ}$. Find $\widehat{NMA}$.
68. Let's take a point $K$ on $BC$ (Fig. 14) such that $\widehat{K A C}=60^{\circ}$, $M K \| A C$. Let $L$ be the intersection point of $A K$ and $M C$; $\triangle A L C$ is equilateral, $\triangle A N C$ is isosceles (calculate the angles), so $\triangle L N C$ is also isosceles, $\widehat{L C N}=20^{\circ}$. Now let's find the angles $N L M$ and $M K N$ - they are both $100^{\circ}$; since $\triangle M K L$ is equilateral, the angles $\widehat{K L N}$ and $\widehat{N K L}$ are both $40^{\circ}$, i.e., $|K N|=|L N|$ and $\triangle M K N = \triangle M L N, \widehat{N M L}=\widehat{K M N}=30^{\circ}$.

Answer: $30^{\circ}$.