11. In the triangle $A B C$ the points $M$ and $N$ lie on the side $A B$ such that $A N=A C$ and $B M=B C$.
We know that $\angle M C N=43^{\circ}$.
Find the size in degrees of $\angle A C B$.
SOlUTION
094

Let $\angle A C M=x^{\circ}$ and $\angle B C N=y^{\circ}$.
Using the base angles property of isosceles triangles $A C N$ and $B C M$, we have $\angle A N C=43+x$ and $\angle B M C=43+y$.
In triangle $C M N, 43+(43+x)+(43+y)=180$.
Therefore, $\angle A C B=x+43+y=94$.