<s>[INST] Finding angle in the isosceles triangle There is the following triangle. I infer that the triangle is isosceles. But I cannot go further.

 [/INST] Without trigonometry
Let $D$ be the symmetric of $C$ with respect to $A$, $H$ and $K$ the orthogonal projections of $A$ and $C$ on $BC$ and $AB$.
$BCD$ is a right triangle. By the midpoint theorem, $CK=2MN=6$ and $MN=MA=4$ so $BK=2$. By Pythagoras theorem, $BC=\sqrt{BK^2+CK^2}=2\sqrt{10}$ and $AH=\sqrt{AB^2-BH^2}=3\sqrt{10}$.
$\dfrac{NC}{ND}=\dfrac{5}{15}=\dfrac{1}{3}$ and $\dfrac{BC}{BD}=\dfrac{BC}{2AH}=\dfrac{1}{3}$ so, by the angle bisector theorem, $BN$ is the interior bisector of the right angle $CBD$. Thus $x=45^\circ$.

With trigonometry
$\cos BAC=4/5$ and $BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos BAC=40$ by the law of cosines.
$NC^2=NB^2+BC^2-2NB\cdot BC\cdot\cos x$ gives $x=45^{\circ}$
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