ans is 75 and 45

Could someone please provide a solution for this?

My friend gave an elegant solution Firstly Join $KN$ and observe that $\triangle KMN $ is equilateral Let $X$ be the mid-point of $AN$ Therefore $AK=AX=XN$ Also observe that $\triangle AKX $ is equilateral $\implies AK=AX=KX=XN$ Note that $\angle KXN =120^{\circ} $ AS $\triangle KXN $ is isosceles with $KX=XN \implies \angle XKN=\angle XNK=30^{\circ}$ Now Note that $\angle XNM=30^{\circ}+60^{\circ}=90^{\circ}$ But in isosceles $\triangle NMC, \angle MNC=90^{\circ} \implies \angle ACM =45^{\circ}$ Which shows that in $\triangle ABC, \angle A=60^{\circ},\angle B=75^{\circ},\angle C=45 ^{\circ}$

How do you know that $\triangle KMN$ is equilateral?

farishafizhan wrote: How do you know that $\triangle KMN$ is equilateral? hopefully a helpful solution i wrote just now: (too lazy to latex it)

official solution !

Attachments:

Let $P$ be midpoint of $AN$. Then, $\triangle AKP$ is isosceles, and so $\angle KAP=60^\circ$ implies that $\triangle AKP$ is in fact equilateral. So, $\triangle PKN$ is isosceles . It follows that it's a $30-120-30$ triangle. Let $\angle B=\beta$, $\angle C=\alpha$, $\angle MKN=\angle MNK=\gamma$. Then we find, \begin{align*}2\beta -\gamma & =90^\circ \\ 2\alpha- \gamma & =30^\circ \\ \alpha+\beta &=2\gamma.\end{align*}Solving we get, $\beta=75^\circ, \alpha=45^\circ$.