Geometry problem - IOQM

Question

In $\triangle ABC$, $AC = BC$, $\angle C=20^{\circ}$, $M$ is on the side $AC$ and $N$ is on the side $BC$, such that $\angle BAN=50^{\circ}, \angle ABM = 60^{\circ}$. Find $\angle NMB$ in degrees.

It is INMO-IOQM level question and I am not able to solve anyhow, please help. I tried by angle chasing but up to no avail.

Answer 1

Trigonometric Solution

Let $\angle NMB=\alpha$

$\dfrac{BN}{\sin\alpha}=\dfrac{MB}{\sin(160-\alpha)}$,$\triangle NMB$,and, $\dfrac{AB}{\sin40}=\dfrac{MB}{\sin 80}$, $\triangle AMB$ (Sine Theorem)

Since $AB=BN$, then

$\dfrac{\sin\alpha}{\sin(160-\alpha)}=\dfrac{\sin40}{\sin 80}=\dfrac{1}{2\cos40}$

$\sin(160-\alpha)=2 \sin\alpha \cos40=\sin(\alpha+40)+ \sin(\alpha-40)$

$\sin(160-\alpha)- \sin(\alpha-40)=\sin(\alpha+40)$

$2\cos60 \sin(100-\alpha)=\sin(\alpha+40)$

$\sin(100-\alpha)=\sin(\alpha+40)$

1. Case

$100-\alpha=\alpha+40 \implies \alpha=30$

2. Case

$100-\alpha+\alpha+40=180 \implies 140\not=180$ , NA

Answer 2

Take a point P on AC between A and M such that $\angle ABP=20$

Hence $BA=BP$

$\angle PMB= \angle PBM= 40$

Hence $PB=PM$

$\angle BAN= \angle BNA= 50$

Hence $BA=BN$

So $\triangle BPN$ is equilateral triangle

$PN=PM$ and $\angle MPN=40$

Finally

$\angle PMN= \angle PNM= 70$

$\angle NMB= 30$