In $\Delta ABC$, $AC = BC$ and $\angle C = 120^\circ$. $M$ is on side $AC$ and $N$ is on side $BC$ such that $\angle BAN = 50^\circ$ and $\angle ABM = 60^\circ$. Find $\angle NMB$ .
What I Tried: Here's a diagram which shows what I have tried :-
You can see that I tried angle-chasing but now I am stuck. Can someone help me out.
Note that this is a possible duplicate of this (only $\angle C = 120^\circ$ here) :- Geometry problem - IOQM
I read the duplicate but didn't make anything out of my question, can someone help me? A solution without trigonometry will be better, but it's not bad to give it.
$MC=BC\cos60^\circ=BC/2$.
Using sine rule in $\triangle ANC$,$$\frac{AC}{\sin100^\circ}=\frac{NC}{\sin20^\circ}$$Using $AC=BC$, we get $NC=BC\sin20^\circ/\sin100^\circ$.
Using sine rule in $\triangle NMC$,$$\begin{align*}\frac{NC}{\sin(60^\circ-x)}&=\frac{MC}{\sin x}\\2\sin x\sin20^\circ&=\sin100^\circ\sin(60^\circ-x)\end{align*}$$
Expand $\sin(60^\circ-x)$ to get $\tan x=\frac{\sqrt 3\sin 100^\circ}{4\sin 20^\circ+\sin 100^\circ}\approx 0.725$, giving $x\approx35.94^\circ$ and $\angle NMB=150^\circ-x\approx 114.06^\circ$.