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.