Is Angle Chasing Useful?

Let A'BC be the equilateral triangle. (A' is in the same side with A over BC).

Because $\angle{NBC} = 50^{\circ} = \angle{C}$ so $NB=NC$ .

Because $\angle{MCB} = 40^{\circ} \text{ so } \angle{NCM} = 10^{\circ} =\angle{NBA'} (=60^{\circ}-50^{\circ})$

Because $\angle{MCB} = 40^{\circ} \text{ so } \angle{CMB} = 70^{\circ} = \angle{B}$ , that leads to $MC=BC=A'B$

From the three above, we have $\bigtriangleup A'BN = \bigtriangleup MCN$ so $\angle{NMC} = \angle{NA'B}$ .

But because $NB=NC$ and A'BC is an equilateral triangle so we can see $\angle{NA'B} = \boxed{30^{\circ}}$

I chased angles via trig. NB and CM turn out to be perpendicular (let's say they meet at O). I gave NC an arbitrary value of 10 (since I wasn't looking for lengths anyway, I was looking for a specific angle). I went counter clockwise through the (4) right triangles that meet at O.

Giving NC = 10, NO = 1.7365 and working thru to MO, MO = 3.0076.

Arctan NO/MO = 30 degrees.

I did this using trigonometry.

Let the lines $BN$ and $CM$ intersect at $O$ . Now note that $\angle OCB = 40^{\circ}$ and $\angle OBC = 50^{\circ}$ . Hence $\angle BOC = 90^{\circ}$ . i.e. lines $MC$ and $NB$ are perpendicular.

Let $\angle NMC = x$ . So we have $\tan x = \frac{NO}{MO}$ , because $\triangle NOM$ is a right triangle.

Similarly, from $\triangle NOC$ we get, $\tan 10^\circ = \frac{NO}{OC}$ , and from $\triangle MOB$ , we get $\tan 20^{\circ} = \frac {MO}{OB}$ . Finally from $\triangle BOC$ , we have $\tan 40^{\circ}= \frac{OB}{OC}$ . Putting this all together we get:

$\frac{\tan 10^\circ}{\tan 20^\circ} = \frac{\frac{NO}{OC}}{\frac{MO}{OB}} = \frac{NO}{MO}\frac{OB}{OC} = \tan x \tan 40^\circ$

Hence, $x = \tan^{-1} \left( \frac{ \tan 10^\circ}{\tan 20^\circ \tan 40^\circ}\right)$

Using a calculator, we get $x = 30^\circ$ .

Let angle NMC=x

Now angle BNC=80 and angle BMC=70

In triangle MNC:

MN/sin(10)=CN/sin(x)---------------------(1)

In triangle MNB:

MN=sin(20).BN/sin(70+x)----------------(2)

In triangle BNC:

BN=sin(50).BC/sin(80)----------------------(3)

From (2) and (3) we get

MN=sin(20).sin(50).BC/sin(80).sin(70+x)-----(4)

Again In triangle BNC:

CN=sin(50).BC/sin(80)-------------------------(5)

Now putting the value of MN and CN in equation (1)

sin(20).sin(50).BC/sin(10).sin(80).sin(70+x)=sin(50).BC/sin(80).sin(x)

or sin(20)/sin(10).sin(70+x)=1/sin(x)

or sin(20).sin(x)=sin(10).sin(70+x)

or 2sin(10).cos(10).sin(x)=sin(10).sin(70+x)

or 2cos(10)sin(x)=sin(70).cos(x)+cos(70).sin(x)

or sin(x)(2cos(10)-cos(70))=sin(70).cos(x)

or sin(x)/cos(x)=sin(70)/(2cos(10)-cos(70))

or tan(x)=0.57735027=1/sqrt(3)

or x=30 degree(ans)

$\angle NBC=\angle BCN, \therefore ~ ~\color{#3D99F6}{NB=NC. ~~~~ \angle CMB=\angle (180-20-50-40)=70.}\\ Applying ~ Sin ~ Law ~ to ~ \Delta s ~~NMC ~ and ~ NMB, ~ we ~ have:-\\ \dfrac{Sin10}{Sin?}=\dfrac{MN}{NC}=\dfrac{MN}{NB}=\dfrac{Sin20}{Sin(70+?)}\\ On ~ expanding ~and ~rearranging, ~\dfrac{Sin20}{Sin10}=Cos70+Sin70Cot?.\\ \therefore ~Cot ? =\Large \dfrac{\frac {Sin20}{Sin10} - Cos70}{Sin70}=1.732050808.\\ \therefore ~ ?= Tan^{-1}\dfrac 1 {1.732050808}=\color{#D61F06}{30^o }$

Masbahul Islam's and my approach is the same. Use of Sin Law. His presentation is detailed and a little round about.