Answer: 83

Small Hint:

The base angles are $37^\circ.$ In triangle $AMC,$ the Law of Sines with $AC = 1$ gives $CM = 2\sin 7^\circ.$

Big Hint:

Apply the Law of Cosines in triangle $BMC,$ using the fact that the cosine of $\angle MCB$ equals $\sin 7^\circ,$ to show $MB = CB$

Solution:

Assume $AC = BC = 1.$ In triangle $AMC,$ the angles at $A$ and $C$ are $7^\circ$ and $23^\circ,$ so $\angle AMC = 150^\circ,$ and the Law of Sines gives $$CM = \frac{\sin 7^\circ}{\sin 150^\circ} = 2\sin 7^\circ.$$

Also $\angle MCB = 106^\circ - 23^\circ = 83^\circ,$ whose cosine is $\sin 7^\circ.$ The Law of Cosines in triangle $BMC$ then gives $$\begin{aligned} MB^2 &= CM^2 + CB^2 \\ &\quad {}- 2 \cdot CM \cdot CB \cos 83^\circ \\ &= 4\sin^2 7^\circ + 1 \\ &\quad {}- 4\sin^2 7^\circ = 1. \end{aligned}$$

So $MB = 1 = CB,$ making triangle $BMC$ isosceles with $\angle CMB = \angle MCB = 83^\circ.$ The answer is $83.$