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Let $\angle CMB = \theta$ . Using Sine Rule, we have:

$\def\arraystretch{2.5} \begin{array}{c}\triangle ACM: & \dfrac {CM}{\sin{\angle MAC}} = \dfrac{AC}{\sin{\angle CMA}} \quad \Rightarrow \dfrac {CM}{\sin{7^\circ}} = \dfrac{AC}{\sin{\angle CMA}} \\ & \Rightarrow CM = \dfrac {AC\sin{7^\circ}}{\sin{150^\circ}} = \dfrac {AC\sin{7^\circ}}{\sin{30^\circ}} = 2AC\sin{7^\circ} \\ \triangle BCM: & \dfrac {CM}{\sin{\angle MBC}} = \dfrac{BC}{\sin{\angle CMB}} \quad \Rightarrow \dfrac {CM}{\sin{(97^\circ-\theta)}} = \dfrac{BC}{\sin{\theta}} \\ & \Rightarrow CM = \dfrac {\color{#3D99F6}{BC}\cos{(\theta-7^\circ)}}{\sin{\theta}} = \dfrac {\color{#3D99F6}{AC}\cos{(\theta-7^\circ)}}{\sin{\theta}} \end{array}$

$\begin{aligned} \Rightarrow 2\sin{7^\circ} & = \frac {\cos{(\theta-7^\circ)}}{\sin{\theta}} \\ 2\sin{7^\circ} \sin{\theta} & = \cos{(\theta-7^\circ)} \\ 2\sin{7^\circ} \sin{\theta} & = \cos{7^\circ} \cos{\theta} + \sin{7^\circ} \sin{\theta} \\ \cos{7^\circ} \cos{\theta} - \sin{7^\circ} \sin{\theta} & = 0 \\ \cos{\theta + 7^\circ} & = 0 \\ \Rightarrow \theta + 7^\circ & = 90^\circ \\ \Rightarrow \angle CMB & = \theta = 90^\circ-7^\circ = \boxed{83^\circ} \end{aligned}$

Let Triangle BCD be an equilateral triangle. Following from AC=CB=CD, the angle of CAD=7(degrees). Therefore, A,M,D are colinear. Because the angle of CBD is twice the angle of AMC, and because BC=BD, B is the circumcenter of triangle MCD. Therefore CB=BM. Therefore the angle of CMB is 73(degrees)