Oh yes!

Let $\angle BMC = \theta$ , $BC = a$ , $AC = b$ and $MC = d$ . By sine rule, we have:

$\begin{cases} \dfrac{a}{b} = \dfrac{\sin \angle A}{\sin \angle B} & = \dfrac{\sin 35^\circ}{\sin 110^\circ} \\ \dfrac{b}{d} = \dfrac{\sin \angle AMC}{\sin \angle MAC} & = \dfrac{\sin 125^\circ}{\sin 30^\circ} \\ \dfrac{a}{d} = \dfrac{\sin \angle MBC}{\sin \angle BMC} & = \dfrac{\sin \theta}{\sin (170^\circ-\theta)} \end{cases}$

$\begin{aligned} \frac{a}{d} & = \frac{a}{b}\times \frac{b}{d} \\ \Rightarrow \frac{\sin \theta}{\sin (170^\circ-\theta)} & = \frac{\sin 35^\circ \sin 125^\circ}{\sin 110^\circ \sin 30^\circ} \\ & = \frac{\cos (90^\circ - 35^\circ) \sin (180^\circ - 125^\circ)}{2\sin 55^\circ \cos 55^\circ \times \frac{1}{2}} \\ & = \frac{\cos 55^\circ \sin 55^\circ}{\sin 55^\circ \cos 55^\circ} \\ & = 1 \\ \Rightarrow \sin \theta & = \sin (170^\circ - \theta) \\ \theta & = 170^\circ - \theta \\ 2 \theta & = 170^\circ \\ \Rightarrow \theta & = \angle BMC = \boxed{85}^\circ \end{aligned}$

Just a tip, you can do degrees with ^{ \circle}