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In triangle $\bigtriangleup ABC$, $\angle BCA = 70, \angle BAC = 30$. $M$ is a point inside the triangle such that $\angle ACM = 30, \angle CAM = 10$. Find $\angle BMC$.

I solved it by Trig-Ceva. I don't wanna write those trigging right now. If anyone solved synthetic, please post ur solu.

Let $O$ be the circumcenter of $\triangle ABC$, Then $\angle ABC =180^{\circ}-30^{\circ} -70^{\circ}=80^{\circ} ,\angle OAC =90^{\circ}-\angle ABC=10^{\circ}.$ So $O \in AM$ .$\angle OBC=\angle ABC -\angle ABO =80^{\circ}-20^{\circ}=60^{\circ} $ .As $OB=OC$ so , $\triangle OBC$ is equilateral .$\angle BOM =2\angle BAO=40^{\circ}=\angle BCM $. So by symmetry $\angle BMO=\angle BMC=\dfrac {1}{2}(360^{\circ} -\angle AMC) =110^{\circ}$