Angle Chasing in geometry in triangle

Question

In triangle $\Delta ABC$, $\angle CAB = 30^{\circ}$ and $\angle ABC = 80^\circ$. The point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find the value of $180^\circ – \angle BMC$ in degrees .

Answer 1

Hint:

Extend lin segment $AM$ and draw line segment $BF$ such that $\angle CBF = 10^0$.

Can you see $ABFC$ is cyclic with $\angle ACF = \angle ABF = 90^0$?

That leads to center of the circle on $AF$.

If $\angle ABG = 20^0$ then can you see $AG = BG = CG$?

You can then show $\triangle BCG$ is equilateral.

Now, as $\angle MGC = \angle MCG = 20^0$, $\angle GBM = \angle CBM = 30^0$.

You also know $\angle BCM = 40^0$.

And adding both $\angle CBM$ and $\angle BCM$, you get $(180^0 - \angle BMC)$.