In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ – \angle BMC)$.
What I Tried: Here is a picture to keep track of the angle-chasing .
The red angle is $30^\circ$ , the green one is $10^\circ$ , the purple one is $20^\circ$ . Now I extend $MC$ to $AB$ such that it meets at $K$. First I noticed that $AK = KC$ but that didn't seem to help. Also the $2$ yellow ones are equal to $40^\circ$ each, the brown one is equal to $80^\circ$ . The light green angle is $60^\circ$ . We have to find the blue angle.
After all these information, it seems like I am still missing something, because from here you actually cannot deduce the value of the blue angle. Can anyone help?
Thank You.
Best Answer
Let $O$ be the circumcenter of $\triangle ABC$. Consider $\triangle OAC$,
$\angle AOC = 2\angle ABC = 160°$
$\angle OAC = \angle OCA = \frac{180° - 160°}{2} = 10°$
Hence, $O$ lies on $AM$.
$\angle MOC = 10° + 10° = \angle MCO \Rightarrow MO = MC$.
$\angle OCB = \angle OBC = \frac{180° - 60°}{2} = \angle COB \Rightarrow BC = OB$.
Hence, $\triangle BMO \cong \triangle BMC$.
$\therefore 180° - \angle BMC = \frac{\angle OBC}{2} + \angle BCO - 20° = 70°$