## Find the angle

### Find the angle

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 $\angle BMC$.

The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

- Charles Caleb Colton

- Atonu Roy Chowdhury
**Posts:**40**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

### Re: Find the angle

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

### Re: Find the angle

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 $\angle BMO=\angle BMC=\dfrac {1}{2}(360^{\circ} -\angle AMC) =110^{\circ}$

The more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein