## Find the angle

For students of class 9-10 (age 14-16)
dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### 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

Atonu Roy Chowdhury
Posts: 40
Joined: Fri Aug 05, 2016 7:57 pm

### 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.

joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

### 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