Too Many Angles, Too Little Information

let x be the desired angle:

$\triangle ABC:\quad \frac { AB }{ CB } =\quad \frac { sin70 }{ sin30 } \\ \triangle ABM:\quad \frac { MB }{ AB } =\quad \frac { sin20 }{ sin(220-x) } \\ \triangle MBC:\quad \frac { CB }{ MB } =\quad \frac { sin(x) }{ sin40 }$

Multiplying the 3 equations, we have:

$\frac { AB }{ CB } \cdot \frac { MB }{ AB } \cdot \frac { CB }{ MB } = 1,$

then:

$\sin70\cdot sin20\cdot sin(x) = sin30\cdot sin(220-x)\cdot sin40$

replacing: $\sin70 = \cos20$ and replacing: $\sin40 = 2\cdot sin20\cdot cos20$

we will get: $\sin(x) = \sin(220-x)$

Solving it, we get x = $\boxed{110 \quad degrees}$

Synthetic one : First, it is easy to see that $\angle{ACB} = 70^{\circ}$ and $\angle{MCB} = 40^{\circ}$

Now, let $O$ be the circumcenter of $\triangle{ABC}$ , then $\angle{AOC} = 2\angle{ABC} = 160^{\circ}$ and so, $\angle{OAC} = \angle{OCA} = 10^{\circ}$ . Then, since $\angle{MAC} = \angle{OAC} = 10^{\circ}$ , $O$ must lies on $AM$ , so we get $\angle{MCO} = \angle{MCA} - \angle{OCA} = 20^{\circ}$ and $\angle{MCO} = \angle{OAC}+\angle{OCA} = 20^{\circ}$ . Hence, $\triangle{MOC}$ is an isosceles triangle, so $MO = MC$ .

Notice again that $\angle{COB} = 2\angle{CAB} = 60^{\circ}$ and so, because $OB = OC$ , $\triangle{COB}$ is an equilateral triangle, so $OC = OB = BC$ . So, $\triangle{OMB} \cong \triangle{CMB}$ . Thus, $\angle{MBC} = \angle{MBO}$ . But clearly, $\angle{OBC} = 60^{\circ}$ because $\triangle{OBC}$ is an equilateral. So, $\angle{MBC} = \angle{MBO} = 30^{\circ}$ . it follows that $\angle{BMC} = 110^{\circ}$ .

Let angle BMC be x degrees.

C=180-B-A=70 degrees MCB=C-MCA=40 degrees MBC=180-40-x=(140-x) degrees

Applying Sine rule in triangle ABC, AC/sin(80) = BC/sin(30) - (1) Applying Sine rule in triangle MAC, AC/sin(140) = MC/sin(10) - (2) Applying Sine rule in triangle MBC, MC/sin(140-x) = BC/sin(x)

Substituting for MC and BC from (1) and (2), AC * sin(10) / (sin(140) * sin(140-x)) = AC * sin(30) / (sin(80) * sin(x))

Rearranging the above equation and solving for x, gives x=110 degrees.

This problem is perfect candidate for applying the trigonometric form of Ceva's Theorem

In this case: $\frac {\sin (20) }{\sin (10)} \frac {\sin (30)}{\sin (40)} \frac {\sin (x)}{\sin (80-x)} = 1 \Rightarrow x = 30$ $\Rightarrow \angle BMC = 110$

You need to find, Angle BCM, Angle BAM and Angle ABC. - Angle BCM is found through taking Angle ACB - Angle ACM = 70 degrees - 30 degrees = 40 degrees. - Angle BAM is found through taking Angle BAC - Angle MAC = 30 degrees - 10 degrees = 20 degrees. - Angle ABC is already stated; which is 80 degrees.

Now, use the concept of a quadrilateral; 4 sided figure diagram. Which adds up to 360 degrees. Kindly, subtract 360 degrees with Angle BCM, BAM & ABC: 360 degrees - 40 degrees - 20 degrees - 80 degrees = 220 degrees.

Last step is just two divide 220 degrees by two because reflex AMC is proportionate to one another: 220 degrees / 2 = 110 degrees :D