23. In $\triangle \mathrm{ABC}, \angle \mathrm{CAB}=30^{\circ}$ and $\angle \mathrm{ABC}=80^{\circ}$. The point $\mathrm{M}$ lies inside the triangle such that $\angle \mathrm{MAC}=10^{\circ}$ and $\angle \mathrm{MCA}=30^{\circ}$. Find $\angle \mathrm{BMC}$ in degrees.
23. Answer: $110^{\circ}$.
Construct a circumcircle of the triangle $\mathrm{ABC}$, with $\mathrm{O}$ as the centre as shown.

Note that $\angle \mathrm{ACB}=70^{\circ}$
Since $\mathrm{OC}=\mathrm{OB}$ and $\angle \mathrm{COB}=60^{\circ} \Rightarrow \angle \mathrm{OCB}=60^{\circ}$
$\triangle \mathrm{COB}$ is an equilateral triangle.
Thus $\angle \mathrm{OCA}=70^{\circ}-60^{\circ}=10^{\circ}=\angle \mathrm{OAC}$
But $\angle \mathrm{MAC}=10^{\circ}$ (given). So AOM lies on a straight line.
$$
\angle \mathrm{AOC}=160^{\circ} \Rightarrow \angle \mathrm{COM}=20^{\circ} \text {. }
$$

Since $\angle \mathrm{MCA}=30^{\circ}$ (given) and $\angle \mathrm{OCA}=10^{\circ}$, thus $\angle \mathrm{MCO}=20^{\circ}$. That means $\triangle \mathrm{MCO}$ is an isosceles triangle and $\angle \mathrm{MCB}=70-30=40^{\circ}$
So BM is the perpendicular bisector of the equilateral triangle OBC. Thus
$$
\begin{array}{l}
\angle \mathrm{OBM}=60^{\circ} \div 2=30^{\circ} \\
\angle \mathrm{BMC}=180^{\circ}-40^{\circ}-30^{\circ}=110^{\circ}
\end{array}
$$