Given a triangle $\triangle ABC$ with an internal point $O$, compute $\angle CBO$ As title suggests, in the given obtuse triangle $\triangle ABC$, the goal is to find the angle measure of $\angle CBO$ called $x$:

I'm going to post my own attempt at the question down as an answer. Please feel free to add more answers, particularly approaches that are unique (geometric or not) and critique my approach if necessary.
 A: My own approach is rather simple. I'll add a brief explanation as well:

This is how I go about it:
Note that I accidentally mislabeled some points on the triangle, therefore the terminology I use will be referring to MY drawing and NOT the one in the question above
1.) Locate point $D$ on $BC$ and connect it with point $A$ and $O$ as well such that $\angle DAB=DBA=40$, therefore $DA=DB$. Notice that $\triangle DAB$ is isosceles and $\angle BDA=100$. Also notice that $\angle BOA=50$ which is half of $\angle BDA$, therefore it follows that $\angle BOA$ is the inscribed angle of $\angle BDA$.
2.) Above implies that, via the converse of inscribed angle theorem, segment $OD=DA=DB$. Notice that $\angle DAO=60$ and, since $OD=DA$, this implies that $\triangle DAO$ is equilateral, therefore $OA=OD=DA=DB$. Note that $\triangle ADC$ is an isosceles triangle $(80-20-80)$ with $AC=DC$. Since $\angle ODA=60$, this means that $\angle ODC= \angle ADC-\angle ODA=80-60=20$.
3.) Because $\angle OAC=\angle ODC=20$, $AC=DC$ and $OA=OD$, it follows that $\triangle OAC$ and $\triangle ODC$ are congruent via the SAS property. This means that $\angle OCA=\angle OCD=x$. Therefore, $\angle ACD=2x=20$. Therefore $x=10$.
A: I confess that I am never able to find such elegant solutions.
But the trigonometric approach leads also to the solution.
Assume OC has unit length.
Applying the sin rule in triangle ACO results in $AO=\frac{sin(x)}{sin(20^{\circ})}$
Applying the sin rule in BOC results in $BO=\frac{sin(20^{\circ}-x)}{sin(10^{\circ})}$
Finally applying the sin rule in ABO results in the equation $\frac{sin(20^{\circ}-x)}{sin(x)}=\frac{sin(10^{\circ})\cdot sin(100^{\circ})}{sin(20^{\circ})\cdot sin(30^{\circ})}=\frac{sin(10^{\circ})\cdot cos(10^{\circ})}{sin(20^{\circ})\cdot \frac{1}{2}}=\frac{2\cdot sin(10^{\circ})\cdot cos(10^{\circ})}{sin(20^{\circ})}=1$
$\cfrac{sin(20^{\circ}-x)}{sin(x)}=1$   Therefore $x=10^{\circ}$
