$\triangle ABC$ is a triangle with internal point $O$. Find $\angle x$. As title states, the triangle in the following figure has 3 equal sides and some given angles, and the goal is to find the measure of $\angle x$. As always, I'll post my own approach here, please share your own approaches as well!


A very simple approach, based on your picture, using the law of sines:
$$\angle CAO=40-x,\angle COA=140^{\circ} \implies \frac{\sin (40-x)^{\circ}}{\sin 140^{\circ}}=\frac {\sin (20+x)^{\circ}}{\sin 40^{\circ}}=\frac{\sin (20+x)^{\circ}}{\sin 140^{\circ}}$$
$$\implies \sin (40-x)^{\circ}=\sin (20+x)^{\circ}\implies 40-x=20+x\implies x=10^{\circ}.$$
