In $\triangle{ABC}$, $\angle{ABC}=20^{\circ}$, $\angle{ACB}=30^{\circ}$, $D$ is a point inside the triangle and $\angle{ABD}=10^{\circ}$, $\angle{ACD}=10^{\circ}$, find $\angle{CAD}$.
Note: I have seen some very similar question with beautiful solution in pure geometric format. I know how to solve this problem in trigonometric format. But I think this problem deserves a beautiful geometric approach as solution, and that's why I post it here.
As request, here is approach applying Ceva's theorem in trigonometric form,
$$\begin{align*} \frac{\sin130}{\cos130+2\cos10}&=\tan(x)\Longrightarrow \frac{\sin120\cos10+\cos120\sin10}{\cos120\cos10-\sin120\sin10+2\cos10}\\ &=\frac{\sqrt{3}\cos10-\sin10}{3\cos10-\sqrt{3}\sin10}.\\ &=\frac{1}{\sqrt{3}}\\ &=\tan30\Longrightarrow x\\ &=\boxed{30}\\ \end{align*}$$