Calculate the angle measure $TBA$ for reference:
On the internal bisector of $\measuredangle C$ of an isosceles triangle $ABC (AB = BC)$ if
externally marks point $T$, in such a way that $\measuredangle CTA = \measuredangle TAB = 30^o$.
Calculate $\measuredangle TBA$ (Answer $20^o$)

My progress:
$\triangle ABC(Isósceles) \rightarrow \measuredangle BAC = \measuredangle C\\
2\alpha+\alpha =120^o \therefore \alpha =40^0\rightarrow \measuredangle ABC = 180^o - 160^o = 20^0\\
\theta +x = 120^o$ $\text{but I can't finish...there's some equation missing}$


Longer method with the use of calculator!
Drop altitude $TD$ to the side $AB$. Then we have:
$$\sin 30^\circ = \frac{DT}{AT} \Rightarrow DT=\frac12AT\\
\begin{cases}\tan x=\frac{DT}{BD}\\ \tan 30^\circ =\frac{DT}{AD}\end{cases}\Rightarrow AD+BD=\frac{DT}{\tan x}+\sqrt3\cdot DT\Rightarrow AB=\frac12AT\left(\frac1{\tan x}+\sqrt3\right) \quad (1)\\
\frac{AT}{\sin 40^\circ}=\frac{AC}{\sin 30^\circ} \quad (2)\\
\frac{AB}{\sin 80^\circ}=\frac{AC}{\sin 20^\circ} \quad (3)$$
From $(2)$ and $(3)$:
$$AB=\frac{AT\cdot \sin80^\circ}{2\sin40^\circ \sin20^\circ} \quad (4)$$
From $(1)$ and $(4)$:
$$\frac1{\tan x}+\sqrt3=\frac{\sin80^\circ}{\sin40^\circ \sin20^\circ}=\frac{1}{\sin 40^\circ \cdot \sin 10^\circ}\Rightarrow \\
x=\arctan\left(\frac{1}{\frac1{\sin 10^\circ\cdot \sin 40^\circ}-\sqrt3}\right)=20^\circ.$$
Wolframalpha answer.
