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

 A: $\triangle ABC$ is isosceles. Therefore,
$\measuredangle BAC = \measuredangle C \implies 2\alpha+\alpha =120^\circ\implies \alpha =40^\circ$
Therefore $\measuredangle ABC = 180^\circ - 160^\circ = 20^\circ$
Draw $AX$ such that $\measuredangle XAK = 30^\circ$.
Draw $TX$.
$\measuredangle CDB
= 120^\circ ~~\therefore \measuredangle DIA = 90^\circ$
Since $CI$ is angle bisector $IA = IX$
$\triangle ITA \cong \triangle IXT ~(L.A.L) \rightarrow \measuredangle AXT = 60^\circ$
Therefore $\triangle ATX$ is equilateral.
$\measuredangle KXB = 180 ^\circ -110^\circ= 70^\circ$
$AK$ is angle bisector and the height of triangle $ATX$
$ \therefore AKT = 90^\circ$
$TK = KX \rightarrow \triangle KTB \cong \triangle KXB~ (L.A.L)$
Therefore $\triangle TBX$ is isosceles.
$\measuredangle B = 180^\circ -140^\circ =40^\circ$
$\therefore \boxed{\color{red}x=40^\circ-20^\circ=20^\circ}$

A: Let $AB\cap TC=\{D\}$.
Thus, since $$\frac{TD}{BD}\cdot\frac{BD}{CD}\cdot\frac{CD}{AD}\cdot\frac{AD}{TD}=1,$$ we obtain:
$$\frac{\sin{x}}{\sin(120^{\circ}-x)}\cdot\frac{\sin40^{\circ}}{\sin20^{\circ}}\cdot\frac{\sin80^{\circ}}{\sin40^{\circ}}\cdot1=1$$ or
$$\sin80^{\circ}\sin{x}=\sin20^{\circ}\sin(60^{\circ}+x)$$ or
$$\cos(80^{\circ}-x)-\cos(x+80^{\circ})=\cos(40^{\circ}+x)-\cos(x+80^{\circ}),$$ which gives $x=20^{\circ}.$
A: 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.
A: Let $X$ be the reflection of $A$ in the line $CT$. $X$ lies on BC.
See that $\triangle AXT$ is equilateral, therefore $X$ is the reflection of $T$ on $AB$ as well. Thus, $\angle ABT = \angle ABC = x$.
From $\triangle ACT$, $60+3\alpha=180^{\circ}$, therefore $\alpha=40^{\circ}$, thence $\angle ABC=20^{\circ}=x$, done.
