Hello, googa!
A slight variation of ticbol's solution . . .
In $\displaystyle \Delta ABC,\:X$ is a point on $\displaystyle AC$ such that $\displaystyle \frac{AX}{XC}=3$
and $\displaystyle \angle AXB = 70^o,\;\;\angle ABC = 110^o,\;\;BC = 6$ cm.
Find the length of $\displaystyle AC.$ Code:
B
o
* **
* * *
* * β* 6
* * *
* * *
* * *
* β 70° * α *
o * * * * * * * o * o
A 3a X a C
We have: .$\displaystyle \angle ABC = 110^o,\;\;\angle AXB = 70^o,\;\;BC = 6$
Let: $\displaystyle a = XC,\;AX = 3a$
Consider $\displaystyle \Delta ABC$ and $\displaystyle \Delta BXC$
. . They share $\displaystyle \angle \alpha = \angle BCX$
Since $\displaystyle \angle AXB = 70^o$, then $\displaystyle \angle BXC = 110^o$
. . They have equal angles: .$\displaystyle \angle ABC \:=\:\angle BXC \:=\:110^o$
Hence, their third angles are equal: .$\displaystyle \angle BAC \:=\:\angle CBX \:=\:\beta$
In $\displaystyle \Delta ABC\!:\;\;\frac{\sin\beta}{6} \:=\:\frac{\sin110^o}{4a} \quad\Rightarrow\quad \sin\beta \:=\:\frac{3\sin110^o}{2a}$ .[1]
In $\displaystyle \Delta BXC\!:\;\;\frac{\sin\beta}{a} \:=\:\frac{\sin110^o}{6} \quad\Rightarrow\quad \sin\beta \:=\:\frac{a\sin110^o}{6}$ .[2]
Equate [1] and [2]: .$\displaystyle \frac{3\sin110^o}{2a} \:=\:\frac{a\sin110^o}{6} \quad\Rightarrow\quad 18 \:=\:2a^2 \quad\Rightarrow\quad a \:=\:3$
Therefore: .$\displaystyle AC \:=\:4a \:=\:\boxed{12\text{ cm}}$