Solve this geometry problem without using any trigonometry

Question

Given:
$ \triangle ABC $
$AC = BC$
$\angle C = 150$
$AB (base) = \sqrt{12}$
Find the radius of the Circumcircle.
I have no idea how to solve this challenge, the answer which is given in the textbook is $2\sqrt{3}$ (which is the base).

P.S. The textbook is for 8th grade, and in my country, sin and cosine law are learned in 9th grade(next year). So proofs without trigonometry are preferred.

Answer 1

From the figure, since $\angle{BCA} = 150$, we have $$\angle{BDA} = 180-150 = 30 \,\,\, (\because \text{Cyclic quadrilateral }ACBD)$$ and therefore $$\angle{BOA} = 2\times 30 = 60 \,\,\, (\because \text{Central angle is twice the inscribed angle})$$ Hence, $\Delta AOB$ is equilateral.

This is because $OA = OB \implies $base angles are equal, which inturn $\implies$ that all angles are $60$, which in turn $\implies$ the triangle is equiangular, which inturn $\implies$ the triangle $AOB$ is also equilateral.

Hence, the circumradius is $AB = \sqrt{12} = 2\sqrt3$.

Answer 2

$$R=\frac{c}{2\sin(150^\circ)}=\frac{2\sqrt{3}}{2\cdot\frac{1}{2}}=2\sqrt{3}$$

Answer 3

Consider $AC$ and $CB$ as two successive sides of a regular dodecagon inscribed in some circle $\gamma$. $AB$ is then the side of a regular hexagon inscribed in $\gamma$, hence equal to the radius of $\gamma$. Since $\gamma$ is also the circumcircle of $\triangle(ABC)$ the claim follows.