Given a triangle $\triangle ABC$, with an internal point $K$, find $\angle AKC$. As title suggests, the question is to find the measure of $\angle AKC$ in the triangle $\triangle ABC$, with an internal point $K$ and some given angles.

This is a pretty challenging problem, I'm going to post my solution below as an answer, but I'm not sure if my answer is correct and if my method is correct. Please share your own answers and approaches. I want to see if my answer is correct and if there are any other ways to arrive at an answer.

We have $\angle ABC = 66^\circ$ and $\angle AKC = 150^\circ$. Assume $BC = 1$, then $AC = \frac{\sin 48^\circ}{\sin 66^\circ} = \frac{\sin 48^\circ}{\cos 24^\circ} = 2 \sin 24^\circ$ and $KC = \frac{\sin 12^\circ}{\sin 150^\circ} = 2\sin 12^\circ$.
Now
$$AK^2 = AC^2 + KC^2 - 2AC \cdot KC \cdot \cos \angle ACK = 4\sin^2 24^\circ + 4\sin^2 12^\circ - 8 \sin 12^\circ \sin 24^\circ \cos 48^\circ.$$
Note that $\sin^2 12^\circ = \frac{(1 - \cos 24^\circ)}{2}$ and $\sin 12^\circ \cos 48^\circ = \frac{(\sin 60^\circ - \sin 36^\circ)}{2}$. Now everything is in $24^\circ$ and $36^\circ$ and the values can be calculated. The result is $\frac{(3 - \sqrt{5})}{2}$. So $AK = \frac{(\sqrt{5} - 1)}{2}$.
$$\cos X = \frac{AK^2 + KC^2 - AC^2}{2AK \cdot KC}$$
Calculate again and get $\cos X = -\frac{\sqrt{7 - \sqrt{5} - \sqrt{30 - 6\sqrt{5}}}}{4}$. Compare this with a cos table and get $X = 102^\circ$.
