What is the value of angle x in the figure, if: AP = 3BQ.

Question

For reference:

My progresss:

$T.Tales: \frac{CB}{AC}=\frac{CQ}{CP}\\ \triangle CPA\sim\triangle CQB: \frac{AP}{BQ} = \frac{AC}{BC} = \frac{PC}{CQ} = 3\\ \triangle FKE \sim \triangle EBC: \\ \frac{KE}{KF}=\frac{BC}{BE}\implies \frac{AB}{AF-AK}=\frac{BC}{BE}\rightarrow \frac{AB}{{AF-BE}}=\frac{BC}{BE}\\ \frac{BE}{AF-BE}=\frac{BC}{AB} =\frac{1}{2}\implies AF = 3BE$

I think the solution will be through a notable triangle (1: 2). I would need to show that $BE = 2CE$ or $BC = 2BQ$

Answer 1

Using your diagram, show $ \small AF = 3BE = 3 r$ (If $r$ is the radius of the smaller circle)
Then $ \small FE = AF + BE = 4 BE = 4r$
$ \small FK = AF - AK = AF - BE = 2 r$
$ \small FK:FE = 1:2$. So $ \small x = \angle FEK = ?$

Answer 2

Let $R$-radius of big circle, $r$-radius of small circle. From similar triangles, $R=3r$.

Also it is easy to prove $2CE=EF$, therefore $CE=\frac{R+r}2=2r$.

As $BE=r$, from $\triangle BEC$ $\implies\sin x=\frac12\:\therefore x=30^\circ$.