$\triangle ABC$ is isosceles where $P$ and $Q$ lie on $AC$. If $AP=2$, $QC=1$, and $\angle PBQ=60^\circ$, find $PQ$ As the title says, in the setup given below, we have an isosceles triangle $\triangle ABC$ where the two base angles are $30^\circ$ each. $P$ and $Q$ are located on $AC$ such that $AP=2$ and $QC=1$, and $\angle PBQ=60^\circ$. The goal is to find the measure of length $PQ$:

This problem was posted by a user on a languages learning app that I came across. The claim is that it is meant for 10th grade students. The user did not share the correct answer.
At first I attempted to "complete the triangle" by forming a large equilateral triangle, but that approach didn't lead anywhere. I will share my only successful approach below as an answer. Please share your own answers as well and let me know if the answer I arrived at is correct.
 A: This is my approach:

1.) Draw a line from $B$ onto point $D$ outside of $\triangle ABC$ such that $\angle ABD =\angle CBQ$ and $BD=BQ$. Connect $A$ and $D$ via $AD$ and notice that $\triangle ABD$ is congruent to $\triangle CBQ$. Also join $D$ and $P$ via $DP$ and $D$ and $Q$ via $DQ$. Notice that $\angle DAP=60^\circ$ and that, $AP=2$ and $AD=CQ=1$, this means that $\triangle DAP$ is a "$30-60-90$" triangle.
2.) Above implies that $DP=\sqrt{3}$ and $\angle APD=30^\circ$. Now notice that $\angle ABP+\angle ABD=60^\circ$, this means that line segment $BP$ is the perpendicular bisector of the isosceles triangle $\triangle DBQ$. This means that $DE=EQ$. But, this also implies that $\triangle DPQ$ is an isosceles triangle. Therefore, $PQ=DP=\sqrt{3}$
A: A trigonometric method:
Let $\angle ABP=\alpha$ and $AB=BC=x$. Then the other angles can be written as in the following diagram.

Using sine law for triangles $ABP$ and $CBQ$ we get,
$$\frac2{\sin\alpha}=\frac x{\sin(30+\alpha)}\tag1$$ and $$\frac1{\sin(60-\alpha)}=\frac x{\sin(90-\alpha)}$$ $$\frac1{\cos(30+\alpha)}=\frac x{\cos\alpha}\tag2$$ respectively.
Dividing (1) by (2) we obtain,
$$2\sin(30+\alpha)\cos(30+\alpha)=\sin\alpha\cos\alpha$$
$$2\sin(60+2\alpha)=\sin2\alpha$$
$$\sqrt3\cos2\alpha+\sin2\alpha=\sin2\alpha$$
$$\cos2\alpha=0$$
$$\alpha=45^\circ$$
Then from (1),
$$x=\frac{2\sin(30+45)}{\sin45}=1+\sqrt3$$
And also we notice $\angle BPC=\angle PBC=75^\circ$ which implies $CB=CP$.
Therefore $PQ=x-1=\sqrt3$.
A: Let's make use of the following elementary formulas relating area, lengths and angles of a triangle.



Call $\ell$ the common length of $[A,B]$ and $[C,B]$; also let $x=PQ$ the unknown length.
• From triangle $ABP$ we get $BP^2=AB^2+AP^2-2AB.AP\cos(30)$, which provides:
$BP^2=\ell^2-2\ell\sqrt3+4$.
• Similarly, from triangle $CBQ$ we get: $BQ^2=\ell^2-\ell\sqrt3+1$.
• From triangle $PBQ$ comes: $x^2=BP^2+BQ^2-2BP.BQ\cos(60)=BP^2+BQ^2-BP.BQ$.
Using the two previous formulas, we express $BP^2$ and $BQ^2$ in terms of $\ell$ to obtain:
$$x^2=2\ell^2-3\ell\sqrt3+5-BP.BQ.$$
• We may express $BP.BQ$ in terms of $PBQ$'s area:
$\displaystyle BP.BQ=\frac{2\mathscr{A}(PBQ)}{\sin(60)}
=\frac{4\mathscr{A}(PBQ)}{\sqrt3}\cdot$
• But the above area can also be calculated as
$$\mathscr{A}(ABC)-\big(\mathscr{A}(ABP)+\mathscr{A}(CBQ)\big)
=\ell^2\frac{\sqrt3}4-\left(\frac\ell2+\frac\ell4\right)
=\ell\frac{\sqrt3}4(\ell-\sqrt3).$$
We thus have $BP.BQ=\ell(\ell-\sqrt3)$, then $\color{firebrick}{x^2=\ell^2-2\ell\sqrt3+5}$.
• It remains to find one last relation between $\ell$ and $x$.
Segment $[A,C]$ has length $\ell\sqrt3$, but also $2+x+1$; hence $\color{firebrick}{\ell=\sqrt3+\frac{x}{\sqrt3}}\cdot$
• Finally, substituting $\ell$ with $\sqrt3+\frac{x}{\sqrt3}$ in the expression of $x^2$ above, we obtain this second degree equation: $x^2=\frac{x^2}3+2$.
Its positive solution is $x=\sqrt3$.
• Comment on the configuration.
Interesting properties can be noticed.
The circles $\mathscr{C}_1$ and $\mathscr{C}_2$ circumscribed to triangle $ABP$ and $CBQ$ respectively are tangent at $B$, and the common tangent is thus the bisector of $[BP)$ and$[BQ)$.
Also $\mathscr{C}$, the circumscribed circle to $ABC$, meets the straight lines $(BP)$ and $(BQ)$ at $P'$ and $Q'$ respectively.
Then $(AP')$ and $(CQ')$ are parallel, and $P'Q'=AC$…

