Mysterious points on an angle bisector

Suppose that $AB = AC = 1$ . Then $BC = 2\sin50^\circ = 2\sin130^\circ$ and so, using the Sine Rule, $PC = 2\sin30^\circ = 1$ . Thus $APC$ is isosceles, and hence $\angle APC = 80^\circ$ , $\angle APB = 150^\circ$ . and hence $\angle PAB = 20^\circ$ . Thus we deduce that triangles $PAB$ and $QCA$ are congruent, and hence $\angle AQC = 150^\circ$ , so that $\theta= \boxed{30^\circ}$ .

Let $AD$ and $PE$ be perpendicular to $BC$ , $PF$ perpendicular to $AD$ , and $BP = 1$ . Then $PE = BP \cdot \sin 30^\circ = \dfrac 12$ and $BE=BP\cdot \cos 30^\circ = \dfrac {\sqrt 3}2$ . And

$\begin{aligned} BC & = BE + EC = \frac {\sqrt 3}2 + PE \cdot \cot 20^\circ = \frac {\sqrt 3}2 + \frac 12 \tan 70^\circ & \small \blue{\text{Let }t=\tan 10^\circ} \\ & = \frac {\sqrt 3}2 + \frac {\sqrt 3 + t}{2(1-\sqrt 3t)} = \frac {\sqrt 3-t}{1-\sqrt 3 t} \\ \implies BD & = \frac {BC}2 = \frac {\sqrt 3-t}{2(1-\sqrt 3t)} \\ PF & = ED = BD - BE = \frac {\sqrt 3-t}{2(1-\sqrt 3t)} - \frac {\sqrt 3}2 = \frac t{1-\sqrt 3 t} \\ AD & = BD \cdot \tan 40^\circ = BD \cdot \cot 50^\circ = \frac {\sqrt 3-t}{2(1-\sqrt 3t)} \cdot \frac {1+\sqrt 3t}{\sqrt 3 - t} = \frac {1+\sqrt 3t}{2(1-\sqrt 3t)} \\ AF & = AD - FD = AD - PE = \frac {1+\sqrt 3t}{2(1-\sqrt 3t)} - \frac 12 = \frac {\sqrt 3t}{1-\sqrt 3t} \\ \tan \angle PAF & = \frac {PF}{AF} = \frac t{\sqrt 3 t} = \frac 1{\sqrt 3} \\ \implies \angle PAF & = 30^\circ \end{aligned}$

This means that $\angle BAP = \angle BAD - \angle PAF = 50^\circ - 30^\circ = 20^\circ$ . Then $\triangle ABP$ and $\triangle ACQ$ are congruent and $\angle CAQ = \angle ABP = 10^\circ$ . And $\angle AQP = \angle CAQ + \angle ACQ = 10^\circ + 20^\circ = \boxed{30}^\circ$ .

This is more so a solution sketch rather than a solution, but I tried to solve it without the use of trigonometry.

The crucial claim is that $\triangle AQC \cong \triangle BPA$ . We have information that $AC = AB$ and $AP = QC$ . If we determine that $\angle BAP = \angle ACQ$ or $BP = AQ$ , then we determine that the triangles are congruent.

If we can prove that these triangles are congruent, we can deduce that $\angle BAQ = 90^\circ$ . Combined with the fact that $\angle APQ = 60^\circ$ , then we can show that $\theta = \boxed{30^\circ}$ .