Answer: 571

Small Hint:

Let $\angle A = \alpha$ and $AP = 1.$ Isosceles triangle $APQ$ gives $\angle AQP = \alpha$ and $AQ = 2\cos\alpha$

Big Hint:

With $BC = 1,$ the law of sines gives $AB = \frac{1}{2\sin(\frac{\alpha}{2})},$ and $AQ + QB = AB$ reduces to $\sin\frac{3\alpha}{2} = \frac{1}{2}$

Solution:

Let $\angle A = \alpha,$ and scale so $AP = PQ = QB = BC = 1.$ In triangle $APQ,$ the equal sides $AP = PQ$ give $\angle AQP = \angle A = \alpha,$ so $\angle APQ = 180^\circ - 2\alpha$ and, by the law of sines, $AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha.$ In triangle $ABC,$ $\angle B = \angle C = 90^\circ - \frac{\alpha}{2},$ so $$\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\frac{\alpha}{2})}{\sin\alpha} = \frac{1}{2\sin(\frac{\alpha}{2})}. \end{aligned}$$

Since $AQ + QB = AB,$ $$\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\frac{\alpha}{2})} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1. \end{aligned}$$ By the product-to-sum identity, $4\sin\frac{\alpha}{2}\cos\alpha$ $= 2\sin\frac{3\alpha}{2}$ $- 2\sin\frac{\alpha}{2},$ so the equation collapses to $\sin\frac{3\alpha}{2} = \frac{1}{2}.$ Then $\alpha = 20^\circ$ or $\alpha = 100^\circ,$ but the latter makes $AQ = 2\cos\alpha$ negative, so $\alpha = 20^\circ.$

Now $\angle ACB = 80^\circ$ and $\angle APQ = 140^\circ,$ so $r = \frac{80}{140} = \frac{4}{7},$ and $\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571.$