Quadrilateral inscribed in semicircle. How to find BC using sin ABD? AB = 3, BD = 5 , tanABD = 0.75
BC is diameter 


Question : How to find BC using sin ABD?

I can find sin cos ABD and lenght AD.
I can find BC using Ptolemy's theorem.

Use $$\measuredangle ABD=\arccos\frac{3}{x}-\arccos\frac{5}{x},$$ where $BC=x=2R$.
Thus, $$\frac{3}{4}=\frac{\sqrt{\frac{x^2}{9}-1}-\sqrt{\frac{x^2}{25}-1}}{1+\sqrt{\frac{x^2}{9}-1}\cdot\sqrt{\frac{x^2}{25}-1}}$$
Can you end it now?
I got $BC=\frac{5}{3}\sqrt{10}.$
Ptolemy also helps.
Indeed, $AD=\sqrt{10},$ $AC=\sqrt{x^2-9}$ and $DC=\sqrt{x^2-25}.$
Thus, since $$AB\cdot DC+AD\cdot BC=AC\cdot BD,$$ we obtain:
$$3\sqrt{x^2-25}+x\sqrt{10}=5\sqrt{x^2-9}$$ and the rest is smooth. 
