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.
 A: By extended sine rule, $BC=\frac{AD}{\sin \angle ABD}$.
A: 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. 
A: Sine rule, brother!
$\dfrac{AD}{\sin \widehat{ABD}}=2r=BC$
You've done $AD$ already? You're halfway there!
A: Draw the other diagonal $AC$. Since $BC$ is a diameter of the circumcircle of the quad $BADC$, $$\angle\, BAC = \angle \, BDC = 90^{\circ}$$
Let $P$ be the intersection point of the diagonals $AC$ and $BD$. Then triangle $ABP$ is right-angled, so 
$$\frac{AP}{AB} = \frac{AP}{3} = \tan(\angle\, ABP) = \tan(\angle\, ABD) = \frac{3}{4}$$ so
$AP = \frac{9}{4}$. By Pythagoras' theorem for $ABP$
$$BP = \sqrt{AB^2 + AP^2} = \sqrt{3^2 + \frac{9^2}{4^2}} = \frac{15}{4}$$ Then $DP = BD - BP = 5 -  \frac{15}{4} =  \frac{5}{4}$
In the cyclic quad $BADC$ the triangles $ABP$ and $DCP$ are simiplar, right-angled triangles, therefore 
$$\frac{DP}{CD} = \frac{5}{4 \, CD} = \tan(\angle \, DCP) = \tan(\angle \, ABP) = \frac{3}{4}$$ so
$$CD = \frac{5}{3}$$ Apply Pythagors' theorem to the right-angled triangle $BCD$
$$BC = \sqrt{BD^2 + CD^2} = \sqrt{5^2 + \frac{5^2}{3^2}} = \frac{5}{3}\sqrt{10}$$
Finally, the triangles $ADP$ and $BCP$ are similar ($BADC$ is cyclic), so 
$$\frac{AD}{BC} = \frac{AP}{BP}$$
$${AD} = \frac{AP}{BP}{BC} = \frac{9\cdot 4}{4 \cdot 15} \cdot \frac{5}{3}\sqrt{10} = \sqrt{10}$$
