Given that $\sin \angle BOC = \frac{4}{5}$, $AC=24$ , find $[\triangle BDE]$ . 
Let $B$ be a point on the circle centered at $O$ with diameter $AC$. Let $D$ and $E$ be the circum-centres of the triangles $OAB$ and $OBC$ respectively. Given that $\sin \angle BOC = \frac{4}{5}$, $AC=24$, find $[\triangle BDE]$ .

What I Tried: Here is a picture in Geogebra :-

From the given information, as I am not that good at Trigonometry I could only conclude that $\sin \angle BOC = \frac{4}{5}$ probably means $\angle BOC$ is around $53^circ$ or so, in considering $\angle BOC$ to be acute. The value however, is not exactly $53^\circ$, but varies.
The other thing which is known is the radius of the circle. I don't seem to find a way to use it, especially I can see that$\angle ABC$ is $90^\circ$ but they come to no use, and maybe to find $[\triangle BDE]$, I have to use the formula like $\frac{1}{2}ac \sin \beta$ or so? I also could not understand if the circum-centres $D$ and $E$ can be used in some way or not.
Can anyone help me here? Thank You.
 A: 
Let $(ABC)=r$, $(BAO)=r_1$ and $(BOC)=r_2$
By sine rule, $$2r_1=\dfrac{r}{\sin \theta} , \quad 2r_2=\dfrac{r}{\sin (\pi/2-\theta)} =\dfrac{r}{\cos \theta}$$
As $\triangle COB$ is isosceles, it can be shown $OE$ is its internal angle bisector and $OD$ exterior angle bisector. So $\angle DOE=90$.
Also $\triangle DOE \cong \triangle DBE \, $ by SSS. $\angle DBE=90$.
$$\therefore [BDE] = \dfrac{1}{2}\cdot BD \cdot BE$$
$$= \dfrac{1}{2}\cdot r_1 \cdot r_2$$
$$= \dfrac{r^2}{4 \sin 2\theta}$$
Given is $r=12$ and $\sin 2\theta = 4/5$
A: $BO$ is common chord of both circumcircles. So $DE$ is perpendicular to $BO$.
$\angle BED = \angle BCO \,$ (as $\angle BEO = 2\angle BCO)$
$\angle BDE = \frac{1}{2} \angle BDO = \angle BAO$
So, $\triangle BDE$ is right angled triangle.
$If \angle BOC = 2 \theta$,
As per extended Sine law,
$\frac{12}{\cos \angle \theta} = 2r$ ($r$ is circumradius of $\triangle BOC$, $\angle BCO = 90^0 - \theta$).
Similarly $\frac{12}{\sin \theta} = 2R$ ($R$ is circumradius of $\triangle AOB$, $\angle BAC = \theta$).
Desired area = $\frac{1}{2} R \times r = \frac{36}{\sin 2\theta} = 45$
