Side length of a quadrilateral incribed on a circle

With sines? We need to find $\angle CBD$. To do so, we can find $\angle BDC$ by the law of sines comparing to $\angle DCB$. Then $\angle BDC = 180^\circ-\angle BDC-\angle DCB$. It's not particularly convenient, but at least it works.

Sine rule:

$${a\over\sin\alpha}={b\over\sin\beta}={c\over\sin\gamma}=2r$$

You may compute $2r$ from previously computed $|BD|$ as

$$2r = {|BD|\over\sin{88^o}}$$

Now (see the picture, where $O$ is the center of the circle, and $\alpha, \beta, \text{ and } \gamma$ have no relation with the same Greek letters used in the sine rule above):

\begin{aligned} \color{green}{\beta} & \color{green}{\;= 2\alpha}\\ \color{red}{\gamma} & \color{red}{\;= 360^o-\beta} \end{aligned}

From the upper (tinted) triangle we may compute angle $\delta$, because $\displaystyle {5\over\sin\delta} =2r$, and then in the bottom triangle we may compute its angle as $\gamma-\delta$.

Now applying the sine rule for this bottom triangle we will obtain

$${|CD| \over\sin (\gamma-\delta)} = 2r \implies \color{red}{|CD|=2r\sin(\gamma-\delta)}$$