Side length of a quadrilateral incribed on a circle I've been doing math for 10 years now, yet every so often I get stumped by a "basic" high school question. This is one of those times.
Here's the question:

Part a is easy; we apply the cosine rule to the angle $92^\circ$.
I don't understand how to use the sine rule to find $CD$. I managed to find it using the cosine rule; draw a line between $BD$ and use this as the "main" length in the cosine rule. The other two lengths are $5cm$ and $CD$. You end up with a quadratic involving $CD$ that can be solved.
But I don't understand how to use the sine rule to do this. Ideas anyone?
 A: 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.
A: 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
$${|CD| \over\sin (\gamma-\delta)} = 2r \implies \color{red}{|CD|=2r\sin(\gamma-\delta)}$$
