# 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?

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.

• I'm amazed at how simple this is. Great answer! – goblin Mar 25 at 4:46

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

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