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:

enter image description here

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.

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

Sine rule:


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}

enter image description here

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.