0
$\begingroup$

The problem is as follows:

In triangle $ABC$, $BC=2$. Point $D$ is on $\overline{AC}$ such that $AD=1$ and $CD=2$. If $m\angle BDC=2m\angle A$, compute $\sin A$.

I tried several ways of making similar triangles, but that didn't work.I couldn't find a simple way to apply the double angle identities either, and I think that's probably where I messed up. Is there a good way to do this without the double angle identities?

$\endgroup$
1
$\begingroup$

Let be $\measuredangle BAC=\alpha$ and $\measuredangle BDC=2\alpha$. Since $CD=BC=2$, we have that $\measuredangle CBD=\measuredangle BDC=2\alpha$. Also,

\begin{eqnarray} \measuredangle BAD+\measuredangle ABD&=&\measuredangle BDC, \\ \measuredangle BAD+\alpha&=&2\alpha, \\ \measuredangle BAD&=&\alpha. \end{eqnarray}

We have $BD=AD=1$ (since $\measuredangle BAD=\measuredangle DAB$, in triangle $ABD$).

Now, from law of cosines (in triangle $BCD$), we get $2^2=2^2+1^2-2*1*2*\cos{2\alpha}$ and $\frac{1}{4}=\cos{2\alpha}=1-2\sin^2{\alpha}$.

Finally, $\sin{\alpha}=\sqrt{\frac{3}{8}}$.

$\endgroup$
  • $\begingroup$ Thank you. I had gotten the first para and those equations, I just didnt have time to apply law of cosines and work it out. $\endgroup$ – Asimov Jun 4 '14 at 20:56

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.