0
$\begingroup$

I was reading over this proof of the law of sines and they say that $\angle CAB = \angle DOB$ because of "basic geometry". I do not get it though, how can you say that the angles are equal?

$\endgroup$
1
  • $\begingroup$ The link is not working for me. Regardless, though, it is preferable to make questions self-contained, whenever possible. Please edit your question to include relevant excerpts and diagrams. $\endgroup$ – Cameron Buie Nov 18 '13 at 20:52
2
$\begingroup$

I think it is referring to a property of inscribed angles:

The measure of the intercepted arc (equal to its central angle) is exactly twice the measure of the inscribed angle.

So $m\angle COB = 2 m\angle CAB$, and they bisected segment $CB$ to halve the angle.

$\endgroup$
1
  • $\begingroup$ Ah, thanks. I knew that inscribed angles could be equal if they mark the same arc, but I didnt know the theorem about the central central angle. $\endgroup$ – David says Reinstate Monica Nov 18 '13 at 20:56
2
$\begingroup$

$\angle CAB$ is an inscribed angle; $\angle DOB$ is half of a central angle through the same arc, so the angles are equal.

See http://www.mathopenref.com/arccentralangletheorem.html for more.

$\endgroup$

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.