Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anybody know how to solve this problem? I am trying to calculate the green sides of this triangle:

enter image description here

I know / have :

the arc length, the arch base, the radius, and the h (distance from the red dot to the center of the circle)

Is it possible :) ?

Thanks a lot!

share|cite|improve this question
duplicated here: – Zev Chonoles Feb 17 '12 at 17:06
up vote 2 down vote accepted

No, this isn't possible. What you have is invariant under rotations, but the lengths of the green sides aren't. You need something that fixes the orientation, ideally the angle that the chord makes with one of the axes or something similar.

share|cite|improve this answer
And say I have the coordinates of the left most black dot? Then it should be possible right? – Jonas Dec 15 '11 at 12:13
Yes. In that case, it would be easiest to translate your coordinate system such that the origin lies at the centre of the circle (i.e. subtract the radius from all coordinates); then you could calculate the angle at which that black dot lies from the centre as the arctangent of the ratio of its coordinates, and then add the arctangent of the ratio of $dB$ to $h$ to get the angle $\alpha$ at which the red dot lies from the centre; then its coordinates (in the translated coordinate system) are $(h\cos\alpha,h\sin\alpha)$. – joriki Dec 15 '11 at 12:23
You'll need to take some care to get the signs right. If your computing environment has an atan2 function, you should use that; otherwise you might have to make some case distinctions. – joriki Dec 15 '11 at 12:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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