Wikipedia:Reference desk/Archives/Mathematics/2012 November 26
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November 26
[edit]Can anyone tell me what the full sine rule (including 2r) is? thanks.
[edit]I'm doing British mathematical olympiad round 1, and the one thing in the booklet of reccomended things you ought to know, but I don't actually know if it mentions the full sine rule-which I knew last summer but now I've forgotten. I presume that r is radius, but radus of what? — Preceding unsigned comment added by 86.176.99.71 (talk) 18:38, 26 November 2012 (UTC)
- Does Law of sines#Relation to the circumcircle give you what you are looking for? Looie496 (talk) 20:29, 26 November 2012 (UTC)
- thanks it does indeed. that was relatively simple after all-if only they could say so in as many words... — Preceding unsigned comment added by 86.176.99.71 (talk) 20:24, 27 November 2012 (UTC)
- Draw a circumcircle on your triangle, then slide the vertex A along the circle's arc to make B a right angle. Based on an inscribed angle theorem the A angle keeps its measure unchanged (blue and green on the picture), so does the BC chord. Now it's clear that sin(A) = a/(2R), isn't it? --CiaPan (talk) 06:32, 28 November 2012 (UTC)
- Thanks from me too, CiaPan! That's a very nice demonstration, new to me even though I've used the result. Duoduoduo (talk) 15:32, 28 November 2012 (UTC)
- Glad to be helpful. However, despite how nice demonstration it is, it's not a valid proof. :( Would be better to ask to shift A so that AC becomes the circle's diameter, then use the Thales' theorem to obtain 'B is a right angle'.
Furthermore the procedure would not work for obtuse triangle with A > 90°, unless we replace A with a new vertex D on the other side of BC and prove
1. D = 180° − A
then
2. sin D = sin A
CiaPan (talk) 06:30, 29 November 2012 (UTC)
- Glad to be helpful. However, despite how nice demonstration it is, it's not a valid proof. :( Would be better to ask to shift A so that AC becomes the circle's diameter, then use the Thales' theorem to obtain 'B is a right angle'.
- Thanks from me too, CiaPan! That's a very nice demonstration, new to me even though I've used the result. Duoduoduo (talk) 15:32, 28 November 2012 (UTC)
