Talk:Law of cosines

Correction for solving angle-side-side triangles

In the section where you use the quadratic equation to make a form of the law of cosines that solves angle-side-side triangles, there is an error on the bounds checking. You say that if c>=b, there are no solutions. However:

- If c=b, there is one solution. It is an isosceles triangle with b=c, B=C. The triangle itself is twice the size as the right triangle formed by c=b*sin(C) & B = 90. B and C are acute.

- If c>b, there is also one solution. It is a scalene triangle with B and C being acute. — Preceding unsigned comment added by 69.159.12.95 (talk) 01:35, 26 January 2020 (UTC)

For future reference, WP:SOFIXIT applies. –Deacon Vorbis (carbon • videos) 01:51, 26 January 2020 (UTC)

Another small addition on the angle-side-side solution

Right now it says, "only one positive solution if c = b sin γ". It should say "only one positive solution if c = b sin γ OR if c>=b" — Preceding unsigned comment added by 69.159.12.95 (talk) 06:06, 26 January 2020 (UTC)

WP:SOFIXIT. Also, please place new comments at the end and sign your posts. See Help:Talk for more info. –Deacon Vorbis (carbon • videos) 12:44, 26 January 2020 (UTC)

Demonstration by Chasles relation and scalar product

Will it make sense to add a demonstration with Chasles relation and scalar product? Even if historically it was after, it is a simple and powerful tool to demonstrate it. — Preceding unsigned comment added by 103.139.171.78 (talk) 17:49, 15 July 2020 (UTC)

Proof by application of dot product

Could there be a valid proof using the definition of a dot product. With triangle having side C as the vector A-B, where A,B are the vectors of the triangle legs: then length |C|^2 could be written as |A|^2+|B|^2-2A(dot)B. A(dot)B is defined as |A||B|cos(theta) giving the law of cosines formula. If other editors think this is valid, I could add a formatted proof. Nikolaih^☎️📖 21:40, 6 August 2020 (UTC)

Absolutely. It should be added --83.56.100.228 (talk) 18:37, 10 October 2020 (UTC)

Of course then you have to show why A dot B = |A| |B| cos (theta) is a reasonable definition, since usually dot products are defined by coordinates. Dot product#Scalar projection and first properties Wqwt (talk) 17:44, 15 June 2024 (UTC)

Requires proof reading?

Perhaps i'm mistaken but angle ABC in Fig 2 appears acute rather than obtuse to me. Proof reading anyone? — Preceding unsigned comment added by 198.24.255.98 (talk) 16:24, 22 September 2020 (UTC)

No, it's definitely obtuse (unless there's some issue with it being displayed in a weird aspect ratio). –Deacon Vorbis (carbon • videos) 16:55, 22 September 2020 (UTC)

The angle ABC in Fig 2 clearly looks acute on my monitor. Just to be clear an acute angle is smaller than 90 degrees and an obtuse angle is greater than 90 degrees but less than 180 degrees. It makes the whole section unreadable, and brings into question the correctness of articles on the site. — Preceding unsigned comment added by 2605:A601:A706:8700:F91E:CE18:C1EC:7244 (talk) 23:19, 27 September 2020 (UTC)

Oh, I think I see the confusion. Angle ABC is acute, but triangle ABC is obtuse, because angle ACB is an obtuse angle. –Deacon Vorbis (carbon • videos) 23:28, 27 September 2020 (UTC)

Right, ABC is an angle not a triangle. 'Triangle ABC' is confusing if not complete nonsense. — Preceding unsigned comment added by 198.24.255.98 (talk) 23:35, 2 October 2020 (UTC)

It's perfectly clear what $\Delta ABC$ means: the triangle defined by the vertices A, B, C. Wqwt (talk) 17:45, 15 June 2024 (UTC)