- #1
Prove It
Gold Member
MHB
- 1,465
- 24
Hello my friends,
I posted this picture as a proof of the Cosine Rule in another thread,
however after having a closer look at it, I believe it is incomplete. It works by drawing a segment from one of the vertices so that this segment is perpendicular to one side of the triangle, and then applying Pythagoras' Theorem.
However, if you have an obtuse-angled triangle, it is impossible to draw a segment from one of the acute vertices to make a right-angle triangle. So how is it possible to prove this relationship:
\(\displaystyle \displaystyle c^2 = a^2 + b^2 - 2\,a\,b\cos{(C)}\)
when C is an obtuse angle?
I posted this picture as a proof of the Cosine Rule in another thread,
![cosinerule_zps33b193fb.jpg](http://i22.photobucket.com/albums/b347/naggindragon/cosinerule_zps33b193fb.jpg)
however after having a closer look at it, I believe it is incomplete. It works by drawing a segment from one of the vertices so that this segment is perpendicular to one side of the triangle, and then applying Pythagoras' Theorem.
However, if you have an obtuse-angled triangle, it is impossible to draw a segment from one of the acute vertices to make a right-angle triangle. So how is it possible to prove this relationship:
\(\displaystyle \displaystyle c^2 = a^2 + b^2 - 2\,a\,b\cos{(C)}\)
when C is an obtuse angle?