MHB Completion of the proof of the Cosine Rule

AI Thread Summary
The discussion centers on the proof of the Cosine Rule, specifically addressing its applicability in obtuse-angled triangles. The initial proof presented relies on constructing a perpendicular segment from a vertex, which is not feasible in obtuse triangles. The challenge is to demonstrate the relationship c² = a² + b² - 2ab cos(C) when angle C is obtuse. The user references trigonometric identities to derive the formula using Pythagoras' Theorem, ultimately confirming the validity of the Cosine Rule for all triangle types. This highlights the importance of considering different triangle configurations in geometric proofs.
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Hello my friends,

I posted this picture as a proof of the Cosine Rule in another thread,

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:

[math]\displaystyle c^2 = a^2 + b^2 - 2\,a\,b\cos{(C)}[/math]

when C is an obtuse angle?
 
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Please refer to the following diagram:

View attachment 867

Note: I have used $$\sin(\pi-\theta)=\sin(\theta)$$ and $$\cos(\pi-\theta)=-\cos(\theta)$$.

By Pythagoras, we now have:

$$(a-b\cos(\theta))^2+(b\sin(\theta))^2=c^2$$

$$c^2=a^2-2ab\cos(\theta)+b^2\cos^2(\theta)+b^2\sin^2(\theta)$$

$$c^2=a^2+b^2-2ab\cos(\theta)$$
 

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