Law of Cosines/Proof 3

Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then $c^2 = a^2 + b^2 - 2ab \cos C$.

Note
Equations are written in terms of a cartesian coordinate system.

Angles are written as if the length of the line sought defines the angle taken implicitly.

Lemma
Pythagoras's Theorem is a special case of the Law of Cosines.

Let $r^2 = x^2 + y^2$.

Then:

Thus $y^2 = r^2 + x^2 - 2xr \cos \theta$.

Method 2 of Lemma 1
Let $r^2=x^2+y^2$

Then:

Lemma 2
Case:The Law of Cosines with constants equal to $\pi$

Let $C=2\pi r$ and $r^2=x^2+y^2$