Law of Cosines/Right Triangle
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Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that:
Let $\triangle ABC$ be a right triangle such that $\angle A$ is right.
Then:
- $c^2 = a^2 + b^2 - 2 a b \cos C$
Proof
Let $\triangle ABC$ be a right triangle such that $\angle A$ is right.
\(\ds a^2\) | \(=\) | \(\ds b^2 + c^2\) | Pythagoras's Theorem | |||||||||||
\(\ds c^2\) | \(=\) | \(\ds a^2 - b^2\) | adding $-b^2$ to both sides and rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 - 2 b^2 + b^2\) | adding $0 = b^2 - b^2$ to the right hand side | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 - 2 a b \left({\frac b a}\right) + b^2\) | multiplying $2 b^2$ by $\dfrac a a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | Definition of Cosine: $\cos C = \dfrac b a$ |
$\blacksquare$