Law of Cosines/Right Triangle

Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that:

$a$ is opposite $A$

$b$ is opposite $B$

$c$ is opposite $C$.

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: $\cos C = \dfrac b a$

$\blacksquare$