Trigonometry/Law of Cosines

Consider this triangle:

It has three sides

• A, length A, opposite angle a at vertex a
• B, length B, opposite angle b at vertex b
• C, length C, opposite angle c at vertex c

The perpendicular, oc, from line ab to vertex c has length h

The Law of Cosines states that:

$A^2=B^2+C^2-2BC\cdot \cos a$

$B^2=A^2+C^2-2AC\cdot \cos b$

$C^2=A^2+B^2-2AB\cdot \cos c$

Proof

The perpendicular, oc, divides this triangle into two right angled triangles, aco and bco.

First we will find the length of the other two sides of triangle aco in terms of known quantities, using triangle bco.

h=A sin b

Side C is split into two segments, total length C.

ob, length C cos b

ao, length C - A cos b

Now we can use Pythagoras to find B, since B² = ao² + h²

$\begin{array}{ccc}{B}^2& =& (C-A\cos b{)}^2+A^2{\sin }^{2}b\\ & =& C^2-2AC\cos b+A^2{\cos }^{2}b+A^2{\sin }^{2}b\\ & =& A^2+C^2-2AC\cos b\end{array}$

The corresponding expressions for A and C can be proved similarly.