Triangle

# The Cosine Law

The cosine law states that in any triangle $ABC$, $1.\;\;\;\cos A=\frac{b^2+c^2-a^2}{2bc}$ $\text{or,}\;\; a^2=b^2+c^2-2bc\cos A$ $2.\;\;\;\cos B=\frac{c^2+a^2-b^2}{2ca}$ $\text{or,}\;\; b^2=c^2+a^2-2ca\cos B$ $3.\;\;\;\cos C=\frac{a^2+b^2-c^2}{2ab}$ $\text{or,}\;\; c^2=a^2+b^2-2ab\cos C$

Let $ABC$ be a triangle placed in the standard position with the vertex $A$ at the origin and the side $AB$ along the positive x-axis. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively.

Draw $BN\perp AC$, then $\cos A=\frac{ON}{OB}=\frac{ON}{c}$ $\therefore ON=c\cos A$ and, $\sin A=\frac{BN}{OB}=\frac{BN}{c}$ $\therefore BN=c\sin A$

Then, the coordinates of $A$, $C$ and $B$ are $(0,0)$, $(b,0)$ and $(c\cos A, c\sin A)$ respectively.

Now, using distance formula to find the length of $BC$, $BC=\sqrt{(c\cos A-b)^2+(c\sin A-0)^2}$ $a=\sqrt{c^2\cos^2A-2bc\cos A+b^2+c^2\sin^2A}$ $a^2=c^2-2bc\cos A+b^2$ $2bc\cos A=b^2+c^2-a^2$ $\therefore\cos A=\frac{b^2+c^2-a^2}{2bc}$

This same formula can also be proved if the triangle $ABC$ lies in the second or third quadrant in which the coordinates of $A$, $C$ and $B$ are $(0,0)$, $(-b,0)$ and $(-c\cos A, \pm c\sin A)$ respectively. And, the similar follows for the fourth quadrant where the coordinates of $A$, $C$ and $B$ are $(0,0)$, $(b,0)$ and $(c\cos A, -c\sin A)$.

Again, by placing the other angles $B$ and $C$ in the standard position, we can prove that $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and, $\cos C=\frac{a^2+b^2-c^2}{2ab}$

[Also see: Vector Method to Prove the Cosine Law]

In particular, if $A$ is right angle, then $\cos 90°=\frac{b^2+c^2-a^2}{2bc}$ $\therefore a^2=b^2+c^2$ This is the Pythagorean theorem.