The Cosine Rule

The Cosine Rule. In $\triangle ABC$: (i) $a^2=b^2+c^2-2bc\cos A$; (ii) $b^2=c^2+a^2-2ca\cos B$; (iii) $c^2=a^2+b^2-2ab\cos C$.

Proof. Take $A$ as the origin, the X-axis along $AB$, and the perpendicular to $AB$ through $A$ as the Y-axis. Then the co-ordinates of $A$, $B$, $C$ are $(0,0)$, $(c,0)$, $(b\cos A,b\sin A)$ respectively

(the co-ordinates of $C$ follow because $AC=b$ and the angle at $A$ between $AC$ and the X-axis is $A$). By the distance formula, $a^2=BC^2=(c-b\cos A)^2+(0-b\sin A)^2=c^2-2bc\cos A+b^2\cos^2A+b^2\sin^2A=c^2+b^2(\cos^2A+\sin^2A)-2bc\cos A=b^2+c^2-2bc\cos A$. So $a^2=b^2+c^2-2bc\cos A$. Placing the origin at $B$ or $C$ in turn and repeating the same argument gives the other two forms, $b^2=c^2+a^2-2ca\cos B$ and $c^2=a^2+b^2-2ab\cos C$. $\blacksquare$

Remark (cosine-solved form). The Cosine Rule can equivalently be written to solve directly for each angle: $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, $\cos B=\dfrac{c^2+a^2-b^2}{2ca}$, $\cos C=\dfrac{a^2+b^2-c^2}{2ab}$.

Ex.(5) In $\triangle ABC$, if $a=2$, $b=3$, $c=4$ then prove that the triangle is obtuse angled. The angle opposite the largest side is the largest angle, so here $C$ (opposite side $AB=c=4$, the largest side) is the largest angle; we check whether $C$ is obtuse. $\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{4+9-16}{2(2)(3)}=\dfrac{-3}{24}=-\dfrac18$. Since $\cos C$ is negative, $C$ is obtuse, so $\triangle ABC$ is an obtuse-angled triangle. …