Cosine Rule

10.2 The Cosine Rule

Theorem. In any $\triangle ABC$,

Equivalently, $a^2=b^2+c^2-2bc\cos A$ and cyclically.

Proof. Place the triangle in coordinates with $A$ at the origin and $B$ on the positive $x$-axis, so $B=(c,0)$ (since $AB=c$). Then $C$ lies at $C=(b\cos A,\,b\sin A)$, because $AC=b$ and the angle between $AC$ and $AB$ is $A$. The side $a=BC$ is the distance between $B$ and $C$:

using $\cos^2A+\sin^2A=1$. Solving for $\cos A$ gives $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. Placing $B$ or $C$ at the origin instead gives the other two forms by the identical argument. $\blacksquare$

Remark — solving a triangle by SAS/SSS. The cosine rule is the tool of choice exactly when the sine rule cannot get started: given two sides and the included angle (SAS), or all three sides (SSS) with no angle at all. In the SAS case it gives the third side directly; in the SSS case it gives $\cos A$ (and hence $A$) directly from the three known sides, after which the sine rule finishes the remaining angles.

Special case. When $A=90^\circ$, $\cos A=0$ and the rule collapses to $a^2=b^2+c^2$, the Pythagorean theorem — the cosine rule is a genuine generalisation of it to non-right triangles.

Worked example. For $a=7,b=8,c=9$: $\cos A=\dfrac{8^2+9^2-7^2}{2\cdot8\cdot9}=\dfrac{96}{144}=\dfrac23$, so $A=\cos^{-1}\!\left(\dfrac23\right)\approx48.19^\circ$.

Detecting the triangle's shape. Because the sign of $\cos A$ is fixed by $b^2+c^2-a^2$, the cosine rule doubles as an angle-classification test: if $a^2<b^2+c^2$ then $A$ is acute; if $a^2=b^2+c^2$ then $A=90^\circ$; if $a^2>b^2+c^2$ then $A$ is obtuse. This is often faster than computing the angle numerically when a problem only asks whether the largest angle is obtuse.

A second worked example. For $a=5,b=6,c=8$: since $c$ is the largest side, check $\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{25+36-64}{60}=\dfrac{-3}{60}=-0.05$. As $\cos C<0$, angle $C$ is obtuse, so this triangle is obtuse-angled at $C$ — found without ever computing $C$ itself. …