The law of cosines

In a triangle, the square of a side is the sum of the squares of the other two sides, less twice their product times the cosine of the angle between them:

$\displaystyle \gamma$ – the angle between $\displaystyle a$ and $\displaystyle b$, opposite $\displaystyle c$. When $\displaystyle {\gamma}={90^{\circ}}$, $\displaystyle {\cos\gamma}={0}$, and this is Pythagoras’ theorem: the law of cosines extends it to every triangle.

The proof below places the triangle on axes, $\displaystyle C$ at the origin and $\displaystyle CA$ on the $\displaystyle x$-axis, and works out $\displaystyle {c}={AB}$ by the distance formula.

• $\displaystyle {A}={(b,0)}$.
• $\displaystyle B$ is at distance $\displaystyle a$ from the origin, at the angle $\displaystyle \gamma$ from the $\displaystyle x$-axis: the point $\displaystyle (\cos\gamma,\sin\gamma)$ of the unit circle, enlarged $\displaystyle a$ times – $\displaystyle {B}={(a\cos\gamma,\ a\sin\gamma)}$.
• $\displaystyle {c^2}={(a\cos\gamma-b)^2+(a\sin\gamma)^2}={a^2\cos^2\gamma-2ab\cos\gamma+b^2+a^2\sin^2\gamma}$.
• $\displaystyle {\cos^2\gamma+\sin^2\gamma}={1}$, so $\displaystyle {c^2}={a^2+b^2-2ab\cos\gamma}$.

The definition on the unit circle holds for every angle, so the proof holds when $\displaystyle \gamma$ is obtuse too.

When all three sides are known, isolate $\displaystyle \cos\gamma$:

The sign of the numerator fixes the kind of angle opposite $\displaystyle c$:

• $\displaystyle c^2<a^2+b^2$ – $\displaystyle \cos\gamma>0$, the angle is acute;
• $\displaystyle {c^2}={a^2+b^2}$ – the angle is right;
• $\displaystyle c^2>a^2+b^2$ – $\displaystyle \cos\gamma<0$, the angle is obtuse.

Every cosine between $\displaystyle -1$ and $\displaystyle 1$ has one angle only between $\displaystyle 0^{\circ}$ and $\displaystyle 180^{\circ}$, so here – unlike the law of sines – there are no two possibilities.

When the known angle is not between the two known sides, the missing side is one of the sides at the angle, and the law of cosines gives a quadratic equation in it.

The example below has, in triangle $\displaystyle ABC$: $\displaystyle {AC}={5}$, $\displaystyle {BC}={7}$, $\displaystyle {\angle A}={60^{\circ}}$, and finds $\displaystyle {AB}={x}$.

• $\displaystyle BC$ is opposite $\displaystyle \angle A$: $\displaystyle {49}={x^2+25-2\cdot5x\cos60^{\circ}}={x^2+25-5x}$;
• $\displaystyle {x^2-5x-24}={0}$, so $\displaystyle {x}={8}$ or $\displaystyle {x}={-3}$;
• a length is positive: $\displaystyle {AB}={8}$.

An equation with two positive solutions means two different triangles with the same data.

Work out the third side $\displaystyle c$.

$\displaystyle {a}={5}$, $\displaystyle {b}={8}$, $\displaystyle {\gamma}={60^{\circ}}$.

$\displaystyle {a}={3}$, $\displaystyle {b}={5}$, $\displaystyle {\gamma}={120^{\circ}}$.

The sides of a triangle are $\displaystyle 5$, $\displaystyle 7$ and $\displaystyle 8$.
Work out the angle opposite the side $\displaystyle 7$, in degrees.

The sides of a triangle are $\displaystyle 4$, $\displaystyle 5$ and $\displaystyle 8$.

Without working out an angle: is the triangle acute, right-angled or obtuse? Explain.

Work out the largest angle of the triangle. Round to two decimal places.

In triangle $\displaystyle ABC$: $\displaystyle {AC}={3}$, $\displaystyle {BC}={7}$, and $\displaystyle {\angle A}={60^{\circ}}$.
$\displaystyle {x}={AB}$. Solve the equation the law of cosines gives.

The sides of a parallelogram are $\displaystyle 6$ and $\displaystyle 10$, and the angle between them is $\displaystyle 60^{\circ}$.
Work out the diagonals, as exact values.