# Cosine Rule (Law of Cosines)

The **cosine rule**, also known as the **law of cosines**, relates all 3 sides of a triangle with an angle of a triangle.

It is most useful for solving for missing information in a triangle. For example, if all three sides of the triangle are known, the cosine rule allows one to find any of the angle measures. Similarly, if two sides and the angle between them is known, the cosine rule allows one to find the third side length.

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## Statement and Proof of the Theorem

Given the following triangle \(ABC\) with corresponding sides length \(a\), \(b\), and \(c:\)

the **law of cosines** states that

\[\begin{align} a^2&=b^2+c^2-2bc \cdot \cos A\\ b^2&=a^2+c^2-2ac \cdot \cos B\\ c^2&=a^2+b^2-2ab \cdot \cos C. \end{align}\]

It can be seen as a generalization of the Pythagorean theorem. Take one arbitrary side of the triangle, for instance, \(a\). Then its square equals the sum of the squares of the other two sides, i.e. \(a^2 = b^2 + c^2\). If you were to plug in 90 degrees, you would be left with the Pythagorean theorem. Since the angle \(\alpha\) that faces our arbitrary side \(a\) is not necessarily \(90°\), we will have to subtract something, as the identity \(a^2 = b^2 + c^2\) does not hold yet. The right side of this equation is still "too big." That something we have to subtract becomes \(2bc \cdot \cos A\).

We'll prove for side \(a\). Let's denote its facing angle as \(\alpha\). The other two equations can be done in a similar way.

By definition we have \[\begin{align} \sin \alpha &= \frac{h}{c} \implies h = c \cdot \sin \alpha \\ \cos \alpha &= \frac{r}{c} \implies r = c \cdot \cos \alpha. \end{align}\] Using the Pythagorean theorem, we get \[a^2 = h^2 + (b-r)^2.\] Substituting for \(h\) and \(r\), we get \[\begin{align} a^2 &= (c \cdot \sin \alpha)^2 + (b - c \cdot \cos \alpha)^2 \\ &= c^2 \cdot \sin^2 \alpha + b^2 - 2bc \cdot \cos \alpha + c^2 \cdot \cos^2 \alpha \\ &= c^2 \cdot \big( \sin^2 \alpha + \cos^2 \alpha \big) + b^2 - 2bc \cdot \cos \alpha\\ &= c^2 + b^2 - 2bc \cdot \cos \alpha.\ _\square \end{align}\]

Notes:

- The identity \(\sin^2 \alpha + \cos^2 \alpha = 1\) is also known as the Pythagorean identity.
- This proof isn't perfect. We should have been worried about angles. This can be avoided by using
directed angles.

Let \( \vec{a} \cdot \vec{b} \) denote the dot product between \(\vec{a}\) and \(\vec{b}\).

Also, let \( \vec{a} = \vec {BC}\), \( \vec{b} = \vec {AC}\), and \( \vec{c} = \vec {AB}\).

Then \[c^2 = \vec{c} \cdot \vec{c} = \big(\vec{b} - \vec{a}\big) \cdot \big(\vec{b} - \vec{a}\big) = b^2 + a^2 - 2\vec{a} \cdot \vec{b} = b^2 + a^2 - 2ab\cos C.\ _\square\]

## Finding Missing Side Lengths and Angles

**Given side-angle-side:**

In triangle \( ABC\), we have \( \angle BAC = \frac {\pi}{4}, \lvert \overline{BC} \rvert = \sqrt{5}\) and \( \lvert \overline{AB} \rvert = 3\). Determine \( \lvert \overline{AC} \rvert\).

Applying the cosine rule on \( \angle BAC\), we get

\[\begin{align} \lvert \overline{AB} \rvert^2 + \lvert \overline{AC} \rvert^2 - 2 \lvert \overline{AB} \rvert \cdot \lvert \overline{AC} \rvert \cdot \cos \frac {\pi}{4} &= \lvert \overline{BC} \rvert^2\\ \lvert \overline{AC} \rvert^2 - 3 \sqrt{2} \lvert \overline{AC} \rvert + 4 &=0 \\ \left(\lvert \overline{AC} \rvert-\sqrt{2}\right)\left(\lvert \overline{AC} \rvert-2\sqrt{2}\right)&=0. \end{align}\]

Hence, \( \lvert \overline{AC} \rvert=\sqrt{2} \text{ or } 2\sqrt{2}\). \( _\square \)

Note: This is similar to the 'ambiguous case' of sine rule, since we have \( 3 \sin \frac {\pi}{4} < \sqrt{5} < 3\), which is the condition \( c \sin \alpha < a < c\).

No problem found with slug "trigonometry-22"

In triangle \( ABC\), \(\angle ACB = 90^\circ \). Points \(A, D, E,\) and \(B\) are consecutive points on side \(AB\) such that \(\overline{AD}=\overline{DE}=\overline{EB}\). If there exists \( \theta\) such that \( \overline{CD} = 5 \cos \theta\) and \(\overline{CE} = 5 \sin \theta \), what is \(\overline{AB}^2?\)

**Given side-side-side:**

In triangle \(ABC\), \(\overline{AB} = \sqrt{6} - \sqrt{2}, \overline{AC} = 2\sqrt{2},\) and \(\overline{BC} = 2\sqrt3.\) Find the measure of \(\angle ABC.\)

Applying the law of cosines on side \(AC,\) we get

\[\begin{align} {\overline{AC}}^2 &= {\overline{AB}}^2 + {\overline{BC}}^2 - 2 \cdot \overline{AB} \cdot \overline{BC} \cdot \cos {\angle ABC} \\\\ {\big(2\sqrt{2}\big)}^2 &= {\big(\sqrt{6} - \sqrt{2}\big)}^2+ {\big(2\sqrt3\big)}^2 - 2\cdot 2\sqrt3 \cdot \big(\sqrt{6} - \sqrt{2}\big) \cdot \cos {\angle ABC}\\\\ 8 &= 6 - 4\sqrt{3} + 2 + 12 - \big(12\sqrt{2} - 4\sqrt{6}\big)\cos{\angle ABC}\\\\ -12 +4\sqrt{3} &= (-12\sqrt{2}+4\sqrt{6})\cos{\angle ABC}\\\\ \cos{\angle ABC} &=\dfrac{-12 +4\sqrt{3}}{-12\sqrt{2}+4\sqrt{6}} \\\\ &=\dfrac{1}{\sqrt2} \cdot \dfrac {-12 +4\sqrt{3}}{-12+4\sqrt{3}} =\dfrac{\sqrt2}{2}\\\\ \Rightarrow \angle ABC &= \dfrac{\pi}{4}.\ _\square \end{align}\]

Try the following problem:

## Applications to Bearing and Heading Problems

When it comes to the applications with regards to the law of cosines, it has its own applications towards vector quantities (and not qualities).

See: Bearing

## Vector Form of Cosine Rule

We have

\[\begin{align} \vec{c}^{2} &= \vec{a}^{2} + \vec{b}^{2} - 2 \cdot \vec{a} \cdot \vec{b}\\ &= \vec{a}^{2} - 2 \cdot \vec{a} \cdot \vec{b} + \vec{b}^{2} \\ &= \big(\vec{a} - \vec{b}\big)^2. \end{align} \]

## Deriving other Laws Using the Cosine Rule

- Triangle Inequality
- Pythagorean Theorem

From the cosine rule, we have

\[c^2 \leq a^2 + b^2 + 2ab = (a+b)^2,\]

and by taking the square root of both sides, we have \(c \leq a + b\), which is also known as the

triangle inequality. One useful application of the triangle inequality is to test if three given lengths can define a triangle.

The Pythagorean theorem applies to right triangles, so let \(\gamma \) be a right angle, i.e., \(\gamma = \frac{\pi}{2}\). Then by the cosine rule,

\[ c^2 = a^2 + b^2 - 2ab \cos \gamma = a^2 + b^2 - 2ab \cos \frac{\pi}{2} = a^2 + b^2 - 0 = a^2 + b^2.\]

## Suppose \( a, b,\) and \( c\) are positive reals such that \( a^2 = b^2+c^2 - bc, b^2 = c^2 + a^2 - ac,\) and \( c^2 = a^2 + b^2 - ab\). Show that \( a = b = c\).

Since \( c^2 = a^2 + b^2 - ab < a^2 + b^2 + 2ab = (a+b)^2\), it follows that \( c < a+b\) and we have similar inequalities for other variables. Hence, the numbers \( a, b\) and \( c\) satisfy the triangle inequality, and there exists a triangle \( ABC\) such that \( \lvert \overline{AB} \rvert=c, \lvert \overline{BC} \rvert=a, \lvert \overline{CA} \rvert=b\). As such

\[ \cos \angle ABC = \frac {a^2 + c^2 - b^2}{2ac} = \frac {a^2 + c^2 - (a^2+c^2-ac)}{2ac} = \frac {1}{2},\]

which gives us that \( \angle ABC = 60^\circ\). Similarily, we have \( \angle BCA = 60^\circ\) and \(\angle CAB = 60^\circ\), which show that triangle \( ABC\) is equilaterial, implying that \( a=b=c\). \( _\square \)

Note:This may also be done directly by summing up the 3 equations to get \( (a-b)^2 + (b-c)^2 + (c-a)^2 =0\).

## Problem Solving

An ant, located in a square field, is \(13\text{ m}\) from one of the corner posts of the field, \(17\text{ m}\) from the corner post diagonally opposite to the first one, and \(20 \text{ m}\) from a third corner post. Assuming the land is flat, find the area of the field \(\big(\)in \(\text{m}^2\big).\)

One big and one small circle share the same center \(O\).

Then \(\triangle ABC\) is constructed such that points \(A\) and \(B\) are on the big circle while point \(C\) is on the smaller circle. \(AB\) intersects the smaller circle at \(D\) and \(E\) and \(AC\) passes through \(F\) and \(O\), as shown above.

If \(AF = 3\) (red segment), \(DE = 6\) (blue segment), and \(BC = 11\) (green side), what is the radius of the smaller, orange circle?

## See Also

**Cite as:**Cosine Rule (Law of Cosines).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/cosine-rule/