10D - The cosine rule

Two paths have the same lengths but meet at different angles. How much can the direct distance between their endpoints change? You will use the cosine rule to find sides and angles, then choose a suitable triangle method in exact calculations and two-dimensional journeys.

Finding an unknown side

Two sides measure $7\,\mathrm{cm}$ and $9\,\mathrm{cm}$ . Is the third side fixed before you know the angle between them?

If the sides meet at a right angle, Pythagoras gives the third side as $\sqrt{7^2+9^2}$ . Closing the angle makes the opposite side shorter; opening it makes that side longer. The cosine rule measures this change.

Cosine rule

In a non-degenerate triangle, with side $a$ opposite angle $A$ ,

a^2=b^2+c^2-2bc\cos A.

Here $b$ and $c$ meet at the included angle $A$ , with $b,c>0$ and $0^\circ<A<180^\circ$ . You can relabel the letters to match the triangle.

On the formulae sheet.

Use it directly for SAS data: two sides and their included angle. Find the squared length first, then take its positive square root. Since $\cos90^\circ=0$ , the formula agrees with Pythagoras for a right triangle. An acute $A$ subtracts a positive amount; an obtuse $A$ subtracts a negative amount.

Worked example 1: an acute included angle

Question. In triangle $ABC$ , $AB=9\,\mathrm{cm}$ , $AC=7\,\mathrm{cm}$ and $A=60^\circ$ . Find $BC$ exactly.

Plan. The known angle lies between the known sides, so use the cosine rule for the opposite side $BC$ .

\begin{aligned} BC^2&=7^2+9^2-2(7)(9)\cos60^\circ &&\text{SAS data}\\ &=130-63=67 &&\text{use }\cos60^\circ=\tfrac12\\ BC&=\sqrt{67} &&\text{a length is positive} \end{aligned}

Answer. $BC=\sqrt{67}\,\mathrm{cm}$ .

Check. Put $A=(0,0)$ and $B=(9,0)$ . Then $C=(7/2,7\sqrt3/2)$ , so the distance squared is $(9-7/2)^2+(7\sqrt3/2)^2=67$ , independently confirming the answer.

Change only the included angle to $120^\circ$ . Its cosine changes sign, so the correction now increases the squared length.

Worked example 2: an obtuse included angle

Question. In triangle $ABC$ , $AB=9\,\mathrm{cm}$ , $AC=7\,\mathrm{cm}$ and $A=120^\circ$ . Find $BC$ exactly.

Plan. Keep the same side pairing and retain the negative value of the cosine.

\begin{aligned} BC^2&=7^2+9^2-2(7)(9)\cos120^\circ\\ &=130-126\left(-\frac12\right)=193\\ BC&=\sqrt{193} \end{aligned}

Answer. $BC=\sqrt{193}\,\mathrm{cm}$ .

Check. This time $C=(-7/2,7\sqrt3/2)$ , so $(9+7/2)^2+(7\sqrt3/2)^2=193$ . Also $\sqrt{193}>\sqrt{130}$ , as expected for an angle larger than a right angle.

The opening paths can therefore have a much larger gap between their endpoints even when their own lengths stay fixed.

Finding an unknown angle

Three side lengths determine the shape of a triangle, so they also determine its angles.

Check first that the lengths are positive and that the longest is less than the sum of the other two. Equality gives a straight, degenerate shape; a larger longest side cannot close the triangle.

Rearranging the supplied rule gives

2bc\cos A=b^2+c^2-a^2,

so

\cos A=\frac{b^2+c^2-a^2}{2bc}.

The side $a$ must be opposite the angle being found. Use inverse cosine to obtain the unique interior angle between $0^\circ$ and $180^\circ$ . Unlike inverse sine, inverse cosine distinguishes acute and obtuse interior angles directly.

Worked example 3: an acute angle from three sides

Question. A triangle has $a=7\,\mathrm{cm}$ , $b=5\,\mathrm{cm}$ and $c=8\,\mathrm{cm}$ . Find $A$ .

Plan. All sides are known, so rearrange the cosine rule for the angle opposite $a$ .

The lengths form a triangle because $8<7+5$ .

\begin{aligned} \cos A&=\frac{5^2+8^2-7^2}{2(5)(8)} &&\text{side }7\text{ opposite }A\\ &=\frac{40}{80}=\frac12\\ A&=\cos^{-1}\left(\frac12\right)=60^\circ \end{aligned}

Answer. $A=60^\circ$ .

Check. The sides adjoining $A$ can have endpoints $(8,0)$ and $(5/2,5\sqrt3/2)$ from a common origin. Their separation squared is $(8-5/2)^2+75/4=49$ , giving the required opposite side $7$ .

Keep the adjoining sides $5$ and $8$ , but lengthen the opposite side. A negative cosine now signals an obtuse angle.

Worked example 4: an obtuse angle from three sides

Question. A triangle has $a=\sqrt{129}\,\mathrm{cm}$ , $b=5\,\mathrm{cm}$ and $c=8\,\mathrm{cm}$ . Find $A$ .

Plan. Keep the exact squared side length when substituting into the angle form.

Since $8<\sqrt{129}<13=5+8$ , these lengths form a triangle.

\begin{aligned} \cos A&=\frac{25+64-129}{80}=-\frac12\\ A&=\cos^{-1}\left(-\frac12\right)=120^\circ \end{aligned}

Answer. $A=120^\circ$ .

Check. With adjoining endpoints $(8,0)$ and $(-5/2,5\sqrt3/2)$ , the distance squared is $(8+5/2)^2+75/4=129$ . Also $129>25+64$ , consistent with an obtuse opposite angle.

The numerator gives a quick classification: $a^2<b^2+c^2$ means $A$ is acute, equality means right, and $a^2>b^2+c^2$ means obtuse. This is a check on the angle opposite $a$ , not necessarily on every angle of the triangle.

Watch out

An angle result needs a genuine triangle. For sides $2,3,5$ , finding the angle opposite side $2$ gives cosine $1$ and angle $0^\circ$ : this is degenerate, not a valid triangle. A cosine outside $[-1,1]$ indicates inconsistent data or an error; do not force an angle from it.

Use the side opposite the requested angle in the subtracted term.

Choosing a triangle method

The useful rule is the one that connects the quantities you actually know.

Start with a labelled sketch. Mark the requested side or angle and identify any known opposite pair or included angle.

Available data and target	Useful starting method
Right triangle with two sides	Pythagoras for the third side; a basic ratio for an angle
Right triangle with a side and an acute angle	Sine, cosine or tangent
Complete opposite pair, plus another side or angle	Sine rule; check both candidates for SSA
Two sides and their included angle, seeking the third side	Cosine rule
Three sides, seeking an angle	Cosine rule
Two sides and their included angle, seeking area	Triangle area formula

Last Accessed

Number

Algebra

Series

Geometry

Trigonometry

Calculus

10D - The cosine rule

Finding an unknown side

Finding an unknown angle

Choosing a triangle method

Continue this lesson.

Learn faster with Nova