Applications of the Cosine Law
Lesson

We often think of math as being confined to the classroom and being separate from our experience of the world. The truth is, part of the beauty in math is that it can explain and measure so much of what is happening in the real and natural world.

#### Example

Scientists can use a set of footprints to calculate an animal's step angle, which is a measure of walking efficiency.  The closer the step angle is to $180^\circ$180°, the more efficiently the animal walked.

The angle marked here highlights the step angle.

If a particular set of footprints has length $AC=304$AC=304 cm, length $BC=150$BC=150 cm and length $AB=182$AB=182 cm, find the step angle.

Think: We have 3 side lengths, and require a missing angle, so we will use the cosine rule.

If angle B is the step angle we are trying to find then:

$b^2=a^2+c^2-2ac\cos B$b2=a2+c22accosB   rearrange to make cos(B) the argument of the equation

$\cos B=\frac{b^2-a^2-c^2}{-2ac}$cosB=b2a2c22ac

Do: Then fill in all the values we know,

$\cos B=\frac{304^2-150^2-182^2}{-2\times150\times182}$cosB=3042150218222×150×182

$\cos B=\frac{36792}{-54600}$cosB=3679254600

$B=\cos^{-1}\left(\frac{-36792}{54600}\right)$B=cos1(3679254600)

$B=132.36$B=132.36 degrees (to 2 decimal places)

Notice that the cosine ratio of the angle is negative. This indicates that the angle will be greater than $90^\circ$90°.

## Using the Law of Cosines

The Law of Cosines is useful when you want to find:

• the third side of a triangle when you know two sides and the angle between them
• the angles of a triangle when you know all three sides

The best way to approach problems involving applications of the cosine rule is to label the angles of the triangle, label the corresponding sides and then apply the cosine rule.

Law of cosines

$a^2=b^2+c^2-2bc\cos A$a2=b2+c22bccosA

$b^2=a^2+c^2-2ac\cos B$b2=a2+c22accosB

$c^2=a^2+b^2-2ab\cos C$c2=a2+b22abcosC

#### Examples

##### Question 1

Find the length of the diagonal, $x$x, in parallelogram $ABCD$ABCD.

##### Question 2

Dave leaves town along a road on a bearing of $169^\circ$169° and travels $26$26 km. Maria leaves the same town on another road with a bearing of $289^\circ$289° and travels $9$9 km.

Find the distance between them, $x$x.