The Law of

Cosines

Let's consider types of triangles with the three pieces of information shown below.

SAS

You may have a side, an angle, and then another side

AAA

You may have all three angles.

SSS

You may have all three sides

This case doesn't determine a triangle because similar triangles have the same angles and shape but "blown up" or "shrunk down"

We can't use the Law of Sines on these because we don't have an angle and a side opposite it. We need another method for SAS and SSS triangles.

AAA

Let's place a triangle on the rectangular coordinate system.

a

b

c

(b, 0)

What is the coordinate here?Drop down a perpendicular line from this vertex and use right triangle trig to find it.

a

ysin

(x, y)

a

xcos

sinay cosax

(a cos , a sin )

Now we'll use the distance formula to find c(use the 2 points shown on graph)

22 sin0cos aabc square both sides and FOIL

222222 sincoscos2 aaabbc 22222 sincoscos2 aabbc factor out a2 This = 1

cos2222 abbac

y

x

rearrange terms This is the Law of Cosines

We could do the same thing if gamma was obtuse and we could repeat this process for each of the other sides. We end up with the following:

LAW OF COSINES

cos2222 abbac

cos2222 accab

cos2222 bccba LAW OF COSINES

ab

cba

2cos

222

ac

bca

2cos

222

bc

acb

2cos

222

Use these to findmissing sides

Use these to find missing angles

Since the Law of Cosines is more involved than the Law of Sines, when you see a triangle to solve you first look to see if you have an angle (or can find one) and a side opposite it. You can do this for ASA, AAS and SSA. In these cases you'd solve using the Law of Sines. However, if the 3 pieces of info you know don't include an angle and side opposite it, you must use the Law of Cosines. These would be for SAS and SSS (remember you can't solve for AAA).

Since the Law of Cosines is more involved than the Law of Sines, when you see a triangle to solve you first look to see if you have an angle (or can find one) and a side opposite it. You can do this for ASA, AAS and SSA. In these cases you'd solve using the Law of Sines. However, if the 3 pieces of info you know don't include an angle and side opposite it, you must use the Law of Cosines. These would be for SAS and SSS (remember you can't solve for AAA).

Solve a triangle where b = 1, c = 3 and = 80°

Draw a picture.

80

a

1

3

Do we know an angle and side opposite it? No so we must use Law of Cosines.

Hint: we will be solving for the side opposite the angle we know.

This is SAS

cos2222 bccba times the cosine of the angle between

those sides

One side squared

2a

sum of each of the other sides

squared

minus 2 times the productof those

other sides

312 80cos22 31

Now punch buttons on your calculator to find a. It will be square root of right hand side.

a = 2.99

CAUTION: Don't forget order of operations: powers then multiplication BEFORE addition and subtraction

We'll label side a with the value we found.

We now have all of the sides but how can we find an angle?

80

2.99

1

3

Hint: We have an angle and a side opposite it.

sin80 sin

2.99 3

3sin80

2.99 80.8

80.8

is easy to find since the sum of the angles is a triangle is 180°

180 80 80.8 19.2

19.2

NOTE: These answers are correct to 2 decimal places for sides and 1 for angles. They may differ with book slightly due to rounding. Keep the answer for in your calculator and use that for better accuracy.

cos2222 abbac

Solve a triangle where a = 5, b = 8 and c = 9

Draw a picture.

5

8

9

Do we know an angle and side opposite it? No, so we must use Law of Cosines.

Let's use largest side to find largest angle first.

This is SSS

times the cosine of the angle between

those sides

One side squared

29

sum of each of the other sides

squared

minus 2 times the productof those

other sides

852 cos22 85 CAUTION: Don't forget order of operations: powers then multiplication BEFORE addition and subtraction

cos808981

80

8cos

3.84

10

1cos 1

84.3

How can we find one of the remaining angles?

5

8

9Do we know an angle and side opposite it?

84.3

sin84.3 sin

9 8

8sin84.3sin

9 1 8sin84.3

sin 62.29

62.2

180 84.3 62.2 33.5

33.5

Yes, so use Law of Sines.

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au