The Law
of
Cosines
Let's consider types of triangles with the three pieces of
information shown below.
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.
SAS
You may have a side, an
angle, and then another
side
SSS
You may have all three
sides
AAA
You may have all three angles.
AAA
This case doesn't determine a
triangle because similar
triangles have the same
angles and shape but "blown
up" or "shrunk down"
Let's place a triangle on the rectangular coordinate system.
(a cos(x,
, a
y)sin )
a
c
y

xb
(b, 0)
Now we'll use
the distance
formula to find c
(use the 2 points
shown on graph)
What is the coordinate here?
Drop down a perpendicular line
from this vertex and use right
triangle trig to find it.
y
sin  
a
y  a sin 
x
cos  
a
x  a cos 
b  a cos    0  a sin  
c
2
2
square both sides and FOIL
c  b  2ab cos   a cos   a sin 
c 2  b 2  2ab cos   a 2 cos 2   sin 2  factor out a2
2
2
2

c  a  b  2ab cos 
2
2
2
2
2
2

This
rearrange
is the Law
terms
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
Use these to find
missing sides
c 2  a 2  b 2  2ab cos 
b  a  c  2ac cos 
2
2
2
a 2  b 2  c 2  2bc cos 
LAW OF COSINES
b  c  a cos   a  c  b
cos  
2ac
2bc
2
2
2
a

b

c
Use these to find
cos  
missing angles
2ab
2
2
2
2
2
2
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.
This is SAS
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.
One side squared
Now punch buttons on your
calculator to find a. It will be
square root of right hand
side.
a = 2.99

3
a

80
1
a  b  c  2bc cos 
2
2
2
sum of each of
the other sides
squared
minus 2 times the
times the cosine of
product the angle
of those between
other
those
sides
sides
a  1  3  213cos 80
2
2
2
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?
Hint: We have an angle and
a side opposite it.
3
2.99
80.8
80
sin 80 sin 

2.99
3
3sin 80
 80.8

2.99

19.2
1
 is easy to find since
the sum of the angles
is a triangle is 180°
180  80  80.8  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.
Solve a triangle where a = 5, b = 8 and c = 9
Draw a picture.
This is SSS
One side squared
5
84.3


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.

9
8
c  a  b  2ab cos 
2
2
2
sum of each of
the other sides
squared
minus 2 times the
times the cosine of
product the angle
of those between
other
those
sides
sides
81  89  80 cos 
1  1 
 8   cos  10   84.3
 
cos  
2
2
2
 80
2 5
9  5 8
 8 cos 
CAUTION: Don't forget order of operations: powers
then multiplication BEFORE addition and subtraction
How can we find one of the remaining angles?
Do we know an angle and
side opposite it?
9

62.2
5
84.3

33.5
8
Yes, so use Law of Sines.
sin 84.3 sin 

9
8
8sin 84.3
 sin 
9
 8sin 84.3 
  sin 
  62.2
9


1
  180  84.3  62.2  33.5
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 Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au