The Law of Cosines

1
The Law of Cosines
2
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
AAA
You may have a side, an angle, and then another
side
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"
SSS
You may have all three sides
3
Let's place a triangle on the rectangular
coordinate system.
What is the coordinate here?
(x, y)
(a cos ?, a sin ?)
Drop down a perpendicular line from this vertex
and use right triangle trig to find it.
a
c
y
?
b
x
(b, 0)
Now we'll use the distance formula to find c(use
the 2 points shown on graph)
square both sides and FOIL
This 1
factor out a2
rearrange terms
This is the Law of Cosines
4
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
Use these to findmissing sides
Use these to find missing angles
5
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).
6
Solve a triangle where b 1, c 3 and ? 80
Draw a picture.
?
This is SAS
3
a
Do we know an angle and side opposite it? No so
we must use Law of Cosines.
?
80
1
Hint we will be solving for the side opposite
the angle we know.
times the cosine of the angle between those sides
minus 2 times the productof those other sides
One side squared
sum of each of the other sides squared
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
7
We'll label side a with the value we found.
We now have all of the sides but how can we find
an angle?
?
3
19.2
2.99
?
80.8
80
Hint We have an angle and a side opposite it.
1
? is easy to find since the sum of the angles is
a triangle is 180
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.
8
Solve a triangle where a 5, b 8 and c 9
Draw a picture.
?
9
This is SSS
5
Do we know an angle and side opposite it? No, so
we must use Law of Cosines.
?
84.3
?
8
Let's use largest side to find largest angle
first.
times the cosine of the angle between those sides
minus 2 times the productof those other sides
One side squared
sum of each of the other sides squared
CAUTION Don't forget order of operations
powers then multiplication BEFORE addition and
subtraction
9
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.
10
Acknowledgement I wish to thank Shawna Haider
from Salt Lake Community College, Utah USA for
her hard work in creating this PowerPoint. www.sl
cc.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
Stephens School Carramar www.ststephens.wa.edu.
au