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6.2 The
Law of
Cosines
Solving an SAS Triangle
• The Law of Sines was good for
– ASA
– AAS
– SSA
- two angles and the included side
- two angles and any side
- two sides and an opposite angle
(being aware of possible ambiguity)
• Why would the Law of Sines not work for an
SAS triangle?
No side opposite
from any angle to
get the ratio
15
26°
12.5
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
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"
Do you see a pattern?
LAW OF COSINES
Use these to find
missing sides
c  a  b  2ab cos C
2
2
2
b 2  a 2  c 2  2ac cos B
a 2  b 2  c 2  2bc cos A
LAW OF COSINES
b  c  a cos B  a  c  b
cos A 
2ac
2bc
2
2
2
a

b

c
Use these to find
cos C 
missing angles
2ab
2
2
2
2
2
2
Deriving the Law of Cosines
h  b  sin A
k  b  cos A
C
b
h
a
• Write an equation
k
c-k
A
using Pythagorean
c
theorem for shaded
triangle.
2
2
2
a   b  sin A    c  b  cos A 
B
a 2  b 2 sin 2 A  c 2  2  c  b  cos A  b 2 cos 2 A
a 2  b 2  sin 2 A  cos 2 A   c 2  2  c  b  cos A
a 2  b 2  c 2  2  c  b  cos A
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).
Solve a triangle where b = 1, c = 3 and A = 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
B
3
a
C
80
1
a  b  c  2bc cos A
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?
3
80
Hint: We have an angle and
a side opposite it.
sin 80 sin B

2.99
1
3 sin 80
B
 80.77
2.99
B
19.23
2.99
C
80.7
1 7
B is easy to find since
the sum of the angles
is a triangle is 180°
180  80  80.7  19.23


or 81.15 and 18.85 If you found C first
Solve a triangle where a = 5, b = 8 and c = 9
Draw a picture.
This is SSS
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.
B
A
5
84.26
C
8
c  a  b  2ab cos C
2
One side squared
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 C
1  1 
 8 C  cos  10   84.26
 
cos C 
2
2
2
 80
 258cos C
9
 5 8
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
B
62.18
5
84.26

33.56
A
8
Yes, so use Law of Sines.
sin 84.26 sin B

9
8
8 sin 84.26
 sin B
9
 8 sin 84.26 
sin 
  62.18
9


1
A  180  84.26  62.18  33.56
Try it on your own! #1
• Find the three angles of the triangle ABC if
a  6, b  8, c  12
C
8
A
6
12
B
A  26.38 , B  36.34 , C  117.28



11
Try it on your own! #2
• Find the remaining angles and side of the
triangle ABC if
b  16, c  12, m  A  80
C

16
A
80
B
12
a  18.26, B  59.67 , C  40.33


12
Wing Span
C
• The leading edge of
each wing of the
B-2 Stealth Bomber
A
measures 105.6 feet
in length. The angle between the wing's
leading edges is 109.05°. What is the wing
span (the distance from A to C)?
• Note these are the actual dimensions!
13
Wing Span
C
b 2  a 2  c 2  2ac cos B
A
b  105.6  105.6  2(105.6)(105.6) cos109.05
2
2
2
b  22302.72  7279.46
2
b  172 ft.
14
H Dub
• 6-2 Pg. 443 #2-16even, 17-22all, and 29
More Practice #1
More Practice #2