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6.2
The Law of
Cosines
Let's consider types of triangles with the
three given measurements:
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
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 different (proportional) sides
Law of Cosines
Use these formulas
to find missing sides
and angles
LAW OF COSINES
c 2  a 2  b 2  2ab cos C
b 2  a 2  c 2  2ac cos B
a 2  b 2  c 2  2bc cos A
C
a
b
A
c
B
Example 1.
Solve a triangle where b = 1, c = 3 and A = 80°
B
3
Draw a picture.
a
A
This is SAS
80 °
1
C
a  b  c  2bc cos A
2
One side squared
Do we know an angle and side
opposite it?
No, so we must use Law of
Cosines.
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 sides those sides
Hint: we will be solving for the
side opposite the angle we
know.
Example 1.
Solve a triangle where b = 1, c = 3 and A = 80°
B
a
3
C
80 °
A
1
a  b  c  2bc cos A
2
One side squared
2
2
minus 2
times the
times the
cosine of
product
the angle
of those
between
other sides those sides
sum of each of
the other sides
squared
a  1  3  213 cos80
2
2
2
a = 2.99
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.
sin80 sinC
=
2.99
3
3sin80
sinC =
2.99
C = 81.15°
A
3
B
18.85°
2.99
81.15°
80°
1
C
Angle B is easy to find
since the sum of the
angles is a triangle is 180°
180 – 80 – 81.15 = 18.85°
B = 18.85°
Example 2.
Solve a triangle where a = 5, b = 8 and c = 9
Draw a picture.
This is SSS
B
Do we know an angle and side
opposite it?
9
5
No, so we must use Law of
Cosines.
84.3
A
Let's use largest side to find
largest angle first.
C
8
c = a + b - 2ab cosC
2
2
2
Example 2.
B
9
C = 84.3°
5
84.3
A
C
8
c  a  b  2ab cos C
2
2
2
sum of each of
the other sides
squared
One side squared
minus 2
times the
times the
cosine of
product
the angle
of those
between
other sides those sides
9  5  8  258 cosC
2
2
2
81  89  80cosC
8
cos C 
80
 1 
C  cos    84.3
 10 
1
Example 2.
B
9
62.2
5
Do we know an angle and
side opposite it?
A
Yes, so use Law of Sines.
sin 84.3 sin B

9
8
8sin 84.3
 sin B
9
1  8sin 84.3 
B  sin 
  62.2
9


B = 62.2°
84.3
33.5
8
C
A  180  84.3  62.2  33.5
A = 33.5°
Another Area Formula (given SSS)
The law of cosines can be used to derive a formula for
the area of a triangle given the lengths of three sides
known as Heron’s Formula.
Heron’s Formula
If a triangle has sides of lengths a, b, and c
1
s  ( a  b  c ),
2
Then the area of the triangle is
A  s ( s a)(s b)(s c).
Solve the ∆ABC given following:
1) b = 40, c = 45, A = 51°
a = 36.87, B = 57.47°, C = 71.53°
2) B= 42°, a= 120, c=160
b = 107.07, A = 48.58°, C = 89.42°
3) a =20, b= 12, c = 28
B = 21.79°, A = 89.79°, C = 120°
4) a= 15, b= 18, c = 17
B = 68.12°, A = 50.66°, C = 61.22°