A Triangle is uniquely determined by
a “Contained Angle”
C
a
b
A
O
c
B
IF b,c and O are all known, then O is
called a “Contained Angle”
C
a
b
A
O
c
B
The Cosine Law can be used to find
the length of the opposite side to O
C
a
b
A
O
c
B
Consider the following Triangle.
Find a
C
a=?
5
A
35o
8
B
Construct a vertical line CN to create
2 right triangles
Label the height “h”, and the bases
in terms of X
C
a=?
5
A
35o
X
h
N
8
8-X
B
By the Pythagorean Theorm:
52 = h2 + X2 and a2 = h2 + (8 – X)2
C
a=?
5
A
35o
X
h
N
8-X
B
By the Pythagorean Theorm:
52 = h2 + X2 and a2 = h2 + (8 - X)2
- X2
- (8 - X)2
52 - X2 = h2 and h2 = a2 - (8 - X)2
Since both equations are solved for h
52 - X2 = a2 - (8 - X)2
+ (8 - X)2 =
+ (8 - X)2
a2 = 52 - X2 + (8 - X)2
a2 = 52 - X2 + (8 - X)2
a2 = 52 - X2 + 64 – 16X + X2
a2 = 25 + 64 – 16X
From the Triangle,
redefine X
5
A
35o
X
Cos 35o = X
5
X = 5Cos35o
a2 = 25 + 64 – 16X
a2 = 25 + 64 – 16(5Cos35o)
a2 = 25 + 64 – 80Cos35o
a2 = 23.4678
a = 4.8 (side length opposite A)
In the interest of efficiency, we hunt for a
pattern that will allow us to generate a
formula….take a few steps back:
If we can create a formula
based on the steps, we
can save a great amount
of work.
Examine the
information
given, and
the structure
of the final
equation….
a2 = 25 + 64 – 80Cos35o
C
a=?
5
A
35o
8
B
In General:
a2 = b2 + c2 – 2bcCosOo
C
a
b
A
O
c
B
For Example: Find a
a2 = b2 + c2 – 2bcCosOo
C
a
8m
A
50o
10m
B
a2 = 82 + 102 – 2(8)(10)Cos50o
a2 = 61.15m
a = 7.8 m
C
a
8m
A
50o
10m
B