Transcript Law of Cosines

Law of Cosines
HOMEWORK: Lesson 12.4/1-14
Who's Law Is It, Anyway?

Murphy's Law:


Anything that can possibly go wrong, will
go wrong (at the worst possible moment).
Cole's Law ??

Finely chopped cabbage
2
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
3
Law of Cosines

Note the pattern
a  b  c  2  c  b  cos A
2
2
2
b  a  c  2  a  c  cos B
2
2
2
c  b  a  2  a  b  cos C
2
2
2
C
b
A
a
c
B
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:
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
a 2  b2  c2
cos  
2ab
2
Use these to find
missing angles
2
2
2
2
2
Applying the Cosine Law

Now use it to solve the triangle we
started with
C
15

Label sides
and angles

Side c first
A
26°
c
12.5
B
c  b  a  2  a  b  cos C
2
2
2
c  152  12.52  2 15 12.5  cos 26
6
Applying the Cosine Law
C
15
A

26°
c = 6.65
12.5
B
Now calculate the angles
2
2
2
b

a

c
 2  a  c  cos B
 use
and solve for B
b2  a 2  c 2
cos B 
2  a  c
2
2
2


b

a

c
1
B  cos 

 2  a  c 
7
Applying the Cosine Law
C
15
A

26°
c = 6.65
12.5
B
The remaining angle
determined by subtraction

180 – 93.75 – 26 = 60.25
8
Solve a triangle where b = 1, c = 3 and  = 80°
Draw a picture.
This is SAS

3
a
Hint: we will be
solving for the side
opposite the angle
we know.

80
Do we know an angle and side
opposite it? No so we must use
Law of Cosines.
1
a  b  c  2bc cos 
2
2
2
a  1  3  213cos 80
2
2
2
a = 2.99
Solve a triangle where a = 5, b = 8 and c = 9
This is SSS
Draw a picture.

9
5
Do we know an angle and
side opposite it? No, so we
must use Law of Cosines.
8
Let's use largest side to
find largest angle first.
9  5
2
84.3

c  a  b  2ab cos 
2
2
2
2
cos 




2
5
8
8
2
81  89  80 cos 
8
1  1 
cos  
  cos    84.3
 80
 10 

9
5
84.3

8
11
Wing Span
C


The leading edge of
each wing of the
B-2 Stealth Bomber
measures 105.6 feet
A
in length. The angle between the
wing's leading edges is 109.05°.
What is the wing span (the distance
from A to C)?
Hint … use the law of cosines!
12
C
b 2  a 2  c 2  2  a  c  cos B
x
B
109.05°
A
13
Using the Cosine Law to Find Area


Recall that
We can use
the value for h
to determine
the area
1
Area  c  b  sin A
2
h  b  sin A
C
b
h
a
A
B
c
14
Using the Cosine Law to Find Area

We can find the area knowing two
sides and the included angle
C
1
Area  a  b  sin C
2
1
 c  a  sin B
2
b
A

a
c
B
Note the pattern
15
Try It Out
Determine the area
12m
127°
24m
16
Determine the area
Missing angle – 180-42.8-76.3 = 60.9°
17.9
60.9°
42.8°
Missing side
sin 76.3 sin 60.9

17.9
a
17
Cost of a Lot

An industrial piece of real estate is
priced at $4.15 per square foot. Find,
to the nearest $1000, the cost of a
triangular lot measuring 324 feet by
516 feet by 412 feet.
516
18
516
19
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