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• Work out problems on board
• Add visuals (ranges of arccos and arcsin) to
show why you use LOC for big angles and
LOS for small angles
6.2 The
Law of
Cosines
Which proved triangles
congruent in Geometry?
•
•
•
•
•
•
SSS
ASA
AAS
SAS
AAA
ASS
The same ones that define a
specific triangle!
•
•
•
•
SSS - congruent
ASA - congruent
AAS – congruent
SAS – congruent
• AAA – not congruent
• ASS – not congruent
Which proved triangles
congruent in Geometry?
•
•
•
•
SSS - congruent
ASA – congruent – Solve w/ Law of Sines
AAS – congruent – Solve w/ Law of Sines
SAS – congruent
• AAA – not congruent
• ASS – not congruent
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?
15
No side opposite
from any angle to
get the ratio
26°
12.5
6
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
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
Deriving the Law of Cosines
• Write an equation using Pythagorean
theorem for shaded triangle that
only includes sides and angles of the
oblique triangle.
C
b
h
k
a
c-k
A
a  h  (c  k )
2
h  b  sin A
k  b  cos A
2
c
2
a   b  sin A    c  b  cos A 
2
B
2
2
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
9
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).
Ex. 1: 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 = 2.9930
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

2.993
1
sin 80
sin B 
2.993
B  19.21
B
19.21
2.993
C
80.79
1
C is easy to find since
the sum of the angles
is a triangle is 180°
180  80 19.21  80.79
When taking arcsin, use 2nd answer on your calculator for accuracy!
Ex. 2: 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 choose to find angle
C 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   84.2608
 
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
Yes, so use Law of Sines.
8
sin 84.26 sin B

9
8
8 sin 84.2608
 sin Bsin 1  8 sin 84.2608   62.1819
9
9


A  180  84.2608  62.1819  33.5573
Too easy, what’s the catch?
• After we use L.O.C. we need to use law of
sines to find the remaining sides and angles.
• The range of arcsin is -90 deg to 90 deg, but
what if the angle is obtuse? Then taking the
arcsin won’t get us the correct angle!
• To avoid this problem – When using
L.O.S. after L.O.C. always find the
smallest angle FIRST The smallest angle
has to be acute since there can’t be more
than one obtuse angle in a triangle.
• Then use the triangle sum thm to find the 3rd
angle.
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



16
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  18.257, B  59.67 , C  40.33


17
Summary – What could we use to
solve the following triangles?
70

80
30
Uh, nothing. It’s AAA
18
Summary – What could we use to
solve the following triangles?
20
16

80
ASA – although we don’t know
an angle and side opposite each
other we can find the 3rd angle
then do law of sines
19
Summary – What could we use to
solve the following triangles?
16
20

80
AAS – law of sines
20
Summary – What could we use to
solve the following triangles?
16
20

80
ASS, we can use law of sines but
need to check for 1, 2 or no
triangles.
21
Summary – What could we use to
solve the following triangles?
16

80
12
SAS – don’t know (and can’t
find) angle and side opposite
Law of Cosines
22
Summary – What could we use to
solve the following triangles?
16
20

12
SSS – don’t know (and can’t
find) angle and side opposite
Law of Cosines
23
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!
24
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.
25
• The direction to a point is stated as the
number of degrees east or west of north or
south. For example, the direction of
• A from O is N30ºE.
B is N60ºW from O.
C is S70ºE from O.
D is S80ºW from O
Navigational Bearings
H Dub
• 6-2 Pg. 443 #2-16even, 17-22all, 29, 34, 35
Practice #1
Practice #2
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
Practice #1
Practice #2