Objectives:
1. To derive and use
the Law of Cosines
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•
•
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Assignment:
P. 443: 1-16 S
P. 443: 17-22 S
P. 443-445: 29-43 S
HW Supplement
You will be able to derive and use the
Law of Cosines
Find the length of AC in acute triangle ABC.
As the previous example
demonstrates, you
cannot always use the
Law of Sines for every
triangle. You need either
two sides and an angle or
two angles and a side in
the following
configurations. What we
need is another law.
AAS
SSA
• Ambiguous
On the same set of trigonometric axis, graph the
following functions.
y  cos   x 
y   cos  x 
In the up-coming activity, the identity below will
prove tremendously useful.
cos   x    cos x
In a right triangle, the
square of the length of
the hypotenuse is
equal to the sum of the
squares of the lengths
of the legs.
c2
a2
b2
If the square of the length of the longest side of
a triangle is equal to the sum of the squares of
the lengths of the other two sides, then it is a
right triangle.
Tell whether a triangle with the given side
lengths is a right triangle.
1. 5, 6, 7
2. 5, 6, 61
3. 5, 6, 8
In the previous
example, the side
lengths 5 and 6 do
not make a right
triangle when paired
with either 7 or 8. So
what kind of triangle
do these side lengths
make?
You can use the Pythagorean formula to
determine whether a triangle is right or not:
Acute:
c  a b
2
2
2
Need to subtract
something to = c2
Right:
c  a b
2
Where c is the
longest side.
2
2
Obtuse:
c  a b
2
2
2
Need to add
something to = c2
To derive the Law of
Cosines, we need an
interesting diagram
like the one shown.
Click the picture for a
Geometer’s
Sketchpad
Demonstration of
this useful equation.
If ΔABC has sides of
length a, b, and c as
shown, then
c 2  a 2  b 2  2ab  cos C
b  a  c  2ac  cos B
2
2
2
a 2  b 2  c 2  2bc  cos A
Simplify the equation below for C = 90°.
c 2  a 2  b 2  2ab  cos C
When C is acute, cos C is positive:
Acute:
c  a b
2
2
2
Need to subtract
something to = c2
c 2  a 2  b 2  2ab  cos C
When C is obtuse, cos C is negative:
Obtuse:
c  a b
2
2
2
Need to add
something to = c2
c 2  a 2  b 2  2ab  cos C
This becomes minus a negative,
so you are actually adding
something
Solve ΔABC.
Solve ΔQED.
Solve the equation below for C.
c 2  a 2  b 2  2ab  cos C
Solve ΔABC.
When using the Law of
Cosines to find a missing
1: Use the Law of Cosines
angle (SSS), it’s a good
idea to find the angle
opposite the longest side
first. This is just in case
the angle turns out to be
obtuse. Regardless of
what type of angle this
turns out to be, use the 2: Use the Law of Sines
Law of Sines and the
Triangle Sum Theorem to
3: Use the Triangle Sum
find the other two angles.
Find the indicated measure.
1. x and y
2.  and 
Law of
Sines
• ASA
• AAS
• SSA
Law of
Cosines
• SAS
• SSS
Given three pieces of
any triangle, you
can use the Law of
Sines or the Law of
Cosines to
completely solve
the triangle.
A sailboat leaves its pier at a bearing of 270 at 8
knots (nautical miles per hour). After 15
minutes, the boat changes course to a bearing
of 344 at 10 knots. Find the sailboat’s bearing
and distance from the pier after 12 minutes
on this course.
Armed with a compass and an odometer, you
set off on a bike trail consisting of two legs.
The first leg takes you 20 miles at a bearing of
050, while on the second leg, you ride 12
miles at a bearing of 125. What is the
distance, as the crow flies, from the starting
point to your destination? What is the
bearing?
A ship travels 40 nm
due east and then
changes direction.
When the ship has
traveled 30 nm at
this heading, it is 56
nm from its point of
departure. Describe
the bearing from B to
C.
Objectives:
1. To derive and use
the Law of Cosines
•
•
•
•
Assignment:
P. 443: 1-16 S
P. 443: 17-22 S
P. 443-445: 29-43 S
HW Supplement