Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Lesson 9: Law of Cosines
Student Outcomes

Students prove the law of cosines and use it to solve problems (G-SRT.D.10).
Lesson Notes
In this lesson, students continue the study of oblique triangles. In the previous lesson, students learned the law of sines.
In this lesson, students add the law of cosines to their repertoire of tools that can be used to analyze the measurements
of triangles. To prove the law of cosines, students build squares on the sides of an oblique triangle and then analyze
their areas. In this way, students build on their prior knowledge of the Pythagorean theorem.
Classwork
Exercises 1–2 (2 minutes)
Exercises
1.
Find the value of 𝒙 in the triangle below.
8
26°
x
𝒙
𝟖
We have 𝐜𝐨𝐬(𝟐𝟔°) = . This means 𝒙 = 𝟖 ∙ 𝐜𝐨𝐬(𝟐𝟔°) ≈ 𝟕. 𝟐.
2.
Explain how the figures below are related. Then, describe 𝒙 in terms of 𝝑.
5
1
θ
θ
cos(θ)
x
The triangles above are similar. We can produce the triangle at the right by dilating the triangle at the left using a
scale factor of 𝟓. This means that 𝒙 = 𝟓 ∙ 𝐜𝐨𝐬(𝝑).
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Exploratory Challenge (9 minutes): Exploring the SAS Case
12
5

The triangle on the left has a right angle, so we can apply the
Pythagorean theorem. The third side satisfies 𝑐 2 = 52 + 122 , which
means that 𝑐 = 13. The triangle on the right does not have a right
angle, so we cannot use this strategy.
 What happens if you open
up the included angle so
that it is as large as
possible? What happens
when you close the
included angle as much as
possible?
Yes, the SAS criterion for triangle congruence says that the three given measurements determine the
remaining measurements in the figure.
Let’s try to get some idea of how large the third side of the triangle can be. If two sides of a triangle are 5 and
12, how big can the third side be? Think about this for a moment, then explain your reasoning to a neighbor.

If we open up the angle between the two given sides so that it approaches a straight angle, the third
side of the triangle approaches a line segment with length 5 + 12 = 17.

If we close the angle so that it approaches the zero angle, the third side of the triangle approaches a
line segment with length 12 − 5 = 7.
12
5

12
Despite the fact that we can’t use the Pythagorean theorem on the triangle at the right, do you think you should
be able to determine the remaining measurements in this triangle? Why or why not? Think about this for a
moment, then explain your reasoning to a neighbor.

MP.3
5
Look at the triangles in the diagram above. What is the length of the third side
of the triangle on the left? How did you determine your answer? Can you use
the same strategy to find the third side of the triangle on the right?


 If students need support
with this task, prompt
them to consider extreme
cases and to represent
them with drawings.
70°
90°

Scaffolding:
12
5
So it appears that the third side takes on values between 7 and 17 and that the length of the third side is a
function of the angle between the two given sides. There is a formula called the law of cosines that makes this
relationship precise. We develop this formula together during the course of the lesson.
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
MP.7

Take a few minutes to explore the triangle below. Can you find the length of the third side? Work together
with a partner, and see how much progress you can make.

Can we use the law of sines? Why or why not?

No, we do not know the length of the side opposite a known angle.
 Where could you add an
auxiliary line to this figure?
70°

 Add an auxiliary line that
would allow you to
capitalize on your
knowledge of right
triangles.
12
5
Scaffolding:
We can draw an altitude to form two right triangles:
70°
12
5

Next, we can use trigonometry to describe the sides of the upper triangle. In particular, the small leg is
5 cos(70°) ≈ 1.7, and the longer leg is 5 sin(70°) ≈ 4.7.

Since the original triangle has a side of length 12, we can also figure out the longer leg of the lower
triangle, namely 12 − 1.7 = 10.3.
1.7
70°
5
Lesson 9:
10.3
4.7
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS

Finally, we can apply the Pythagorean theorem to the lower triangle. The hypotenuse of this triangle is
a number 𝑐 which satisfies 4.72 + 10.32 = 𝑐 2 , so we must have 𝑐 ≈ 11.3. So the third side of the
original triangle is about 11.3 units long.
70°
12
5
11.3

Let’s do another example of this kind, and then we’ll look for a way to generalize our work.
Exercise 3 (5 minutes)
Instruct students to solve the following problems, and then have them compare their work with a partner. Select two
students to write their work on the board to share with the class.
3.
Find the length of side ̅̅̅̅
𝑨𝑪 in the triangle below.
C
6
64°
A
B
9
We have 𝟔 ∙ 𝐜𝐨𝐬(𝟔𝟒°) ≈ 𝟐. 𝟔 and 𝟔 ∙ 𝐬𝐢𝐧(𝟔𝟒°) ≈ 𝟓. 𝟒. Then we have 𝟗 − 𝟐. 𝟔 = 𝟔. 𝟒. Finally, we have
√𝟔. 𝟒𝟐 + 𝟓. 𝟒𝟐 ≈ 𝟖. 𝟒.
C
8.4
5.4
6
64°
A
Lesson 9:
6.4
Law of Cosines
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2.6
B
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Discussion (5 minutes): Proving the Law of Cosines

In the examples on the previous page, what information was given to you? What information were you able to
find?


We were given two sides and the included angle. We determined the length of the third side.
Let’s try to develop a general formula that handles problems of this type. In the diagram below, find a way to
describe the side 𝑐 in terms of the other three measurements in the figure. Take a few minutes to work
together with the students around you on this task.
γ
b
a
c

First, we draw in an altitude to create right triangles, and then we describe the segments as before.
γ
a
a∙cos(γ)
b - a∙cos(γ)
a∙sin(γ)
c

The third side 𝑐 must satisfy 𝑐 2 = (𝑎 ∙ sin(𝛾))2 + (𝑏 − 𝑎 ∙ cos(𝛾))2 .

Let’s do some algebra: We have 𝑐 2 = (𝑎 ∙ sin(𝛾))2 + 𝑏 2 − 2𝑎𝑏 ∙ cos(𝛾) + (𝑎 ∙ cos(𝛾))2 .

We also see in the upper triangle that (𝑎 ∙ sin(𝛾))2 + (𝑎 ∙ cos(𝛾))2 = 𝑎2 , so we can rearrange and
replace terms as



𝑐 2 = [(𝑎 ∙ sin(𝛾))2 + (𝑎 ∙ cos(𝛾))2 ] + 𝑏 2 − 2𝑎𝑏 ∙ cos(𝛾)

𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 ∙ cos(𝛾)
We can use this relation to find 𝑐 if we are given 𝑎, 𝑏, and 𝛾.
Since this relation features the cosine function, it is known as the law of cosines.
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Discussion (13 minutes): Geometric Interpretation

Now that we have the law of cosines, let’s explore its geometric meaning. How do all of the symbols in the
formula 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 ∙ cos(𝛾) relate to the parts of the triangle? This is a complex question! Let’s
begin by reviewing what we know about right triangles.

Do you recall the geometric interpretation of the Pythagorean theorem? In particular, what do the quantities
𝑎2 , 𝑏 2 , and 𝑐 2 represent, and how are these three quantities related?
MP.2

The symbols 𝑎, 𝑏, and 𝑐 represent the sides of a right triangle, with 𝑐 being the hypotenuse. The
quantities 𝑎2 , 𝑏 2 , and 𝑐 2 represent the areas of the squares that are built on the sides of the triangle,
as shown in the figure below:
b2
a2
a2
c2

b2
a2
b2
The equation 𝑎2 + 𝑏 2 = 𝑐 2 tells us that the area of the large square is equal to the sum of the areas of
the two smaller squares.

The picture on the right even suggests a proof of this fact and is closely related to the work you’ll do below.
But our main interest today is in the study of oblique triangles, so let’s turn our attention back to those.

Altitudes played a key role in our work above. In particular, they allowed us to split the figure into right
triangles, which allowed us to utilize our knowledge of right triangle trigonometry. The key to understanding
the law of cosines from a geometric point of view is to draw squares on the sides of the triangle and then to
divide those squares by drawing in the altitudes as shown below:
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS

Can you find a way to describe each of the rectangles in the figure? Take several minutes to work with the
students around you on this task.

Let’s begin by choosing labels for each of the components of an oblique triangle.
γ
a1
γ
b2
b
a
b1
a2
β
β
α
c

α
c1
c2
Now we will attempt to describe each of the small segments in the figure.
𝑎1
𝑎2
𝑏1
𝑏2
𝑐1
𝑐2
𝑏 ∙ cos(𝛾)
𝑐 ∙ cos(𝛽)
𝑐 ∙ cos(𝛼)
𝑎 ∙ cos(𝛾)
𝑎 ∙ cos(𝛽)
𝑏 ∙ cos(𝛼)

We can use the information in the table above to analyze the areas in the figure.
VI
V
I
II
III
𝐼
𝐼𝐼
𝐼𝐼𝐼
𝐼𝑉
𝑉
𝑉𝐼
𝑎 ∙ 𝑎1
𝑎 ∙ 𝑎2
𝑐 ∙ 𝑐1
𝑐 ∙ 𝑐2
𝑏 ∙ 𝑏1
𝑏 ∙ 𝑏2
𝑎𝑏 ∙ cos(𝛾)
𝑎𝑐 ∙ cos(𝛽)
𝑐𝑎 ∙ cos(𝛽)
𝑐𝑏 ∙ cos(𝛼)
𝑏𝑐 ∙ cos(𝛼)
𝑏𝑎 ∙ cos(𝛾)


IV
From this table, it’s clear that the pink regions have equal areas, as do the blue and green regions. So
the equation 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 ∙ cos(𝛾) says that the area of the lower square is equal to the sum of
the areas of the upper squares minus the area of the two blue rectangles.
Fascinating, isn’t it? The law of cosines truly is a generalization of the Pythagorean theorem.
Lesson 9:
Law of Cosines
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M4
PRECALCULUS AND ADVANCED TOPICS


Note that if we shift our focus to a different square in the picture, we get a different version of this formula.
Here are all three versions:

𝑎2 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏 ∙ cos(𝛾)

𝑏 + 𝑐 2 = 𝑎2 + 2𝑏𝑐 ∙ cos(𝛼)

𝑐 2 + 𝑎2 = 𝑏 2 + 2𝑐𝑎 ∙ cos(𝛽)
Let’s do a few examples to put all of our hard work in this lesson to use!
Example (3 minutes)
In this example, students use the law of cosines to find an unknown side of a triangle, and then they use the law of
cosines to find an unknown angle in a triangle.

Let’s revisit the triangle you encountered at the start of the lesson. This time, find the missing side length by
directly applying the law of cosines.
70°
12
5
α

β
52 + 122 = 𝑐 2 + 2(5)(12) ∙ cos(70°)
25 + 144 = 𝑐 2 + 120 ∙ cos(70°)
169 = 𝑐 2 + 120 ∙ cos(70°)
𝑐 2 = 169 − 120 ∙ cos(70°)
𝑐 = √169 − 120 ∙ cos(70°)
𝑐 ≈ 11.3

The law of cosines can also be used to find the missing angles in the triangle. Give it a try!

52 + 11.32 = 122 + 2(5)(11.3) ∙ cos(𝛼)
25 + 127.69 = 144 + 113 ∙ cos(𝛼)
152.69 = 144 + 113 ∙ cos(𝛼)
8.69 = 113 ∙ cos(𝛼)
8.69
cos(𝛼) =
113
𝑚∠𝛼 ≈ 86°
Now that we know two of the angles, we can easily find the third: 𝛽 = 180° − 70° − 86° = 24°.
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Exercises 4–5 (3 minutes)
Instruct students to solve the problems below and then to compare their answers with a partner.
4.
Points 𝑩 and 𝑪 are located at the edges of a large body of water. Point 𝑨 is 𝟔 𝐤𝐦 from point 𝑩 and 𝟏𝟎 𝐤𝐦 from
point 𝑪. The angle formed between ̅̅̅̅
𝑩𝑨 and ̅̅̅̅
𝑨𝑪 is 𝟏𝟎𝟖°. How far apart are points 𝑩 and 𝑪?
B
6
108°
A
C
10
𝟔𝟐 + 𝟏𝟎𝟐 = 𝑩𝑪𝟐 + 𝟐(𝟔)(𝟏𝟎) ∙ 𝐜𝐨𝐬(𝟏𝟎𝟖°)
𝟏𝟑𝟔 = 𝑩𝑪𝟐 + 𝟏𝟐𝟎 ∙ 𝐜𝐨𝐬(𝟏𝟎𝟖°)
𝑩𝑪𝟐 = 𝟏𝟑𝟔 − 𝟏𝟐𝟎 ∙ 𝐜𝐨𝐬(𝟏𝟎𝟖°)
𝑩𝑪 ≈ 𝟏𝟑. 𝟐
5.
Use the law of cosines to find the value of 𝝑 in the triangle below.
Scaffolding:
 Be aware that it is common for
students to make order-ofoperations errors in this context.
13.9
6.4
θ
10.5
𝟏𝟎. 𝟓𝟐 + 𝟔. 𝟒𝟐 = 𝟏𝟑. 𝟗𝟐 + 𝟐(𝟏𝟎. 𝟓)(𝟔. 𝟒) ∙ 𝐜𝐨𝐬(𝝑)
−𝟒𝟐 = 𝟏𝟑𝟒. 𝟒 ∙ 𝐜𝐨𝐬(𝝑)
𝒎∠𝝑 ≈ 𝟏𝟎𝟖°
Lesson 9:
Law of Cosines
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 For example, students may write
121 − 96 ∙ cos(𝜗) = 25 ∙ cos(𝜗).
 To address this, ask students to
solve the equation 10 = 16 − 2𝑥
and then check to be sure that their
solution is correct. If they write
10 = 14𝑥, they will find that their
answer does not satisfy the
equation.
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Lesson Extension (optional)
γ
γ
12
5
b
a
c
c

At the start of the lesson, we observed that 7 < 𝑐 < 17. More generally, the triangle inequality tells us that
𝑏 − 𝑎 < 𝑐 < 𝑏 + 𝑎. Does the law of cosines support these constraints on the value of 𝑐?

What does the law of cosines have to say about the triangle on the right? Write an equation.


𝑎2 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏 ∙ cos(𝛾)
Find the minimum and maximum values of 𝑐 using what you know about the cosine function.

The cosine function has a maximum value of 1. In this case, we get the following:
𝑎2 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏
𝑎2 − 2𝑎𝑏 + 𝑏 2 = 𝑐 2
(𝑎 − 𝑏)2 = 𝑐 2
𝑐 =𝑎−𝑏
The cosine function reaches a value of 1 when the input is 0, so this corresponds to a zero angle
between the two sides of the triangle (i.e., they form a straight line).

We also know that the cosine function has a minimum value of −1. This time our analysis is as follows:
𝑎2 + 𝑏 2 = 𝑐 2 − 2𝑎𝑏
𝑎2 + 2𝑎𝑏 + 𝑏 2 = 𝑐 2
(𝑎 + 𝑏)2 = 𝑐 2
𝑎+𝑏 =𝑐
The cosine function reaches a value of −1 when the input is 180, so this corresponds to a straight angle
between the two sides of the triangle (i.e., they form a straight line here as well).

Let’s try another special case. What does the law of cosines have to say when the angle involved is a right
angle?

The cosine of 90° is 0, so in this case we have the following:
𝑎2 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏 ∙ cos(90°)
𝑎2 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏 ∙ 0
𝑎2 + 𝑏 2 = 𝑐 2 + 0
𝑎2 + 𝑏 2 = 𝑐 2

So the law of cosines is truly a generalization of the Pythagorean theorem. It’s nice to know that this formula
does what it is supposed to even in these special cases!
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Closing (2 minutes)

State the law of cosines.



𝑎2 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏 ∙ cos(𝛾)
Interpret each quantity in the formula geometrically.

The expressions 𝑎2 , 𝑏 2 , and 𝑐 2 represent the areas of squares.

The expression 2𝑎𝑏 ∙ cos(𝛾) represents the combined area of two rectangles with equal areas.
What is the law of cosines used for?

You can use the law of cosines to find unknown parts of an oblique triangle.
Exit Ticket (3 minutes)
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
Name
Date
Lesson 9: Law of Cosines
Exit Ticket
Use the law of cosines to solve the following problems.
1.
Josie wishes to install a new television that will take up 15° of her vertical field of view. Using the angle she wants,
she uses a laser measure and finds the distances from the wall to her couch are 8 ft. and 12 ft., but she does not
have any way to mark the spots on the wall. How tall is the television that she wants?
2.
Given the figure shown, find the height of the evergreen tree. Round your answers to the nearest thousandths.
Lesson 9:
Law of Cosines
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 9
M4
PRECALCULUS AND ADVANCED TOPICS
Exit Ticket Sample Solutions
Use the law of cosines to solve the following problems.
1.
Josie wishes to install a new television that will take up 𝟏𝟓° of her vertical field of view. Using the angle she wants,
she uses a laser measure and finds the distances from the wall to her couch are 𝟖 𝐟𝐭. and 𝟏𝟐 𝐟𝐭., but she does not
have any way to mark the spots on the wall. How tall is the television that she wants?
√𝟖𝟐 + 𝟏𝟐𝟐 − 𝟐 ⋅ 𝟖 ⋅ 𝟏𝟐 ⋅ 𝐜𝐨𝐬(𝟏𝟓°) ≈ 𝟒. 𝟕𝟒𝟖
Josie’s television will be more than 𝟒. 𝟕𝟒𝟖 𝐟𝐭. tall.
2.
Given the figure shown, find the height of the evergreen tree. Round your answers to the nearest thousandths.
√𝟏𝟐𝟓𝟐 + 𝟏𝟔𝟖𝟐 − 𝟐 ∙ 𝟏𝟐𝟓 ∙ 𝟏𝟔𝟖 ∙ cos(𝟐𝟕°) ≈ 𝟖𝟎. 𝟏𝟔𝟕
The height of the tree is approximately 𝟖𝟎. 𝟏𝟔𝟕 𝐟𝐭.
Problem Set Sample Solutions
The first few problems reframe the optional lesson extension as problems for the students to explore. If there was not
enough time to complete the lesson extension in class, then use these problems as an alternative. If there was time to
cover the material in class, then students may want to start with Problem 5, or Problems 1–4 can be used to emphasize
the extension.
1.
Consider the case of a triangle with sides 𝟓, 𝟏𝟐, and the angle between them 𝟗𝟎°.
a.
What is the easiest method to find the missing side?
Pythagorean theorem
b.
What is the easiest method to find the missing angles?
Right triangle trigonometry
c.
Can you use the law of cosines to find the missing side? If so, perform the calculations. If not, show why not.
Yes, this is an example of SAS. The missing side is 𝟏𝟑.
d.
Can you use the law of cosines to find the missing angles? If so, perform the calculations. If not, show why
not.
Yes. The measures of the missing angles are 𝟐𝟐. 𝟔𝟐° and 𝟔𝟕. 𝟑𝟖°.
Lesson 9:
Law of Cosines
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
e.
Consider a triangle with sides 𝒂, 𝒃, and the angle between them 𝟗𝟎°. Use the law of cosines to prove a wellknown theorem. State the theorem.
𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 − 𝟐𝒂𝒃 𝐜𝐨𝐬(𝟗𝟎°)
𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 − 𝟐𝒂𝒃 ⋅ 𝟎
𝒄𝟐 = 𝒂 𝟐 + 𝒃 𝟐
The Pythagorean theorem
f.
Summarize what you have learned in parts (a) through (e).
The law of cosines appears to be a more general form of the Pythagorean theorem, one that can work with
any angle between two sides of a triangle, not just a right angle.
2.
Consider the case of ̅̅̅̅
𝑪𝑨 and ̅̅̅̅
𝑪𝑩 of lengths 𝟓 and 𝟏𝟐, respectively, with 𝒎∠𝑪 = 𝟏𝟖𝟎°.
a.
Is 𝑨𝑩𝑪 a triangle?
̅̅̅̅ is a line segment containing 𝑪. △ 𝑨𝑩𝑪 does not exist.
𝑨𝑩
b.
What is the easiest method to find the distance between 𝑨 and 𝑩?
Add the lengths of the two line segments partitioning ̅̅̅̅
𝑨𝑩: 𝟓 + 𝟏𝟐 = 𝟏𝟕.
c.
Can you use the law of cosines to find the distance between 𝑨 and 𝑩? If so, perform the calculations. If not,
show why not.
Yes. 𝟓𝟐 + 𝟏𝟐𝟐 − 𝟐 ⋅ 𝟓 ⋅ 𝟏𝟐 𝐜𝐨𝐬(𝟏𝟖𝟎°) = 𝟐𝟓 + 𝟏𝟒𝟒 + 𝟏𝟐𝟎 = 𝟐𝟖𝟗 = 𝟏𝟕𝟐
The distance between 𝑨 and 𝑩 is 𝟏𝟕 units.
d.
Summarize what you have learned in parts (a) through (c).
The law of cosines can work even when the line segments are a linear pair and do not form a triangle.
3.
Consider the case of ̅̅̅̅
𝑪𝑨 and ̅̅̅̅
𝑪𝑩 of lengths 𝟓 and 𝟏𝟐, respectively, with 𝒎∠𝑪 = 𝟎°.
a.
Is 𝑨𝑩𝑪 a triangle?
̅̅̅̅ is a line segment containing 𝑨. △ 𝑨𝑩𝑪 does not exist.
𝑪𝑩
b.
What is the easiest method to find the distance between 𝑨 and 𝑩?
Subtract the two line segments: 𝟏𝟐 − 𝟓 = 𝟕.
c.
Can you use the law of cosines to find the distance between 𝑨 and 𝑩? If so, perform the calculations. If not,
show why not.
Yes, the law of cosines still applies. 𝟓𝟐 + 𝟏𝟐𝟐 − 𝟐 ⋅ 𝟓 ⋅ 𝟏𝟐 𝐜𝐨𝐬(𝟎°) = 𝟐𝟓 + 𝟏𝟒𝟒 − 𝟏𝟐𝟎 = 𝟒𝟗 = 𝟕𝟐
The distance between 𝑨 and 𝑩 is 𝟕 units.
d.
Summarize what you have learned in parts (a) through (c).
The law of cosines applies even when the line segments have the same starting point and one lies on top of
the other.
Lesson 9:
Law of Cosines
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 9
M4
PRECALCULUS AND ADVANCED TOPICS
4.
̅̅̅̅ and 𝑪𝑩
̅̅̅̅ of lengths 𝟓 and 𝟏𝟐, respectively, with 𝒎∠𝑪 > 𝟏𝟖𝟎°.
Consider the case of 𝑪𝑨
a.
Is the law of cosines consistent in being able to calculate the length of ̅̅̅̅
𝑨𝑩 even using an angle this large? Try
it for 𝒎∠𝑪 = 𝟐𝟎𝟎°, and compare your results to the triangle with 𝒎∠𝑪 = 𝟏𝟔𝟎°. Explain your findings.
Yes, the law of cosines is still able to calculate the length of ̅̅̅̅
𝑨𝑩:
𝒄𝟐 = 𝟓𝟐 + 𝟏𝟐𝟐 − 𝟐 ⋅ 𝟓 ⋅ 𝟏𝟐 𝐜𝐨𝐬(𝟐𝟎𝟎°)
≈ 𝟐𝟖𝟏. 𝟕𝟔
Thus, 𝒄 ≈ 𝟏𝟔. 𝟕𝟖𝟔, which represents the length of line segment 𝑨𝑩. We get the same result for 𝐜𝐨𝐬(𝟏𝟔𝟎°),
which makes sense since it is true that 𝐜𝐨𝐬(𝜽) = 𝐜𝐨𝐬(𝟑𝟔𝟎° − 𝜽).
b.
Consider what you have learned in Problems 1–4. If you were designing a computer program to be able to
measure sides and angles of triangles created from different line segments and angles, would it make sense
to use the law of cosines or several different techniques depending on the shape? Would a computer
program created from the law of cosines have any errors based on different inputs for the line segments and
angle between them?
The benefits of using the law of cosines would be that there would be no need for logic involving different
cases to be programmed, and there would be no exceptions to the formula. The law of cosines would work
even when the shape formed was not a triangle or if the shape was formed using an angle greater than 𝟏𝟖𝟎°.
The triangle inequality theorem would need to be used to verify whether the side lengths could represent
those of a triangle. There would be no errors for two line segments and the angle between them.
5.
Consider triangles with the following measurements. If two sides are given, use the law of cosines to find the
measure of the third side. If three sides are given, use the law of cosines to find the measure of the angle between
𝒂 and 𝒃.
a.
𝒂 = 𝟒, 𝒃 = 𝟔, 𝒎∠𝑪 = 𝟑𝟓°
𝒄 ≈ 𝟑. 𝟓𝟔
b.
𝒂 = 𝟐, 𝒃 = 𝟑, 𝒎∠𝑪 = 𝟏𝟏𝟎°
𝒄 ≈ 𝟒. 𝟏𝟒
c.
𝒂 = 𝟓, 𝒃 = 𝟓, 𝒎∠𝑪 = 𝟑𝟔°
𝒄 ≈ 𝟑. 𝟎𝟗
d.
𝒂 = 𝟕. 𝟓, 𝒃 = 𝟏𝟎, 𝒎∠𝑪 = 𝟗𝟎°
𝒄 ≈ 𝟏𝟐. 𝟓
e.
𝒂 = 𝟒. 𝟒, 𝒃 = 𝟔. 𝟐, 𝒎∠𝑪 = 𝟗°
𝒄 ≈ 𝟏. 𝟗𝟖
f.
𝒂 = 𝟏𝟐, 𝒃 = 𝟓, 𝒎∠𝑪 = 𝟒𝟓°
𝒄 ≈ 𝟗. 𝟏𝟕
g.
𝒂 = 𝟑, 𝒃 = 𝟔, 𝒎∠𝑪 = 𝟔𝟎°
𝒄 ≈ 𝟓. 𝟐
Lesson 9:
Law of Cosines
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from PreCal-M4-TE-1.3.0-10.2015
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
h.
𝒂 = 𝟒, 𝒃 = 𝟓, 𝒄 = 𝟔
𝒎∠𝑪 ≈ 𝟖𝟐. 𝟖𝟐°
i.
𝒂 = 𝟏, 𝒃 = 𝟏, 𝒄 = 𝟏
𝒎∠𝑪 = 𝟔𝟎°
j.
𝒂 = 𝟕, 𝒃 = 𝟖, 𝒄 = 𝟑
𝒎∠𝑪 ≈ 𝟐𝟏. 𝟕𝟗°
k.
𝒂 = 𝟔, 𝒃 = 𝟓. 𝟓, 𝒄 = 𝟔. 𝟓
𝒎∠𝑪 ≈ 𝟔𝟖. 𝟔𝟖°
l.
𝒂 = 𝟖, 𝒃 = 𝟓, 𝒄 = 𝟏𝟐
𝒎∠𝑪 ≈ 𝟏𝟑𝟑. 𝟒𝟑°
m.
𝒂 = 𝟒. 𝟔, 𝒃 = 𝟗, 𝒄 = 𝟏𝟏. 𝟗
𝒎∠𝑪 ≈ 𝟏𝟏𝟖. 𝟒𝟓°
6.
A trebuchet launches a boulder at an angle of elevation of 𝟑𝟑° at a force of 𝟏, 𝟎𝟎𝟎 𝐍. A strong gale wind is blowing
against the boulder parallel to the ground at a force of 𝟑𝟒𝟎 𝐍. The figure is shown below.
a.
What is the force in the direction of the boulder’s path?
Since the wind is blowing in the opposite direction, finding the sum of the two vectors is similar to finding the
difference between two vectors if the wind is in the same direction. Thus, we are finding the third side of a
triangle (the diagonal of the parallelogram).
𝒄𝟐 = 𝟏𝟎𝟎𝟎𝟐 + 𝟑𝟒𝟎𝟐 − 𝟐 ⋅ 𝟏𝟎𝟎𝟎 ⋅ 𝟑𝟒𝟎 𝐜𝐨𝐬(𝟑𝟑°)
≈ 𝟕𝟑𝟖. 𝟒𝟓
The boulder is traveling with an initial force of 𝟕𝟑𝟖. 𝟒𝟓 𝐍.
b.
What is the angle of elevation of the boulder after the wind has influenced its path?
The angle between the original trajectory and the new trajectory is 𝟏𝟒. 𝟓𝟐°, so the new angle of elevation is
𝟒𝟕. 𝟓𝟐°.
Lesson 9:
Law of Cosines
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This file derived from PreCal-M4-TE-1.3.0-10.2015
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Lesson 9
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
PRECALCULUS AND ADVANCED TOPICS
7.
Cliff wants to build a tent for his son’s graduation party. The tent is a regular pentagon, as illustrated below. How
much guy-wire (show in blue) does Cliff need to purchase to build this tent? Round your answers to the nearest
thousandths.
𝑨𝑩 = √𝟐𝟎𝟐 + 𝟏𝟓𝟐 − 𝟐 ∙ 𝟐𝟎 ∙ 𝟏𝟓 ∙ 𝐜𝐨𝐬(𝟕𝟓°) ≈ 𝟐𝟏. 𝟔𝟕𝟑
𝑬𝑪 = √𝟒𝟎𝟐 + 𝟑𝟎𝟐 = 𝟓𝟎
𝟒𝟎
𝒎∠𝑪𝑬𝑫 = 𝐚𝐫𝐜𝐭𝐚𝐧 ( ) = 𝟓𝟓. 𝟏𝟑𝟎°
𝟑𝟎
𝒎∠𝑩𝑬𝑪 = 𝟏𝟖𝟎° − 𝟕𝟓° − 𝟓𝟓. 𝟏𝟑𝟎° = 𝟒𝟗. 𝟖𝟕𝟎°
𝑩𝑪 = √𝟏𝟓𝟐 + 𝟓𝟎𝟐 − 𝟐 ∙ 𝟏𝟓 ∙ 𝟓𝟎 ∙ 𝐜𝐨𝐬(𝟒𝟗. 𝟖𝟕𝟎°) ≈ 𝟒𝟏. 𝟗𝟑𝟏
𝑨𝑩 + 𝑩𝑪 = 𝟔𝟑. 𝟔𝟎𝟒
(𝟓)(𝟔𝟑. 𝟔𝟎𝟒) = 𝟑𝟏𝟖. 𝟎𝟐𝟎
Total guy-wire is 𝟑𝟏𝟖. 𝟎𝟐𝟎 feet long.
8.
A roofing contractor needs to build roof trusses for a house. The side view of the truss is shown below. Given that
𝑮 is the midpoint of ̅̅̅̅
𝑨𝑩, 𝑬 is the midpoint of ̅̅̅̅
𝑨𝑮, 𝑰 is the midpoint of ̅̅̅̅
𝑮𝑩, 𝑨𝑩 = 𝟑𝟐 𝐟𝐭., 𝑨𝑫 = 𝟔 𝐟𝐭., 𝑭𝑪 = 𝟓 𝐟𝐭., and
𝒎∠𝑨𝑮𝑪 = 𝟗𝟎°. Find 𝑫𝑬, 𝑬𝑭, and 𝑭𝑮. Round your answers to the nearest thousandths.
𝟖
𝒎∠𝑪𝑨𝑮 = 𝐚𝐫𝐜𝐭𝐚𝐧 ( ) = 𝟐𝟔. 𝟓𝟔𝟓°
𝟏𝟔
𝑫𝑬 = √𝟔𝟐 + 𝟖𝟐 − 𝟐 ∙ 𝟔 ∙ 𝟖 ∙ 𝐜𝐨𝐬(𝟐𝟔. 𝟓𝟔𝟓°) ≈ 𝟑. 𝟕𝟔𝟎
𝒎∠𝑨𝑪𝑮 = 𝟏𝟖𝟎° − 𝟗𝟎° − 𝟐𝟔. 𝟓𝟔𝟓° = 𝟔𝟑. 𝟒𝟑𝟓°
𝑭𝑮 = √𝟓𝟐 + 𝟖𝟐 − 𝟐 ∙ 𝟓 ∙ 𝟖 ∙ 𝐜𝐨𝐬(𝟔𝟑. 𝟒𝟑𝟓°) ≈ 𝟕. 𝟐𝟗𝟓
𝟖𝟐 + 𝟕. 𝟐𝟗𝟓𝟐 − 𝟓𝟐
𝒎∠𝑭𝑮𝑪 = 𝐚𝐫𝐜𝐜𝐨𝐬 (
) ≈ 𝟑𝟕. 𝟖𝟎𝟖°
𝟐 ∙ 𝟖 ∙ 𝟕. 𝟐𝟗𝟓
𝒎∠𝑭𝑮𝑬 = 𝟗𝟎° − 𝟑𝟕. 𝟖𝟎𝟖° = 𝟓𝟐. 𝟏𝟗𝟐°
𝑬𝑭 = √𝟖𝟐 + 𝟕. 𝟐𝟗𝟓𝟐 − 𝟐 ∙ 𝟖 ∙ 𝟕. 𝟐𝟗𝟓 ∙ 𝐜𝐨𝐬(𝟓𝟐. 𝟏𝟗𝟐°) ≈ 𝟔. 𝟕𝟓𝟖
𝑫𝑬 is approximately 𝟑. 𝟕𝟔𝟎 feet long, 𝑭𝑮 is approximately 𝟕. 𝟐𝟗𝟓 feet long, and 𝑬𝑭 is approximately 𝟔. 𝟕𝟓𝟖 feet
long.
Lesson 9:
Law of Cosines
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This file derived from PreCal-M4-TE-1.3.0-10.2015
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