The Law of Cosines

```The Law of Cosines
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Intro
Law of Sines vs. Cosines
Motivation
Proof
My Video
Information and Application
Examples
Quiz
Extra Help and References
Introduction/Orientation

In addition to the Law of Sines, the Law
of Cosines is a trigonometric method,
used for solving oblique (non-right)
triangles.
Introduction/Orientation (cont.)
The objective of each problem is to
solve for all side and angle measures of
a triangle.
 At the end, a quiz will be given, so be
sure to ask questions and take notes
along the way!
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What Makes The Law of Cosines
different from the Law of Sines?
The Law of Sines (used in cases of
AAS, ASA, and SSA) may involve an
ambiguous case (SSA), where the final
result could be a possibility of 1 triangle,
2 triangles, or even no triangles in the
problem.
 A problem involving The Law of Cosines
guarantees only one triangle, regardless
of the case in which it is used.
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What will you be doing with the
Law of Cosines?
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Using the Law of Cosines, you will be
able to apply this formula to solve for
oblique triangles. This will go more indepth than right-triangle trigonometric
problems, as they truly come down to
knowing the formulas, rather than a
simple ratio to be used.
What will you be doing with the
Law of Cosines? (cont.)
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In order to work with applications of
word problems, dealing with geometric
measurements are relevant to your
knowledge, especially if you go into any
kind of math, science, or engineering.
The Law of Cosines is a good way to
start learning.
Law of Cosines Proof
The Law of Cosines- Any given triangle
with side lengths, a, b, and c, with
opposing angles, A, B, and C.
 Take your triangle and divide it into two
right triangles. From there, since we
have right triangles, we can use the
Pythagorean Theorem and prove our
formula.
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Information
Formulas
 a2 = b2 + c2 – 2bc(cos(A))
 b2 = a2 + c2 – 2ac(cos(B))
 c2 = a2 + b2 – 2ab(cos(C))
Formulas (cont.)
cos A = (b2 + c2 – a2) / (2bc)
 cos B = (a2 - b2 + c2) / (2ac)
 cos C = (a2 + b2 - c2) / (2ab)
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NOTE
Once you find two angles, sum them
and then take 180 and subtract the sum
of the two angles that you have.
 In addition, lower-case letters represent
side measures, and upper-case for
angle measures.
 In Slides 17 and 18, you may click on
the image in the middle of the page that
will take you to an example problem.
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Application
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There are two cases in which the Law of
Cosines can be used.
Side-Side-Side (SSS)
Side-Angle-Side (SAS)
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Let us take a look at some example
problems. When you answer a question
correctly, please click the triangle on the
left-hand side of the “correct answer”
slide for a solution.
Example 1
Given a triangle with side measures, a = 4,
b = 5, and c = 7, Solve for the angles
(NOTE: Angles are measured in degrees).
A. A = 34.0, B = 44.4, C = 101.6
B. A = 0.60, B = 0.80, C = 178.6
C. A = 0.8, B = 0.7, C = 178.5
NO, TRY AGAIN.

NO, TRY AGAIN.
You forgot to take the inverse
cosine of the first angle to solve
for the unknown. Hence, it threw off your
entire problem.
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Solution
A = arccos((25+49-16)/((2)(5)(7))) = 34.0
(approximately)
 B = arccos((16-25+49)/((2)(4)(7))) =
44.4 (approximately)
 C = 180-34.0-44.4 = 101.6
(approximately)
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Example 2
Given a triangle ABC with b = 5, c = 7, and
A = 120, solve for the triangle.
A. a = 109, B = 0.72, C = 59.28
B. a = 10.4, B = 24.8, C = 35.3
C. a = 4.1, B = 1, C = 57
NO, TRY AGAIN.
For the missing side, you forgot to
take the square root of the
In addition, when solving for B, you forgot
to take the inverse cosine of your fraction.
NO, TRY AGAIN.
Your calculator is NOT in degree
mode.
Solution
a = sqrt(25 + 49 – 70 cos(120)) =
sqrt(109) = 10.4
 B = arccos(((10.4)^2 – 25 +
49)/(2*7*10.4)) = 24.8
 C = 180 – 24.8 – 120 = 35.2

Example 3
A person leaves home and walks 5 mi
east, then goes 3 mi northeast. How far is
he from home?
A. 7.43 mi
B. 7.99 mi
C. 55.21 mi
NO, TRY AGAIN.
Calculator is in the wrong mode.
NO, TRY AGAIN.
You forgot to take the square root
of both sides.
Solution
Going NE, thinking about it from the
perspective of a compass, as from North
to East, the right angle is bisected.
Therefore angle C = 135. Now that the
angle is found, we can plug in everything
we need to solve for the missing side.
sqrt(9+25-30cos(135)) = 7.43 mi
Example 4
Two airplanes leave an airport, and the angle
between their flight paths is 40. An hour later,
one plane has traveled 300 mi, while the
other has traveled 200 mi. How far apart are
the planes at this time?
A. 38,075 mi
B. 458 mi
C. 195 mi
D. None of the Above
NO, TRY AGAIN.
You forgot to take the square root
of both sides.
NO, TRY AGAIN.
Use the formulas given earlier,
and apply the numbers you know
into the problem. Then you can punch
everything into your calculator and solve
Solution
This is similar to the previous word
problem. We are given a side of 200, and
another side of 300. In addition, we have
the angle of 40 in between the two sides.
So, in order to find the distance, simply
solve for the missing side.
sqrt(40,000+90,000-120,000(cos(40))) =
195 mi
1. When using your calculator, make
sure that it is in Degree mode, and NOT
 2. IMPORTANT TO NOTE!!! The
Pythagorean Theorem CANNOT be
used as these problems will not deal
with right triangles, so do not do it!
 3. In most cases, sketching a diagram of
the problem helps.
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Quiz
There is no time limit. However, you must
pass this quiz with 100% accuracy. You
may take this quiz as many times as you
wish. Good Luck!
Question 1
Triangle ABC with side measures, a = 9, b
= 12, c = 13.5. Solve for the angle
measures.
A. A = 40.8, B = 60.6, C = 78.6
B. A = 1.18, B = 1.06, C = 177.76
NO, TRY AGAIN.
Your Calculator is in the Wrong
Mode!
Question 2
Two ships leave the same port at noon. One
ship travels at a bearing of N 53 W at 12 mph,
and the other ship travels at a bearing of S 67
W at 16 mph. How far apart will the ships be
at 3 p.m.?
A. 43 mi
B. 1,872 mi
C. 95 mi
D. 100 mi
NO, TRY AGAIN.
Do not forget to take the square
Root of both sides!
NO, TRY AGAIN.
Your Calculator is in the Wrong
Mode!
NO, TRY AGAIN.
calculator, using the formulas
given earlier in the presentation.
Question 3
You are given ASA. Can you use the Law
of Cosines?
A. Yes
B. No
NO, TRY AGAIN.
Can ASA be used for the Law of
Cosines?
Question 4
Given a = 4, b = 5, and C = ((31π)/(180)),
find c.
A. 2.10
B. No Solution
C. 2.59
NO, TRY AGAIN.
degree mode!
NO, TRY AGAIN.
again!
Question 5
Besides the Law of Cosines, what is the
other way to solve an oblique triangle?
The Law of _________.
A. Tangents
B. Secants
C. Cotangents
D. Sines
E. Cosecants
NO, TRY AGAIN.
Look back here.
Question 6
a = 19, b = 26, and C = 42. Solve the
triangle.
A. c = 37.8, A = 0.36, B = 137.1
B. c = 17.4, A = 20.8, B = 117.2
C. c = 302.1, A = 0.9, B = No Solution
NO, TRY AGAIN.
Again, pay attention to your
calculator’s mode.
NO, TRY AGAIN.
Use the formulas given earlier.
Then punch in the given
Question 7
a = 20 m, b = 30 m, and c = 40 m. Solve
for the triangle.
A. A = 0.51, B = 0.81, C = 178.68
B. A = 30, B = 30, C = 120
C. A = 29, B = 46.6, C = 104.4
NO, TRY AGAIN.
NO, TRY AGAIN.
Use the formulas given earlier.
Then punch in the given
Question 8
Given b = 12, c = 20, and A = (π/4),
Manipulate the formula to solve for a.
A. sqrt(144+400-480cos(π/4))
B. sqrt(144+400-480cos(45))
C. A. and B. are the same answer
NO, TRY AGAIN.
Hint: Remember to convert the
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embedded videos on this PowerPoint.
References
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mathworld.wolfram.com/LawofCosines.html
public.asu.edu/~kamman/notes/trigonometry/Law-of-Cosines-r10-14-2007.html
kkuniyuk.com/M1410602.pdf
mathproofs.blogspot.com/2006/06/law-of-cosines.html
regentsprep.org/Regents/math/algtrig/ATT12/lawcosinespractice.htm