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Law of Cosine Ee

The Law of Cosines is used to find the length of any side or angle of a triangle when two sides and the included angle or when all three sides are known. The basic formula is c^2 = a^2 + b^2 - 2abcos(C), where a and b are the lengths of two sides and C is the angle between them, or with the letters changed to refer to the other sides and angles. It can be used in cases where two sides and the included angle (SAS) or all three sides (SSS) are known. The appropriate trigonometric function to use depends on what information is known about the triangle.

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Myko Brian Santillan
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66 views17 pages

Law of Cosine Ee

The Law of Cosines is used to find the length of any side or angle of a triangle when two sides and the included angle or when all three sides are known. The basic formula is c^2 = a^2 + b^2 - 2abcos(C), where a and b are the lengths of two sides and C is the angle between them, or with the letters changed to refer to the other sides and angles. It can be used in cases where two sides and the included angle (SAS) or all three sides (SSS) are known. The appropriate trigonometric function to use depends on what information is known about the triangle.

Uploaded by

Myko Brian Santillan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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6.

2 Law of Cosines

fguilbert
In any triangle, the
Law of Cosines says:
c2 = a2 + b 2 - 2ab Cos C
C

A B
Side a is opposite < A.
Side b is opposite < B
Sice c is opposite < C
2 2 2
c = a + b - 2ab Cos C
C
b a
A c B
fguilbert
When is the Law of Cos
used?

C
b a
A c
fguilbert B
CASE 1 Two sides and
the included angle are
known (SAS).
CASE 2: Three sides are
known (SSS).
The basic form of the
Law of Cosines is :

c2 = a 2 + b 2 - 2ab Cos C
but other letters are changed
Although this is < C, you
can use the formula for A or B
c  a  b  2 ab cos C
2 2 2

b  a  c  2 ac cos 
2 2 2

a  b  c  2bc cos A
2 2 2
fguilbert
Find C,B and a

3 C a

40
A B
4
𝟐 𝟐 𝟐
𝒂 = 𝒃 + 𝒄 − 𝟐𝒃𝒄 𝒄𝒐𝒔𝑨
Cont…
𝟐 𝟐 𝟐
𝒂 = 𝒃 + 𝒄 − 𝟐𝒃𝒄 𝒄𝒐𝒔𝑨
𝒂𝟐 = 𝟑𝟐+ 𝟒 𝟐
− 𝟐(𝟑)(𝟒) 𝒄𝒐𝒔40
𝟐
𝒂 = 𝟗 + 𝟏𝟔 − 𝟐𝟒 𝒄𝒐𝒔40
𝟐
𝒂 = 𝟐𝟓 − 𝟐𝟒 𝒄𝒐𝒔40
a= 𝟐𝟓 − 𝟐𝟒 𝒄𝒐𝒔40
a= 𝟐. 𝟓𝟕
3 C a=2.572

40
A B
4
Solving for C
𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 − 𝟐𝒂𝒃 𝒄𝒐𝒔C
𝟒𝟐 = 𝟐. 𝟓𝟕𝟐𝟐 + 𝟑𝟐 − 𝟐(𝟐. 𝟓𝟕𝟐)(𝟑) 𝒄𝒐𝒔C
16= 𝟔. 𝟔𝟏 + 𝟗 − 𝟏𝟓. 𝟒𝟐𝒄𝒐𝒔C
𝟔.𝟔𝟐+𝟗−𝟏𝟔
cos C=
𝟏𝟓.𝟒𝟑𝟐
−𝟎. 𝟑𝟖
𝑪 = 𝒄𝒐𝒔−𝟏 = 𝟗𝟏. 𝟒𝟏
𝟏𝟓. 𝟒𝟑𝟐
Solving for B
𝑩 = 𝟏𝟖𝟎 − 𝟒𝟎 − 𝟗𝟏. 𝟒𝟏
𝑩 = 𝟒𝟖. 𝟓𝟗
Find <A A
17 29
B 35 C

c2 2 2
= + - 2ab Cos C
a b
2 2 2
a = b + c - 2bc Cos A
35 =29 +17
2 2 2 - 2(29)(17)cosA
Find <A A
17 29
B 35 C
35 =29 +17
2 2 2 - 2(29)(17)cosA
1225= 1130 - 986cosA
95 = - 986cosA
Find <A A
17 29
B 35 C
95
___ = - 986cosA
___
-986 -986
-.0963 = cos A
0
95.5 = A fguilbert
Find <A A
17 29
B 35 C

To find B or C now, it would be


quicker to use the Sine Law.
Both B and C must be acute.
How to Choose the right one.

SAS Law COS


SSS Law COS
ASA Law SINES
SSA Law SINES (0,1,2
triangles are possible)
thought for the day

One who lacks


the courage
to start,
is already
finished .

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