Transcript 8.2 The Law of Cosines

CHAPTER 8:
Applications of Trigonometry
8.1
8.2
8.3
8.4
8.5
8.6
The Law of Sines
The Law of Cosines
Complex Numbers: Trigonometric Form
Polar Coordinates and Graphs
Vectors and Applications
Vector Operations
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8.2
The Law of Cosines


Use the law of cosines to solve triangles.
Determine whether the law of sines or the law of
cosines should be applied to solve a triangle.
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Law of Cosines
The Law of Cosines
In any triangle ABC,
B
a  b  c  2bc cos A
2
2
2
b  a  c  2ac cos B
2
2
c
2
c 2  a 2  b 2  2ab cosC
a
A
b
C
Thus, in any triangle, the square of a side is the sum of
the squares of the other two sides, minus twice the
product of those sides and the cosine of the included
angle. When the included angle is 90º, the law of
cosines reduces to the Pythagorean theorem.
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Slide 8.2 - 4
When to use the Law of Cosines
The Law of Cosines is used to solve triangles given
two sides and the included angle (SAS) or given three
sides (SSS).
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Slide 8.2 - 5
Example
In !ABC, a = 32, c = 48, and B = 125.2º. Solve the
triangle.
Solution:
Draw and label a triangle.
A?
a  32
B  125.2º b  ?
C ?
c  48
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Slide 8.2 - 6
Example
Solution continued
Use the law of cosines to find the third side, b.
b2  a2  c2  2accos B
b 2  322  482  2(32)(48)cos125.2º
b2  5098.8
b  71
We need to find the other two angle measures. We can
use either the law of sines or law of cosines. Using the
law of cosines avoids the possibility of the ambiguous
case. So use the law of cosines.
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Slide 8.2 - 7
Example
Solution continued
Find angle A.
a2  b2  c2  2bccos A
32  71  48  2  71 48cos A
1024  5041 2304  6816 cos A
2
2
2
6321  6816cos A
cos A  0.9273768
A  22.0º
Now find angle C.
C ≈ 180º – (125.2º + 22º)
C ≈ 32.8º
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Thus,
A  22.0º
B  125.2º
C  32.8º
a  32
b  71
c  48
Slide 8.2 - 8
Example
Solve !RST, r = 3.5, s = 4.7, and t = 2.8.
Solution:
Draw and label a triangle.
R?
S?
T ?
r  3.5
s  4.7
t  2.8
s 2  r 2  t 2  2rt cosS
4.7   3.5   2.8 
2
2
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2
 2  3.5  2.8 cosS
Slide 8.2 - 9
Example
Solution continued
2
2
2
3.5   2.8   4.7 

cos S 
2 3.5 2.8 
cosS  0.1020408
S  95.86º
Similarly, find angle R.
r 2  s 2  t 2  2st cos R
3.5   4.7   2.8   2  4.7  2.8 cos R
2
2
2
4.7   2.8   3.5 

cos S 
2 4.7 2.8 
2
2
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2
Slide 8.2 - 10
Example
Solution continued
cos R  0.6717325
R  47.80º
Now find angle T.
T ≈ 180º – (95.86º + 47.80º) ≈ 36.34º
Thus,
R  47.80º
S  95.86º
r  3.5
s  4.7
T  36.34º
t  2.8
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Slide 8.2 - 11
Example
Knife makers know that the bevel of the
blade (the angle formed at the cutting edge
of the blade) determines the cutting
characteristics of the knife. A small bevel
like that of a straight razor makes for a keen
edge, but is impractical for heavy-duty
cutting because the edge dulls quickly and is
prone to chipping. A large bevel is suitable
for heavy-duty work like chopping wood.
Survival knives, being universal in
application, are a compromise between
small and large bevels. The diagram
illustrates the blade of a hand-made Randall
Model 18 survival knife. What is its bevel?
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Slide 8.2 - 12
Example
Solution:
Use the law of cosines to find angle A.
a2  b2  c2  2bccos A
0.5   2   2 
2
2
2
 2  2  2 cos A
4  4  0.25
cos A 
8
cos A  0.96875
A  14.36º
The bevel is approximately 14.36º.
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Slide 8.2 - 13