Transcript Section 3.5 - Canton Local

Chapter 7
Applications of
Trigonometric
Functions
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All rights reserved
© 2010
2011 Pearson Education, Inc. All rights reserved
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SECTION 7.3
The Law of Cosines
OBJECTIVES
1
2
3
Derive the Law of Cosines.
Use the Law of Cosines to solve SAS
triangles.
Use the Law of Cosines to solve SSS
triangles.
THE LAW OF COSINES
In triangle ABC, with sides of lengths a, b, and c,
a 2  b 2  c 2  2bc cos A,
b  c  a  2ca cos B,
and
2
2
2
c  a  b  2ab cosC.
2
2
2
In words, the square of any side of a triangle is equal
to the sum of the squares of the length of the other
two sides less twice the product of the lengths of the
other sides and the cosine of their included angle.
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DERIVATION OF THE LAW OF COSINES
a  d C , B 
a 2  d C , B 
2
a 2  b cos A  c   b sin A  0
2
2
a  b cos A  2bc cos A  c  b sin A
2
2
2
2
2
2
a 2  b 2  c 2  2bc cos A
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SOLVING SAS TRIANGLES
Step 1
Use the appropriate form of the Law
of Cosines to find the side opposite
the given angle.
Step 2
Use the Law of Sines to find the
angle opposite the shorter of the two
given sides. Note that this angle is
always an acute angle.
Step 3
Use the angle sum formula to find
the third angle.
Step 4
Write the solution.
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EXAMPLE 1
Solving SAS Triangles
Solve triangle ABC with a = 15 inches,
b = 10 inches, and C = 60º. Round each answer
to the nearest tenth.
Solution
Step 1 Find side c opposite the given angle C.
c 2  a 2  b 2  2ab cos C
c  15   10   2 15 10  cos 60°
2
2
2
1
c  225  100  2 15 10     175
2
2
c  175  13.2
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EXAMPLE 1
Solving the SAS Triangles
Solution continued
Step 2 Find the angle B opposite side b.
sin B sin C

b
c
b sin C
sin B 
c
10sin 60°
sin B 
175
1  10sin 60° 
B  sin 
  40.9°
175 

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EXAMPLE 1
Solving the SAS Triangles
Solution continued
Step 3 Use the angle sum formula to find the
third angle.
A  180°  60°  40.9°  79.1°
Step 4 The solution of triangle ABC is
A ≈ 79.1°
a = 15 inches
B ≈ 40.9°
b = 10 inches
C = 60°
c ≈ 13.2 inches
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EXAMPLE 2
Using the Law of Cosines
Suppose that a Boeing 747 is flying over
Disney World headed due south at 552 miles
per hour. Twenty minutes later, an F-16 passes
over Disney World with a bearing of N 37º E
at a speed of 1250 mi/hr.
Find the distance between the two planes three
hours after the F-16 passes over Disney World.
Round the answer to the nearest tenth.
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EXAMPLE 2
Using the Law of Cosines
Solution
Suppose the F-16 has been
traveling for t hours after
passing over Disney World.
Then, because the Boeing 747
had a head start of
1
20 minutes = hour, the
3
Boeing 747 has been
 1
traveling  t   hours
 3
due south.
The distance between the two planes is d.
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EXAMPLE 2
Using the Law of Cosines
Solution continued
Using the Law of Cosines in triangle FDB, we have
d  1250t 
2
2
2

 1 
 1
 552  t     2 1250t   552  t   cos143º.
 3 
 3

Substitute t = 3.
d  28,469,270.04
2
d  5335.7 miles
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SOLVING SSS TRIANGLES
Step 1
Use the Law of Cosines to find the
angle opposite the given side.
Step 2
Use the Law of Sines to find either
of the two remaining acute angles.
Step 3
Use the angle sum formula to find
the third angle.
Step 4
Write the solution.
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EXAMPLE 3
Solving the SSS Triangles
Solve triangle ABC with a = 3.1 feet,
b = 5.4 feet, and c = 7.2 feet. Round answers to
the nearest tenth.
Solution
Step 1 Because c is the longest side, find C.
c 2  a 2  b 2  2ab cos C
2ab cos C  a 2  b 2  c 2
3.1   5.4    7.2 

a b c
cos C 

 0.39
2ab
2  3.1 5.4 
2
2
2
2
2
2
C  cos 1  0.39   113°
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EXAMPLE 3
Solving the SSS Triangles
Solution continued
Step 2 Find B.
sin B sin C

b
c
b sin C
sin B 
c
1  b sin C 
B  sin 

 c 
1  5.4sin113° 
 sin 
  43.7°
7.2


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EXAMPLE 3
Solving the SSS Triangles
Solution continued
Step 3 A ≈ 180º − 43.7º − 113º ≈ 23.3º
Step 4 Write the solution.
A ≈ 23.3°
a = 3.1 feet
B ≈ 43.7°
b = 5.4 feet
C ≈ 113°
c = 7.2 feet
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EXAMPLE 4
Solving an SSS Triangle
Solve triangle ABC with a = 2 meters,
b = 9 meters, and c = 5 meters. Round each answer
to the nearest tenth.
Solution
Find B, the angle opposite the longest side.
b 2  c 2  a 2  2ca cos B
2
2
2
c2  a 2  b2
5  2 9
cos B 

2ca
25 2
cos B  2.6
The range of the cosine function is [–1, 1]; there is no
angle B with cos B = −2.6; the triangle cannot exist.
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