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LAW OF SINES
Sec. 5.5
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Deriving the Law of Sines
A
A
B
B
C
Cc
c
a
ab
b
h
h
In either triangle: sinh
Ab
In the top triangle: sinh
Ba
In the bottom triangle: sinh
Ba
But, sin sinB B so each of these last twoexpressions are equal!!!
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Deriving the Law of Sines
A
A
B
B
C
Cc
c
a
ab
b
h
h
sinh
Ab
sinh
Ba
Solve for h:
sinh b A sinh a B
Set equal:
sin sinb A a BWhich is equivalent to:
sin sinA B
a b
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Law of SinesIn any triangle with angles A, B, and Copposite sides a, b, and c, respectively, the followingequation is true:
ABC
sin sin sinA B C
a b c
The Law of Sines works most easily withthese two triangle cases: AAS, ASA
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Guided PracticeSolve , given the following.ABC
36A 48B 8a
A B
C
c
8b
36 48
180 36 48C 96 sin sinA B
a b
sin 36 sin 48
8 b
8sin 48
sin36b
10.115
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Guided PracticeSolve , given the following.ABC
36A 48B 8a
A B
C
c
8b
36 48
sin sinA C
a c
sin 36 sin 96
8 c
8sin96
sin36c
13.536
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The Ambiguous Case (SSA)We wish to construct ABC given angle A, side AB, & side BC.
1. Suppose angle A is obtuse and that side AB is as shownbelow. To complete the triangle, side BC must determine apoint on the dotted horizontal line (which extends infinitely tothe left). Explain from the picture why a unique triangle ABCis determined if BC > AB, but no triangle is determined ifBC < AB.
A
B
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The Ambiguous Case (SSA)We wish to construct ABC given angle A, side AB, & side BC.
2. Suppose angle A is acute and that side AB is as shownbelow. To complete the triangle, side BC must determine apoint on the dotted horizontal line (which extends infinitely tothe right). Explain from the picture why a unique triangle ABCis determined if BC = h, but no triangle is determined if BC < h.
A
B
h
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The Ambiguous Case (SSA)We wish to construct ABC given angle A, side AB, & side BC.
3. Suppose angle A is acute and that side AB is as shownbelow. If AB > BC > h, then we can form a triangle as shown.Find a second point C on the dotted horizontal line that givesa side BC of the same length, but determines a differenttriangle. (This is the “ambiguous case.”)
A
B
h
C C
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The Ambiguous Case (SSA)We wish to construct ABC given angle A, side AB, & side BC.
4. Explain why sin(C) is the same in both triangles in theambiguous case. (This is why the Law of Sines is alsoambiguous in this case.)
A
B
h
C C
5. Explain from the figure below why a unique triangle isdetermined if BC > AB.
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Guided PracticeState whether the given measurements determine zero, one, ortwo triangles.
B
(a) B = 82 , b = 17, c = 15
82
15 17
A
C
h
Solve for h:
sin8215
h
15sin82h 14.9
Because h < c < b, one triangle is formed.
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Practice ProblemsState whether the given measurements determine zero, one, ortwo triangles.
A
(b) A = 73 , a = 24, b = 28
73
28 24
C
B
h
Solve for h:
sin 7328
h
28sin 73h 26.8
Because a < h, no triangle is formed.
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Practice ProblemsState whether the given measurements determine zero, one, ortwo triangles.
C
(c) C = 31 , a = 17, c = 10
31
17 10
B
A
h
Solve for h:
sin 3117
h
17sin31h 8.756
Because h < c < a, two triangles are formed.
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Guided Practice Two triangles can be formed using the given measurements.Solve both triangles.
B
(a) B = 38 , b = 21, c = 25
38
2521
A
C
Here, C is acute:
1sinsin 38
21 25
C
C
211
1
25sin 38sin
21C
47.1
B38
2521
A
C
1 1180 94.9A B C 1
1
21sin
sin 38
Aa
1
1
sinsin 38
21
A
a
33.987
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Guided Practice Two triangles can be formed using the given measurements.Solve both triangles.
B
(a) B = 38 , b = 21, c = 25
38
2521
A
C
What if C is obtuse?
C
21
2 1180 132.9C C
22
21sin
sin 38
Aa
2
2
sinsin 38
21
A
a
5.414
B38
25
A
C
21
2 2180 9.1A B C
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Whiteboard PracticeSolve , given the following.ABC
50A 62B 4a
A B
C
c
4b
50
68C
62
4.610b4.841c
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Whiteboard PracticeSolve , given the following.ABC
16B 103C 12c
A B
C
12
ab 103
61A
16
10.772a 3.395b
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Whiteboard PracticeSolve , given the following.ABC
49A 32a 28b
A B
C
c
3228
49
sin 49 sin
32 28
B
1 28sin 49sin32
B
41.3
180 49 41.3C 89.7 sin89.7 sin 49
32c
32sin89.7
sin 49c
42.400
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Whiteboard PracticeSolve , given the following.ABC
103C 46b 61c
A
B
C
61
a
46103
47.3B 29.7A 31.029a
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Whiteboard ProblemsTwo triangles can be formed using the given measurements.Solve both triangles.
B
(b) B = 57 , a = 11, b = 10
57
1110
C
A
Here, A is acute:
1sinsin 57
10 11
A
A
101
1
11sin 57sin
10A
67.3
B57
1110
C
A
1 1180 55.7C A B 1
1
10sin
sin 57
Cc
1
1
sinsin 57
10
C
c
9.850
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Whiteboard ProblemsTwo triangles can be formed using the given measurements.Solve both triangles.
What if A is obtuse?
2 1180 112.7A A
22
10sin
sin 57
Cc
2
2
sinsin 57
10
C
c
2.132
B57
11
C
A
10
2 2180 10.3C B A
B57
1110
C
AA
10
(b) B = 57 , a = 11, b = 10