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Sullivan Algebra and
Trigonometry: Section 8.2
The Law of Sines
Objectives of this Section
• Solve SAA or ASA Triangles
• Solve SSA Triangles
• Solve Applied Problems
If none of the angles of a triangle is a right
angle, the triangle is called oblique.
All angles are acute
Two acute angles, one
obtuse angle
To solve an oblique triangle means to find the
lengths of its sides and the measurements of its
angles.
FOUR CASES
CASE 1: One side and two angles are
known (SAA or ASA).
CASE 2: Two sides and the angle
opposite one of them are known (SSA).
CASE 3: Two sides and the included
angle are known (SAS).
CASE 4: Three sides are known (SSS).
The Law of Sines is used to solve triangles in
which Case 1 or 2 holds. The Law of Sines is
used to solve SAA, ASA or SSA triangles.
Law of Sines
For a triangle with sides a, b, c and opposite
angles  ,  , , respectively,
sin  sin  sin 


a
b
c
      180
Solve the triangle:  = 30 ,   70 , a  5 (SAA)
     180
30
30  70    180
c
b

70
5
  80
sin30 sin70

5
b

sin30 sin80

5
c

5 sin 70
5 sin 80
b
 9.40 c 
 9.85


sin 30
sin 30
Solve the triangle:  = 20 ,   60 , c  12 (ASA)
20
b
12

60
a
      180
20  60    180
  100
sin20 sin100

a
12
sin60 sin100

b
12
12 sin 20 
12 sin 60
a
 4.17 b 
 10.55


sin 100
sin100
Solve the triangle: b  5, c  3,   30 (SSA)
30
sin 30 sin 

5
3
3

a

5
3sin 30
sin  
 0.3
5


 2  162 .5
 1  17.5
   2  30  162.5  192.5  180
  180 30 17.5  132.5




sin 132.5 sin 30

a
5

5 sin 132.5
a
 7.37

sin 30
a  7 .37 , b  5, c  3
  132 .5 ,   30 ,   17 .5



Solve the triangle: b  8, c  10,   45 (SSA)
sin 45 sin 

8
10

10 sin 45
sin  
 0.88
8


 1  62 .1 or  2  117 .9
45  62.1  180
45  117.9  180
Two triangles!!
Triangle 1:  1  62 . 1
1  180  45  62.1  72.9 

sin 72.9 sin 45

a1
8
8 sin 72 . 9
a1 
 10 . 81

sin 45
a1  10.81, b  8, c  10
1  72.9 ,   45 ,  1  62.1






117
.
9
Triangle 2: 2
 2  180  45 117.9  17.1




sin 17.1 sin 45

a2
8

8 sin 17.1
a2 
 3.33

sin 45
a2  3.33, b  8, c  10
2  17.1 ,   45 ,  2  117.9



Solve the triangle: c  5, b  3,   50 (SSA)
sin 50 sin 

3
5
3
5 sin 50
sin  
3
5
50
a
sin   1 . 28
No triangle with the
given measurements!
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