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Chapter 9 section 2
Law of Sines
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).
A
S
A
ASA
S
A
A
SAA
CASE 1: ASA or SAA
S
A
S
CASE 2: SSA
S
A
S
CASE 3: SAS
S
S
S
CASE 4: SSS
The Law of Sines is used to solve
triangles in which Case 1 or 2
holds. That is, the Law of Sines
is used to solve SAA, ASA or SSA
triangles.
Theorem 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


sin 30 sin 70

5
b



sin 30 sin 80

5
c

5 sin 80
5 sin 70
c


9
.
85
b

9
.
40


sin 30
sin 30
Solve the triangle:  = 20 ,   60 , c  12 (ASA)


      180

20
20  60    180

b
12

60

a



  100



sin 20 sin 100

a
12



sin 60 sin 100

b
12

12 sin 20
12 sin 60
a

4
.
17
b


10
.
55


sin 100
sin 100
Solve the triangle: b  5, c  3,   30 (SSA)

sin 30 sin 

5
3

3 sin 30
sin  
 0.3
5


30
3

a

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




Triangle 2:  2  117.9

 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

5 sin 50
sin  
3

3
5

50
a
sin   128
.
No triangle with the given
measurements!