Concept: The Ambiguous Case
Information:
The Law of Sines works very nicely when
you are given AAS or ASA.
When you are given SAS or SSA it gets tricky.
You could have 1, 2, or no triangles.
If < A is acute and the side opposite the
given angle is greater than the other side . . .
C
15
12
A
28º
c
B
1 triangle exists. Solve using Law of
Sines.
If < A is acute and the side opposite the
given angle is less than the other side . . .
b
b
a
a
and
A
A
b
a
or
A
or no triangles exists.
If < A is acute and the side opposite the
given angle is less than the other side . . .
C
sinA = h
Why?
b
h = bsinA
b
h
A
B
Check a against bsinA
If a = bsinA, 1 Rt. triangle
If a > bsinA, 2 triangles
If a < bsinA, no triangles
If a = bsinA
C
b
a = h = bsinA
A
B
1 Rt. triangle
If a > bsinA
C
b
a
A
B1
a1
1
B2
2 triangles exist
If a < bsinA
C
b
a
A
No triangles can be formed.
(We like these . . . no work!!)
When < A is obtuse . . .
a
b
A
a
b
B
If a < b
No triangle exists
A
B
If a > b
1 triangle exists
When < A is a Rt. angle . . .
C
b
A
C
a
b
B
If a > b
1 triangle exists
A
a
B
If a < b
No triangle exists
Read this carefully!!!
If you get caught up in the A’s , a’s
and b’s of it all you will never get it.
Draw a picture
• Put the given angle at the bottom left
• Put the side with the same letter
opposite that angle
• Draw the other side up from the left
Understanding:
Solve RST
< R = 37.4º
r = 11.6
t = 9.4
S
9.4
11.6
37.4º
R
< R is acute and r > t
Solve using the Law of Sines
Find < T then < S and side s
T
Solve LMN
< M = 35.18º
n = 17.8
m = 11.46
L
11.46
17.8
h
35.18º
M
< M is acute and m < n
Check m against nsinM
m > nsinM
2 triangles exist
Draw them separately
N
h = 17.8sin35.18
= 10.2554. . .
Still solving LMN
L2
L1
17.8
M
35.18º
l1
17.8
11.46
N1
11.46
M
35.18º
l2
N2
Solve them using the Law of Sines
N1 = 63.49º
L1 = 81.33º
l1 = 19.66
N2 = 116.51º
L2 = 28.31º
l2 = 9.43
Solve JKL
< K = 71.4º
k = 45.3
j = 51.4
< K is acute and k < j
Check k against jsinK
k < jsinK
No triangles exist
L
45.3
51.4
h
71.4º
K
J
h = 51.4sin71.4
= 48.71529. . .
Determine the Number of Solutions
< M = 132º
n = 96
m = 105
< K = 29º
k = 23
j = 31
23
31
105
96
h
29º
K
132º
J
< K is acute and k < j
k > 31sin29º
2 triangles exist
M
N
< M is obtuse and m > n
1 triangle exists
Homework Section 8.2:
Page 527 #19-33, 39, 41 ODDS