#### Transcript Ambiguous Case Triangles - Parkway C-2

```Ambiguous Case Triangles
Can given numbers make a triangle? Can
a different triangle be formed with the same
information?
Conditions for Unique Triangles
ASA
two angles must sum to less than 180º
SSS
AAS
two angles must sum to less than 180º
SAS
two shortest sides are longer than the third side
Any set of data that fits these conditions will result in one unique triangle.
Ambiguous Triangle
Case
b
a
A
This diagram is deceiving -- side-side-angle
data may result in two different triangles.
SSA
Side a is given but it might be possible to ‘swing’ it to either
of two positions depending on the other given values.
An acute or an obtuse triangle may be possible.
Example (2 triangles)
Given information
mA = 17º
a = 5.8
b = 14.3
Set up Law of Sines
5.8
14.3

sin 17 sin B
mB  46º
Solve for sin B
14.3 sin 17
sin B 
5 .8
mB  180 – 46 = 134º
Find mC
Find mC
mC  (180 – 17 – 46)  117º
mC  (180 – 17 – 134)  29º
Both values of C are possible, so 2 triangles are possible
Example (1 triangle)
Given information
mA = 58º
a = 20
b = 10
Set up Law of Sines
20
10

sin 58 sin B
mB  25º
Solve for sin B
10 sin 58
sin B 
20
mB  180 – 25 = 155º
Find mC
Find mC
mC  (180 – 58 – 25)  97º
mC  (180 – 58 – 155)  -33º
Only one value of C is possible, so only 1 triangle is possible
Example (0 triangles)
Given information
mA = 71º
a = 12
b = 17
Set up Law of Sines
12
17

sin 71 sin B
Solve for sin B
17 sin 71
sin B 
12
sin B  1.3395
No value of B is possible, so no triangles are possible
Law of Sines Method
1) Use Law of Sines to find angle B
-If there is no value of B (for example, sin B = 2),
then there are no triangles
Remember, sin x is positive in both quadrant I and II
2) Determine value of B in quadrant II
(i.e. 180 – quadrant I value)
3) Figure out the missing angle C for both values of
angle B by subtracting angles A and B from 180
4) If it is possible to find angle C for
-both values of B, then there are 2 triangles
-only the quadrant I value of B, then
only 1 triangle is possible
Solve a triangle where a = 5, b = 8 and c = 9
Draw a picture.
This is SSS
One side squared
5
84.3


Do we know an angle and
side opposite it? No, so we
must use Law of Cosines.
Let's use largest side to
find largest angle first.

9
8
c  a  b  2ab cos 
2
2
2
sum of each of
the other sides
squared
minus 2 times the
times the cosine of
product the angle
of those between
other
those
sides
sides
81  89  80 cos 
1  1 
 8   cos  10   84.3
 
cos  
2
2
2
 80
2 5
9  5 8
 8 cos 
CAUTION: Don't forget order of operations: powers
then multiplication BEFORE addition and subtraction
How can we find one of the remaining angles?
Do we know an angle and
side opposite it?
9

62.2
5
84.3

33.5
8
Yes, so use Law of Sines.
sin 84.3 sin 

9
8
8sin 84.3
 sin 
9
 8sin 84.3 
  sin 
  62.2
9


1
  180  84.3  62.2  33.5
OR…..
•
If you have SSA: (h = b Sin A)
– No triangles exist if:
•
•
a<h
a<b
-- One triangle exists if:
•
•
a=h
a>b
– Two triangles exist if:
•
h<a<b
```