Post on 31-Dec-2015
description
Aim: How do we handle the ambiguous case?
Do Now: In ∆ABC, mA = 30° , a = 6 and c = 10. Find C to the nearest degree.
We can use the Law of Sine to answer the problem.
,sin
10
30sin
6
C ,56,
6
5sin,5sin6 CCC
HW: p.574 # 6b,8b,10b,14b,16,18
Notice that sin C is positive 56° which can be in quadrant I or quadrant II.
If angle C is in quadrant I then it is 56°.
If 56° is in quadrant II then it is 180° – 56° = 124°
Which angle measurement is correct for angle C?
We call this situation as ambiguous case that means there are different choices for C
We have to use the basic idea about the sum of the interior angles of a triangle to figure out.
If C = 56°, then B = 180° – (56 + 30) = 94°
If C = 124°, then B = 180° – (124° + 30) = 26°
Both angle measurements satisfy the basic requirement of interior angles of a triangle. Therefore, we can say that there are two different triangles can be made with the given condition.
How do we deal with this ambiguous case?
We are given A = 30
Let’s see another example:
In 30,15, AmaABC and c = 12
a) Find mC
b) How many triangles can be drawn?
24,5
2
15
6sin,6sin15,
sin
12
30sin
15CCC
C
or 15624180
If ,24Cm then 126)3024(180Bm
If 156Cm
18615630CmAm
Two interior angles of the ∆ABC are already over 180°
Therefore, mC =156° is invalid.
The answer for a) is 24° only
Therefore, there is only one triangle can be made with the given condition.
In ΔABC, b = 18, c = 10 and B = 70. How many triangles can be constructed?
B
AC
11
10
70
,sin
11
70sin
10
C
70sin11sin10 C
,10
70sin11sin
C
C undefined, therefore no triangle can be constructed
Application:
1. If side a = 16, side b = 20, mB = 30°, how many distinct triangles can be constructed?
2. If mA = 68°, side a = 10 and side b = 24, how many distinct triangles can be constructed?
1 triangle
0 triangle
3. If side a = 18, side b = 10 and mC = 70°, how many distinct triangles can be constructed?
2 triangles