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