5.7

The Ambiguous Case for the Law of Sines

AMBIGUOUS

• Open to various interpretations

• Having double meaning

• Difficult to classify, distinguish, or comprehend /ctr

RECALL:

• Opposite sides of angles of a triangle

• Interior Angles of a Triangle Theorem

• Triangle Inequality Theorem

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RECALL:

• Oblique Triangles

Triangles that do not have right angles

(acute or obtuse triangles)

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RECALL:

• LAW OF SINE

– 1 sin 1

c

Csin

b

Bsin

a

Asin

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RECALL:

• Sine values of supplementary angles are equal.

Example:

Sin 80o = 0.9848

Sin 100o = 0.9848/ctr

Law of Sines: The Ambiguous Case

Given:

lengths of two sides and the angle opposite one of them (S-S-A)

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Possible Outcomes

Case 1: If A is acute and a < b

A

C

B

ba

c

h = b sin A

a. If a < b sinA

A

C

B

b

a

c

h

NO SOLUTION

Possible Outcomes

Case 1: If A is acute and a < b

A

C

B

b a

c

h = b sin A

b. If a = b sinA

A

C

B

b= a

c

h

1 SOLUTION

Possible Outcomes

Case 1: If A is acute and a < b

A

C

B

b a

c

h = b sin A

b. If a > b sinA

A

C

B

b

c

h

2 SOLUTIONS

a a

B

180 -

Possible Outcomes

Case 2: If A is obtuse and a > bC

A B

a

b

c

ONE SOLUTION

Possible Outcomes

Case 2: If A is obtuse and a ≤ bC

A B

a

b

c

NO SOLUTION

Determine the number of possible solutions for each triangle.

• i) A=30deg a=8 b=10

• ii) b=8 c = 10 B = 118 deg

Find all solutions for each triangle.

• i) a = 4 b = 3 A = 112 degrees

• ii) A = 51 degrees a = 40 c = 50

Given: ABC where

a = 22 inches

b = 12 inches

mA = 42o

EXAMPLE 1

Find m B, m C, and c.(acute)

a>b

mA > mBSINGLE–SOLUTION CASE

sin A = sin B a b

Sin B 0.36498 mB = 21.41o or 21o

Sine values of supplementary angles are equal.

The supplement of B is B2. mB2=159o

mC = 180o – (42o + 21o) mC = 117o

sin A = sin C a c

c = 29.29 inches

SINGLE–SOLUTION CASE

sin A = sin B a b

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Sin B 1.66032 mB = ?

Sin B > 1 NOT POSSIBLE !

Recall: – 1 sin 1

NO SOLUTION CASE

Given: ABC where

b = 15.2 inches

a = 20 inches

mB = 110o

EXAMPLE 3

Find m B, m C, and c.(obtuse)

b < a

NO SOLUTION CASE

sin A = sin B a b

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Sin B 1.23644 mB = ?

Sin B > 1 NOT POSSIBLE !

Recall: – 1 sin 1

NO SOLUTION CASE

Given: ABC where

a = 24 inches

b = 36 inches

mA = 25o

EXAMPLE 4

Find m B, m C, and c.(acute)

a < b

a ? b sin A 24 > 36 sin 25o

TWO – SOLUTION CASE

sin A = sin B a b

Sin B 0.63393 mB = 39.34o or 39o

The supplement of B is B2. mB2 = 141o

mC1 = 180o – (25o + 39o) mC1 = 116o mC2 = 180o – (25o+141o) mC2 = 14o

sin A = sin C a c1

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c1 = 51.04 inches

sin A = sin C a c2

c = 13.74 inches

Final Answers:

mB1 = 39o

mC1 = 116o

c1 = 51.04 in.

EXAMPLE 3

TWO – SOLUTION CASE

mB2 = 141o

mC2 = 14o

C2= 13.74 in.

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SEATWORK: (notebook)

Answer in pairs.

Find m B, m C, and c, if they exist.

 1) a = 9.1, b = 12, mA = 35o

 2) a = 25, b = 46, mA = 37o

3) a = 15, b = 10, mA = 66o  /ctr

Answers:

 1)Case 1:

mB=49o,mC=96o,c=15.78

Case 2:  

mB=131o,mC=14o,c=3.84

2)No possible solution.

3)mB=38o,mC=76o,c=15.93  

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