LAW OF SINES:

THE AMBIGUOUS CASE

Review Identify if the given oblique triangle can be

solved using the Law of Sines or the Law of Cosines

1. X = 210, Z = 650 and y = 34.7

2. s = 73.1, r = 93.67 and T = 650

3. a = 78.3, b = 23.5 and c = 36.8

Law of Sines

Law of Cosines

Law of Cosines

Law of Sines: The Ambiguous Case

Given:

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

AMBIGUOUS

• Open to various interpretations

• Having double meaning

• Difficult to classify, distinguish, or comprehend

Always set your triangle up this way…

Given angle here

Side opposite hereOther g

iven si

de here

Possible OutcomesCase 1: If is acute and the side opposite the given angle < the other given side.

A

C

B

b a

c

h = b sin A

a. If a < h

A

C

B

b

a

c

h

NO SOLUTION

Possible OutcomesCase 1: If is acute and the side opposite the given angle < the other given side.

A

C

B

b a

c

h = b sin A

b. If a = h

A

C

B

b= a

c

h

1 SOLUTION

Possible OutcomesCase 1: If is acute and the side opposite the given angle < the other given side.

A

C

B

b a

c

h = b sin A

b. If a > h

A

C

B

b

c

h

2 SOLUTIONS

a a

B180 -

Possible OutcomesCase 2: If is obtuse and the side opposite the given angle > the other given side.

C

AB

ab

c

ONE SOLUTION

Possible OutcomesCase 2: If is obtuse and the side opposite the given angle the other given side.

C

A B

a

b

c

NO SOLUTION

SUMMARY is acute or is obtuse

Side opposite < other side

Side opposite > other side

Side opposite > other side

Side opposite < other side

FIND HEIGHT: h = other side sin

Side opposite < h:

Side opposite = h:

Side opposite > h:

No Solution

1 Solution

2 Solutions

1 Solution

1 Solution

No Solution

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

Given: ABC where

c = 15 inches

b = 25 inches

mC = 85o

EXAMPLE 2

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

c < b

c ? b sin C 15 < 25 sin 85o

NO SOLUTION CASE

sin A = sin B a bSin 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

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

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.

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  

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