LAW OF SINES: THE AMBIGUOUS CASE MENTAL DRILL Identify if the given oblique triangle can be solved...
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Transcript of LAW OF SINES: THE AMBIGUOUS CASE MENTAL DRILL Identify if the given oblique triangle can be solved...
MENTAL DRILL
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/ctr
Law of Sines
Law of Cosines
Law of Cosines
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
/ctr
RECALL:
• Oblique Triangles
Triangles that do not have right angles
(acute or obtuse triangles)
/ctr
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)
/ctr
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 -
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
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 b
/ctr
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
/ctr
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
Final Answers:
mB1 = 39o
mC1 = 116o
c1 = 51.04 in.
EXAMPLE 3
TWO – SOLUTION CASE
mB2 = 141o
mC2 = 14o
C2= 13.74 in.
/ctr
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
/ctr
MENTAL DRILL
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/ctr
Law of Sines
Law of Cosines
Law of Cosines
RECALL:
• Opposite sides of angles of a triangle
• Interior Angles of a Triangle Theorem
• Triangle Inequality Theorem
/ctr
RECALL:
• Oblique Triangles
Triangles that do not have right angles
(acute or obtuse triangles)
/ctr
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)
/ctr
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 -
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
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 b
/ctr
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
/ctr
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
Final Answers:
mB1 = 39o
mC1 = 116o
c1 = 51.04 in.
EXAMPLE 3
TWO – SOLUTION CASE
mB2 = 141o
mC2 = 14o
C2= 13.74 in.
/ctr
Final Answers:
mB1 = 39o
mC1 = 116o
c1 = 51.04 in.
EXAMPLE 3
TWO – SOLUTION CASE
mB2 = 141o
mC2 = 14o
C2= 13.74 in.
/ctr