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LAW OF SINES:. THE AMBIGUOUS CASE. MENTAL DRILL. Identify if the given oblique triangle can be solved using the Law of Sines or the Law of Cosines. 1. X = 21 0 , Z = 65 0 and y = 34.7. Law of Sines. 2. s = 73.1, r = 93.67 and T = 65 0. Law of Cosines.
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LAW OF SINES: THE AMBIGUOUS CASE
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 Law of Sines 2. s = 73.1, r = 93.67 and T = 650 Law of Cosines 3. a = 78.3, b = 23.5 and c = 36.8 Law of Cosines /ctr
AMBIGUOUS • Open to various interpretations • Having double meaning • Difficult to classify, distinguish, or comprehend
RECALL: • Opposite sides of angles of a triangle • Interior Angles of a Triangle Theorem • Triangle Inequality Theorem
RECALL: • Oblique Triangles • Triangles that do not have right angles • (acute or obtuse triangles)
RECALL: • LAW OF SINE • – 1 sin 1
RECALL: • Sine values of supplementary angles are equal. • Example: • Sin 80o = 0.9848 • Sin 100o = 0.9848
Law of Sines: The Ambiguous Case Given: lengths of two sides and the angle opposite one of them (S-S-A)
Possible Outcomes Case 1: If A is acute and a < b C a. If a < b sinA a C b a h = b sin A b B A h c A B c NO SOLUTION
C a b B A c Possible Outcomes Case 1: If A is acute and a < b b. If a = b sinA C h = b sin A b = a h A c B 1 SOLUTION
C b a h = b sin A B A c Possible Outcomes Case 1: If A is acute and a < b b. If a > b sinA C b a h a 180 - A B B c 2 SOLUTIONS
Possible Outcomes Case 2: If A is obtuse and a > b C a b A c B ONE SOLUTION
Possible Outcomes Case 2: If A is obtuse and a ≤ b C a b A c B NO SOLUTION
a>b mA > mB EXAMPLE 1 Given:ABC where a= 22 inches b = 12 inches mA = 42o SINGLE–SOLUTION CASE (acute) Find mB, mC, and c.
sin A = sin B ab 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 Cac c= 29.29 inches SINGLE–SOLUTION CASE
c < b EXAMPLE 2 Given:ABC where c= 15 inches b = 25 inches mC = 85o 15 < 25 sin 85o c ? b sin C NO SOLUTION CASE (acute) Find mB, mC, and c.
sin A = sin B ab Sin B 1.66032 mB = ? Sin B > 1 NOT POSSIBLE ! Recall:– 1 sin 1 NO SOLUTION CASE
b < a EXAMPLE 3 Given:ABC where b= 15.2 inches a = 20 inches mB = 110o NO SOLUTION CASE (obtuse) Find mB, mC, and c.
sin A = sin B ab Sin B 1.23644 mB = ? Sin B > 1 NOT POSSIBLE ! Recall:– 1 sin 1 NO SOLUTION CASE
a < b EXAMPLE 4 Given:ABC where a= 24 inches b = 36 inches mA = 25o a ? b sin A 24 > 36 sin 25o TWO – SOLUTION CASE (acute) Find mB, mC, and c.
sin A = sin B ab 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 Cac1 c1 = 51.04 inches sin A = sin Cac2 c = 13.74 inches
EXAMPLE 3 Final Answers: mB1 = 39o mC1 = 116o c1 = 51.04 in. mB2 = 141o mC2 = 14o C2= 13.74 in. TWO – SOLUTION CASE
SEATWORK: (notebook) Answer in pairs. Find mB, mC, 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
LAW OF SINES: THE AMBIGUOUS CASE
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 Law of Sines 2. s = 73.1, r = 93.67 and T = 650 Law of Cosines 3. a = 78.3, b = 23.5 and c = 36.8 Law of Cosines /ctr
RECALL: • Opposite sides of angles of a triangle • Interior Angles of a Triangle Theorem • Triangle Inequality Theorem
RECALL: • Oblique Triangles • Triangles that do not have right angles • (acute or obtuse triangles)
RECALL: • LAW OF SINE • – 1 sin 1
RECALL: • Sine values of supplementary angles are equal. • Example: • Sin 80o = 0.9848 • Sin 100o = 0.9848
Law of Sines: The Ambiguous Case Given: lengths of two sides and the angle opposite one of them (S-S-A)
Possible Outcomes Case 1: If A is acute and a < b C a. If a < b sinA a C b a h = b sin A b B A h c A B c NO SOLUTION
C a b B A c Possible Outcomes Case 1: If A is acute and a < b b. If a = b sinA C h = b sin A b = a h A c B 1 SOLUTION
C b a h = b sin A B A c Possible Outcomes Case 1: If A is acute and a < b b. If a > b sinA C b a h a 180 - A B B c 2 SOLUTIONS
Possible Outcomes Case 2: If A is obtuse and a > b C a b A c B ONE SOLUTION
Possible Outcomes Case 2: If A is obtuse and a ≤ b C a b A c B NO SOLUTION
a>b mA > mB EXAMPLE 1 Given:ABC where a= 22 inches b = 12 inches mA = 42o SINGLE–SOLUTION CASE (acute) Find mB, mC, and c.
sin A = sin B ab 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 Cac c= 29.29 inches SINGLE–SOLUTION CASE
c < b EXAMPLE 2 Given:ABC where c= 15 inches b = 25 inches mC = 85o 15 < 25 sin 85o c ? b sin C NO SOLUTION CASE (acute) Find mB, mC, and c.
sin A = sin B ab Sin B 1.66032 mB = ? Sin B > 1 NOT POSSIBLE ! Recall:– 1 sin 1 NO SOLUTION CASE
b < a EXAMPLE 3 Given:ABC where b= 15.2 inches a = 20 inches mB = 110o NO SOLUTION CASE (obtuse) Find mB, mC, and c.
sin A = sin B ab Sin B 1.23644 mB = ? Sin B > 1 NOT POSSIBLE ! Recall:– 1 sin 1 NO SOLUTION CASE
a < b EXAMPLE 4 Given:ABC where a= 24 inches b = 36 inches mA = 25o a ? b sin A 24 > 36 sin 25o TWO – SOLUTION CASE (acute) Find mB, mC, and c.
sin A = sin B ab 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 Cac1 c1 = 51.04 inches sin A = sin Cac2 c = 13.74 inches