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 = s = 73.1, r = and T = a = 78.3, b = 23.5 and c = 36.8 /ctr Law of Sines 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: LAW OF SINE – 1 sin 1 /ctr
RECALL: Sine values of supplementary angles are equal. Example: Sin 80 o = Sin 100 o = /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 b a 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 AC 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 AC B ba c h = b sin A b. If a > b sinA A C B b c h 2 SOLUTIONS aa B
Possible Outcomes Case 2: If A is obtuse and a > b C A B a b c ONE SOLUTION
Possible Outcomes Case 2: If A is obtuse and a ≤ b C A B a b c NO SOLUTION
Given: ABC where a = 22 inches b = 12 inches m A = 42 o EXAMPLE 1 Find m B, m C, and c. (acute) a>b m A > m B SINGLE–SOLUTION CASE
sin A = sin B a b Sin B m B = o or 21 o Sine values of supplementary angles are equal. The supplement of B is B 2. m B 2 =159 o
m C = 180 o – (42 o + 21 o ) m C = 117 o sin A = sin C a c c = inches SINGLE–SOLUTION CASE
Given: ABC where c = 15 inches b = 25 inches m C = 85 o EXAMPLE 2 Find m B, m C, and c. (acute) c < b c ? b sin C 15 < 25 sin 85 o NO SOLUTION CASE
sin A = sin B a b /ctr Sin B m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE
Given: ABC where b = 15.2 inches a = 20 inches m B = 110 o EXAMPLE 3 Find m B, m C, and c. (obtuse) b < a NO SOLUTION CASE
sin A = sin B a b /ctr Sin B m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE
Given: ABC where a = 24 inches b = 36 inches m A = 25 o EXAMPLE 4 Find m B, m C, and c. (acute) a < b a ? b sin A24 > 36 sin 25 o TWO – SOLUTION CASE
sin A = sin B a b Sin B m B = o or 39 o The supplement of B is B 2. m B 2 = 141 o m C 1 = 180 o – (25 o + 39 o ) m C 1 = 116 o m C 2 = 180 o – (25 o +141 o ) m C 2 = 14 o
sin A = sin C a c 1 /ctr c 1 = inches sin A = sin C a c 2 c = inches
Final Answers: m B 1 = 39 o m C 1 = 116 o c 1 = in. EXAMPLE 3 TWO – SOLUTION CASE m B 2 = 141 o m C 2 = 14 o C 2 = in. /ctr
SEATWORK: (notebook) Answer in pairs. Find m B, m C, and c, if they exist. 1 ) a = 9.1, b = 12, m A = 35 o 2) a = 25, b = 46, m A = 37 o 3) a = 15, b = 10, m A = 66 o /ctr
Answers: 1 )Case 1: m B=49 o,m C=96 o,c=15.78 Case 2: m B=131 o,m C=14 o,c=3.84 2)No possible solution. 3)m B=38 o,m C=76 o,c=15.93 /ctr
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 = s = 73.1, r = and T = a = 78.3, b = 23.5 and c = 36.8 /ctr Law of Sines 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: LAW OF SINE – 1 sin 1 /ctr
RECALL: Sine values of supplementary angles are equal. Example: Sin 80 o = Sin 100 o = /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 b a 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 AC 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 AC B ba c h = b sin A b. If a > b sinA A C B b c h 2 SOLUTIONS aa B
Possible Outcomes Case 2: If A is obtuse and a > b C A B a b c ONE SOLUTION
Possible Outcomes Case 2: If A is obtuse and a ≤ b C A B a b c NO SOLUTION
Given: ABC where a = 22 inches b = 12 inches m A = 42 o EXAMPLE 1 Find m B, m C, and c. (acute) a>b m A > m B SINGLE–SOLUTION CASE
sin A = sin B a b Sin B m B = o or 21 o Sine values of supplementary angles are equal. The supplement of B is B 2. m B 2 =159 o
m C = 180 o – (42 o + 21 o ) m C = 117 o sin A = sin C a c c = inches SINGLE–SOLUTION CASE
Given: ABC where c = 15 inches b = 25 inches m C = 85 o EXAMPLE 2 Find m B, m C, and c. (acute) c < b c ? b sin C 15 < 25 sin 85 o NO SOLUTION CASE
sin A = sin B a b /ctr Sin B m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE
Given: ABC where b = 15.2 inches a = 20 inches m B = 110 o EXAMPLE 3 Find m B, m C, and c. (obtuse) b < a NO SOLUTION CASE
sin A = sin B a b /ctr Sin B m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE
Given: ABC where a = 24 inches b = 36 inches m A = 25 o EXAMPLE 4 Find m B, m C, and c. (acute) a < b a ? b sin A24 > 36 sin 25 o TWO – SOLUTION CASE
sin A = sin B a b Sin B m B = o or 39 o The supplement of B is B 2. m B 2 = 141 o m C 1 = 180 o – (25 o + 39 o ) m C 1 = 116 o m C 2 = 180 o – (25 o +141 o ) m C 2 = 14 o
sin A = sin C a c 1 /ctr c 1 = inches sin A = sin C a c 2 c = inches
Final Answers: m B 1 = 39 o m C 1 = 116 o c 1 = in. EXAMPLE 3 TWO – SOLUTION CASE m B 2 = 141 o m C 2 = 14 o C 2 = in. /ctr
Final Answers: m B 1 = 39 o m C 1 = 116 o c 1 = in. EXAMPLE 3 TWO – SOLUTION CASE m B 2 = 141 o m C 2 = 14 o C 2 = in. /ctr
Answers: 1 )Case 1: m B=49 o,m C=96 o,c=15.78 Case 2: m B=131 o,m C=14 o,c=3.84 2)No possible solution. 3)m B=38 o,m C=76 o,c=15.93 /ctr