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LAW OF SINES: THE AMBIGUOUS CASE

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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 2. s = 73.1, r = 93.67 and T = 65 0 3. a = 78.3, b = 23.5 and c = 36.8 /ctr Law of Sines Law of Cosines

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AMBIGUOUS Open to various interpretations Having double meaning Difficult to classify, distinguish, or comprehend /ctr

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RECALL: Opposite sides of angles of a triangle Interior Angles of a Triangle Theorem Triangle Inequality Theorem /ctr

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RECALL: Oblique Triangles Triangles that do not have right angles (acute or obtuse triangles) /ctr

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RECALL: LAW OF SINE – 1 sin 1 /ctr

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RECALL: Sine values of supplementary angles are equal. Example: Sin 80 o = 0.9848 Sin 100 o = 0.9848 /ctr

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Law of Sines: The Ambiguous Case Given: lengths of two sides and the angle opposite one of them (S-S-A) /ctr

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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

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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

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Possible Outcomes Case 1: If A is acute and a < b AC B ba c h = b sin A c. If a > b sinA A C B b c h 2 SOLUTIONS aa B 180 -

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Possible Outcomes Case 2: If A is acute and a > b C B h 1 SOLUTION ba A AC B ba c h = b sin A

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Possible Outcomes Case 3: If A is obtuse and a > b C A B a b c ONE SOLUTION

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Possible Outcomes Case 3: If A is obtuse and a ≤ b C A B a b c NO SOLUTION

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First, determine what case you have. (ASA, AAS, SSA, etc.) Law of Sines (“plug & chug”) determine # of solutions ASA, AAS SSA use the Law of Sines to find them.

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For SSA Triangles: 1. If A < 90° a.a < b 1.a < b(sin A)No Solution 2.a = b(sin A)1 Solution 3.a > b(sin A)2 Solution b.a ≥ b1 Solution 2. If A ≥ 90° a.a ≤ bNo Solution b.a > b1 Solution

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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

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sin A = sin B a b Sin B 0.36498 m B = 21.41 o or 21 o Sine values of supplementary angles are equal. The supplement of B is B 2. m B 2 =159 o

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m C = 180 o – (42 o + 21 o ) m C = 117 o sin A = sin C a c c = 29.29 inches SINGLE–SOLUTION CASE

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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

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sin A = sin B a b /ctr Sin B 1.66032 m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE

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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

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sin A = sin B a b /ctr Sin B 1.23644 m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE

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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

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sin A = sin B a b Sin B 0.63393 m B = 39.34 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

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sin A = sin C a c 1 /ctr c 1 = 51.04 inches sin A = sin C a c 2 c = 13.74 inches

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Final Answers: m B 1 = 39 o m C 1 = 116 o c 1 = 51.04 in. EXAMPLE 3 TWO – SOLUTION CASE m B 2 = 141 o m C 2 = 14 o C 2 = 13.74 in. /ctr

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CLASSWORK: (notebook) 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

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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

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9-3 L AW OF S INES. L AW OF S INES A B Given an oblique triangle (no right angle) we can draw in the altitude from vertex B Label the altitude k and find.

9-3 L AW OF S INES. L AW OF S INES A B Given an oblique triangle (no right angle) we can draw in the altitude from vertex B Label the altitude k and find.

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