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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 = 21 0,

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Presentation on theme: "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 = 21 0,"— Presentation transcript:

1 LAW OF SINES: THE AMBIGUOUS CASE

2 Review 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 Law of Sines Law of Cosines

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

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

5 Always set your triangle up this way…  Given angle here Side opposite here Other given side here

6 Possible Outcomes Case 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

7 Possible Outcomes Case 1: If  is acute and the side opposite the given angle < the other given side. AC B b a c h = b sin A b. If a = h A C B b = a c h 1 SOLUTION

8 Possible Outcomes Case 1: If  is acute and the side opposite the given angle < the other given side. AC B b a c h = b sin A b. If a > h A C B b c h 2 SOLUTIONS aa B   180 - 

9 Possible Outcomes Case 2: If  is obtuse and the side opposite the given angle > the other given side. C A B a b c ONE SOLUTION

10 Possible Outcomes Case 2: If  is obtuse and the side opposite the given angle  the other given side. C A B a b c NO SOLUTION

11 SUMMARY  is acute or  is obtuse 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 No Solution

12 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

13 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

14 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

15 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

16 sin A = sin B a b Sin B  1.66032 m  B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1  sin   1 NO SOLUTION CASE

17 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

18 sin A = sin B a b Sin B  1.23644 m  B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1  sin   1 NO SOLUTION CASE

19 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

20 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

21 sin A = sin C a c 1 c 1 = 51.04 inches sin A = sin C a c 2 c = 13.74 inches

22 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.

23 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

24 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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