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LAW OF SINES:

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 , Z = 65 0 and y = 34.7. Law of Sines. 2. s = 73.1, r = 93.67 and T = 65 0. Law of Cosines. 3. a = 78.3, b = 23.5 and c = 36.8.

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LAW OF SINES:

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  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 = 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

  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… Other given side here Side opposite here  Given angle here

  6. Possible Outcomes Case 1: If  is acute and the side opposite the given angle < the other given side. C a. If a < h C a b a h = b sin A b B A h c A B c NO SOLUTION

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

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

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

  11. SUMMARY  is acute or  is obtuse Side opposite <other side Side opposite >other side FIND HEIGHT: h = other side  sin 1 Solution Side opposite <h: No Solution Side opposite =h: 1 Solution Side opposite >h: 2 Solutions Side opposite >other side Side opposite <other side No Solution 1 Solution

  12. a>b mA > mB EXAMPLE 1 Given:ABC where a= 22 inches b = 12 inches mA = 42o SINGLE–SOLUTION CASE (acute) Find mB, mC, and c.

  13. sin A = sin B ab Sin B  0.36498 mB = 21.41o or 21o Sine values of supplementary angles are equal. The supplement of B is B2.  mB2=159o

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

  15. c < b EXAMPLE 2 Given:ABC where c= 15 inches b = 25 inches mC = 85o 15 < 25 sin 85o c ? b sin C NO SOLUTION CASE (acute) Find mB, mC, and c.

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

  17. b < a EXAMPLE 3 Given:ABC where b= 15.2 inches a = 20 inches mB = 110o NO SOLUTION CASE (obtuse) Find mB, mC, and c.

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

  19. a < b EXAMPLE 4 Given:ABC where a= 24 inches b = 36 inches mA = 25o a ? b sin A 24 > 36 sin 25o TWO – SOLUTION CASE (acute) Find mB, mC, and c.

  20. sin A = sin B ab Sin B  0.63393 mB = 39.34o or 39o The supplement of B is B2.  mB2 = 141o mC1 = 180o – (25o + 39o) mC1 = 116o mC2 = 180o – (25o+141o) mC2 = 14o

  21. sin A = sin Cac1 c1 = 51.04 inches sin A = sin Cac2 c = 13.74 inches

  22. EXAMPLE 3 Final Answers: mB1 = 39o mC1 = 116o c1 = 51.04 in. mB2 = 141o mC2 = 14o C2= 13.74 in. TWO – SOLUTION CASE

  23. SEATWORK: (notebook) Answer in pairs. Find mB, mC, and c, if they exist.  1) a = 9.1, b = 12, mA = 35o  2) a = 25, b = 46, mA = 37o 3) a = 15, b = 10, mA = 66o

  24. Answers:  1)Case 1: mB=49o,mC=96o,c=15.78 Case 2: mB=131o,mC=14o,c=3.84 2)No possible solution. 3)mB=38o,mC=76o,c=15.93

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