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The Law of

The Law of. Sines. There is no doubt that the 3 PTRs are extremely useful when solving problems modeled on a right triangle. Unfortunately, the world does not consist only of right triangles…. As a matter of fact, right triangles end up being more of a rarity than commonplace.

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The Law of

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  1. The Law of Sines

  2. There is no doubt that the 3 PTRs are extremely useful when solving problems modeled on a right triangle. Unfortunately, the world does not consist only of right triangles…

  3. As a matter of fact, right triangles end up being more of a rarity than commonplace. Does that mean when we come across a situation that can only be modeled with a non-right triangle that we abandon our pursuit?….

  4. No Way!!!! There exists 2 Laws of Trigonometry that allow one to solve problems that involve non-right Triangles: Law of Sines Law of Cosines

  5. A triangle is uniquely determined by two angles and a particular side A C b a O2 O1 B c

  6. If a corresponding angle and side are known, they form an “opposing pair” A C b a O2 O1 B c

  7. The Sine Law can be used to determine an unknown side or angle given an “opposing pair” A C b a O2 O1 B c

  8. Find the length of b A C b 5 65o 30o B c

  9. Construct CN with height h A C b 5 h 30o 65o B c N

  10. By the right triangle SIN ratio Sin 30o = h Sin 65o = h b 5 A C b 5 h h 30o 30o 65o 65o B c N

  11. Solve both equations for h Sin 30o = h Sin 65o = h X 5 X b b 5 bSin30o = h h = 5Sin65o Because the equations are equal bSin30o = 5Sin65o

  12. bSin30o = 5Sin65o b = 5Sin65o Sin30o b = 9.1 Consider the general case:

  13. C b a h A B N c Sin A = h Sin B = h b a bSinA = aSinB

  14. bSinA = aSinB a a bSinA = SinB a bSinA = SinB ab b SinA = SinB a b

  15. Extend this to all 3 sides of a triangle, and the Sine Law is generated! SinA = SinB = SinC a b c

  16. Find the length of a C a = 24 a Sin73o Sin57o 57o a = 27.4 c 73o N A 24

  17. h Find h 5.9O 10.3O 2.9 km

  18. Find h 1. Find O O = 180O – 5.9O – 10.3O = 163.8O O 5.9O 10.3O 2.9 km

  19. X 2.9 = Find X SIN 10.3O SIN163.8O X = 1.86km 163.8O X 5.9O 10.3O 2.9 km

  20. h SIN 5.9O = h Find h 1.86 km h = 191.2 m 1.86 km 5.9O 10.3O 2.9 km

  21. The Ambiguous Case

  22. Find A SinA Sin48o 11 = 11 9 9 A = 65.3o Does that make sense? 48o A No Way!!!

  23. Side 9 can also be drawn as: Could A be 65o in this case? 11 9 48o A

  24. This type of discrepancy is called the “Ambiguous Case”Be sure to check the diagram to see which answer fits:O, or 180o - O

  25. Page 366

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