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Law of Sines

Law of Sines. Section 6.1. So far we have learned how to solve for only one type of triangle Right Triangles Next, we are going to be solving oblique triangles Any triangle that is not a right triangle. In general:. C. a. b. A. B. c.

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Law of Sines

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  1. Law of Sines Section 6.1

  2. So far we have learned how to solve for only one type of triangle • Right Triangles • Next, we are going to be solving oblique triangles • Any triangle that is not a right triangle

  3. In general: C a b A B c

  4. To solve an oblique triangle, we must know 3 pieces of information: • 1 Side of the triangle • Any 2 other components • Either 2 sides, an angle and a side, and 2 angles

  5. C • AAS • ASA • SSA • SSS • SAS Law of Sines a b B A c

  6. Law of Sines • If ABC is a triangle with sides a, b, and c, then:

  7. ASA or AAS C A = 49º 27.4 102.3º a = 43.06 28.7º B c = A

  8. ASA or AAS C A = 49º 27.4 102.3º a = 43.06 28.7º B c = 55.75 A

  9. Solve the following Triangle: • A = 123º, B = 41º, and a = 10 C = 16º C 10 b 123º 41º c

  10. C C = 16º 10 b = 7.8 b 123º 41º c

  11. C C = 16º 10 b = 7.8 b c = 3.3 123º 41º c

  12. Solve the following Triangle: • A = 60º, a = 9, and c = 10 How is this problem different? C What can we solve for? 9 b 60º B 10

  13. C C = 74.2º 9 b 60º B 10

  14. C C = 74.2º 9 b B = 45.8º c = 7.5 60º B 10

  15. What we covered: • Solving right triangles using the Law of Sines when given: • Two angles and a side (ASA or AAS) • One side and two angles (SSA) • Tomorrow we will continue with SSA

  16. SSA The Ambiguous Case

  17. Yesterday • Yesterday we used the Law of Sines to solve problems that had two angles as part of the given information. • When we are given SSA, there are 3 possible situations. • No such triangle exists • One triangle exists • Two triangles exist

  18. Consider if you are given a, b, and A Can we solve for h? a b h h = b Sin A A If a < h, no such triangle exists

  19. Consider if you are given a, b, and A a b h A If a = h, one triangle exists

  20. Consider if you are given a, b, and A a b h A If a > h, one triangle exists

  21. Consider if you are given a, b, and A a b A If a ≤ b, no such triangle exists

  22. Consider if you are given a, b, and A a b A If a > b, one such triangle exists

  23. Hint, hint, hint… • Assume that there are two triangles unless you are proven otherwise.

  24. Two Solutions • Solve the following triangle. a = 12, b = 31, A = 20.5º 31 12 20.5º

  25. 2 Solutions First Triangle Second Triangle B’ = 180 – B = 115.2º C’ = 44.3º C’ = 23.93 • B = 64.8º • C = 94.7º • c = 34.15

  26. Problems with SSA • Solve the first triangle (if possible) • Subtract the first angle you found from 180 • Find the next angle knowing the sum of all three angles equals 180 • Find the missing side using the angle you found in step 3.

  27. A = 60º; a = 9, c = 10 First Triangle Second Triangle C’ = 105.8º B’ = 14.2º b ’ = 2.6 • C = 74.2º • B = 48.8º • b = 7.5

  28. One Solution • Solve the following triangle. What happens when you try to solve for the second triangle? a = 22; b = 12; A = 42º

  29. a = 22; b = 12; A = 42º First Triangle Second Triangle B’ = 158.6º C’ = -20.6º • B = 21.4º • C = 116.6º • c = 29.4

  30. No Solution • Solve the following triangle. a = 15; b = 25; A = 85º Error → No such triangle

  31. Law of Sines Section 6.1

  32. Warm Up • Solve the following triangles (if possible) • A = 58º; a = 20; c = 10 • B = 78º; b = 207; c = 210 • A = 62º; a = 10; b = 12

  33. A = 58º; a = 20; c = 10 C = 25º B= 97º b= 23.4

  34. B = 78º; b = 207; c = 210 C = 82.9º A= 19.1º b= 69.2

  35. B = 78º; b = 207, c = 210 First Triangle Second Triangle C’ = 97.1º A’ = 4.9º a’ = 18.1 • C = 82.9º • A= 19.1º • a= 69.2

  36. A = 62º; a = 10; b = 12 No such triangle

  37. Area of an oblique triangle

  38. What is the area of a triangle? a b = c b h h = b Sin A A

  39. Area of an Oblique Triangle • The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle. • i.e.: Area = ½ bc Sin A = ½ ac Sin b = ½ ab Sin B

  40. Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102º. Area = ½ (Side) (Side) Sine (angle) Area = ½ (90) (52) Sine (102º) Area = 1,189 m²

  41. Find the area of the triangle: • A = 35º, a = 14.7, b = 14.7

  42. The course for a boat race starts at a point A and proceeds in the direction S 52º W to point B, then in the direction of S 40º E to point C, and finally back to A. Point C lies 8 miles directly south of point A. How far was the race? A 52 B 40 C

  43. An airplane left an airport and flew east for 169 miles. Then it turned northward to N 32º E. When it was 264 miles from the airport, there was an engine problem and it turned to take the shortest route back to the airport. Find θ, the angle through which the airplane turned.

  44. A farmer has a triangular field with sides that are 450 feet, 900 feet, and an included angle of 30º. He wants to apply fall fertilizer to the field. If it takes one 40 pound bag of fertilizer to cover 6,000 square feet, how many bags does he need to cover the entire field?

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