1. There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle. Albert Einstein (1879 – 1955) (1879 – 1955) a German-born theoretical physicist who developed the theory of general relativity, effecting a revolution in physics.

2. Chapter 6 Additional Topics in Trigonometry

3. Day I. Law of Sines (6.1) Part 1

4. 6.1 GOAL 1 How to use the Law of Sines to solve oblique triangles.

5. Why should you learn it? You can use the Law of Sines to solve real-life problems, for example to determine the distance from a ranger station to a forest fire

6. -- a triangle that has no right angles -- a triangle that has no right angles Oblique Triangle

7. you need To Solve

8. 2 angles and any side 1-Today A V AS A S A

9. 2 sides and an angle opposite one of them 2-Next Time A SS SSASSASSASSA

10. 3 sides 3-Later SSS

11. 2 sides and their included angle 4-Later SAS

12. Students know the Law of Sines. Standard 13.1

13. Law of Sines If ABC is a triangle with sides a, b, and c, then a b c a b c sin A sin B sin C = =

14. Law of Sines Alternate Form sin A sin B sin C a b c a b c = =

15. Students can apply the Law of Sines to solve problems. Standard 13.2

16. Finding a Measurement Example 1

17. Find, to the nearest meter, the distance across Perch Lake from point A to point B. The length of AC, or b, equals 110 m, and measures of the angles of the triangle are as shown.

18. A B C c a b sin B = h hc OR h = c sin B sin C = hb OR h = b sin C

19. A B C c a b h c sin B = b sin C c b sin C sin B =

20. A B C 40 67 110 m h

21. c b sin C sin B = 110 67 40 Solve.

22. 110  sin 67  sin 40  = 110  sin 67  = c  sin 40  sin 67  sin 40  c 110 c = c  158 meters

23. Given Two Angles and One Side - AAS Example 2

24. Using the given information, solve the triangle. A 25 35 3.5 c b B C

25. What does it mean “to solve a triangle”? Find all unknowns

26. 3.5  sin 35  sin 25  = b  sin 25  = 3.5  sin 35  sin 25  sin 35  3.5 b b = b  4.8

27. Using the given information, solve the triangle. A 25 35 3.5 c  4.8 B C

28.  C = 180  – (25  + 35  )  C = 120  How do we find  C?

29. Using the given information, solve the triangle. A 25 35 3.5 c  4.8 B C 120

30. Can I use the Pythagorean Theorem to find c? Why or why not? NOT A RIGHT TRIANGLE!

31. 3.5  sin 120  sin 25  = c  sin 25  = 3.5  sin 120  sin 25  sin 120  3.5 c c = c  7.2

32. Using the given information, solve the triangle. A 25 35 3.5  7.2  4.8 B C 120

33. Your Turn

34. Using the given information, solve the triangle. A 135 10 a 45 b B C

35. 45  sin 10  sin 135  = b  sin 135  = 45  sin 10  sin 135  sin 10  45 b b = b  11.0

36. Using the given information, solve the triangle. A 135 10 a 45 11.0 B C

37.  A = 180  – (135  + 10  )  A = 35 

38. Using the given information, solve the triangle. A 135 10 a 45 11.0 B C 35

39. 45  sin 35  sin 135  = a  sin 135  = 45  sin 35  sin 135  sin 35  45 a a = a  36.5

40. Using the given information, solve the triangle. A 135 10 36.5 45 11.0 B C 35

41. Finding a Measurement Example 3

42. A pole tilts away from the sun at an 8° angle from the vertical, and it casts a 22 foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is 43°. How tall is the pole?

43. A B C 8888 43 22’ What do we need to find in order to use AAS or ASA? p  CBA

44. A B C 8888 43 22’ p = 82   CBA = 90  - 8  82  BCA = 180  - (82 + 43) = 55  55

45. 22  sin 43  sin 55  = p  sin 55  = 22  sin 43  sin 55  sin 43  22 p p = p  18.3 ft

46. What was the psychiatrist’s reply when a patient exclaimed, “I’m a teepee. I’m a wigwam!”?

47. “Relax… You’re too tense!”

48. Given Two Angles and One Side - ASA Example 4

49. Using the given information, solve the triangle. A = 102.4 , C = 16.7 , and b = 21.6

50. B C A c 102.4 16.7 21.6 a  B=180 – (102.4 + 16.7) = 60.9  60.9

51. 21.6  sin 102.4  sin 60.9  = 21.6  sin 102.4  = a  sin 60.9  sin 102.4  sin 60.9  a 21.6 a = a  24.1

52. B C A c 102.4 16.7 21.6 24.1 60.9

53. 21.6  sin 16.7  sin 60.9  = 21.6  sin 16.7  = c  sin 60.9  sin 16.7  sin 60.9  c 21.6 c = c  7.1

54. B C A 7.1 102.4 16.7 21.6 24.1 60.9

55. Your Turn

56. Using the given information, solve the triangle. A = 12.4 , C = 86.4 , and b = 22.5

57. B A C  22.7 86.4 12.4 22.5  4.9 81.2

58. 6.1 GOAL 2 How to find the areas of oblique triangles.

59. Students determine the area of triangle, given one angle and the two adjacent sides. Standard 14.0

60. A B C c sin B = h hc OR h = c sin B a A = ½bh a csin B

61. 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. That is, Area= ½bc sin A = ½ab sin C = ½ac sin B

62. Finding an Area of a Triangle Example 5

63. Find the area if C = 84  30’, a = 16, b = 20. Area = ½ab sin C = ½(16)(20)sin 84.5  Area ≈ 159.3 units 2

64. Your Turn

65. Find the area if A = 5  15’, b = 4.5, c = 22. Area = ½bc sin A = ½(4.5)(22)sin 5  15’ Area ≈ 4.5 units 2

66. Finding an Angle Example 6

67. Find  C. The area is 262 ft 2, and a = 86’ and b = 11’. Area = ½ab sin C 262 = ½(86)(11)sin C

68. sin C = 262/473 262 = 473 sin C  C ≈ 33.6   C = sin -1 (262/473)

69. Your Turn

70. Find  B. The area is 1492 ft 2, and a = 202’ and c = 66’. Area = ½ac sin B 1492 = ½(202)(66)sin B

71. sin B = 1492/6666 1492 = 6666 sin B  B ≈ 12.9   B = sin -1 (1492/6666)

72. Application Example 7

73. The bearing from Pine Knob fire tower to the Colt Station fire tower is N 65  E and the two towers are 30 km apart.

74. A fire spotted by rangers in each tower has a bearing of N 80  E from Pine Knob and S 70  E from Colt Station.

75. Find the distance of the fire from each tower.

76. N PK CS 65 N 70 80 30 km  42.4 km  15.5 km