1. Some Basic Geometric Facts of Triangles

2. Some Basic Geometric Facts of Triangles Standard labeling of a triangle

5. (Triangle Inequality) The sum of the lengths of any two sides is more than the length of third side.Standard labeling of a triangle

7. (Triangle Inequality) The sum of the lengths of any two sides is more than the length of third side.

8. The longest side is opposite to the largest angle.The medium side is opposite to the medium angle. The shortest side is opposite to the smallest angle. Standard labeling of a triangle

10. (Triangle Inequality) The sum of the lengths of any two sides is more than the length of third side.

12. Sine and Cosine Laws There are two important theorems concerning the sides and the trig-values of the angles of a given triangle.

13. Sine and Cosine Laws There are two important theorems concerning the sides and the trig-values of the angles of a given triangle. They are the Sine Law and the Cosine Law.

14. Sine and Cosine Laws There are two important theorems concerning the sides and the trig-values of the angles of a given triangle. They are the Sine Law and the Cosine Law. Sine Law: For any triangle,

15. Sine and Cosine Laws There are two important theorems concerning the sides and the trig-values of the angles of a given triangle. They are the Sine Law and the Cosine Law. Sine Law: For any triangle, c a b = = sin(A) sin(C) sin(B) or sin(A) sin(B) sin(C) = = c a b

16. Sine and Cosine Laws There are two important theorems concerning the sides and the trig-values of the angles of a given triangle. They are the Sine Law and the Cosine Law. Sine Law: For any triangle, c a b = = sin(A) sin(C) sin(B) or sin(A) sin(B) sin(C) = = c a b When we find all the angles and sides of a triangle, we say we "solve the triangle".

17. Sine and Cosine Laws There are two important theorems concerning the sides and the trig-values of the angles of a given triangle. They are the Sine Law and the Cosine Law. Sine Law: For any triangle, c a b = = sin(A) sin(C) sin(B) or sin(A) sin(B) sin(C) = = c a b When we find all the angles and sides of a triangle, we say we "solve the triangle". When given two angles of a triangle (whose sum is less than 180o) and the side between them, we use the Sine Law to solve the triangle.

18. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label.

19. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. 16

20. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. A 29 16

21. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. C A 29 74 16

22. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. A C 29 74 16

23. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: B c a A C 29 74 16

24. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 B c a A 29 74 16

25. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) B c a A C 29 74 16

26. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) 16sin(29) so a = sin(77) B c a A C 29 74 16

27. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) 16sin(29)  7.96 so a = sin(77) B c a A C 29 74 16

28. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) 16sin(29)  7.96 so a = sin(77) B c 16 = c a sin(74) sin(77) A C 29 74 16

29. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) 16sin(29)  7.96 so a = sin(77) B c 16 = c a sin(74) sin(77) A C 29 74 16 16sin(74) so c = sin(77)

30. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) 16sin(29)  7.96 so a = sin(77) B c 16 = c a sin(74) sin(77) A C 29 74 16 16sin(74) 15.8 so c = sin(77)

31. Sine and Cosine Laws Example A: Solve for the triangle given that <A=29o, b=16, <C=74o. Draw and label. To solve the triangle: <B = 180 – 29 – 74 = 77 a 16 = sin(29) sin(77) 16sin(29)  7.96 so a = sin(77) B c 16 = c a sin(74) sin(77) A C 29 74 16 16sin(74) 15.8 so c = sin(77) The Cosine Law gives the needed adjustment to the Pythygorean Theorem for non-right-triangles.

32. Sine and Cosine Laws Cosine Law:

33. Sine and Cosine Laws Cosine Law: For any triangle,

34. Sine and Cosine Laws Cosine Law: For any triangle, c2 = a2 + b2

35. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C)

36. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C) Similarly, b2 = a2 + c2 – 2*a*c*cos(B) a2 = b2 + c2 – 2*b*c*cos(A)

37. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C) Similarly, b2 = a2 + c2 – 2*a*c*cos(B) a2 = b2 + c2 – 2*b*c*cos(A) Example B: Given that a=29, <B=16o, c=74. draw and find side b.

38. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C) Similarly, b2 = a2 + c2 – 2*a*c*cos(B) a2 = b2 + c2 – 2*b*c*cos(A) Example B: Given that a=29, <B=16o, c=74. draw and find side b.

39. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C) Similarly, b2 = a2 + c2 – 2*a*c*cos(B) a2 = b2 + c2 – 2*b*c*cos(A) Example B: Given that a=29, <B=16o, c=74. draw and find side b. Use the Cosine Law to find b.

40. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C) Similarly, b2 = a2 + c2 – 2*a*c*cos(B) a2 = b2 + c2 – 2*b*c*cos(A) Example B: Given that a=29, <B=16o, c=74. draw and find side b. Use the Cosine Law to find b. b2 = 292 + 742 – [ 2(29)(74)cos(16) ]

41. Sine and Cosine Laws Cosine Law: Adjustment to the Pythagorean Theorem For any triangle, c2 = a2 + b2 – 2*a*b*cos(C) Similarly, b2 = a2 + c2 – 2*a*c*cos(B) a2 = b2 + c2 – 2*b*c*cos(A) Example B: Given that a=29, <B=16o, c=74. draw and find side b. Use the Cosine Law to find b. b2 = 292 + 742 – [ 2(29)(74)cos(16) ] b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8

42. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions.

43. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions.

44. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value.

45. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, ….

46. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, …. We select π/6 as the inverse valuesof sine for ½.

47. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, …. We select π/6 as the inverse valuesof sine for ½. We write it as sin-1(1/2) = π/6.

48. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, …. We select π/6 as the inverse valuesof sine for ½. We write it as sin-1(1/2) = π/6. In general we select the inverse values of sine to be the angles between –π/2 and π/2.

49. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, …. We select π/6 as the inverse valuesof sine for ½. We write it as sin-1(1/2) = π/6. In general we select the inverse values of sine to be the angles between –π/2 and π/2. Specifically, Given the number a such that -1 < a <1,

50. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, …. We select π/6 as the inverse valuesof sine for ½. We write it as sin-1(1/2) = π/6. In general we select the inverse values of sine to be the angles between –π/2 and π/2. Specifically, Given the number a such that -1 < a <1, we define * sin-1(a) =  where sin() = a and –π/2 <  < π/2.

51. Inverse Trigonometric Functions To find angles, we need theinverse trig-functions. But trig-functions are not one to one, therefore the inverse procedure is not functions. We have to select one value as the inverse value. For example, if sin() = 1/2, then  = -11π/6, π/6, 13π/6, …. We select π/6 as the inverse valuesof sine for ½. We write it as sin-1(1/2) = π/6. In general we select the inverse values of sine to be the angles between –π/2 and π/2. Specifically, Given the number a such that -1 < a <1, we define * sin-1(a) =  where sin() = a and –π/2 <  < π/2. (sin-1(a) is read as "sine-inverse of a")

52. Inverse Trigonometric Functions In a similar manner, we define the inverse cosine.

53. Inverse Trigonometric Functions In a similar manner, we define the inverse cosine. Given the number a such that -1 < a <1, cos-1(a)= where cos() = a and 0 <  < π (cos-1(a) is read as "cosine-inverse of a")

54. Inverse Trigonometric Functions In a similar manner, we define the inverse cosine. Given the number a such that -1 < a <1, cos-1(a)= where cos() = a and 0 <  < π (cos-1(a) is read as "cosine-inverse of a") Since tangent may output any real numbers, the domain of the inverse tangent is the set of all numbers.

55. Inverse Trigonometric Functions In a similar manner, we define the inverse cosine. Given the number a such that -1 < a <1, cos-1(a)= where cos() = a and 0 <  < π (cos-1(a) is read as "cosine-inverse of a") Since tangent may output any real numbers, the domain of the inverse tangent is the set of all numbers. Given the number a such that -1 < a <1, tan-1(a)= where tan() = a and 0 <  < π (tan-1(a) is read as "tangent-inverse of a")

56. Inverse Trigonometric Functions In a similar manner, we define the inverse cosine. Given the number a such that -1 < a <1, cos-1(a)= where cos() = a and 0 <  < π (cos-1(a) is read as "cosine-inverse of a") Since tangent may output any real numbers, the domain of the inverse tangent is the set of all numbers. Given the number a such that -1 < a <1, tan-1(a)= where tan() = a and 0 <  < π (tan-1(a) is read as "tangent-inverse of a") We also write arcsin(a), arccos(a) and arctan(a), for sin-1(a), cos-1(a) and tan-1(a).

57. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.

58. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) sin-1(a) =  ,  is between π/2 and π/2

59. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between -π/2 and π/2

60. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between -π/2 and π/2 tan-1(a) =  ,  is between -π/2 and π/2

61. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between -π/2 and π/2 tan-1(a) =  ,  is between -π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

62. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between π/2 and π/2 tan-1(a) =  ,  is between π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

63. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between π/2 and π/2 tan-1(a) =  ,  is between π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

64. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between π/2 and π/2 tan-1(a) =  ,  is between π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

65. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between π/2 and π/2 tan-1(a) =  ,  is between π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

66. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between π/2 and π/2 tan-1(a) =  ,  is between π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

67. Inverse Trigonometric Functions Following are the pictures of the inverse trig-functions.  = sin-1(a) a cos-1(a) =  ,  is between 0 and π sin-1(a) =  ,  is between π/2 and π/2 tan-1(a) =  ,  is between π/2 and π/2 Example C: a. sin-1(1/2) = π/6 b. cos-1(-1/2) = 2π/3 c. cos-1(-1) = π d. sin-1(2) = UND e. cos-1(0.787)  38.1o f. tan-1(3) = π/3

68. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74.

69. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8.

70. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8. We need to solve for the angles.

71. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8. We need to solve for the angles. Note that <C might be more than 90o, hence we can't solve it with the sine-inverse (why?).

72. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8. We need to solve for the angles. Note that <C might be more than 90o, hence we can't solve it with the sine-inverse (why?). Therefore, we solve for <A first via the Sine Law .

73. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8. We need to solve for the angles. Note that <C might be more than 90o, hence we can't solve it with the sine-inverse (why?). Therefore, we solve for <A first via the Sine Law . sin(A) sin(16) = 29 46.8

74. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8. We need to solve for the angles. Note that <C might be more than 90o, hence we can't solve it with the sine-inverse (why?). Therefore, we solve for <A first via the Sine Law . 29sin(16) sin(A) sin(16) sin(A)  = =  0.171 46.8 29 46.8

75. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8. We need to solve for the angles. Note that <C might be more than 90o, hence we can't solve it with the sine-inverse (why?). Therefore, we solve for <A first via the Sine Law . 29sin(16) sin(A) sin(16) sin(A)  = =  0.171 46.8 29 46.8 Hence A = sin-1(0.171)  9.83o

76. Inverse Trigonometric Functions Example D (Continued with B): Solve the triangle with a=29, <B=16o, c=74. We have from example B that b  46.8 We need to solve for the angles. Note that <C might be more than 90o, hence we can't solve it with the sine-inverse (why?). Therefore, we solve for <A first via the Sine Law . 29sin(16) sin(A) sin(16) sin(A)  = =  0.171 46.8 29 46.8 Hence A = sin-1(0.171)  9.83o and C = 180 – 16 – 9.83  154.2o

77. Inverse Trigonometric Functions We may draw right triangles that represent the inverse trig-values.

78. Inverse Trigonometric Functions We may draw right triangles that represent the inverse trig-values. Example D: Draw the angle cos-1(2/a) and find tan(cos-1(2/a) ).

79. Inverse Trigonometric Functions We may draw right triangles that represent the inverse trig-values. Example D: Draw the angle cos-1(2/a) and find tan(cos-1(2/a) ). Let α = cos-1(2/a), the following triangle represents α. a α 2

80. Inverse Trigonometric Functions We may draw right triangles that represent the inverse trig-values. Example D: Draw the angle cos-1(2/a) and find tan(cos-1(2/a) ). Let α = cos-1(2/a), the following triangle represents α. Hence the third side is a2 – 4 a a2 – 4 α 2

81. Inverse Trigonometric Functions We may draw right triangles that represent the inverse trig-values. Example D: Draw the angle cos-1(2/a) and find tan(cos-1(2/a) ). Let α = cos-1(2/a), the following triangle represents α. Hence the third side is a2 – 4 Thererfore, tan(cos-1(2/a)) = tan(α) a a2 – 4 α 2

82. Inverse Trigonometric Functions We may draw right triangles that represent the inverse trig-values. Example D: Draw the angle cos-1(2/a) and find tan(cos-1(2/a) ). Let α = cos-1(2/a), the following triangle represents α. Hence the third side is a2 – 4 Thererfore, tan(cos-1(2/a)) = tan(α) = a a2 – 4 α 2 a2 – 4 2