six-trigonometric ratio grade nine .pptx

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1. R ATIO S Grade 9 Mathematics SIX TRIGONOMETRIC

2. S ides of a Triangle Hypotenuse Opposite Adjacent

3. S ides of a Triangle Hypotenuse Opposite Adjacent ᶿ

4. S ides of a Triangle Hypotenuse Opposite Adjacent ᶿ h ypotenuse

5. S ides of a Triangle Hypotenuse Opposite Adjacent ᶿ h ypotenuse o pposite

6. S ides of a Triangle Hypotenuse Opposite Adjacent ᶿ h ypotenuse opposite

7. S ides of a Triangle Hypotenuse Opposite Adjacent ᶿ h ypotenuse o pposite

8. S ides of a Triangle Hypotenuse Opposite Adjacent ᶿ h ypotenuse o pposite Adjacent

9. F ind the reciprocal of the given terms 1 . = 2. = 3. = 4. 16 = 5. 25 =  

10. F ind the reciprocal of the given terms 1 . = 2. = 3. = 4. 16 = 5. 25 =  

11. F ind the reciprocal of the given terms 1 . = 2. = 3. = 4. 16 = 5. 25 =  

12. F ind the reciprocal of the given terms 1 . = 2. = 3. = 4. 16 = 5. 25 =  

13. F ind the reciprocal of the given terms 1 . = 2. = 3. = 4. 16 = 5. 25 =  

14. F ind the reciprocal of the given terms 1 . = 2. = 3. = 4. 16 = 5. 25 =  

15. The three primary trigonometric ratios are sine, cosine and tangent Trigonometric ratio Abbreviation Ratio of Lengths sine theta sin ᶿ cosine theta cos ᶿ tangent theta tan ᶿ Trigonometric ratio Abbreviation Ratio of Lengths sine theta sin ᶿ cosine theta cos ᶿ tangent theta tan ᶿ The three secondary trigonometric ratios cosecant, secant and cotangent Trigonometric ratio Abbreviation Ratio of Lengths cosecant theta csc ᶿ secant theta sec ᶿ cotangent theta cot ᶿ Trigonometric ratio Abbreviation Ratio of Lengths cosecant theta csc ᶿ secant theta sec ᶿ cotangent theta cot ᶿ

16. The three primary trigonometric ratios are sine, cosine and tangent Trigonometric ratio Abbreviation Ratio of Lengths sine theta sin ᶿ cosine theta cos ᶿ tangent theta tan ᶿ Trigonometric ratio Abbreviation Ratio of Lengths sine theta sin ᶿ cosine theta cos ᶿ tangent theta tan ᶿ The three secondary trigonometric ratios cosecant, secant and cotangent Trigonometric ratio Abbreviation Ratio of Lengths cosecant theta csc ᶿ secant theta sec ᶿ cotangent theta cot ᶿ Trigonometric ratio Abbreviation Ratio of Lengths cosecant theta csc ᶿ secant theta sec ᶿ cotangent theta cot ᶿ SOH CAH TOA

17. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A = C A 5 7 6 B csc A = sec A = cot A =

18. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A = C A 5 7 6 B csc A = sec A = cot A = hypotenuse

19. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A = C A 5 7 6 B csc A = sec A = cot A = opposite hypotenuse

20. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A = C A 5 7 6 B csc A = sec A = cot A = opposite hypotenuse adjacent

21. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A =   C A 5 7 6 adjacent hypotenuse opposite B csc A = sec A = cot A =

22. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A =   C A 5 7 6 adjacent hypotenuse opposite B csc A = sec A = cot A =

23. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A =   C A 5 7 6 adjacent hypotenuse opposite B csc A = sec A = cot A =

24. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A =   C A 5 7 6 adjacent hypotenuse opposite B csc A = sec A = cot A =  

25. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A =   C A 5 7 6 adjacent hypotenuse opposite B csc A = sec A = cot A =  

26. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin A = cos A = tan A =   C A 5 7 6 adjacent hypotenuse opposite B csc A = sec A = cot A =  

27. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B = C A 5 7 6 B csc B = sec B = cot B =

28. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B = C A 5 7 6 B csc B = sec B = cot B = hypotenuse

29. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B = C A 5 7 6 B csc B = sec B = cot B = hypotenuse opposite

30. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B = C A 5 7 6 B csc B = sec B = cot B = hypotenuse adjacent opposite

31. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B =   C A 5 7 6 B csc B = sec B = cot B = hypotenuse adjacent opposite

32. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B =   C A 5 7 6 B csc B = sec B = cot B = hypotenuse adjacent opposite

33. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B =   C A 5 7 6 B csc B = sec B = cot B = hypotenuse adjacent opposite

34. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B =   C A 5 7 6 B csc B = sec B = cot B =   hypotenuse adjacent opposite

35. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B =   C A 5 7 6 B csc B = sec B = cot B =   hypotenuse adjacent opposite

36. EXAMPLE 1 FIND THE TRIGONOMETRIC RATIOS OF EACH OF THE ACUTE ANGLE OF THE GIVEN RIGHT TRIANGLE sin B = cos B = tan B =   C A 5 7 6 B csc B = sec B = cot B =   hypotenuse adjacent opposite

37. EXAMPLE 2 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle θ 4 3 r

38. EXAMPLE 2 θ 4 3 r Find the length of the missing side using Pythagorean Theorem.   FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

39. EXAMPLE 2 θ 4 3 r Find the length of the missing side using Pythagorean Theorem.     FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

40. EXAMPLE 2 θ 4 3 r Find the length of the missing side using Pythagorean Theorem.     FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

41. EXAMPLE 2 θ 4 3 r Find the length of the missing side using Pythagorean Theorem.     FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

42. EXAMPLE 2 θ 4 3 r Find the length of the missing side using Pythagorean Theorem.     FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

43. EXAMPLE 2 θ 4 3 r= 5 Find the length of the missing side using Pythagorean Theorem.   r   FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

44. EXAMPLE 2 csc θ = sec θ = cot θ = sin θ = cos θ = tan θ = θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

45. EXAMPLE 2 csc θ = sec θ = cot θ = sin θ = cos θ = tan θ =   θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

46. EXAMPLE 2 csc θ = sec θ = cot θ = sin θ = cos θ = tan θ =   θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

47. EXAMPLE 2 csc θ = sec θ = cot θ = sin θ = cos θ = tan θ =   θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

48. EXAMPLE 2 csc θ = sec θ = cot θ =   sin θ = cos θ = tan θ =   θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

49. EXAMPLE 2 csc θ = sec θ = cot θ =   sin θ = cos θ = tan θ =   θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

50. EXAMPLE 2 csc θ = sec θ = cot θ =   sin θ = cos θ = tan θ =   θ 4 3 r= 5 FIND THE VALUE OF SIX TRIDONOMETRIC RATIOS OF θ in a triangle

51. EXAMPLE 4 Given θ , an angle in a right triangle and sin θ = , find the remaining trigonometric ratios of θ   θ

52. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = sin X= cos X = csc Y= sec Y =

53. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = Find the length of the missing side using Pythagorean Theorem.

54. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X =   Find the length of the missing side using Pythagorean Theorem.

55. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X =   Find the length of the missing side using Pythagorean Theorem.

56. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X =   Find the length of the missing side using Pythagorean Theorem.

57. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X =   Find the length of the missing side using Pythagorean Theorem.

58. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 x   Find the length of the missing side using Pythagorean Theorem.

59. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X= csc Y= sec Y =

60. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = csc Y= sec Y = csc Y= sec Y =

61. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = csc Y= sec Y= csc Y= sec Y=

62. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = csc Y= sec Y = csc Y= sec Y =

63. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = = csc Y= sec Y = csc Y= sec Y =

64. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = = csc Y= sec Y = csc Y= sec Y = CHO csc = SHA sec = CAO cot = CHO SHA CAO

65. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = = csc Y= sec Y = CHO csc = SHA sec = CAO cot = CHO SHA CAO

66. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = = csc Y= = sec Y = CHO csc = SHA sec = CAO cot = CHO SHA CAO

67. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = = csc Y= = sec Y= CHO csc = SHA sec = CAO cot = CHO SHA CAO

68. EXAMPLE 3 Use triangle XYZ to find the four trigonometric ratio. X Y Z 10 8 X = 6 sin X= cos X = = csc Y= = sec Y= = CHO csc = SHA sec = CAO cot = CHO SHA CAO

69. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. SOH-CAH-TOA

70. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA

71. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle

72. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse

73. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse y is the adjacent side to ∠X

74. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse y is the adjacent side to ∠X Use CAH, that is cos θ =  

75. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse y is the adjacent side to ∠X cos X =   Use CAH, that is cos θ =  

76. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse y is the adjacent side to ∠X cos X = cos 54 ° =   Use CAH, that is cos θ =  

77. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. a. Solve for y . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse y is the adjacent side to ∠X y = 12cos54 ° cos X = cos 54 ° =   Use CAH, that is cos θ =  

78. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. b. Solve for x . SOH-CAH-TOA

79. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. b. Solve for x . SOH-CAH-TOA ∠X is an acute angle

80. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. b. Solve for x . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse

81. EXAMPLE 4 Y Z X y x z=12 54 ° Determine the formula to find the missing term of the triangle. b. Solve for x . SOH-CAH-TOA ∠X is an acute angle z is the hypotenuse x is the opposite side to ∠X