2. RATIOS IN RIGHT TRIANGLES OPPOSITE SIDE? ADJACENT SIDE? HYPOTENUSE? SINE? COSINE? TANGENT? INVERSE OF TRIGONOMETRIC RATIOS PROBLEM 1a PROBLEM 1b PROBLEM 3c PROBLEM 2c PROBLEM 1c PROBLEM 2b PROBLEM 3a PROBLEM 3b PROBLEM 2a Standards 15, 18, 19 PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved

3. Standard 15: Students use the pythagoream theorem to determine distance and find missing lengths of sides of right triangles. Los estudiantes usan el teorema de Pitágoras para determinar distancia y encontrar las longitudes de los lados de teoremas rectángulos. Standard 18: Students know the definitions of the basic trigonometric functions defined by the angles of a right triangle. They also know and are able to use elementary relationships between them, (e.g., tan(x)=sin(x)/cos(x), etc.) Los estudiantes conocen las definiciones de las funciones básicas trigonométricas definidas para los ángulos de triángulos rectángulos. Ellos también conocen y son capaces de usar relaciones básicas entre ellos. (ej., tan(x)=sin(x)/cos(x), etc.) Standard 19: Students use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. Los estudiantes usan funciones trigonométricas para resolver para una longitud desconocida de un triángulo rectángulo, dado un ángulo y la longitud de un lado.

4. ADJACENT SIDE HYPOTENUSE Standard 18

5. OPPOSITE SIDE Standard 18

6. ADJACENT SIDE HYPOTENUSE Standard 18

7. OPPOSITE SIDE Standard 18

8. ADJACENT SIDE HYPOTENUSE Standard 18

9. OPPOSITE SIDE Standard 18

10. ADJACENT SIDE HYPOTENUSE Standard 18

11. OPPOSITE SIDE Standard 18

12. ADJACENT SIDE HYPOTENUSE Standard 18

13. OPPOSITE SIDE Standard 18

14. ADJACENT SIDE HYPOTENUSE Standard 18

15. OPPOSITE SIDE Standard 18

16. b a c C B A The ADJACENT side to A is side “b” or CA Standard 18

17. b a c C B A The OPPOSITE side to A is side “a” or BC Standard 18

18. k m l L K M What is the adjacent side to M? The ADJACENT side to M is side “k” or LM Standard 18

19. What is the opposite side to M? The OPPOSITE side to M is side “ m ” or KL k m l L K M Standard 18

20. r t s S R T What is the adjacent side to R? The ADJACENT side to R is side “t” or RS Standard 18

21. What is the opposite side to R? The OPPOSITE side to R is side “r” or ST r t s S R T Standard 18

22. z x y Y Z X What is the adjacent side to X? The ADJACENT side to X is side “y” or XZ Standard 18

23. What is the opposite side to X? The OPPOSITE side to X is side “x” or ZY z x y Y Z X Standard 18

24. b a c R Q S What is the adjacent side to S? The ADJACENT side to S is side “b” or RS Standard 18

25. b a c What is the opposite side to S? The OPPOSITE side to S is side “a” or QR R Q S Standard 18

26. o u i C B A What is the adjacent side to C? The ADJACENT side to C is side “o” or CA Standard 18

27. C B A What is the opposite side to C? The OPPOSITE side to C is side “i” or AB o u i Standard 18

28. e i u Q R S What is the adjacent side to Q? The ADJACENT side to Q is side “e” or QS Standard 18

29. What is the opposite side to Q? The OPPOSITE side to Q is side “u” or RS e i u Q R S Standard 18

30. o u i C B A Sin C= Hypotenuse Opposite side SINE u i Standard 18

31. o u i C B A Cos C= Hypotenuse Adjacent side COSINE u o Standard 18

32. o u i C B A Tan C= Adjacent side Opposite side TANGENT i o Standard 18

33. o u i C B A Tan C= i o Adjacent side Opposite side TANGENT Sin C= i u Hypotenuse Opposite side SINE Cos C= o u Hypotenuse COSINE Adjacent side Standard 18

34. b a c C B A Sin A= a Opposite side c Hypotenuse Standard 18

35. b a c C B A Cos A= c Hypotenuse b Adjacent side Standard 18

36. b a c C B A Tan A= a Opposite side Adjacent s. b Standard 18

37. b a c C B A Sin B= c Hypotenuse b Opposite side Standard 18

38. b a c C B A Cos B= c Hypotenuse a Adjacent side Standard 18

39. b a c C B A b Opposite side Adjacent c Tan B= Standard 18

40. k m l L K M Sin M= k Hypotenuse m Opposite side Standard 18

41. k m l L K M Cos M= k Hypotenuse l Adjacent side Standard 18

42. k m l L K M Tan M= m Opposite side Adjacent l Standard 18

43. k m l L K M Sin L= k Hypotenuse l Opposite side Standard 18

44. k m l L K M Cos L= k Hypotenuse m Adjacent side Standard 18

45. k m l L K M Tan L= Adjacent m l Opposite side Standard 18

46. r t s S R T Sin T= s Hypotenuse t Opposite side Standard 18

47. r t s S R T Cos T= s Hypotenuse r Adjacent side Standard 18

48. r t s S R T Tan T= t Opposite side Adjacent r Standard 18

49. r t s S R T Sin R= s Hypotenuse r Opposite side Standard 18

50. r t s S R T Cos R= s Hypotenuse t Adjacent side Standard 18

51. r t s S R T Tan R= Adjacent t r Opposite side Standard 18

52. 4 3 5 Y Z X Sin Z= 5 Hypotenuse 4 Opposite side Sin Z= 0.8 Standard 18

53. Y Z X Cos Z= 5 Hypotenuse 3 Adjacent side 4 3 5 Cos Z= 0.6 Standard 18

54. Y Z X Tan Z= 4 Opposite side Adjacent 3 4 3 5 Tan Z= 1.3 Standard 18

55. Y Z X Sin X= 5 Hypotenuse 4 3 5 3 Opposite side Sin X=0.6 Sin ( ) = 36.86° m X = Sin ( ) m X = 36.86° 0.6 36.86° 0.6 Standard 18

56. Y Z X 4 3 5 Cos X= 5 Hypotenuse 4 Adjacent side Cos X= 0.8 Cos ( ) = 36.86° m X = Cos ( ) m X = 36.86° 0.8 36.86° 0.8 Standard 18

57. Y Z X 4 3 5 Tan X= 4 Adjacent 3 Opposite Side Tan X=.75 Tan ( )= 53.14° 36.86° m X = Tan( ) m X = 36.86° 90°- = m Z=.75 36.86°.75 36.86° 53.14° Standard 18

58. 8 17 15 Q R S Sin Q= 17 Hypotenuse 15 Opposite side Sin Q=.8829 61.93° 28.07° m Q = Sin( ) m Q = 90°- = m R=.8829 61.93° 28.07° Standard 18

59. 9 41 40 Q R S Cos Q= 41 Hypotenuse 9 Adjacent side Cos Q=.2195 77.31° 12.69° m Q = Cos( ) m Q = 90°- = m R=.2195 77.31° 12.69° Standard 18

60. 9 15 12 Q R S Tan R= 12 Adjacent 9 Opposite Side Tan R=.75 53.14° 36.86° m R = Tan ( ) m R = 90°- = m Q=.75 36.86° 53.14° 36.86° Standard 18

61. 24 51 45 Q R S Sin Q= 51 Hypotenuse 45 Opposite side Sin Q=.8829 61.93° 28.07° m Q = Sin( ) m Q = 90°- = m R=.8829 61.93° 28.07° 61.93° Standard 18

62. 21 75 72 Q R S Cos R= 75 Hypotenuse 72 Adjacent side Cos R=.96 73.31° 16.26° m R = Cos( ) m R = 90°- = m Q=.96 16.26° 73.31° 16.26° Standard 18

63. 36 i 48 Q R S Tan R= 48 Adjacent 36 Opposite Side Tan R=.75 53.14° 36.86° m R = Tan ( ) m R = 36 + 48 = i 22 2 i = 3600 2 2 1296 + 2304 = i 2 |i|=60 i=60 and i=-60 90°- = m Q=.75 36.86° 53.14° 36.86° SOLVE QRS: Standards 15, 18, 19

64. Tan F= Tan F=.7446 m F = Tan ( ) m F = u = 3434 2 2 1225 + 2209 = u 2 |u|=58.6 u=58.6 and u=-58.6 90°- = m H=.7446 36.67° SOLVE FGH: 47 35 u G F H 35 + 47 = u 22 2 47 Adjacent 35 Opposite Side 36.67° 53.32° Standards 15, 18, 19

65. 30 i u Q R S 35° SOLVE SRQ: Sin ( )= u 30 Sin S= Opposite side Sin 35°= u 30 (30) u=30 Sin 35° u=30( ) u=17.2 55° Cos 35°= i Adjacent side Cos 35°= i 30 (30) i=30 Cos 35° i=30( ) i=24.57 90°- = m Q= Hypotenuse 30 Hypotenuse 35°.5736 35° 55°.8192 Standards 15, 18, 19

66. 45 i u K L M 35° SOLVE LMK: Sin( )= i 45 Sin K= Opposite side Sin 55°= i 45 (45) i=45 Sin 55° i=45( ) i=36.86 55° Cos 55°= u Adjacent side Cos 55°= u 45 (45) u=45 Cos 55° u=45( ) u=25.81 90°- = m M= Hypotenuse 45 Hypotenuse 55°.8191 55° 35°.5735 Standards 15, 18, 19

67. e i 9 Q R S 30° Tan ( ) = 9 i 30° = Tan ( 30° ) 9 i 1 i = 9 Tan ( 30° ) i = 9.5774 i =15.58 e = 324 2 2 243 + 81 = e 2 |e|=18 e=18 and e=-18 15.58 + 9 = e 2 2 2 90°- = m Q= 30° 60° SOLVE QRS: i Tan(30°) = 9 Tan(30°) Standards 15, 18, 19

68. Y Z X i 6 a 36° Tan ( ) = 36° = Tan ( 36° ) 6 i 1 i = 6 Tan ( 36° ) i = 6.7265 i =8.26 90°- = m Z= 36° 54° a = 104.2 2 2 68.2 + 36 = a 2 |a|=10.2 a=10.2 and a=-10.2 i 8.26 + 6 = a 2 2 2 6 54° i Tan(36°) = 6 Tan(36°) Standards 15, 18, 19

69. Tan F= Tan F=.7391 m F = Tan ( ) m F = u = 818 2 2 289 + 529 = u 2 |u|=28.6 u=28.6 and u=-28.6 90°- = m H=.7391 36.46° SOLVE FGH: 23 17 u G F H 17 + 23 = u 22 2 23 Adjacent 17 Opposite Side 36.46° 53.53° Standards 15, 18, 19

70. 25 i u K L M 33° SOLVE LMK: Sin( )= i 25 Sin K= Opposite side Sin 57°= i 25 (25) i=25 Sin 57° i=25( ) i=20.96 57° Cos 57°= u Adjacent side Cos 57°= u 25 (25) u=25 Cos 57° u=25( ) u=13.61 90°- = m M= Hypotenuse 25 Hypotenuse 57°.8386 57° 33°.5446 Standards 15, 18, 19

71. Y Z X i 7 a 29° Tan ( ) = 29° = Tan ( 29° ) 7 i 1 i = 7 Tan ( 29° ) i = 7.5543 i =12.62 90°- = m Z= 29° 61° a = 208.5 2 2 159.5 + 49 = a 2 |a|=14.4 a=14.4 and a=-14.4 i 12.6 + 7 = a 2 2 2 7 61° i Tan(29°) = 7 Tan(29°) Standards 15, 18, 19