6 trigonometric functions sohcahtoa-nat

trigonometric functions sohcahtoa

1. SOHCAHTOA

2. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA

3. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . θ O pposite A djacent H ypotenuse

4. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . We define the following ratios: θ O pposite A djacent H ypotenuse

5. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . S in( θ ) = opposite hypotenuse = O H We define the following ratios: θ O pposite A djacent H ypotenuse

6. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . S in( θ ) = opposite hypotenuse = O H C os( θ ) = adjacent hypotenuse = A H We define the following ratios: θ O pposite A djacent H ypotenuse

7. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . S in( θ ) = opposite hypotenuse = O H C os( θ ) = adjacent hypotenuse = A H T an( θ ) = adjacent = O A opposite We define the following ratios: θ O pposite A djacent H ypotenuse

8. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . S in( θ ) = opposite hypotenuse = O H C os( θ ) = adjacent hypotenuse = A H T an( θ ) = adjacent = O A opposite We define the following ratios: Hence the SOHCAHTOA . θ O pposite A djacent H ypotenuse

9. Right triangles are differentiated in shape by the ratios of their sides. This gives us the SOHCAHTOA definition of the trigonometric functions. SOHCAHTOA Given the angle θ in a right triangle as shown, the three sides may be labeled as the opposite, the adjacent, or the hypotenuse with respect to θ . S in( θ ) = opposite hypotenuse = O H C os( θ ) = adjacent hypotenuse = A H T an( θ ) = adjacent = O A opposite We define the following ratios: Hence the SOHCAHTOA . Note that these ratios depend on A but not on the size of the right triangle. θ O pposite A djacent H ypotenuse

10. Example A. Given the following right triangles, a . find sin( θ ) and tan( θ ). b. find sin(B) and tan(A) in terms of b. SOHCAHTOA θ O pposite A djacent H ypotenuse S ( θ ) = O H C ( θ ) = A H T( θ ) = O A θ 2 3 A b B 4

11. Example A. Given the following right triangles, a . find sin( θ ) and tan( θ ). b. find sin(B) and tan(A) in terms of b. SOHCAHTOA θ O pposite A djacent H ypotenuse S ( θ ) = O H C ( θ ) = A H T( θ ) = O A θ 2 3 x =√ 5 By the Pythagorean Theorem x 2 + 2 2 = 3 2 so x 2 = 5 or x = √ 5 A b B 4

12. Example A. Given the following right triangles, a . find sin( θ ) and tan( θ ). b. find sin(B) and tan(A) in terms of b. SOHCAHTOA θ O pposite A djacent H ypotenuse S ( θ ) = O H C ( θ ) = A H T( θ ) = O A θ 2 3 x =√ 5 By the Pythagorean Theorem x 2 + 2 2 = 3 2 so x 2 = 5 or x = √5 Hence sin( θ ) = 2/3 and tan( θ ) = 2/√ 5 . A b B 4

13. Example A. Given the following right triangles, a . find sin( θ ) and tan( θ ). b. find sin(B) and tan(A) in terms of b. SOHCAHTOA θ O pposite A djacent H ypotenuse S ( θ ) = O H C ( θ ) = A H T( θ ) = O A θ 2 3 x =√ 5 By the Pythagorean Theorem x 2 + 2 2 = 3 2 so x 2 = 5 or x = √5 Hence sin( θ ) = 2/3 and tan( θ ) = 2/√ 5 . A b B 4 We have that c 2 = b 2 + 4 2 so c = √ b 2 + 16. (√b 2 +16)

14. Example A. Given the following right triangles, a . find sin( θ ) and tan( θ ). b. find sin(B) and tan (A ) in terms of b. SOHCAHTOA θ O pposite A djacent H ypotenuse S ( θ ) = O H C ( θ ) = A H T( θ ) = O A θ 2 3 x =√ 5 By the Pythagorean Theorem x 2 + 2 2 = 3 2 so x 2 = 5 or x = √5 Hence sin( θ ) = 2/3 and tan( θ ) = 2/√ 5 . A b B 4 We have that c 2 = b 2 + 4 2 so c = √ b 2 + 16. Therefore sin(B) = b/ √ b 2 + 16 and tan (A ) = 4/b. (√b 2 +16)

15. SOHCAHTOA Applying the SOHCAHTOA to the 30–60 and 45–45 triangles, we have the following important trig-values. They are to be memorized! sin( π /4 ) =  2/2 ≈ 0.707 cos( π /4 ) =  2/2 tan( π /4 ) = 1 sin( π /6 ) = 1/2 cos( π /6 ) =  3/2 ≈ 0.866 tan( π /6 ) = 1/  3 ≈ 0.577 sin( π /3 ) =  3 /2 cos( π /3 ) = 1 /2 tan( π /3 ) =  3 ≈ 1.73 sin( ) = sin( π /2 ) = 1 cos( ) = 1 cos( π /2 ) =

16. Given the angle A and one side of a right triangle, we can solve for the other two sides . SOHCAHTOA

17. 34 o 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA A

18. <B = 90 – 34 = 56 o 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA

19. <B = 90 – 34 = 56 o tan(56 o ) = b/23  23tan(56 o ) = b Hence b  34.1” 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA

20. <B = 90 – 34 = 56 o tan(56 o ) = b/23  23tan(56 o ) = b Hence b  34.1” sin(34 o ) = 23/c  c = 23/sin(34 o ) Hence c  41.1” 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA

21. <B = 90 – 34 = 56 o tan(56 o ) = b/23  23tan(56 o ) = b Hence b  34.1” sin(34 o ) = 23/c  c = 23/sin(34 o ) Hence c  41.1” 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA π /3 1 x π /6 b. Given the following right-triangle , find x.

22. <B = 90 – 34 = 56 o tan(56 o ) = b/23  23tan(56 o ) = b Hence b  34.1” sin(34 o ) = 23/c  c = 23/sin(34 o ) Hence c  41.1” 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA π /3 1 x π /6 (h) Let h be as shown then h/1 = tan( π /3 ) = √3. b. Given the following right-triangle , find x.

23. <B = 90 – 34 = 56 o tan(56 o ) = b/23  23tan(56 o ) = b Hence b  34.1” sin(34 o ) = 23/c  c = 23/sin(34 o ) Hence c  41.1” 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. SOHCAHTOA π /3 1 x π /6 (h) Let h be as shown then h/1 = tan( π /3 ) = √3. On the other hand sin( π /6 ) = h/x = ½ b. Given the following right-triangle , find x.

24. <B = 90 – 34 = 56 o tan(56 o ) = b/23  23tan(56 o ) = b Hence b  34.1” sin(34 o ) = 23/c  c = 23/sin(34 o ) Hence c  41.1” 34 o A 23 b c B Given the angle A and one side of a right triangle, we can solve for the other two sides . Example B. a. Given the following information, find the missing angle and sides. b. Given the following right-triangle , find x. SOHCAHTOA π /3 1 x π /6 (h) Let h be as shown then h/1 = tan( π /3 ) = √3. On the other hand sin( π /6 ) = h/x = ½ so x = 2h = 2 √3.

25. From a low point, the angle of looking up, measured against the horizon, is called the angle of elevation. Angles of Elevation and Depression angle of elevation

26. From a low point, the angle of looking up, measured against the horizon, is called the angle of elevation. Angles of Elevation and Depression angle of elevation angle of depression From a high point, the angle of looking down, measured against the horizon, is called the angle of depression .

27. From a low point, the angle of looking up, measured against the horizon, is called the angle of elevation. Angles of Elevation and Depression angle of elevation angle of depression From a high point, the angle of looking down, measured against the horizon, is called the angle of depression . angle of elevation = angle of depression = Note:

28. Many distance problems reduce to measuring right triangles. Some problems of Distances Example C. A girl with eye-level at 5-ft is looking up at a painting hung on a wall that is 12 ft away. The angle of elevation to the bottom of the painting is 12 o . The angle of elevation to the top of the painting is 52 o . How high is the bottom of the painting off the floor and what is the height of the painting? Draw.

29. Many distance problems reduce to measuring right triangles. Some problems of Distances 5 Example C. A girl with eye-level at 5-ft is looking up at a painting hung on a wall that is 12 ft away. The angle of elevation to the bottom of the painting is 12 o . The angle of elevation to the top of the painting is 52 o . How high is the bottom of the painting off the floor and what is the height of the painting? Draw.

30. Many distance problems reduce to measuring right triangles. Some problems of Distances 5 Example C. A girl with eye-level at 5-ft is looking up at a painting hung on a wall that is 12 ft away. The angle of elevation to the bottom of the painting is 12 o . The angle of elevation to the top of the painting is 52 o . How high is the bottom of the painting off the floor and what is the height of the painting? Draw.

31. Many distance problems reduce to measuring right triangles. Some problems of Distances 5 h Example C. A girl with eye-level at 5-ft is looking up at a painting hung on a wall that is 12 ft away. The angle of elevation to the bottom of the painting is 12 o . The angle of elevation to the top of the painting is 52 o . How high is the bottom of the painting off the floor and what is the height of the painting? Draw. x Let x and h be as shown in the figure.

32. 5 h x Some problems of Distances Find the x first.

33. 5 h x Some problems of Distances Find the x first. tan(12) = x/12,

34. 5 h x Some problems of Distances Find the x first. tan(12) = x/12 , so x = 12tan(12)  2.55 ft

35. 5 h x Some problems of Distances Find the x first. tan(12) = x/12 , so x = 12tan(12)  2.55 ft So the painting is 5 + 2.55 = 7.55 ft off the ground .

36. 5 h x Some problems of Distances Find the x first. tan(12) = x/12 , so x = 12tan(12)  2.55 ft So the painting is 5 + 2.55 = 7.55 ft off the ground . We also have that ( h + 2.55)/ 12 = tan(52) or h + 2.55 = 12tan( 52)

37. 5 h x Some problems of Distances Find the x first. tan(12) = x/12 , so x = 12tan(12)  2.55 ft So the painting is 5 + 2.55 = 7.55 ft off the ground . We also have that ( h + 2.55)/ 12 = tan(52) or h + 2.55 = 12tan( 52) or that h = 12tan(52) – 2.55

38. 5 h x Some problems of Distances Find the x first. tan(12) = x/12 , so x = 12tan(12)  2.55 ft So the painting is 5 + 2.55 = 7.55 ft off the ground . We also have that ( h + 2.55)/ 12 = tan(52) or h + 2.55 = 12tan( 52) or that h = 12tan(52) – 2.55 So h  12.8 ft is the height of the painting .

39. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. Some problems of Distances 46 o

40. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? Some problems of Distances 46 o

41. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o Some problems of Distances 46 o 20 ’

42. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o Some problems of Distances 46 o h=? 20 ’

43. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. x 20 ’

44. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)* x x 20 ’

45. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20 ) x 20 ’

46. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20) Equating these two, tan(56)x = tan(46)(x + 20 ) x 20 ’

47. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20) Equating these two, tan(56)x = tan(46)(x + 20) tan(56)x = tan(46)x + 20tan(46 ) x 20 ’

48. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20) Equating these two, tan(56)x = tan(46)(x + 20) tan(56)x = tan(46)x + 20tan(46) tan(56)x – tan(46)x = 20tan(46 ) x 20 ’

49. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20) Equating these two, tan(56)x = tan(46)(x + 20) tan(56)x = tan(46)x + 20tan(46) tan(56)x – tan(46)x = 20tan(46) x(tan(56) – tan(46)) = 20tan(46 ) x 20 ’

50. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20) Equating these two, tan(56)x = tan(46)(x + 20) tan(56)x = tan(46)x + 20tan(46) tan(56)x – tan(46)x = 20tan(46) x(tan(56) – tan(46)) = 20tan(46 ) hence x  46.3’ 20tan(46) x = tan(56) – tan(46) 20 ’

51. Example D. A drone spots a whale straight ahead in the ocean with an 46 o angle of depression. It flew 20 feet straight ahead, the angle of depression changes to 56 o . How high is the drone flying? 5 6 o h=? Some problems of Distances 46 o 20 ’ Label the x as shown. h/x = tan(56) so h = tan(56)*x h/(x + 20) = tan(46), hence h = tan(46)(x + 20) Equating these two, tan(56)x = tan(46)(x + 20) tan(56)x = tan(46)x + 20tan(46) tan(56)x – tan(46)x = 20tan(46) x(tan(56) – tan(46)) = 20tan(46 ) hence x  46.3’ and h = tan(56 ) x  68.7’. 20tan(46) x = tan(56) – tan(46)

52. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Some problems of Distances 72 o 120 ’ 86 o

53. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . Some problems of Distances 72 o 120 ’ 86 o

54. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances 72 o 120 ’ 86 o

55. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. 72 o 120 ’ 86 o h x

56. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. Find x first. 120/x = tan(72 ) 72 o 120 ’ 86 o h x 72 o

57. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. Find x first. 120/x = tan(72) so x = 120/tan(72)  39 72 o 120 ’ 86 o h x 72 o

58. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. Find x first. 120/x = tan(72) so x = 120/tan(72)  39 To find h, h/x = tan(86 ) 72 o 120 ’ 86 o h x 72 o

59. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. Find x first. 120/x = tan(72) so x = 120/tan(72)  39 To find h, h/x = tan(86) so h = tan(86) x 72 o 120 ’ 86 o h x 72 o

60. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. Find x first. 120/x = tan(72) so x = 120/tan(72)  39 To find h, h/x = tan(86) so h = tan(86) x or that h = tan(86) 120/tan(72)  558 ’ 72 o 120 ’ 86 o h x 72 o

61. Example E. A person at the top of a 120-foot building is looking down at the base of the building across the chasm with angle of depression 72 o . Another person at the base of the same building is looking up at the top of the building across the chasm at an angle of elevation of 86 o . How tall is the building across the chasm? Some problems of Distances Let x and h be as shown. Find x first. 120/x = tan(72) so x = 120/tan(72)  39 To find h, h/x = tan(86) so h = tan(86) x or that h = tan(86) 120/tan(72)  558 ’ 72 o 120 ’ 86 o h x 72 o

62. Solving for Angles of Right Triangles In many practical problems, we may reduce the context to finding the height h of a wall with the following setup.

63. Solving for Angles of Right Triangles In many practical problems, we may reduce the context to finding the height h of a wall with the following setup. A slide of is leaning against a wall with an angle A of elevation as shown. h slide floor A

64. Solving for Angles of Right Triangles In many practical problems, we may reduce the context to finding the height h of a wall with the following setup. A slide of is leaning against a wall with an angle A of elevation as shown. h slide floor A tan(A)* floor = h t hen tan(A)* floor = h

65. Solving for Angles of Right Triangles In many practical problems, we may reduce the context to finding the height h of a wall with the following setup. h slide floor A t hen tan(A)* floor = h sin(A)* slide = h tan(A)* floor = h sin(A)* slide = h A slide of is leaning against a wall with an angle A of elevation as shown.

66. Solving for Angles of Right Triangles For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o .

67. Given the triangle as shown, w e have that sin(A) = 2/5, Solving for Angles of Right Triangles A 2 5 For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o .

68. Given the triangle as shown, w e have that sin(A) = 2/5, so to find the angle A, apply the sine–inverse to 2/5 so A = sin –1 (2/5) ≈ 23.6 o . Solving for Angles of Right Triangles A 2 5 For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o .

69. Given the triangle as shown, w e have that sin(A) = 2/5, so to find the angle A, apply the sine–inverse to 2/5 so A = sin –1 (2/5) ≈ 23.6 o . Solving for Angles of Right Triangles A 2 5 For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o . Similarly, for 0 ≤ s ≤ 1, cos –1 (s ) = B , i.e. the cosine–inverse of s is B, if cos(B) = s and 0 o ≤ B ≤ 90 o .

70. Given the triangle as shown, w e have that sin(A) = 2/5, so to find the angle A, apply the sine–inverse to 2/5 so A = sin –1 (2/5) ≈ 23.6 o . Solving for Angles of Right Triangles A 2 5 For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o . Similarly, for 0 ≤ s ≤ 1, cos –1 (s ) = B , i.e. the cosine–inverse of s is B, if cos(B) = s and 0 o ≤ B ≤ 90 o . B 2 5 Hence B = cos –1 (2/5) ≈ 66.4 o

71. Given the triangle as shown, w e have that sin(A) = 2/5, so to find the angle A, apply the sine–inverse to 2/5 so A = sin –1 (2/5) ≈ 23.6 o . Solving for Angles of Right Triangles A 2 5 For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o . Similarly, for 0 ≤ s ≤ 1, cos –1 (s ) = B , i.e. the cosine–inverse of s is B, if cos(B) = s and 0 o ≤ B ≤ 90 o . B 2 5 Hence B = cos –1 (2/5) ≈ 66.4 o since cos( 66.4 o ) = 2/5 and o ≤ 66.4 o ≤ 90 o .

72. Given the triangle as shown, w e have that sin(A) = 2/5, so to find the angle A, apply the sine–inverse to 2/5 so A = sin –1 (2/5) ≈ 23.6 o . Solving for Angles of Right Triangles A 2 5 For any number 0 ≤ s ≤ 1, the sine–inverse of s, sin –1 (s ) = A is the angle where sin(A) = s with o ≤ A ≤ 90 o . Similarly, for 0 ≤ s ≤ 1, cos –1 (s ) = B , i.e. the cosine–inverse of s is B, if cos(B) = s and 0 o ≤ B ≤ 90 o . B 2 5 Hence B = cos –1 (2/5) ≈ 66.4 o since cos( 66.4 o ) = 2/5 and o ≤ 66.4 o ≤ 90 o . We will extend these inverse formulas to the inverses of negative trig–values later.

73. SOHCAHTOA Applying the SOHCAHTOA to the 30–60 and 45–45 triangles, we have the following important trig-values. They are to be memorized! sin( π /4 ) =  2/2 ≈ 0.707 cos( π /4 ) =  2/2 tan( π /4 ) = 1 sin( π /6 ) = 1/2 cos( π /6 ) =  3/2 ≈ 0.866 tan( π /6 ) =  3/3 ≈ 0.577 sin( π /3 ) =  3 /2 cos( π /3 ) = 1 /2 tan( π /3 ) =  3 ≈ 1.73 sin( ) = sin( π /2 ) = 1 cos( ) = 1 cos( π /2 ) =

74. Ex. A. G iven the following right triangles, find sin(A), cos(A), tan(A), sin(B), cos(B) and tan(B). SOHCAHTOA A 1. B A 5 A x A B 9 A x A x 2 3 7. 8. 6. 5. 4. 3. 2. 9. A A B B B B B B B B 1 6 5 4 1 4 4 2 4 3 4

75. 34 o A 23 Ex. B. Find all sides and angles of the following right triangles. SOHCAHTOA 1. 64 o A 23 B 54 o A 15 82 74 o A 64 o x 82 74 o A A 29 64 o A 64 o x A x 55 o x – 1 x+5 2x – 1 7. 8. 6. 5. 4. 3. 2. 9. For 7, 8 and 9, find x.

76. Ex. C. Solve for x. Give the answer in the exact form and its approximate value. SOHCAHTOA 1. 3. 2. π /4 1 x π /6 π /4 1 x π /6 π /4 1 x π /6 4 . 6 . 5. π /3 1 x π /6 π /3 1 x π /6 π /3 1 x π /6 7 . 9 . 8 . π /3 1 x π /4 π /3 1 x π /4 π /3 1 x π /4

77. Ex. D. Solve for x and h. Give the answers in the exact form and its approximate values. SOHCAHTOA 1. 3. 2. 46 o h x 5 2 o 30 46 o h x 5 2 o 30 46 o h x 5 2 o 30 41 o h x 32 o 30 41 o h x 32 o 30 41 o h x 32 o 30 4. 6. 5. 7 . 9 . 8 . 20 x 20 x x 71 o 49 o 71 o 49 o 71 o 49 o h h 20 h

78. Ex. E . Solve. Give the answers in the exact form and its approximate values. SOHCAHTOA 1. A flagpole casts a 42-foot shadow. If the angle of elevation of the sun is 67  , what is the height of the flagpole? 2. An airplane is flying at 4000 feet above the ground. If the angle of depression from the airplane to the beginning of the runway is 9 .5  , what is the horizontal distance to the nearest tenth of a kilometer of the airplane to the beginning of the runway? 3. A group of Boy Scouts on a straight trail headed N41  E found the trail led through a landslide. They chose to walk 56 meters due east along the south edge of the landslide , then due north along the east edge of the landslide back to the trail. How much do they have to walk north to get back on the trail? 4. In the High Mountains lookout station Running Deer is 28 kilometers due west of station Lazy Bear. The bearing from Running Deer to a fire directly south of Lazy Bear is S 41.9  E. To the nearest tenth of a mile, how far is the fire from Running Deer? From Lazy Bear?   5. A ship is 15 miles due west of lighthouse A on the island of Hawaii. If lighthouse B is 6.2 miles due south of lighthouse A, what is the bearing to the nearest tenth of a degree of the ship from lighthouse B? To the nearest tenth of a mile, how far is the ship from lighthouse B?

79. Ex. E . Solve. Give the answers in the exact form and its approximate values. SOHCAHTOA 6. A balloon flying at 5000 feet above the ocean measures the angles of depression of each end of an island directly in front of the balloon to be 65.8  and 12.6  . What is the length of the island to the nearest tenth of a mile? 7. An observer standing on a cliff 320 feet above the ocean measured angles of depression of the near and far sides of an island to be 16.5  and 10.5  respectively. How long is the island to the nearest foot ?   8. Two observers, 8150 feet apart, observe a balloonist flying overhead between them. Their measures of the angles of elevation at 3:30pm are 27.6  and 48.2  respectively. What is the height of the balloonist above the ground? 10. An observer stands 450 feet away from a church and measures the angles of elevation of the top and bottom of the steeple to be 24.4  and 26.2  respectively. What is the height of the steeple to the nearest meter?   9. A 40-foot flagpole stands on the top of a hill. To measure the height of the hill, a surveyor standing at the base of the hill finds the measure of the angles of elevation of the top and bottom of the flagpole to be 49.7  and 39.3  respectively. Find the height of the hill to the nearest foot.

80. SOHCAHTOA A. 1. sin(A) =1/3, cos(A) = √8/3, tan(A) = 1/√8, sin(B) =√8/3, cos(B) = 1/3, tan(B) = √ 8 3 . sin(A) = 4/ √41, cos(A) = 5/ √41, tan(A) = 4/5, sin(B) = 5/√41, cos(B ) = 4 /√41, tan(B) = 5/4 5 . sin(A) = 1 /2, cos(A) = √3/2, tan(A) = 1 /√3, sin(B) = √3/2 , cos(B ) = 1/2 , tan(B) = √3 7 . sin(A) = √4 – x 2 /2, cos(A ) = x/2, tan(A) = √4 – x 2 /x, sin(B) = x/2 , cos(B ) = √4 – x 2 /2 , tan(B) = x/√4 – x 2 9 . sin(A) = x/√16 + x 2 , cos(A ) = 4/√ 16 + x 2 , tan(A ) = x/4, sin(B) = √16 + x 2 , x/2, cos(B ) = x/√16 + x 2 , tan(B ) = 4/x Answers

81. Answers SOHCAHTOA C . 1. x = 1/(√3 – 1) ≈ 1.37 3. x = ( √3 – 1 )/2 ≈ 0.366 5. x = 3/2 7. x ≈ 2.37 9. x = √2/2 ≈ 0.707 D . 1. x ≈ 18.7 h ≈ 26.1 3. x ≈ 63.6 h ≈ 56.6. 5. x ≈ 33.8 h ≈ 26.7 7. x ≈ 13.1 h ≈ 38 9. x ≈ 77 h ≈ 31.1 E . 1. ≈ 98.9 ft 3. ≈ 64.4 m 5. ≈ 112° 7. ≈ 646 ft 9. ≈ 90.8 ft B . 1. a = 23, b ≈ 34.1, c ≈ 41.1, B = 56 o 3 . a = 15, b ≈ 10.9, c ≈ 18.5, B = 36 o 5 . a ≈ 23.5, b = 82, c ≈ 85.3, B = 16 o 7 . x ≈ 44.4 9. x ≈ 3.34