Measurement - Trigonometry One Notes

introduction

Math 10C - Trigonometry OneTrigonometry Math Animations
Transcript of: Trigonometry One.

Trigonometry One: Introduction

 

  1. In this lesson, we'll investigate right triangles using trigonometry. In part a of the introduction, we are asked to label the sides of each triangle, relative to the given angle.
  2. The hypotenuse is opposite the right angle.
  3. The opposite side is directly across from the angle.
  4. The adjacent side is the remaining side.
  5. Now we'll label the sides of a second triangle.
  6. The hypotenuse is opposite the right angle.
  7. The opposite side is directly across from the angle.
  8. The adjacent side is the remaining side.
  9. In part b, we'll calculate the ratio of the opposite side to the adjacent side for each similar triangle.
  10. Let's calculate opposite/adjacent ratio for the first triangle.
  11. The opposite side to the angle is 1 cm, and the adjacent side is 2 cm.
  12. Divide to get 0.5. The units of cm cancel, so 0.5 is a unitless number.
  13. Now we'll move on to the second triangle.
  14. The opposite side is 2 cm, and the adjacent side is 4 cm.
  15. Divide to get 0.5.
  16. Now we'll move on to the third triangle.
  17. The opposite side is 4 cm, and the adjacent side is 8 cm.
  18. Divide to get 0.5.
  19. Notice that the opposite/adjacent ratio is identical for similar triangles.
  20. In part c, we'll define the tangent ratio.
  21. The tangent ratio is the fraction opposite/adjacent.
  22. We can express tangent as an equation using tan(theta) = opposite/adjacent.
  23. Theta is a Greek letter commonly used to represent an angle.
  24. In part d we'll calculate the ratio of the opposite side to the hypotenuse for each similar triangle.
  25. Let's calculate opposite/hypotenuse ratio for the first triangle.
  26. The opposite side is 2 cm, and the hypotenuse is 5 cm.
  27. Divide to get 0.4.
  28. Now we'll move on to the second triangle.
  29. The opposite side is 4 cm, and the hypotenuse is 10 cm.
  30. Divide to get 0.4.
  31. Now we'll move on to the third triangle.
  32. The opposite side is 8 cm, and the hypotenuse is 20 cm.
  33. Divide to get 0.4.
  34. Notice that the opposite/hypotenuse ratio is identical for similar triangles.
  35. In part e, we'll define the sine ratio.
  36. The sine ratio is the fraction opposite/hypotenuse.
  37. We can express sine as an equation using sin(theta)=opposite/hypotenuse. Note: We write sine with an "e" when using it as a word, but we drop the "e" when writing mathematical equations.
  38. In part f we'll calculate the ratio of the adjacent side to the hypotenuse for each similar triangle.
  39. Let's calculate the adjacent/hypotenuse ratio for the first triangle.
  40. The adjacent side is 3 cm, and the hypotenuse is 4 cm.
  41. This divides to 0.75.
  42. Now we'll move on to the second triangle.
  43. The adjacent is 8 cm, and the hypotenuse is 6 cm.
  44. This divides to 0.75.
  45. Now we'll move on to the third triangle.
  46. The adjacent is 12 cm, and the hypotenuse is 16 cm.
  47. This divides to 0.75.
  48. Notice that the adjacent/hypotenuse ratio is identical for similar triangles.
  49. In part g, we'll define the cosine ratio.
  50. The cosine ratio is the fraction adjacent over hypotenuse.
  51. We can express cosine as an equation using cos(theta)=adjacent/hypotenuse.
  52. In part h, we are asked "What is a useful memorization tool to remember the trigonometric ratios?"
  53. A lot of students find it useful to memorize the phrase SOH CAH TOA. The order of the letters matches the trigonometric ratios. Finally, keep in mind that SOH CAH TOA can only be used in right triangles. Non-right triangles have different rules that you will learn about in future courses.
  54. sine is opposite over hypotenuse.
  55. cosine is adjacent over hypotenuse.
  56. tan is opposite over adjacent.

 

Example 1: Trigonometric Ratios.

 

  1. For each triangle, calculate each trigonometric ratio.
  2. Let's label each side of the triangle in part a.
  3. Now we'll calculate sinθ.
  4. sinθ is opposite over hypotenuse.
  5. The opposite side is 4 cm, and the hypotenuse is 5 cm.
  6. Divide to get 0.8.
  7. Next we'll calculate cosθ.
  8. cosθ = adjacent/hypotenuse.
  9. The adjacent side is 3 cm, and the hypotenuse is 5 cm.
  10. Divide to get 0.6.
  11. Next we'll calculate tanθ.
  12. tanθ = opposite/adjacent.
  13. The opposite side is 4 cm, and the adjacent side is 3 cm.
  14. tanθ = 1.3 repeating.
  15. Now we'll move on to the triangle in part b.
  16. Label each side of the triangle.
  17. Now we'll calculate sinθ.
  18. sinθ is opposite/hypotenuse.
  19. The opposite side is 21 cm, and the hypotenuse is 29 cm.
  20. Divide to get 0.7241.
  21. Next we'll calculate cosθ.
  22. cosθ is adjacent over hypotenuse.
  23. The adjacent side is 20 cm, and the hypotenuse is 29 cm.
  24. Divide to get 0.6897.
  25. Next we'll calculate tanθ.
  26. tanθ is opposite over adjacent.
  27. The opposite is 21 cm, and the adjacent is 20 cm.
  28. Divide to get 1.05.
  29. Now we'll move on to part c.
  30. Label each side of the triangle.
  31. First we'll calculate sinθ.
  32. sinθ is opposite over hypotenuse.
  33. The opposite side is 63 cm, and the hypotenuse is 65 cm.
  34. Divide to get 0.9692.
  35. Next we'll evaluate cosθ.
  36. cosθ is adjacent over hypotenuse.
  37. The adjacent is 16 cm, and the hypotenuse is 65 cm.
  38. The value of cosθ is 0.2462.
  39. Next we'll evaluate tanθ.
  40. tanθ is opposite over adjacent.
  41. The opposite side is 63 cm, and the adjacent side is 16 cm.
  42. Divide to get 3.9375.
  43. Now we'll move on to part d.
  44. Label each side of the triangle.
  45. First we'll calculate sinθ.
  46. sinθ is opposite over hypotenuse.
  47. The opposite side is 12 cm, and the hypotenuse is 13 cm.
  48. Divide to get 0.9231.
  49. Next we'll calculate cosθ.
  50. cosθ is adjacent over hypotenuse.
  51. The adjacent side is 5 cm, and the hypotenuse is 13 cm.
  52. Divide to get 0.3846.
  53. Next we'll evaluate tanθ.
  54. tanθ is opposite over adjacent.
  55. The opposite side is 12, and the adjacent side is 5.
  56. Divide to get 2.4.

 

Example 2: Finding the Angles of a Triangle.

 

  1. Calculate the angle θ in each triangle.
  2. Label the sides of the triangle.
  3. We know the adjacent and hypotenuse, so let's use the cosine ratio to find the unknown angle.
  4. The adjacent side is 36 cm, and the hypotenuse is 47 cm.
  5. Divide to get 0.7660.
  6. In order to get the angle θ, use the inverse cosine feature of your calculator. On a TI-83/84, this is the 2nd function of the cosine button. If you type in 2nd -> cos -> 0.7660, you will get the angle 40°. Note: The calculator must be in degree mode for this to work. Press the MODE button and look at the third row. Make sure Degree is highlighted.
  7. The answer is 40 degrees.
  8. Now we'll move on to part b.
  9. Label the sides of the triangle.
  10. We know the opposite and adjacent sides, so let's use the tangent ratio to solve for the unknown angle.
  11. The opposite side is 87 cm, and the adjacent side is 26 cm.
  12. Divide to get 3.3462.
  13. Use inverse tan to get the angle 73 degrees.
  14. Now we'll move on to part c.
  15. All of the angles in a triangle must add up to 180 degrees. Since we know two of the three angles in the triangle, we can use this as a shortcut.
  16. The unknown angle, plus 90 degrees, plus 73.74 degrees equals 180 degrees.
  17. Take 73.74 degrees and 90 degrees to the other side of the equation and change their signs.
  18. Perform the arithmetic to get the angle 16.26 degrees.
  19. Now we'll move on to part d.
  20. Label each side in the triangle.
  21. We know the opposite and hypotenuse, so let's use the sine ratio to get the unknown angle.
  22. The opposite side is 50 cm, and the hypotenuse is 68 cm.
  23. Divide to get 0.7353.
  24. Use inverse sine to get the angle 47 degrees.

 

Example 3: Finding the Side Lengths of a Triangle.

 

  1. Calculate the missing side of each triangle using two methods.
  2. Let's find the unknown side in part a using the Pythagorean Theorem.
  3. The Pythagorean Theorem is a2 + b2 = c2.
  4. Plug in 7 for a and 24 for b.
  5. This gives c2 = 625.
  6. Square root both sides to get c = 25.
  7. Now we'll try part a again, but this time we'll find the side using a trigonometric ratio.
  8. Label the sides of the triangle.
  9. We need to find the hypotenuse. We already know both the adjacent and opposite sides, so we can use either cosθ or sinθ. In this example,we'll use cosθ.
  10. The cosine ratio is adjacent over hypotenuse.
  11. The adjacent side is 7 cm, and the hypotenuse is the unknown h.
  12. Cross-multiply.
  13. Divide both sides by cos73.74 to isolate h.
  14. Evaluate to get 25 cm. This is the same hypotenuse length we found in part a, using the Pythagorean Theorem.
  15. Now we'll move on to part b.
  16. First we'll find the unknown side using the Pythagorean Theorem.
  17. The Pythagorean Theorem is a2 + b2 = c2.
  18. Plug in 48 for b and 73 for c. Recall that the hypotenuse is always c, while the other two sides can be labelled in either order.
  19. Square the numbers.
  20. Subtract 2304 from both sides to get a2 = 3025.
  21. Square root both sides to get a = 55 cm.
  22. Now we'll try this again, using a trigonometric ratio to solve for the unknown side.
  23. Label all the sides of the triangle.
  24. We are trying to find the opposite side, and we know the adjacent and hypotenuse. We can use either tanθ or sinθ. In this example, we'll use tanθ.
  25. The tangent ratio is opposite over adjacent.
  26. The opposite side is x, and the adjacent side is 48.
  27. Cross-multiply.
  28. Evaluate to get 55 cm. This is the same side length we obtained earlier when we used the Pythagorean Theorem.

 

Example 4: Solving a Triangle (find all unknown side lengths and angles).

 

  1. Solve each triangle. Round answers to the nearest tenth. Note that "solving" a triangle means finding all unknown sides and angles of the triangle.
  2. First we'll solve for x.
  3. Relative to 41 degrees, we know the hypotenuse and want the adjacent. Use the cosine ratio.
  4. The cosine ratio is adjacent over hypotenuse.
  5. The angle is 41 degrees, so plug that in for theta. On the right side, the adjacent side is x, and the hypotenuse is 70.3 cm.
  6. Cross-multiply.
  7. The value of x is approximately 53 cm.
  8. Now solve for y.
  9. Relative to 41 degrees, we know the hypotenuse and want the opposite. Use the sine ratio. Note: We could alternatively use tanθ to solve for y.
  10. The sine ratio is opposite over hypotenuse.
  11. The angle is 41 degrees, so plug that in for theta. On the right side, the opposite side is y, and the hypotenuse is 70.3 cm.
  12. Cross-multiply.
  13. The value of y is approximately 46 cm.
  14. Now solve for the angle m.
  15. In a triangle, all angles add up to 180 degrees, so we have 90 degrees, plus 41 degrees, plus m = 180 degrees.
  16. Take 90 degrees and 41 degrees to the other side of the equation and change their signs.
  17. Perform the arithmetic to get 49 degrees.
  18. Now we'll solve the triangle in part b.
  19. First we'll solve for y.
  20. Relative to 8°, we know the adjacent and want to find the opposite. Use tanθ. Note: We could alternatively use sinθ to solve for h.
  21. The tangent ratio is opposite over adjacent.
  22. The angle is 8 degrees, so plug that in for theta. On the right side, the opposite side is y, and the adjacent side is 23.8.
  23. Cross-multiply.
  24. The value of y is approximately 3 cm.
  25. Next, solve for h.
  26. Relative to 8°, we know the adjacent and want the hypotenuse. Use cosθ.
  27. The cosine ratio is adjacent over hypotenuse.
  28. The angle is 8 degrees, so plug that in for theta. On the right side, the adjacent is 23.8 cm and the hypotenuse is h.
  29. Cross-multiply.
  30. Divide both sides by the cosine of 8 degrees.
  31. The hypotenuse is approximately 24 cm.
  32. Finally, solve for the angle m.
  33. All angles in a triangle add up to 180 degrees, so we have 90 degrees, plus 8 degrees, plus m, equals 180 degrees.
  34. Take 90 degrees and 8 degrees to the other side of the equation and change their signs.
  35. Angle m is 82 degrees.
  36. Now we'll solve the triangle in part c.
  37. First solve for the hypotenuse.
  38. We don't have an angle that will help us find the hypotenuse. Use the Pythagorean Theorem.
  39. The Pythagorean Theorem is a2 + b2 = c2.
  40. Plug in 16.4 cm for a, 17.2 cm for b, and h for c.
  41. Square each number.
  42. This gives us h2 = 564.78.
  43. The hypotenuse is approximately 24 cm.
  44. Now solve for angle m.
  45. Relative to angle m, we know all sides of the triangle. Any of the trig ratios can be used to calculate angle m. In this example, we'll use tanθ.
  46. The tangent ratio is opposite over hypotenuse.
  47. The opposite side is 16.4 cm, and the adjacent side is 17.2 cm.
  48. Divide to get 0.9535.
  49. The angle m is approximately 44 degrees.
  50. Finally, solve for angle n.
  51. All angles in a triangle add up to 180 degrees. This gives us 90 degrees, plus 44 degrees, plus n = 180 degrees.
  52. Bring 90 degrees and 44 degrees to the other side of the equation and change their signs.
  53. Angle n is 46 degrees.
  54. Now we'll solve the triangle in part d.
  55. First we'll find x.
  56. There are no angles we can use to find x, so we'll use the Pythagorean Theorem.
  57. The Pythagorean Theorem is a2 + b2 = c2.
  58. Plug in x for a, 22.3 for b, and 28.9 for the hypotenuse.
  59. Square each number.
  60. This gives us x2 = 337.92.
  61. Square root both sides to get an x value of approximately 18 cm.
  62. Next we'll find the angle m.
  63. Relative to angle m, we know all sides of the triangle. Any of the trig ratios can be used to calculate angle m. In this example, we'll use cosθ.
  64. The cosine ratio is adjacent over hypotenuse.
  65. Relative to angle m, the adjacent side is 22.3 cm, and the hypotenuse is 28.9 cm.
  66. Divide to get 0.7716.
  67. Use inverse cosine to get approximately 39.5 degrees as the value of angle m.
  68. Finally, find angle n.
  69. All angles in a triangle add up to 180 degrees, so we have 90 degrees, plus 39.5 degrees, plus n, equals 180 degrees.
  70. Bring 90 degrees and 39.5 degrees to the other side of the equation and change their signs.
  71. Angle n is 50.5 degrees.

 

Example 5: Trigonometry and Surveying.

 

  1. The sketch on the right was drawn by a surveyor who is trying to determine the distance between two trees across a river. Using the information in the sketch, calculate the distance between the trees.
  2. First we'll complete the triangle by drawing the line x, across the river.
  3. Relative to 46°, we know the adjacent and want to find the opposite. Use tanθ to find the opposite side.
  4. The tangent ratio is opposite over adjacent.
  5. The angle is 46 degrees, so plug that in for theta. The opposite side is x, and the adjacent side is 80 m.
  6. Cross-multiply.
  7. The distance across the river is approximately 83 m.
  8. Now we'll move on to part b. A 16 ft. ladder is leaning against the roof of a house. The angle between the ladder and the ground is 62°. How high above the ground is the roof?
  9. Let's clearly draw the triangle we are investigating. We'll label the height of the house y.
  10. Relative to 62°, we are trying to find the opposite side and we know the hypotenuse. Use sinθ to find the side.
  11. The sine ratio is opposite over hypotenuse.
  12. Plug in 62 degrees for the angle, y for the opposite side, and 16 ft. For the hypotenuse.
  13. Cross-multiply.
  14. The height of the base of the roof is approximately 14 ft.