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Engineering math Review

This detailed engineering math review covers essential topics including trigonometry, systems of equations, and vector operations. You'll revisit the core principles of trigonometry, including Pythagorean triples, sine, cosine, and tangent. Explore the Law of Sines and the Law of Cosines with practical applications in right triangles. Additionally, gain insights into vector addition, subtraction, and multiplication, ensuring a strong foundation in mathematical concepts crucial for engineering studies.

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Engineering math Review

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  1. Engineering math Review

    Trigonometry Systems of Equations Vectors Vector Addition and Subtraction Vector Multiplication

  2. Trigonometry Review

    Pythagorean Triples Sine, Cosine, and Tangent revisited. Law of Sines, Law of Cosines
  3. Starting With Right Triangles c a b
  4. Starting With Right Triangles c a b Recall the Pythagorean Theorem
  5. Starting With Right Triangles c a b
  6. Pythagorean Triples There are a few sets of numbers that satisfy the Pythagorean Theorem using only whole numbers for a, b, and c. c a b
  7. Sine c a b
  8. Cosine c a b
  9. tangent c a b
  10. Inverse Trig Functions Recall that inverse trig functions are used to solve for the angle, , after setting up the trig equations c a b For example: so then
  11. Law of Sines B c a A C b Because of the properties of the sine function between 0 and 180, the Law of Sines may yield 2 answers. If possible, its best to use the Law of Sines when you know whether an angle is acute (less than 90) or obtuse (greater than 90)
  12. Law of Cosines The Law of Cosines can be used for obtuse angles (angles greater than 90 degrees) as well as acute angles B c a A C b
  13. Law of Cosines The Law of Cosines can be used for obtuse angles (angles greater than 90 degrees) as well as acute angles B c a A C b
  14. Law of Cosines The Law of Cosines can be used for obtuse angles (angles greater than 90 degrees) as well as acute angles B c a A C b
  15. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 A C b = 18
  16. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Start by using Law of Cosines to determine angle B. This will be the largest angle in the triangle because it is opposite to the longest side. A C b = 18
  17. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Our goal is to calculate angle B. To do this, we’ll need to isolate the expression with angle B in it. We start by squaring both sides of the equation A C b = 18
  18. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Subtract the a2 and c2terms next. A C b = 18
  19. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Divide both sides by -2ac A C b = 18
  20. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Take the inverse cosine of both sides of the equation to isolate the angle of interest, angle B. A C b = 18
  21. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Substitute the known lengths of the sides of the triangle and solve for angle B A C b = 18
  22. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Substitute the known lengths of the sides of the triangle and solve for angle B A C b = 18
  23. Example Problem Solve for all angles of the triangle pictured to the left B c = 12 a = 9 Substitute the known lengths of the sides of the triangle and solve for angle B A C b = 18
  24. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Now that we know one of the angles, and we know that the remaining angles have to be less than 90°, we can use Law of Sines to determine angle A A C b = 18
  25. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Multiplying both sides by ‘a’ isolates the term with angle A in it A C b = 18
  26. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Taking the inverse sine of both sides of the equation solves in terms of angle A A C b = 18
  27. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Substitute known values to solve for angle A A C b = 18
  28. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Substitute known values to solve for angle A A C b = 18
  29. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 We can use Law of Sines to determine angle C or capitalize on the fact that A + B + C = 180°. We’ll use Law of Sines, and then check the results using A + B + C = 180° A = 26.4° C b = 18
  30. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Multiplying both sides by ‘c’ isolates the term with angle C in it A = 26.4° C b = 18
  31. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Taking the inverse sine of both sides of the equation solves in terms of angle C A = 26.4° C b = 18
  32. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Substitute known values to solve for angle C A = 26.4° C b = 18
  33. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Substitute known values to solve for angle C A = 26.4° C b = 18
  34. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Now, we can check our answer and see if A + B + C = 180° A = 26.4° C = 36.3° b = 18
  35. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Now, we can check our answer and see if A + B + C = 180° A = 26.4° C = 36.3° b = 18
  36. Example Problem Solve for all angles of the triangle pictured to the left B = 117.3° c = 12 a = 9 Everything checks out! A = 26.4° C = 36.3° b = 18
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