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The Law of Cosines

The Law of Cosines. Main Menu. Intro Law of Sines vs. Cosines Motivation Proof My Video Information and Application Examples Quiz Extra Help and References. Introduction/Orientation.

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The Law of Cosines

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  1. The Law of Cosines

  2. Main Menu • Intro • Law of Sines vs. Cosines • Motivation • Proof • My Video • Information and Application • Examples • Quiz • Extra Help and References

  3. Introduction/Orientation • In addition to the Law of Sines, the Law of Cosines is a trigonometric method, used for solving oblique (non-right) triangles.

  4. Introduction/Orientation (cont.) • The objective of each problem is to solve for all side and angle measures of a triangle. • At the end, a quiz will be given, so be sure to ask questions and take notes along the way!

  5. What Makes The Law of Cosines different from the Law of Sines? • The Law of Sines (used in cases of AAS, ASA, and SSA) may involve an ambiguous case (SSA), where the final result could be a possibility of 1 triangle, 2 triangles, or even no triangles in the problem. • A problem involving The Law of Cosines guarantees only one triangle, regardless of the case in which it is used.

  6. What will you be doing with the Law of Cosines? • Using the Law of Cosines, you will be able to apply this formula to solve for oblique triangles. This will go more in-depth than right-triangle trigonometric problems, as they truly come down to knowing the formulas, rather than a simple ratio to be used.

  7. What will you be doing with the Law of Cosines? (cont.) • In order to work with applications of word problems, dealing with geometric measurements are relevant to your knowledge, especially if you go into any kind of math, science, or engineering. The Law of Cosines is a good way to start learning.

  8. Law of Cosines Proof • The Law of Cosines- Any given triangle with side lengths, a, b, and c, with opposing angles, A, B, and C. • Take your triangle and divide it into two right triangles. From there, since we have right triangles, we can use the Pythagorean Theorem and prove our formula.

  9. My Videos

  10. If there is an issue with viewing the videos on this PowerPoint, then please use these links. Thank you. • http://www.youtube.com/watch?v=hfPIw9WfpmQ • http://www.youtube.com/watch?v=4Bfk6Wc9vNM&context=C41709ffADvjVQa1PpcFNBdTOOCAoTJc1fS0NQ5YddXCQ0wuD5yyc=

  11. Information Formulas • a2 = b2 + c2 – 2bc(cos(A)) • b2= a2+ c2 – 2ac(cos(B)) • c2= a2+ b2– 2ab(cos(C))

  12. Formulas (cont.) • cos A = (b2 + c2 – a2) / (2bc) • cosB = (a2 - b2 + c2) / (2ac) • cos C = (a2 + b2- c2) / (2ab)

  13. NOTE • Once you find two angles, sum them and then take 180 and subtract the sum of the two angles that you have. • In addition, lower-case letters represent side measures, and upper-case for angle measures. • In Slides 17 and 18, you may click on the image in the middle of the page that will take you to an example problem.

  14. Application • There are two cases in which the Law of Cosines can be used.

  15. Side-Side-Side (SSS)

  16. Side-Angle-Side (SAS)

  17. Let us take a look at some example problems. When you answer a question correctly, please click the triangle on the left-hand side of the “correct answer” slide for a solution.

  18. Example 1 Given a triangle with side measures, a = 4, b = 5, and c = 7, Solve for the angles (NOTE: Angles are measured in degrees). A. A = 34.0, B = 44.4, C = 101.6 B. A = 0.60, B = 0.80, C = 178.6 C. A = 0.8, B = 0.7, C = 178.5

  19. That is the correct Answer.

  20. NO, TRY AGAIN. • Your calculator is in radians.

  21. NO, TRY AGAIN. • You forgot to take the inverse cosine of the first angle to solve for the unknown. Hence, it threw off your entire problem.

  22. Solution • A = arccos((25+49-16)/((2)(5)(7))) = 34.0 (approximately) • B = arccos((16-25+49)/((2)(4)(7))) = 44.4 (approximately) • C = 180-34.0-44.4 = 101.6 (approximately)

  23. Example 2 Given a triangle ABC with b = 5, c = 7, and A = 120, solve for the triangle. A. a = 109, B = 0.72, C = 59.28 B. a = 10.4, B = 24.8, C = 35.3 C. a = 4.1, B = 1, C = 57

  24. NO, TRY AGAIN. For the missing side, you forgot to takethe square root of the numericalvalue inside the radical. In addition, when solving for B, you forgot to take the inverse cosine of your fraction.

  25. That is the correct Answer.

  26. NO, TRY AGAIN. Your calculator is NOT in degree mode.

  27. Solution • a = sqrt(25 + 49 – 70 cos(120)) = sqrt(109) = 10.4 • B = arccos(((10.4)^2 – 25 + 49)/(2*7*10.4)) = 24.8 • C = 180 – 24.8 – 120 = 35.2

  28. Example 3 A person leaves home and walks 5 mi east, then goes 3 mi northeast. How far is he from home? A. 7.43 mi B. 7.99 mi C. 55.21 mi

  29. That Is The Correct Answer.

  30. NO, TRY AGAIN. Calculator is in the wrong mode.

  31. NO, TRY AGAIN. You forgot to take the square root of both sides.

  32. Solution Going NE, thinking about it from the perspective of a compass, as from North to East, the right angle is bisected. Therefore angle C = 135. Now that the angle is found, we can plug in everything we need to solve for the missing side. sqrt(9+25-30cos(135)) = 7.43 mi

  33. Example 4 Two airplanes leave an airport, and the angle between their flight paths is 40. An hour later, one plane has traveled 300 mi, while the other has traveled 200 mi. How far apart are the planes at this time? A. 38,075 mi B. 458 mi C. 195 mi D. None of the Above

  34. NO, TRY AGAIN. You forgot to take the square root of both sides.

  35. NO, TRY AGAIN. Use the formulas given earlier, and apply the numbers you know into the problem. Then you can punch everything into your calculator and solve for the correct answer.

  36. That Is The Correct Answer.

  37. Solution This is similar to the previous word problem. We are given a side of 200, and another side of 300. In addition, we have the angle of 40 in between the two sides. So, in order to find the distance, simply solve for the missing side. sqrt(40,000+90,000-120,000(cos(40))) = 195 mi

  38. Pieces of Advice/Warnings • 1. When using your calculator, make sure that it is in Degree mode, and NOT in Radians. • 2. IMPORTANT TO NOTE!!! The Pythagorean Theorem CANNOT be used as these problems will not deal with right triangles, so do not do it! • 3. In most cases, sketching a diagram of the problem helps.

  39. Quiz There is no time limit. However, you must pass this quiz with 100% accuracy. You may take this quiz as many times as you wish. Good Luck!

  40. Question 1 Triangle ABC with side measures, a = 9, b = 12, c = 13.5. Solve for the angle measures. A. A = 40.8, B = 60.6, C = 78.6 B. A = 1.18, B = 1.06, C = 177.76

  41. That Is The Correct Answer.

  42. NO, TRY AGAIN. Your Calculator is in the Wrong Mode!

  43. Question 2 Two ships leave the same port at noon. One ship travels at a bearing of N 53 W at 12 mph, and the other ship travels at a bearing of S 67 W at 16 mph. How far apart will the ships be at 3 p.m.? A. 43 mi B. 1,872 miC. 95 mi D. 100 mi

  44. That Is The Correct Answer.

  45. NO, TRY AGAIN. Do not forget to take the square Root of both sides!

  46. NO, TRY AGAIN. Your Calculator is in the Wrong Mode!

  47. NO, TRY AGAIN. Throw your numbers into your calculator, using the formulas given earlier in the presentation.

  48. Question 3 You are given ASA. Can you use the Law of Cosines? A. Yes B. No

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