1. 1 Right Triangle Trigonometry Pre-Calculus Monday, April 20

2. Today’s Objective Review right triangle trigonometry from Geometry and expand it to all the trigonometric functions Begin learning some of the Trigonometric identities 2

3. Evaluate trigonometric functions of acute angles. Use fundamental trigonometric identities. Use a calculator to evaluate trigonometric functions. Use trigonometric functions to model and solve real-life problems. What You Should Learn

4. 4 Plan Questions from last week? Notes! Guided Practice Homework

5. 5 Right Triangle Trigonometry Trigonometry is based upon ratios of the sides of right triangles. The ratio of sides in triangles with the same angles is consistent. The size of the triangle does not matter because the triangles are similar (same shape different size).

6. 6 The six trigonometric functions of a right triangle, with an acute angle, are defined by ratios of two sides of the triangle. The sides of the right triangle are:  the side opposite the acute angle  the side adjacent to the acute angle,  and the hypotenuse of the right triangle. opp adj hyp θ θ θ θ

7. 7 The trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. opp adj hyp θ Trigonometric Functions Sin = cos = tan = Csc = sec = cot = opp hyp adj hyp adj opp adj Note: sine and cosecant are reciprocals, cosine and secant are reciprocals, and tangent and cotangent are reciprocals. θ θ θ θ θ θ

8. 8 Reciprocal Functions Another way to look at it… sin  = 1/csc  csc  = 1/sin  cos  = 1/sec  sec  = 1/cos  tan  = 1/cot  cot  = 1/tan 

9. Given 2 sides of a right triangle you should be able to find the value of all 6 trigonometric functions. Example: 9 5 12 

10. 10 Calculate the trigonometric functions for . The six trig ratios are 4 3 5  Sin = Tan = Sec = Cos = Cot = Csc = Example: Six Trig Ratios cos α = sin α = cot α =tan α = csc α = sec α = What is the relationship of α and θ? They are complementary (α = 90 – θ) Calculate the trigonometric functions for . θ θ θ θ θ θ

11. 11 Note : These functions of the complements are called cofunctions. Note sin  = cos(90   ), for 0 <  < 90 Note that  and 90   are complementary angles. Side a is opposite θ and also adjacent to 90 ○ – θ. a hyp b θ 90 ○ – θ sin  = and cos (90   ) =. So, sin  = cos (90   ). Example: Using Trigonometric Identities

12. 12 Cofunctions sin  = cos (90   ) cos  = sin (90   ) sin  = cos (π/2   ) cos  = sin (π/2   ) tan  = cot (90   ) cot  = tan (90   ) tan  = cot (π/2   ) cot  = tan (π/2   ) sec  = csc (90   ) csc  = sec (90   ) sec  = csc (π/2   ) csc  = sec (π/2   )

13. 13 Trigonometric Identities are trigonometric equations that hold for all values of the variables. Example: Using Trigonometric Identities We will learn many Trigonometric Identities and use them to simplify and solve problems.

14. 14 Quotient Identities Sin = cos = tan = hyp adj opp adj hyp θ The same argument can be made for cot… since it is the reciprocal function of tan. θ θ θ

15. 15 Quotient Identities

16. Pythagorean Identities Three additional identities that we will use are those related to the Pythagorean Theorem: 16 Pythagorean Identities sin 2  + cos 2  = 1 tan 2  + 1 = sec 2  cot 2  + 1 = csc 2 

17. 17 Some old geometry favorites… Let’s look at the trigonometric functions of a few familiar triangles…

18. 18 Geometry of the 45-45-90 triangle Consider an isosceles right triangle with two sides of length 1. 1 1 45 The Pythagorean Theorem implies that the hypotenuse is of length. Geometry of the 45-45-90 Triangle Remember a 2 + b 2 = c 2

19. 19 Calculate the trigonometric functions for a 45 ° angle. Examp le: Trig Functi ons for  45  1 1 45 csc 45 ° = = = opp hyp sec 45 ° = = = adj hyp cos 45 ° = = = hyp adj sin 45 ° = = = cot 45 ° = = = 1 opp adj tan 45 ° = = = 1 adj opp

20. 20 60 ○ Consider an equilateral triangle with each side of length 2. The perpendicular bisector of the base bisects the opposite angle. The three sides are equal, so the angles are equal; each is 60 °. Geometry of the 30-60-90 triangle 22 2 11 30 ○ Use the Pythagorean Theorem to find the length of the altitude,. Geometry of the 30-60-90 Triangle

21. 21 Calculate the trigonometric functions for a 30  angle. 1 2 30 Example: Trig Functions for  30  csc 30 ° = = = 2 opp hyp sec 30 ° = = = adj hyp cos 30 ° = = hyp adj tan 30 ° = = = adj opp cot 30 ° = = = opp adj sin 30 ° = =

22. 22 Calculate the trigonometric functions for a 60  angle. 1 2 60 ○ Example: Trig Functions for  60  csc 60  = = = opp hyp sec 60  = = = 2 adj hyp cos 60  = = hyp adj tan 60  = = = adj opp cot 60  = = = opp adj sin 60  = =

23. Some basic trig values SineCosineTangent 30 0  /6 45 0  /4 1 60 0  /3 23

24. IDENTITIES WE HAVE REVIEWED SO FAR… 24

25. 25 Fundamental Trigonometric Identities Co function Identities sin  = cos(90   ) cos  = sin(90   ) sin  = cos (π/2   ) cos  = sin (π/2   ) tan  = cot(90   ) cot  = tan(90   ) tan  = cot (π/2   ) cot  = tan (π/2   ) sec  = csc(90   ) csc  = sec(90   ) sec  = csc (π/2   ) csc  = sec (π/2   ) Reciprocal Identities sin  = 1/csc  cos  = 1/sec  tan  = 1/cot  cot  = 1/tan  sec  = 1/cos  csc  = 1/sin  Quotient Identities tan  = sin  /cos  cot  = cos  /sin  Pythagorean Identities sin 2  + cos 2  = 1 tan 2  + 1 = sec 2  cot 2  + 1 = csc 2  Fundamental Trigonometric Identities for

26. 26 Example: Given sec  = 4, find the values of the other five trigonometric functions of . Use the Pythagorean Theorem to solve for the third side of the triangle. tan  = = cot  =sin  = csc  = = cos  = sec  = = 4 θ 4 1 Draw a right triangle with an angle  such that 4 = sec  = =. adj hyp Example: Given 1 Trig Function, Find Other Functions

27. Using the calculator Function Keys Reciprocal Key Inverse Keys 27

28. Using Trigonometry to Solve a Right Triangle A surveyor is standing 115 feet from the base of the Washington Monument. The surveyor measures the angle of elevation to the top of the monument as 78.3 . How tall is the Washington Monument? Figure 4.33

29. Applications Involving Right Triangles The angle you are given is the angle of elevation, which represents the angle from the horizontal upward to an object. For objects that lie below the horizontal, it is common to use the term angle of depression.

30. Solution where x = 115 and y is the height of the monument. So, the height of the Washington Monument is y = x tan 78.3   115(4.82882)  555 feet.

31. 31 Homework 4-2 Practice 1 1-17 ODD

32. Tuesday, April 21, 2015 BENCHMARK TOMORROW 32

33. Kahoot all day long! Here’s the deal… 1. You need a PENCIL and PAPER 2. You will be GRADED for participating so make sure your Kahoot name is your real name 3. If you get kicked out, you must log back in 4. If you do not have an ipad/smart phone, you may work in teams of TWO or do your work on a piece of paper and turn that in 5. THIS IS REQUIRED. 33

34. Review Exponents and Log Converting Logs and Exponents https://play.kahoot.it/#/k/68a4661b-b90b- 4987-820e-956c7e5af6bd https://play.kahoot.it/#/k/68a4661b-b90b- 4987-820e-956c7e5af6bd Log and Inverses https://play.kahoot.it/#/k/45a02d91-aa36- 485f-ba52-436768d981f7 https://play.kahoot.it/#/k/45a02d91-aa36- 485f-ba52-436768d981f7 34

35. Review Intro to Trig 35 Right Triangle Trig and Angle Measures https://play.kahoot.it/#/k/8858d7ed-b092- 46c6-bc5d-b16c044665c0 https://play.kahoot.it/#/k/8858d7ed-b092- 46c6-bc5d-b16c044665c0 Radians, Degrees, Arc Length https://play.kahoot.it/#/k/63ac7b20-59ce- 4fd4-9266-954797b1b39a https://play.kahoot.it/#/k/63ac7b20-59ce- 4fd4-9266-954797b1b39a

36. Wednesday BENCHMARK! 36

37. Thursday, April 23 37

38. Do Now – How was the benchmark? What can you improve?

39. Benchmark Data 39

40. Benchmark Data 1 st Period 40

41. Analysis By Question 41

42. Analysis By Question 42

43. Analysis By Question 43

44. Analysis By Question 44

45. Analysis By Question 45

46. Benchmark Data 2 nd Period 46

47. Analysis By Question 47

48. Analysis By Question 48

49. Analysis By Question 49

50. Analysis By Question 50

51. Analysis By Question 51

52. Special Angle Names Angle of Elevation From Horizontal Up Angle of Depression From Horizontal Down

53. Angle of Elevation and Depression The angle of elevation is measured from the horizontal up to the object. Imagine you are standing here.

54. Angle of Elevation and Depression The angle of depression is measured from the horizontal down to the object. Constructing a right triangle, we are able to use trig to solve the triangle.

55. Guided Practice! Follow along on your handout!

56. Lighthouse & Sailboat Suppose the angle of depression from a lighthouse to a sailboat is 5.7 o. If the lighthouse is 150 ft tall, how far away is the sailboat? Construct a triangle and label the known parts. Use a variable for the unknown value. 5.7 o 150 ft. x

57. Lighthouse & Sailboat Suppose the angle of depression from a lighthouse to a sailboat is 5.7 o. If the lighthouse is 150 ft tall, how far away is the sailboat? 5.7 o x Set up an equation and solve. 150 ft.

58. Lighthouse & Sailboat Remember to use degree mode! x is approximately 1,503 ft. 5.7 o x 150 ft.

59. River Width A surveyor is measuring a river’s width. He uses a tree and a big rock that are on the edge of the river on opposite sides. After turning through an angle of 90° at the big rock, he walks 100 meters away to his tent. He finds the angle from his walking path to the tree on the opposite side to be 25°. What is the width of the river? Draw a diagram to describe this situation. Label the variable(s)

60. River Width We are looking at the “opposite” and the “adjacent” from the given angle, so we will use tangent Multiply by 100 on both sides

61. Subway The DuPont Circle Metrorail Station in Washington DC has an escalator which carries passengers from the underground tunnel to the street above. If the angle of elevation of the escalator is 52° and a passenger rides the escalator for 188 ft, find the vertical distance between the tunnel and the street. In other words, how far below street level is the tunnel?

62. Subway We are looking at the “opposite” and the “hypotenuse” from the given angle so we will use sine Multiply by 188 on each side

63. Building Height A spire sits on top of the top floor of a building. From a point 500 ft. from the base of a building, the angle of elevation to the top floor of the building is 35 o. The angle of elevation to the top of the spire is 38 o. How tall is the spire? Construct the required triangles and label. 500 ft. 38 o 35 o

64. Building Height Write an equation and solve. Total height (t) = building height (b) + spire height (s) 500 ft. 38 o 35 o Solve for the spire height. t b s Total Height

65. Building Height Write an equation and solve. 500 ft. 38 o 35 o Building Height t b s

66. Write an equation and solve. 500 ft. 38 o 35 o The height of the spire is approximately 41 feet. t b s Total height (t) = building height (b) + spire height (s)

67. Mountain Height A hiker measures the angle of elevation to a mountain peak in the distance at 28 o. Moving 1,500 ft closer on a level surface, the angle of elevation is measured to be 29 o. How much higher is the mountain peak than the hiker? Construct a diagram and label. 1 st measurement 28 o. 2 nd measurement 1,500 ft closer is 29 o.

68. Mountain Height Adding labels to the diagram, we need to find h. 28 o 29 o 1500 ftx ft h ft Write an equation for each triangle. Remember, we can only solve right triangles. The base of the triangle with an angle of 28 o is 1500 + x.

69. Mountain Height Now we have two equations with two variables. Solve by substitution. Solve each equation for h. Substitute.

70. Mountain Height Solve for x. Distribute. Get the x’s on one side and factor out the x. Divide.

71. Mountain Height However, we were to find the height of the mountain. Use one of the equations solved for “h” to solve for the height. The height of the mountain above the hiker is 19,562 ft.