1. Copyright © 2005 Pearson Education, Inc.

2. Chapter 2 Acute Angles and Right Triangles

3. Copyright © 2005 Pearson Education, Inc. 2.1 Trigonometric Functions of Acute Angles

4. Copyright © 2005 Pearson Education, Inc. Slide 2-4 Right-triangle Based Definitions of Trigonometric Functions For any acute angle A in standard position.

5. Copyright © 2005 Pearson Education, Inc. Slide 2-5 Example: Finding Trig Functions of Acute Angles Find the values of sin A, cos A, and tan A in the right triangle shown. A C B 52 48 20

6. Copyright © 2005 Pearson Education, Inc. Slide 2-6 Cofunction Identities For any acute angle A, sin A = cos(90   A)csc A = sec(90   A) tan A = cot(90   A)cos A = sin(90   A) sec A = csc(90   A)cot A = tan(90   A)

7. Copyright © 2005 Pearson Education, Inc. Slide 2-7 Example: Write Functions in Terms of Cofunctions Write each function in terms of its cofunction. a) cos 38  cos 38  = sin (90   38  ) = sin 52  b) sec 78  sec 78  = csc (90   78  ) = csc 12 

8. Copyright © 2005 Pearson Education, Inc. Slide 2-8 Example: Solving Equations Find one solution for the equation Assume all angles are acute angles.

9. Copyright © 2005 Pearson Education, Inc. Slide 2-9 Example: Comparing Function Values Tell whether the statement is true or false. sin 31  > sin 29  In the interval from 0  to 90 , as the angle increases, so does the sine of the angle, which makes sin 31  > sin 29  a true statement.

10. Copyright © 2005 Pearson Education, Inc. Slide 2-10 Special Triangles 30-60-90 Triangle 45-45-90 Triangle

11. Copyright © 2005 Pearson Education, Inc. Slide 2-11 Function Values of Special Angles 2 60  11 45  2 30  csc  sec  cot  tan  cos  sin 

12. Copyright © 2005 Pearson Education, Inc. 2.2 Trigonometric Functions of Non-Acute Angles

13. Copyright © 2005 Pearson Education, Inc. Slide 2-13 Answers to homework Sincostan 1. 21/2920/2921/20 2. 45/5328/5345/28 3. n/pm/pn/m 4. k/zy/zk/y 5. C 6. H 9. E 7. B 10. A 8. G

14. Copyright © 2005 Pearson Education, Inc. Slide 2-14 Homework answers SinCosTanCscSeccot 11.12/135/1312/513/1213/55/12 125/33/5 136/77/6 147/1212/7

15. Copyright © 2005 Pearson Education, Inc. Slide 2-15 Reference Angles A reference angle for an angle  is the positive acute angle made by the terminal side of angle  and the x-axis.

16. Copyright © 2005 Pearson Education, Inc. Slide 2-16 Example: Find the reference angle for each angle. a) 218  Positive acute angle made by the terminal side of the angle and the x- axis is 218   180  = 38 . 1387  Divide 1387 by 360 to get a quotient of about 3.9. Begin by subtracting 360 three times. 1387  – 3(360  ) = 307 . The reference angle for 307  is 360  – 307  = 53 

17. Copyright © 2005 Pearson Education, Inc. Slide 2-17 Example: Finding Trigonometric Function Values of a Quadrant Angle Find the values of the trigonometric functions for 210 . Reference angle: 210  – 180  = 30  Choose point P on the terminal side of the angle so the distance from the origin to P is 2.

18. Copyright © 2005 Pearson Education, Inc. Slide 2-18 Example: Finding Trigonometric Function Values of a Quadrant Angle continued The coordinates of P are  x = y =  1r = 2

19. Copyright © 2005 Pearson Education, Inc. Slide 2-19 Finding Trigonometric Function Values for Any Nonquadrantal Angle  Step 1If  > 360 , or if  < 0 , then find a coterminal angle by adding or subtracting 360  as many times as needed to get an angle greater than 0  but less than 360 . Step 2Find the reference angle  '. Step 3Find the trigonometric function values for reference angle  '. Step 4Determine the correct signs for the values found in Step 3. (Use the table of signs in section 5.2, if necessary.) This gives the values of the trigonometric functions for angle .

20. Copyright © 2005 Pearson Education, Inc. Slide 2-20 Example: Finding Trig Function Values Using Reference Angles Find the exact value of each expression. cos (  240  ) Since an angle of  240  is coterminal with and angle of  240  + 360  = 120 , the reference angles is 180   120  = 60 , as shown.

21. Copyright © 2005 Pearson Education, Inc. Slide 2-21 Homework Pg 59-60 # 1-6, 10-13

22. Copyright © 2005 Pearson Education, Inc. Slide 2-22 Example: Evaluating an Expression with Function Values of Special Angles Evaluate cos 120  + 2 sin 2 60   tan 2 30 . Since cos 120  + 2 sin 2 60   tan 2 30  =

23. Copyright © 2005 Pearson Education, Inc. Slide 2-23

24. Copyright © 2005 Pearson Education, Inc. Slide 2-24

25. Copyright © 2005 Pearson Education, Inc. Slide 2-25 Example: Using Coterminal Angles Evaluate each function by first expressing the function in terms of an angle between 0  and 360 . cos 780   cos 780  = cos (780   2(360  ) = cos 60  =

26. Copyright © 2005 Pearson Education, Inc. 2.3 Finding Trigonometric Function Values Using a Calculator

27. Copyright © 2005 Pearson Education, Inc. Slide 2-27 Function Values Using a Calculator Calculators are capable of finding trigonometric function values. When evaluating trigonometric functions of angles given in degrees, remember that the calculator must be set in degree mode. Remember that most calculator values of trigonometric functions are approximations.

28. Copyright © 2005 Pearson Education, Inc. Slide 2-28 Example: Finding Function Values with a Calculator a) Convert 38  to decimal degrees. b) cot 68.4832  Use the identity cot 68.4832 .3942492

29. Copyright © 2005 Pearson Education, Inc. Slide 2-29 Angle Measures Using a Calculator Graphing calculators have three inverse functions. If x is an appropriate number, then gives the measure of an angle whose sine, cosine, or tangent is x.

30. Copyright © 2005 Pearson Education, Inc. Slide 2-30 Example: Using Inverse Trigonometric Functions to Find Angles Use a calculator to find an angle in the interval that satisfies each condition. Using the degree mode and the inverse sine function, we find that an angle having sine value.8535508 is 58.6. We write the result as

31. Copyright © 2005 Pearson Education, Inc. Slide 2-31 Example: Using Inverse Trigonometric Functions to Find Angles continued Use the identity Find the reciprocal of 2.48679 to get Now find using the inverse cosine function. The result is 66.289824

32. Copyright © 2005 Pearson Education, Inc. Slide 2-32 Page 64 # 22-29 22. 57.99717206 23. 55.84549629 24. 81.166807334 25. 16.16664145 26. 30.50274845 27. 38.49157974 28. 46.17358205 29. 68.6732406

33. Copyright © 2005 Pearson Education, Inc. 2.4 Solving Right Triangles

34. Copyright © 2005 Pearson Education, Inc. Slide 2-34 Significant Digits for Angles A significant digit is a digit obtained by actual measurement. Your answer is no more accurate then the least accurate number in your calculation. Tenth of a minute, or nearest thousandth of a degree 5 Minute, or nearest hundredth of a degree4 Ten minutes, or nearest tenth of a degree3 Degree2 Angle Measure to Nearest:Number of Significant Digits

35. Copyright © 2005 Pearson Education, Inc. Slide 2-35 Example: Solving a Right Triangle, Given an Angle and a Side Solve right triangle ABC, if A = 42  30' and c = 18.4. B = 90  42  30' B = 47  30' A C B c = 18.4 42  30'

36. Copyright © 2005 Pearson Education, Inc. Slide 2-36 Example: Solving a Right Triangle Given Two Sides Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm. B = 90  24.35  B = 65.65  A C B c = 27.82 a = 11.47

37. Copyright © 2005 Pearson Education, Inc. Slide 2-37 Solve the right triangle A= 28.00 o, and c = 17.4 ft

38. Copyright © 2005 Pearson Education, Inc. Slide 2-38 Homework Page 73 # 9-14

39. Copyright © 2005 Pearson Education, Inc. Slide 2-39 Definitions Angle of Elevation: from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint X. Angle of Depression: from point X to point Y (below) is the acute angle formed by ray XY and a horizontal ray with endpoint X.

40. Copyright © 2005 Pearson Education, Inc. Slide 2-40 Solving an Applied Trigonometry Problem Step 1Draw a sketch, and label it with the given information. Label the quantity to be found with a variable. Step 2Use the sketch to write an equation relating the given quantities to the variable. Step 3Solve the equation, and check that your answer makes sense.

41. Copyright © 2005 Pearson Education, Inc. Slide 2-41 Example: Application The length of the shadow of a tree 22.02 m tall is 28.34 m. Find the angle of elevation of the sun. Draw a sketch. The angle of elevation of the sun is 37.85 . 22.02 m 28.34 m B

42. Copyright © 2005 Pearson Education, Inc. 2.5 Further Applications of Right Triangles

43. Copyright © 2005 Pearson Education, Inc. Slide 2-43 Bearing Other applications of right triangles involve bearing, an important idea in navigation.

44. Copyright © 2005 Pearson Education, Inc. Slide 2-44 Example An airplane leave the airport flying at a bearing of N 32  E for 200 miles and lands. How far east of its starting point is the plane? The airplane is approximately 106 miles east of its starting point. e 200 32º

45. Copyright © 2005 Pearson Education, Inc. Slide 2-45 Example: Using Trigonometry to Measure a Distance A method that surveyors use to determine a small distance d between two points P and Q is called the subtense bar method. The subtense bar with length b is centered at Q and situated perpendicular to the line of sight between P and Q. Angle  is measured, then the distance d can be determined. a) Find d when  = and b = 2.0000 cm b) Angle  usually cannot be measured more accurately than to the nearest 1 second. How much change would there be in the value of d if  were measured 1 second larger?

46. Copyright © 2005 Pearson Education, Inc. Slide 2-46 Example: Using Trigonometry to Measure a Distance continued Let b = 2, change  to decimal degrees. b) Since  is 1 second larger, use 1.386944. The difference is.0170 cm.

47. Copyright © 2005 Pearson Education, Inc. Slide 2-47 Example: Solving a Problem Involving Angles of Elevation Sean wants to know the height of a Ferris wheel. From a given point on the ground, he finds the angle of elevation to the top of the Ferris wheel is 42.3. He then moves back 75 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 25.4. Find the height of the Ferris wheel.

48. Copyright © 2005 Pearson Education, Inc. Slide 2-48 Example: Solving a Problem Involving Angles of Elevation continued The figure shows two unknowns: x and h. Since nothing is given about the length of the hypotenuse, of either triangle, use a ratio that does not involve the hypotenuse, (the tangent ratio). In triangle ABC, In triangle BCD, x C B h DA 75 ft

49. Copyright © 2005 Pearson Education, Inc. Slide 2-49 Example: Solving a Problem Involving Angles of Elevation continued Since each expression equals h, the expressions must be equal to each other. Solve for x. Distributive Property Factor out x. Get x-terms on one side. Divide by the coefficient of x.

50. Copyright © 2005 Pearson Education, Inc. Slide 2-50 We saw above that Substituting for x. tan 42.3 =.9099299 and tan 25.4 =.4748349. So, tan 42.3 - tan 25.4 =.9099299 -.4748349 =.435095 and  The height of the Ferris wheel is approximately 74.48 ft. Example: Solving a Problem Involving Angles of Elevation continued