1. 6.7 Applications and Models

2. 2 What You Should Learn Solve real-life problems involving right triangles. Solve real-life problems involving directional bearings. Solve real-life problems involving harmonic motion.

3. 3 Applications Involving Right Triangles

4. 4 The three angles of a right triangle are denoted by the letters A, B and C (where C is the right angle), and the lengths of the sides opposite these angles by the letters a, b and c (where c is the hypotenuse).

5. 5 Example – Solving a Right Triangle Solve the right triangle shown in Figure 4.77 for all unknown sides and angles. Solution: Because C = 90 , it follows that A + B = 90  and B = 90  – 34.2  = 55.8 . Figure 4.77

6. 6 Example – Solution To solve for a, use the fact that a = b tan A. So, a = 19.4 tan 34.2   13.18. Similarly, to solve for c, use the fact that  23.46. cont’d

7. 7 Find the remaining parts of the triangle. Your Turn – Solving a Right Triangle

8. 8 Find the remaining parts of the triangle. Your Turn – Solving a Right Triangle

9. 9 Find the remaining parts of the triangle. Your Turn – Solving a Right Triangle

10. 10 Find the remaining parts of the triangle. For Angle B Your Turn – Solving a Right Triangle

11. 11 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' Your Turn:

12. 12 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 Your Turn:

13. 13 Solving an Applied Trigonometry Problem Step 1Draw a sketch, and label it with the given information. Label the quantity to be 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.

14. 14 Applications of Right Triangles Angle of Elevation Angle of Depression

15. 15      Angles of Elevation and Depression

16. 16 Example – Finding a Side of a Right Triangle A safety regulation states that the maximum angle of elevation for a rescue ladder is 72 . A fire department’s longest ladder is 110 feet. What is the maximum safe rescue height? Solution: A sketch is shown in Figure 4.79. From the equation sin A = a / c, it follows that a = c sin A Figure 4.79

17. 17 Example 2 – Solution = 110 sin 72   104.6. So, the maximum safe rescue height is about 104.6 feet above the height of the fire truck. cont’d

18. 18 Example - Find of an Object in the Distance Finding the height of a tree on a mountain.

19. 19 Finding the height of a tree on a mountain. Example - Find of an Object in the Distance

20. 20 Finding the height of a tree on a mountain. Example - Find of an Object in the Distance

21. 21 Your Turn: A swimming pool is 20 meters long and 12 meters wide. The bottom of the pool is slanted so that the water depth is 1.3 meters at the shallow end and 4 meters at the deep end, as shown. Find the angle of depression of the bottom of the pool. 7.69°

22. 22 Your Turn: 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

23. 23 If the ladder is 40’ long and the angle of elevation is 42 ̊, how high up is the fireman? 40’ h Your Turn:

24. 24 At a point 200 feet from the base of a building, the angle of elevation to the bottom of a smokestack is 35 ̊, while the angle to the top is 53 ̊. Find the height, s, of the smokestack alone. 200 ft. s x 265.41 -140.04 125.37 Your Turn:

25. 25 Your Turn: Problem: The angle of elevation of the Sun is 35.1 ̊ at the instant it casts a shadow 789 feet long of the Washington Monument. Use this information to calculate the height of the monument. Answer: 554.52 ft

26. Bob is at the top of a vertical cliff, 80m high. He sees a boat out at sea. The angle of depression from Bob to the boat is 34 o. How far from the cliff base is the boat? 80m 34 o x Your Turn:

27. From a point on a riverside, a bridge arch is 57m away and has an angle of elevation of 3 o, and the top of a yacht mast 14m away has an angle of elevation of 12 o. Will the yacht fit under the bridge? 57m 14m 3o3o 12 o x y Yes! But only just. Your Turn:

28. 28 Trigonometry and Bearings

29. 29 Bearings - Types In compass bearing, all bearings are measured from the north or from the south in an acute angle to the east or to the west. e.g.The compass bearing of B from A is N35  E. The compass bearing of A from B is S35  W. In true bearing, all bearings are measured from the north in a clockwise direction. A three-digit integer is used to represent the integral part of true bearing. e.g.The true bearing of B from A is 035 . The true bearing of A from B is 215 .

30. 30 Compass Bearings In surveying and navigation, directions are generally given in terms of bearings. A bearing measures the acute angle a path or line of sight makes with a fixed north-south line, as shown in Figure 4.81. For instance, the bearing of S 35  E in Figure 4.81(a) means 35  degrees east of south. (c)(b)(a) Figure 4.81

31. 31 Compass Bearing - Examples Name the Bearing

32. 32 Examples: A bearing measures the acute angle that a path or line of sight makes with a fixed north- south line. N S 35° EW N S 70° EW

33. 33 Your Turn: You try. Draw a bearing of: N80 0 W S30 0 E N S EW N S EW

34. 34 Solution You try. Draw a bearing of: N80 0 W S30 0 E

35. Caution A correctly labeled sketch is crucial when solving bearing applications. Some of the necessary information is often not directly stated in the problem and can be determined only from the sketch.

36. A ship leaves port at noon and heads due west at 20 knots, or 20 nautical miles (nm) per hour. At 2 P.M. the ship changes course to N 54 o W. Find the ship’s bearing and distance from the port of departure at 3 P.M. 54 o 20 nmph for 2 hrs 40 nm 20 nmph for 1 hr 20 nm a b 20sin(36 o ) 36 o θ d 20cos(36 o ) 78.181 o Bearing: N 78.181 o W Example – Finding Directions Using Bearings

37. A ship leaves port at noon and heads due west at 20 knots, or 20 nautical miles (nm) per hour. At 2 P.M. the ship changes course to N 54 o W. Find the ship’s bearing and distance from the port of departure at 3 P.M. 54 o 20 nmph for 2 hrs 40 nm 20 nmph for 1 hr 20 nm a b 20sin(36 o ) 36 o θ d 20cos(36 o ) 78.181 o Bearing: N 78.181 o W

38. Two lookout towers are 50 kilometers apart. Tower A is due west of tower B. A roadway connects the two towers. A dinosaur is spotted from each of the towers. The bearing of the dinosaur from A is N 43 o E. The bearing of the dinosaur from tower B is N 58 o W. Find the distance of the dinosaur to the roadway that connects the two towers. AB 47 o 32 o x50– x h 43 o 58 o

39. AB 47 o 32 o x50– x h 19.741 km

40. Two lookout towers spot a fire at the same time. Tower B is Northeast of Tower A. The bearing of the fire from tower A is N 33 o E and is calculated to be 45 km from the tower. The bearing of the fire from tower B is N 63 o W and is calculated to be 72 km from the tower. Find the distance between the two towers and the bearing from tower A to tower B. A B 33 o 63 o 45 72 a b c d s a + c b – d 45sin(33 0 ) + 72sin(63 0 ) 45cos(33 0 ) – 72sin(63 0 )

41. A B 33 o 63 o 45 72 a b c d s a + c b – d 45sin(33 0 ) + 72sin(63 0 ) 45cos(33 0 ) – 72sin(63 0 ) 88.805 km

42. A B 33 o 63 o 45 72 a b c d s a + c b – d 45sin(33 0 ) + 72sin(63 0 ) 45cos(33 0 ) – 72sin(63 0 ) θ 88.805 km

43. 43 Your Turn: Radar stations A and B are on an east-west line, 3.7 km apart. Station A detects a plane at C, on a bearing of 61°. Station B simultaneously detects the same plane, on a bearing of 331°. Find the distance from A to C.

44. 44 Your Turn: The bearing from A to C is S 52° E. The bearing from A to B is N 84° E. The bearing from B to C is S 38° W. A plane flying at 250 mph takes 2.4 hours to go from A to B. Find the distance from A to C. To draw the sketch, first draw the two bearings from point A. Choose a point B on the bearing N 84° E from A, and draw the bearing to C, which is located at the intersection of the bearing lines from A and B.

45. Solution: (cont.) The distance from A to B is about 430 miles.

46. 46 Harmonic Motion

47. 47 Terminology Harmonic Motion – Simple vibration, oscillation, rotation, or wave motion. It can be described using the sine and cosine functions. Displacement – Distance from equilibrium.

48. 48 Simple Harmonic Motion Equilibrium: Rest position (B). Amplitude: Distance from rest position to greatest displacement (A-B or B-C). Period: Length of time to complete one vibration (A back to A or C back to C).

49. 49 Simple Harmonic Motion Simple harmonic motion is related to circular motion.

50. 50 Simple Harmonic Motion A point that moves on a coordinate line is in simple harmonic motion if its distance d from the origin at time t is given by where a and ω are real numbers (ω>0) and frequency is number of cycles per unit of time.

51. 51 Simple Harmonic Motion Frequency of an object in simple harmonic motion: Number of oscillations per unit time Frequency f is reciprocal of period

52. 52 Harmonic Motion

53. 53 Simple Harmonic Motion 10 cm 0 cm 10 cm For example, consider a ball that is bobbing up and down on the end of a spring, as shown in below.

54. 54 Harmonic Motion Suppose that 10 centimeters is the maximum distance the ball moves vertically upward or downward from its equilibrium (at-rest) position. Suppose further that the time it takes for the ball to move from its maximum displacement above zero to its maximum displacement below zero and back again is t = 4 seconds. Assuming the ideal conditions of perfect elasticity and no friction or air resistance, the ball would continue to move up and down in a uniform and regular manner.

55. 55 Harmonic Motion From this spring you can conclude that the period (time for one complete cycle) of the motion is Period = 4 seconds its amplitude (maximum displacement from equilibrium) is Amplitude = 10 centimeters and its frequency (number of cycles per second) is Frequency = cycle per second. Motion of this nature can be described by a sine or cosine function, and is called simple harmonic motion.

56. 56 Example – Simple Harmonic Motion Write the equation for the simple harmonic motion of the ball illustrated in below, where the period is 4 seconds. What is the frequency of this motion? 10 cm 0 cm 10 cm

57. 57 Example – Solution Because the spring is at equilibrium (d = 0) when t = 0, you use the equation d = a sin  t. Moreover, because the maximum displacement from zero is 10 and the period is 4, you have the following. Amplitude = | a | = 10 = 4 Consequently, the equation of motion is d = 10

58. 58 Example – Solution Note that the choice of a = 10 or a = –10 depends on whether the ball initially moves up or down. The frequency is Frequency cycle per second. cont’d

59. 59 Example – Simple Harmonic Motion Given this equation for simple harmonic motion Find: a)Maximum displacement b)Frequency c)Value of d at t=4 d)The least positive value of t when d=0

60. 60 A mass attached to a spring vibrates up and down in simple harmonic motion according to the equation Find: Maximum displacement Frequency Value of d at 2 values of t for which d=0 Your Turn:

61. 61 A weight attached to the end of a spring is pulled down 5 cm below its equilibrium point and released. It takes 4 seconds to complete one cycle of moving from 5 cm below the equilibrium point to 5 cm above the equilibrium point and then returning to its low point. Find the sinusoidal function that best represents the motion of the moving weight. Find the position of the weight 9 seconds after it is released. Example:

62. 62 A buoy oscillates in simple harmonic motion as waves go past. At a given time it is noted that the buoy moves a total of 6 feet from its low point to its high point, returning to its high point every 15 seconds. Write a sinusoidal function that describes the motion of the buoy if it is at the high point at t=0. Find the position of the buoy 10 seconds after it is released. Your Turn:

63. 63 Homework: Section 6.7 Notes Worksheet Sec. 6.7, Pg. 521 – 523; #1 – 39 odd, 55 – 61 odd.