1. Problem 1 The Escalator Click on the link below to view the lesson starter. Stuck on an Escalator

2. In order to save our friends, the rescue team must determine the total length of the escalator. If the vertical rise of the escalator has been measured at 195 ft., 9.5 in. at an angle of elevation of 10.4°, what is the length of the escalator to the nearest foot? Problem 1 (Part 1)

3. What’s your strategy? 1.Draw the figure. 2.Convert measurement to desired unit (feet). 3.Calculate the length of the escalator.

4. 1. Draw the figure. 10.4° 195 ft., 9.5 in. M E O x

5. 2. Convert measurement to desired unit (feet). 195 ft., 9.5 in. 9.5 in. x 1 ft. = 9.5 ft. = 0.792 ft. 12 in. 12 195 ft. + 0.792 ft. = 195.792 ft. 195.792 ft. 10.4° M EO x

6. 3. Calculate the length of the escalator. Sin O° = ME MO Sin 32.4° = 90.792 ft. x x (Sin 32.4°) = 90.792 ft. x = 90.792 ft. Sin 32.4° x = 169.443 ft. MO = 169.443 ≈ 169 ft.

7. What’s your strategy? 1.Calculate starting distance from the bottom of the escalator. 2.Calculate distance traveled along the incline. 3.Use trigonometry to calculate horizontal distance.

8. Problem 1 (Part 2) The rescue team determines our friends are stuck a third of the length of the escalator from the bottom when the escalator begins to move again. If the escalator moves our friends for just 10 seconds before stopping again, along the incline at a speed of 3 feet per second, what is the horizontal distance traveled to the nearest tenth?

9. 1.Calculate starting distance from the bottom of the escalator. …our friends are stuck a third of the length of the escalator from the bottom… Current location = (1/3) (169 ft.) = 56.33 ft. 10.4° M EO 56.33 ft.

10. 2. Calculate distance traveled along the incline. …the escalator moves our friends for 10 seconds along the incline at a speed of 3 feet per second… d = r (t) d = 3 (10) = 30 ft. 10.4° M EO 56.33 ft. 30 ft.

11. 3. Use trigonometry to calculate horizontal distance. 10.4° M EO 86.33 ft. T J y Cos O° = TO OJ Cos (10.4°) = y 86.33 ft. y = (Cos 10.4°) (86.33) ≈ 84.9 ft.

12. Problem 1 (Part 3) After this 10 seconds of movement, would it be shorter for the stranded riders to walk the rest of the way up or revert back down to the bottom of the escalator? Explain why.

13. Problem 2 Jessica observed a mountain climber reaching the summit, which is known to be at 2,358 ft. If she sighted the climber standing 1500 ft. from the base, at what angle did Jessica sight the mountain climber to the nearest degree?

14. What’s your strategy? 1.Draw a figure to represent the problem. 2.Determine which trigonometric ratio to use. 3.Calculate the angle of elevation.

15. 1. Draw a figure to represent the problem. 2,358 ft. 1,500 ft. x° M S C

16. 2. Determine which trigonometric ratio to use. 2,358 ft. 1,500 ft. x° Opposite Adjacent Tangent M S C

17. 3. Calculate the Angle of Elevation. Tan C° = MS SC Tan x° = 2,358 ft. 1,500 ft. x = Tan -1 2,358 ft. 1,500 ft. x = 57.538° C≈ 58° ∨

18. Problem 3 A rescue helicopter pilot sights a life raft at an angle of depression of 26 o. The helicopter is 3 km above the water. What is the pilot’s surface distance from the raft to the nearest km?

19. What’s your strategy? 1.Draw a figure to represent the problem. 2.Determine which trigonometric ratio to use. 3.Calculate the surface distance.

20. 1. Draw a figure to represent the problem. 3 km x km H W R 26° Alternate Interior Angles

21. 2. Determine which trigonometric ratio to use. 3 km x km H W R 26° Opposite Adjacent Tangent

22. 3. Calculate the surface distance. Tan R° = HW WR Tan 26° = 3 km x km x (Tan 26°) = 3 km x = 3 km Tan 26° x = WR ≈ 6.1509 km

23. Kevin is standing at the back of the cruise ship and observes two sea turtles following each other, swimming in a straight line in the opposite direction of the ship. Kevin’s position is 206 meters above sea level and the angles of depression to the two sea turtles are 43° and 47°. Calculate the distance between the two sea turtles to the nearest meter. K 206m 47° 43° S TO Problem 4

24. What’s your strategy? 1.Separate and re-draw the two triangles. 2.Calculate individual horizontal distances. 3.Calculate the difference between the two horizontal distances.

25. K 206m 47° 43° S T O x 206m O y 47° 43° K 1. Separate and re-draw the two triangles.

26. 2. Calculate individual horizontal distances. Tan S° = KO SO Tan 43° = 206 m x x (Tan 43°) = 206 m x = 206 m Tan 43° x = 220.90795 SO = 220.90795 m Tan T° = KO TO Tan 47° = 206 m y x (Tan 47°) = 206 m x = 206 m Tan 47° x = 193.03062 SO = 193.03062 m

27. 3. Calculate the differences between the two horizontal distances. ST = SO – TO ST = 220.90795 m – 193.03062 m ST = 27.8773 m ST ≈ 28 m