1. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

2. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

3. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

4. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

5. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

6. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

7. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

8. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

9. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

10. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

11. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

12. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

13. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

14. Mathematics Curriculum Centre eye - level Mr. Florben G. Mendoza

15. Mathematics Curriculum Centre eye - level The angles are equal – they are alternate interior angles Mr. Florben G. Mendoza

16. Mathematics Curriculum Centre 1. From a point 48 meters from the base of a flagpole, the angle of elevation to the top of a flagpole is 34°. Find the height of the flagpole. Angle of Elevation h 48 m 34° Mr. Florben G. Mendoza

17. Mathematics Curriculum Centre 1. From a point 48 meters from the base of a flagpole, the angle of elevation to the top of a flagpole is 34°. Find the height of the flagpole. Angle of Elevation Mr. Florben G. Mendoza h 48 m 34° SOH-CAH-TOA tan θ = O A 1 h = 48 (tan 34°) h = 32.38 meters 48 tan 34°h = (O) (A)

18. Mathematics Curriculum Centre 2. From the top of a tower which is 175 ft tall, the angle of depression to a car is 13°. How far is the car from the tower? Angle of Depression 175 ft 13° d Mr. Florben G. Mendoza

19. Mathematics Curriculum Centre Angle of Depression 175 ft 13° d (O) (A) SOH-CAH-TOA tan θ = O A d tan 13° 175 = 1 d (tan 13°) = 175 tan 13° d = 758.01 ft Mr. Florben G. Mendoza

20. Mathematics Curriculum Centre Angle of Elevation & Depression 3. From a window of one building, the angle of depression to the base of the second building is 29° 30 ’, and the angle of elevation to the top is 47° 50 ’. If the buildings are 200 m apart, how high is the second building? 29° 30’ 47° 50’ 200 m h Mr. Florben G. Mendoza

21. Mathematics Curriculum Centre Angle of Elevation & Depression 29° 30’ 47° 50’ 200 m h1h1 47° 50’ H (O) (A) 200 tan 47 50° h1h1 = 1 h 1 = 200 (tan 47° 50’) h 1 = 220.83 meters h 2 (O) 200 tan 29° 30’ h2h2 = 1 h 2 = 200 (tan 29° 30’) h 2 = 113.15 meters H = h 1 + h 2 H = 220.83 + 113.15 H = 333.98 meters 29° 30’ Mr. Florben G. Mendoza

22. Mathematics Curriculum Centre 4. When you stand at a certain spot near a building, you get an elevation of 70°. You then walk 60 feet away from the building and get a 50° angle of elevation. How tall is the building? Multi-Step Problem 70° 50° 60 ft h Mr. Florben G. Mendoza

23. Mathematics Curriculum Centre h= x (tan 70°) h= (60 + x)(tan 50°) by Substitution: h= (60 + x)(tan 50°) x(tan 70°)= (60) (tan 50°) + x (tan 50°) x(tan 70°) - x (tan 50°) = (60) (tan 50°) x(tan 70° - tan 50°) = (60) (tan 50°) (tan 70° - tan 50°) (tan 70° - tan 50°) x = 45.96 ft h= x (tan 70°) h= (45.96 ft) (tan 70°) h= 126.27 ft Mr. Florben G. Mendoza 70° 50 ° 60 ft h

24. Mathematics Curriculum Centre Inscribed Polygon 5. Find the length of the side of a regular pentagon inscribed in a circle of radius 23 inches. 23 in x Mr. Florben G. Mendoza

25. Mathematics Curriculum Centre 70° 50° 60 ft x 60 + x h 70° h x 50° h 60 + x x tan 70° h = h= x (tan 70°) 60 +x tan 50° h = h= (60 + x)(tan 50°) Mr. Florben G. Mendoza

26. Mathematics Curriculum Centre 23 in x 360° 5 =72° 23 in 36° 23 in x y y y (O) (H) sin θ = O H 23 sin 36° y = 1 y = 23 (sin 36°) y = 13.52 inches x = 2y x = 27.04 inches Inscribed Polygon Mr. Florben G. Mendoza

27. Mathematics Curriculum Centre Circumscribed Polygon 6. Find the length of the radius of a circumscribed regular hexagon if the length of one side measures 72 inches. 72 inches r Mr. Florben G. Mendoza

28. Mathematics Curriculum Centre Circumscribed Polygon 6. Find the length of the radius of a circumscribed regular hexagon if the length of one side measures 72 inches. 72 inches r 30° 36 in tan θ = O A r tan 30° 36 = 1 r r(tan 30°) = 36 tan 30° r = 62.35 in 30° 36 in r r (O) (A) Mr. Florben G. Mendoza

29. Mathematics Curriculum Centre DIRECTION – Direction is given in terms of acute angle made with the NS (North – South) direction. All directions start with an N for North or an S for South, then the angle between 0° & 90° and end with an E or W for East and West, respectively. N S EW 50° N S EW N S EW N S EW N 50° ES 50° E S 50° W N 50° W Mr. Florben G. Mendoza

30. Mathematics Curriculum Centre N S EW 40° N S EW 55° N S EW 70° N S EW 60° N 40° E S 55° E S 70° W N 60° W Mr. Florben G. Mendoza

31. Mathematics Curriculum Centre N S EW 60° N S EW 50° N S EW 25° N S EW 45° N 30° E S 40° E S 65° W N 45° W 30° 40° 65° 45° Mr. Florben G. Mendoza

32. Mathematics Curriculum Centre 7. At a certain instant, a ship was 5 km north of a lighthouse. The ship was traveling eastward and after 20 minutes, its direction was N 20° E. What was the speed of the ship per hour? 5 km 20° d Mr. Florben G. Mendoza

33. Mathematics Curriculum Centre 5 km 20° d (O) (A) tan θ= O A tan 20°= d 5 d = 5 (tan 20°) d = 1.82 km d = rt r= d t 20 min 1 hr 60 min x = 1 3 hour r= 1.82 km 1 / 3 hour = 5.46 kph Mr. Florben G. Mendoza

34. Mathematics Curriculum Centre The bearing to a point is the angle measured in a clockwise direction from the north line. N S EW O 50° N S EW O 120° N S EW O 27° Direction: N50°E 050° S60°E 120° N27°W 333° Bearing: Direction: Bearing: Direction: Bearing: Mr. Florben G. Mendoza

35. Mathematics Curriculum Centre N S EW 40° N S EW 30° N S EW 50° N S EW 110° 040° 330° 230 ° 070° Mr. Florben G. Mendoza

36. Mathematics Curriculum Centre 8. A cyclist travels 10 km south, then 8 km east. Find the cyclist's bearing from her starting point to the nearest degree. N S E W (O) 10 km 8 km (A) SOH-CAH-TOA tan θ = O A 10 tan θ 8 = θ = 38.66° tan θ = 0.8 θ = tan -1 0.8 B = 180° - 38.66° B = 141.34° Mr. Florben G. Mendoza