1. Trigonometry Pythagoras Theorem & Trigo Ratios of Acute Angles

2. Pythagoras Theorem a + b = c a 2 + b 2 = c 2 where c is the hypotenuse while a and b are the lengths of the other two sides. c b a

3. Trigo Ratios of Acute angles O P Q hypotenuse adjacent opposite Hypotenuse = side opposite right angle/longest side Adjacent = side touching theta Opposite= side opposite theta

4. Trigo Ratios of Acute angles Hypotenuse = AB Adjacent = AC Opposite= BC A B C X Y Z Hypotenuse = XZ Adjacent = XY Opposite= YZ

5. Trigo Ratios of Acute angles O P Q hypotenuse adjacent opposite Tangent ratio Cosine ratio Sine ratio tan cossin

6. Trigo Ratios of Acute angles O P Q hypotenuse adjacent opposite

7. Trigo Ratios of Acute angles O P Q hypotenuse adjacent opposite TOA CAH SOH

8. Exercise 1 5 13 12 4 3 5

9. Exercise 1 9 15 12 51 45 24

10. Exercise 2 4 3 5 8 15 17 20 21 29

11. Exercise 3

12. Exercise 4 sin (  2) = sin   2 sin (30°  2) = 0.2588… sin 30°  2 = 0.25 cos 2 = 2  cos  tan (10° + 30°) = 0.839… tan 10° + tan 30° = 0.753… tan (A + B) = tan A + tan B cos (2× 30°) = 0.5 2× cos 30° = 1.732…

13. Exercise 5 sin  = 0.4537  = sin -1 0.4537 = 26.981≈27.0° cos  = 0.3625  = cos -1 0.3625 = 68.746≈68.7° tan  = 4.393  = tan -1 4.393 = 77.176≈77.2°

14. Exercise 5 sin  = 0.8888  = sin -1 0.8888 = 62.722≈62.7° cos  = 0.9999  = cos -1 0.9999 = 0.8102≈0.8° tan  = 0.5177  = tan -1 0.5177 = 27.370≈27.4°

15. B AC 7 cm 8 cm D E 54.8° In the diagram, BCE is a straight line, angle ECD = 54.8° and angle CDE = angle ACB = 90°. BC = 7 cm and AC = CE = 8 cm. Calculate angle CED, angle DCB, angle BAC, the length of ED, the length of AE, Further Examples 1

16. B AC 7 cm 8 cm D E 54.8° In the diagram, BCE is a straight line, angle ECD = 54.8° and angle CDE = angle ACB = 90°. BC = 7 cm and AC = CE = 8 cm. Calculate angle CED = 180° − 90° − 54.8° = 35.2° angle CED?

17. Further Examples 1 B AC 7 cm 8 cm D E 54.8° In the diagram, BCE is a straight line, angle ECD = 54.8° and angle CDE = angle ACB = 90°. BC = 7 cm and AC = CE = 8 cm. Calculate angle DCB = 180° − 54.8° = 125.2° angle DCB?

18. Further Examples 1 B AC 7 cm 8 cm D E 54.8° Let angle BAC be . angle BAC?

19. Further Examples 1 B AC 7 cm 8 cm D E 54.8° the length of ED?

20. Further Examples 1 B AC 7 cm 8 cm D E 54.8° the length of AE?

21. Further Examples 2 A 16 m ladder is leaning against a house. It touches the bottom of a window that is 12 m above the ground. What is the measure of the angle that the ladder forms with the ground? Let the angle be . 16 m 12 m

22. Further Examples 3 A 16 m ladder is leaning against a house. It touches the bottom of a window that is 12 m above the ground. What is the measure of the angle that the ladder forms with the ground? Let the angle be . 16 m 12 m

23. Exercise 6 30° 50° A B C D 4 cm In the diagram, angle ADC = 30°, angle ACB = 50°, angle ABD = 90° and BC = 4 cm. Calculate (a) angle DAC

24. Applications – Angle of elevation and Angle of depression

26. Example 1

27. Example 2 A surveyor is 100 meters from the base of a dam. The angle of elevation to the top of the dam measures. The surveyor's eye-level is 1.73 meters above the ground. Find the height of the dam.

28. Trigonometric Ratios of Special Angles: 30°, 45° and 60°. 1 1 11 2

29. Trigonometric Ratios of Complementary Angles. b P Q R a c

30. At the point P, a boat observes that the angle of elevation of the cliff at point T is 32 o, and the distance PT is 150m. It sails for a certain distance to reach point Q, and observes that the angle of elevation of the point T becomes 48 o. T RQP 48 o 32 o 150m (i)Calculate the height of the cliff. (ii) Calculate the distance the boat is from the cliff at point Q. (iii) Calculate the distance travelled by the boat from point P to point Q.

31. T RQP 48 o 32 o 150m Let the height of the cliff = TR

32. T RQP 48 o 32 o 150m Let the distance the boat is from the cliff at point Q = QR

33. T RQP 48 o 32 o 150m Let the distance travelled by the boat from point P to point Q = PQ

34. Q2 1.3 m 3 m Let the angle be . 

35. Q3 In 15 Secs, distance travelled = 140 x 15 = 2100 m Plane 10 ° altitude 2100 m Let the altitude be a.

36. Q4 65 m Let the height of cliff be h. 37°

37. Q5 60 m 53° 65° cliff tower Let the height of cliff be h. Let the height of cliff and tower be x. Let the height of tower be t.

38. Q6 30 m 67° h kite Let the height of kite be h.

39. Q7 Danny 75° 30 m balloon Let the distance be d. d

40. Q8 25 m 23° d Let the distance be d. Buoy

41. Q9 h 60 o 50 o 1000m Let the height be h. x

42. Q10 18 m 46° 58° h x Let the height be h.

43. Q11 h 33 o 22 o 20 m Let the height be h. x

44. Q12 h 58 o 39 o 35 m Let the height be h. xAB

45. Q13(a) xoxo xoxo 40 60 E B Let the angle of depression be x.

46. Q13(b) 40 100 60 70 o B A C D E F