1. © T Madas Trigonometric Calculations

2. © T Madas x 16 m 35° tanθ = Opp Adj c tan35° = x 16 c x = c x ≈ 11.2 m x tan35° Trigonometric Calculations S O H C A H T O A

3. © T Madas cos65° h 65° 9 m cosθ = Adj Hyp c cos65° = 9 h c h x cos65° c 9 = h = 9 cos65° c h ≈ 21.3 m S O H C A H T O A = Trigonometric Calculations

4. © T Madas x 14 m 38° sinθ = Opp Hyp c sin38° = x 14 c x = c x ≈ 8.62 m x sin38° S O H C A H T O A Trigonometric Calculations

5. © T Madas x 19 m 31° cosθ = Adj Hyp c cos31° = x 19 c x = c x ≈ 16.29 m x cos31° S O H C A H T O A Trigonometric Calculations

6. © T Madas 40 m x tan55° 55° tanθ = Opp Adj c tan55° = 40 x c x x tan55° c 40 = x = tan55° c x ≈ 28.0 m S O H C A H T O A = Trigonometric Calculations

7. © T Madas sin25° h 25° 8 m sinθ = Opp Hyp c sin25° = 8 h c h x sin25° c 8 = h = 8 sin25° c h ≈ 18.93 m S O H C A H T O A = Trigonometric Calculations

8. © T Madas Inverse Trigonometric Calculations

9. © T Madas 0.5 16 m Inverse Trigonometric Calculations ? 8 m θ sinθ = Opp Hyp c sinθ = 8 16 c sinθ = 0.5 c θ = sin -1 ( ) c θ = 30° S O H C A H T O A

10. © T Madas ( ) 0.5454 11 m ? 6 m θ cosθ = Adj Hyp c cosθ = 6 11 c cosθ ≈ 0.5454 c θ = cos -1 c θ = 56.9° [1 d.p.] S O H C A H T O A Inverse Trigonometric Calculations

11. © T Madas ( ) 2 16 m ? 8 m θ tanθ = Opp Adj c tanθ = 16 8 c tanθ = 2 c θ = tan -1 c θ = 63.4° [1 d.p.] S O H C A H T O A Inverse Trigonometric Calculations

12. © T Madas …more Sine & Cosine Calculations

13. © T Madas a 6.93 m 12 m S O H C A H T O A 30° More Trigonometric Calculations Calculate the sides of this triangle b tanθ = Opp Adj c tan30° = a 12 c a = c a ≈ 6.93 m x tan30°

14. © T Madas 6.93 m 12 m S O H C A H T O A 30° More Trigonometric Calculations Calculate the sides of this triangle b Why should we try to avoid using 6.93 m? cos30° cosθ = Adj Hyp c cos30° = 12 b c b x cos30° c 12 = b = cos30° c b ≈ 13.9 m =

15. © T Madas 6.93 m 12 m 30° More Trigonometric Calculations Calculate the sides of this triangle b We could use Pythagoras but we must make sure we have the answer of 6.93 to greater accuracy 12 2 + 6.9282 2 = b 2= b 2 c 144+ 48b 2 = c 192b 2 = c 192b = c b ≈ 13.9 m

16. © T Madas …further Sine & Cosine Calculations

17. © T Madas b a 8 cm Calculate the area of this triangle 50° 60° h S O H C A H T O A sinθ = Opp Hyp c sin60° = h 8 c h = 8 c h ≈ 6.928 cm x sin60° h ≈ 6.928

18. © T Madas b a 8 cm Calculate the area of this triangle 50° 60° h S O H C A H T O A h ≈ 6.928 cosθ = Adj Hyp c cos60° = a 8 c a = 8 c a = 4 cm x cos60° a ≈ 4

19. © T Madas b a 8 cm Calculate the area of this triangle 50° 60° h S O H C A H T O A h ≈ 6.928 a ≈ 4 tan50° tanθ = Opp Adj c tan50° = h b c b x tan50° c 6.928 = b = tan50° c b ≈ 5.813 cm b = 5.813 =

20. © T Madas b a 8 cm Calculate the area of this triangle 50° 60° h S O H C A H T O A h ≈ 6.928 a = 4 b = 5.813

21. © T Madas Exam Question

22. © T Madas An A4 size sheet of card is a rectangle 297 mm long by 209 mm wide. (a) Calculate the length of its diagonal, to 3 s.f. (b) Find the acute angle formed by the diagonal and the longer of the two sides of the rectangle, to 2 s.f. A B C D 297 mm 209 mm d 363 mm 209 2 + 297 2 c = d 2= d 2 43681+ 88209 c = d 2= d 2 131890 c = d 2= d 2 c = d= d 363 mmd ≈

23. © T Madas A B C D 297 mm 209 mm 363 mm θ We can use trig to find θ Which trig ratio can we use? Why should we avoid using a ratio which involves the length of 363 mm? An A4 size sheet of card is a rectangle 297 mm long by 209 mm wide. (a) Calculate the length of its diagonal, to 3 s.f. (b) Find the acute angle formed by the diagonal and the longer of the two sides of the rectangle, to 2 s.f.

24. © T Madas An A4 size sheet of card is a rectangle 297 mm long by 209 mm wide. (a) Calculate the length of its diagonal, to 3 s.f. (b) Find the acute angle formed by the diagonal and the longer of the two sides of the rectangle, to 2 s.f. tan -1 A B C D 297 mm 209 mm 363 mm θ tanθ = Opp adj tanθ = 209 297 θ = θ = θ ≈ 35° 35° 209 297 c c c

25. © T Madas

26. x 12.9 cm A cylindrical glass has a diameter of 6 cm and a straw is resting on the glass as shown, the straw protruding above the rim of the glass by 5 cm. The straw forms an angle of 65° with the base of the glass. 1. Calculate the height of glass, correct to 1 decimal place. 2. Calculate the length of the straw, correct to 1 decimal place. 6 cm 5 cm tanθ = Opp Adj c tan65° = x 6 c x = 6 c x ≈ 12.9 cm x tan65° 65 °

27. © T Madas 12.9 cm A cylindrical glass has a diameter of 6 cm and a straw is resting on the glass as shown, the straw protruding above the rim of the glass by 5 cm. The straw forms an angle of 65° with the base of the glass. 1. Calculate the height of glass, correct to 1 decimal place. 2. Calculate the length of the straw, correct to 1 decimal place. 6 cm 5 cm 65 ° We can find y by using Pythagoras Theorem or trigonometry. If we use Pythagoras Theorem we must obtain the value of 12.9 to greater accuracy. We are going to use trigonometry instead so we can use the length of 6 cm. y

28. © T Madas 12.9 cm A cylindrical glass has a diameter of 6 cm and a straw is resting on the glass as shown, the straw protruding above the rim of the glass by 5 cm. The straw forms an angle of 65° with the base of the glass. 1. Calculate the height of glass, correct to 1 decimal place. 2. Calculate the length of the straw, correct to 1 decimal place. 6 cm 5 cm 65 ° y cos65° cosθ = Adj Hyp c cos65° = 6 y c y x cos65° c y = y = 6 cos65° c y ≈ 14.2 cm = The straw is 19.2 cm long

29. © T Madas

30. The diagram below shows an alley between two houses. A ladder 3.6 m long is placed against house B as shown. The angle between the ground and the ladder is 70°. Calculate how wide is the alley, giving your answer correct to 3 significant figures. A B 3.6 m 70° x S O H C A H T O A x 3.6 = cos70° c x 3.6 = x cos70° c x 3.6 ≈ x 0.342 c x 1.23 m = [3 s.f.] 1.23 m

31. © T Madas The diagram below shows an alley between two houses. A different ladder is placed against house A as shown. This ladder is 4.2 m long. Calculate the angle the ladder makes with the wall, giving your answer correct to the nearest degree. A B 4.2 m θ S O H C A H T O A 1.23 4.2 = sinθ c 0.293 ≈ c θ sin -1 [ ] = 0.293 c θ 17° = [nearest degree] 1.23 m

32. © T Madas

33. 2.8 4.5 The figure below shows the cross section of a barn. AE = AB and all lengths are in metres [not to scale] 1.Calculate the length of AB. 2.Calculate the size of the angle ABE, giving your answer to the nearest degree. A B C D E 11 9 8.2 2.8 2 = x 2 + 4.5 2 c 7.84+ 20.25= x 2 c x 2x 2 = 28.09 c x =28.09 By Pythagoras Theorem: c x = 5.3 m x 5.3 lengths in metres

34. © T Madas 2.8 4.5 The figure below shows the cross section of a barn. AE = AB and all lengths are in metres [not to scale] 1.Calculate the length of AB. 2.Calculate the size of the angle ABE, giving your answer to the nearest degree. A B C D E 11 9 8.2 5.3 lengths in metres θ Since we know the exact values of all the sides of the right angled triangle we can use any one of the three trigonometric ratios: S O H C A H T O A 2.8 5.3 = sinθ c 0.528 ≈ c θ sin -1 [ ] = 0.528 c θ 32° = [nearest degree]

35. © T Madas

36. 2.7 6 5.6 The figure below shows a right angled trapezium ABCD. All lengths are in cm [not to scale] 1.Calculate the length of BC. 2.Calculate the size of the angle DCB, giving your answer correct to 1 decimal place. A B C D lengths in cm 5.6 3.3 x 3.3 2 = x 2 + 5.6 2 c 10.89+ 31.36= x 2 c x 2x 2 = 42.25 c x =42.25 By Pythagoras Theorem: c x = 6.5 cm 6.5

37. © T Madas 2.7 6 5.6 The figure below shows a right angled trapezium ABCD. All lengths are in cm [not to scale] 1.Calculate the length of BC. 2.Calculate the size of the angle DCB, giving your answer correct to 1 decimal place. A B C D lengths in cm 5.6 3.3 6.5 Since we know the exact values of all the sides of the right angled triangle we can use any one of the three trigonometric ratios: S O H C A H T O A 5.6 3.3 = tanθ c 1.697 ≈ c θ tan -1 [ ] = 1.697 c θ 59.5° = [1 d.p.] θ

38. © T Madas

39. 5.66 cm ABCD is a quadrilateral with R DCB = R ADB = 90°, R DAB = 40°, DB = 9 cm and CB = 7 cm. Calculate to 3 significant figures: 1. The length of CD 2. The length of AB 3. The size of R CBD A B C D 41° 9 cm 7 cm x 7 27 2 + x 2 = 9 2= 9 2 c 49+ x 2 = 81 c – 49 x 2x 2 = 81 c x 2x 2 = 32 c x ≈5.66 cm By Pythagoras Theorem: x =32 c [3 s.f.]

40. © T Madas 13.7 cm 5.66 cm ABCD is a quadrilateral with R DCB = R ADB = 90°, R DAB = 40°, DB = 9 cm and CB = 7 cm. Calculate to 3 significant figures: 1. The length of CD 2. The length of AB 3. The size of R CBD A B C D 41° 9 cm 7 cm S O H C A H T O A y 9 y = sin41° c y 9 = x sin41° c y 9 sin41° = c y 13.7 cm = [3 s.f.] y 9 0.656 = c

41. © T Madas 13.7 cm 5.66 cm ABCD is a quadrilateral with R DCB = R ADB = 90°, R DAB = 40°, DB = 9 cm and CB = 7 cm. Calculate to 3 significant figures: 1. The length of CD 2. The length of AB 3. The size of R CBD A B C D 41° 9 cm 7 cm S O H C A H T O A 7 9 = cosθ c 0.777 ≈ c θ cos -1 [ ] = 0.777 c θ 38.9° = [3 s.f.] θ Why should we avoid using a trig ratio which involves the length CD ?

42. © T Madas

43. The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 1.The vertical height of B above the level of A 2.The vertical height of D above the level of A 3.The length EC 4.The height of the water level if the tank was to be stood upright with AD horizontal. 40 cm 30 cm 40°

44. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 1.The vertical height of B above the level of A 40 cm 30 cm 40° x S O H C A H T O A x 40 = sin40° c x 40 = x sin40° c x 25.7 cm = [3 s.f.]

45. © T Madas 1.The vertical height of B above the level of A 2.The vertical height of D above the level of A 3.The length EC 4.The height of the water level if the tank was to be stood upright with AD horizontal. The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 40° x

46. © T Madas 1.The vertical height of B above the level of A 2.The vertical height of D above the level of A The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 40° x S O H C A H T O A y 30 = sin50° c y 30 = x sin50° c y 23.0 cm = [3 s.f.] y 50°

47. © T Madas α 1.The vertical height of B above the level of A 2.The vertical height of D above the level of A 3.The length EC 4.The height of the water level if the tank was to be stood upright with AD horizontal. The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 40° x y 50° 40° 30 cm

48. © T Madas α The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 40° x y 50° 40° 30 cm S O H C A H T O A 30 α = tan40° c α 30 = x tan40° c α 30 tan40° = c α 35.8 cm = [3 s.f.] α 30 0.839 = c 35.8cm

49. © T Madas 1.The vertical height of B above the level of A 2.The vertical height of D above the level of A 3.The length EC 4.The height of the water level if the tank was to be stood upright with AD horizontal. The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 35.8cm

50. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 35.8cm

51. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 35.8cm

52. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 35.8cm

53. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 35.8cm

54. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A B C D E 40 cm 30 cm 35.8cm

55. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A C B D E 40 cm 30 cm 35.8cm The area of the rectangle (above the water level) must be equal to the area of the triangle BCE 1212 x 30 x 35.8 = 30 h x h 17.9 cm = h 22.1cm

56. © T Madas The figure below shows the rectangular cross-section ABCD of a fish tank with AB = CD = 40 cm and BD = DA = 30 cm and the side AB tilted at an angle of 40° to the horizontal. The water inside the tank is level with point B. Calculate to 3 significant figures: A C B D E 40 cm 30 cm 35.8cm h 22.1cm 1.The vertical height of B above the level of A 2.The vertical height of D above the level of A 3.The length EC 4.The height of the water level if the tank was to be stood upright with AD horizontal.

57. © T Madas