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Trigonometry

Trigonometry. Pythagoras Theorem & Trigo Ratios of Acute Angles. Pythagoras Theorem. a 2 + b 2 = c 2. where c is the hypotenuse while a and b are the lengths of the other two sides. c. a. b. P. O. Q. Trigo Ratios of Acute angles. hypotenuse. opposite. adjacent.

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Trigonometry

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  1. Trigonometry Pythagoras Theorem & Trigo Ratios of Acute Angles

  2. Pythagoras Theorem a2 + b2 = c2 where c is the hypotenuse while a and b are the lengths of the other two sides. c a b

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

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

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

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

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

  8. 13 5 5 3 12 4 Exercise 1

  9. 24 12 9 45 15 51 Exercise 1

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

  11. Exercise 3

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

  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 7 cm A C 8 cm 54.8° D 8 cm E Further Examples 1 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,

  16. B 7 cm A C 8 cm 54.8° D 8 cm E Further Examples 1 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 CED = 180° − 90° − 54.8° = 35.2°

  17. B 7 cm A C 8 cm 54.8° D 8 cm E Further Examples 1 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? angle DCB = 180° − 54.8° = 125.2°

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

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

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

  21. 65 m h GROUND Further Examples 2 John is flying a kite whose string is making a angle with the ground. The kite string is 65 meters long. How far is the kite above the ground? Let the height be h.

  22. 65 m ? GROUND Further Examples 2 John is flying a kite whose string is making a angle with the ground. The kite string is 65 meters long. How far is the kite above the ground? Let the height be h.

  23. 192 Inches 16 feet 150 inches 12 feet 6 inches 1 foot = 12 inches Further Examples 3 A 16 feet ladder is leaning against a house. It touches the bottom of a window that is 12 feet 6 inches above the ground. What is the measure of the angle that the ladder forms with the ground? Let the angle be .

  24. 192 Inches 150 inches 1 foot = 12 inches Further Examples 3 A 16 feet ladder is leaning against a house. It touches the bottom of a window that is 12 feet 6 inches above the ground. What is the measure of the angle that the ladder forms with the ground? Let the angle be .

  25. 4 cm A 50° 30° D B C Exercise In the diagram, angle ADC = 30°, angle ACB = 50°, angle ABD = 90° andBC = 4 cm. Calculate (a) angle DAC

  26. Applications – Angle of elevation and Angle of depression

  27. Applications – Angle of elevation and Angle of depression

  28. Example 1

  29. 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.

  30. T 150m 48o 32o P Q R At the point P, a boat observes that the angle of elevation of the cliff at point T is 32o, 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 48o. (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 150m 48o 32o P Q R Let the height of the cliff = TR

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

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

  34. 2 1 1 1 1 Trigonometric Ratios of Special Angles: 30°, 45° and 60°.

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

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