![Simplify sin(4x) / cos(2x)
sin 4𝑥 = sin[2(2𝑥)]
Let 𝐴 = 2𝑥
sin 2 2𝑥 = sin(2𝐴)
From double angles,
sin(2𝐴) = 2 sin 𝐴 cos 𝐴
Going back to 𝐴 = 2𝑥,
2 sin 𝐴 cos 𝐴 = 2 sin 2𝑥 cos(2𝑥)
Therefore,
sin 4𝑥 = 2 sin 2𝑥 cos(2𝑥)
sin(4𝑥)
cos(2𝑥)
=
2 sin 2𝑥 cos(2𝑥)
cos(2𝑥)
= 2 sin(2𝑥)
For double angles,
sin 2𝑥 = 2 sin(𝑥) cos(𝑥)
= 2 sin(2𝑥)
= 2[2 sin(𝑥) cos(𝑥)]
= 4 sin 𝑥 cos(𝑥)](https://image.slidesharecdn.com/ppt-2trigonometry1-250324000042-183f5bfe/85/Correlation-Powerpoint-2-Trigonometry-1-pdf-57-320.jpg)
Correlation: Powerpoint 2- Trigonometry (1).pdf
- 2. SUB-TOPIC 1 Circular and Trigonometric Functions
- 3. • To discuss the unit circle and angles • To solve problems involving trigonometric functions
- 4. • The study of triangles applying the relationship between sides and angles • Word origin: • “TRIGONON” meaning triangle • “METRIA” meaning measurement
- 7. Median • Line joining a vertex to the midpoint of the opposite side. • Intersection: Centroid
- 8. Altitude • Line through a vertex and perpendicular to the opposite side. • Intersection: Orthocenter
- 9. Perpendicular Bisector • Create a right angle at the midpoint of the segment • Intersection: Circumcenter
- 10. Angle Bisector • Split the internal angle into two congruent angles. • Intersection: Incenter
- 11. EULER LINE • the line joining the centroid, circumcenter, and the incenter.
- 12. Trigonometric Functions sin Θ = 𝑜 ℎ cos Θ = 𝑎 ℎ tan Θ = 𝑜 𝑎 csc Θ = ℎ 𝑜 sec Θ = ℎ 𝑎 cot Θ = 𝑎 𝑜
- 13. Pythagorean Theorem 𝑎2 + 𝑜2 = ℎ2 “In a right triangle, the sum of the squares of the sides is equal to the square of the hypotenuse” -Pythagoras, 500 BC
- 14. Coterminal Angles • Two angles have the same initial side and the same terminal side, but different amounts of rotation • To find coterminal angles for any angle, add or subtract 2π or 360° to the given angle
- 15. Trigonometric Cofunctions sin 900 − 𝜃 = cos 𝜃 cos 900 − 𝜃 = sin 𝜃 tan 900 − 𝜃 = cot 𝜃 cot 900 − 𝜃 = tan 𝜃 sec 900 − 𝜃 = c𝑠𝑐 𝜃 csc(900 − 𝜃) = sec 𝜃 Trigonometric Functions for Special Triangles 450 450 1 1 2 600 300 1 2 3
- 16. Unit Circle Positive sign convention for terminal sides falling in region: A = all trigonometric functions are positive C = cosine function is positive T = tangent function is positive S = sine function is positive 00 900 1800 2700 1,0 0,1 -1,0 0,-1 A C T S
- 17. If the value of cos A is a negative fraction, then angle A terminates in what quadrant? c. 3rd and 4th d. 1st and 4th a. 1st and 2nd b. 2nd and 3rd
- 18. If the value of cos A is a negative fraction, then angle A terminates in what quadrant? 𝐶𝑜𝑠𝑖𝑛𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑛𝑣𝑜𝑙𝑣𝑒𝑠 𝑥 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒 𝑎𝑛𝑑 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 cos 𝐴 = 𝑥 𝑟 Thus, for it to be a negative fraction, the x-coordinate must be negative, and it is located at 2nd and 3rd quadrants.
- 19. The angle between 180 degrees and 270 degrees has: c. Negative secant and tangent d. Negative sine and cosine a. Negative cotangent and cosecant b. Negative sine and tangent
- 20. The angle between 180 degrees and 270 degrees has: If the angle is between 1800 and 2700, it is located at the third quadrant. At this quadrant, a point has a negative x-coordinate and negative y-coordinate. Since both x and y are negative, tangent and cotangent functions are always positive in 3rd quadrant. Thus, choices involving tangent and cotangent are wrong.
- 21. Simplify: (sinθ + cosθtanθ)/(cosθ) c. sin θ d. 2tan θ a. tan θ b. 2cot θ
- 22. Simplify: (sinθ + cosθtanθ)/(cosθ) sin 𝜃 + cos 𝜃 tan 𝜃 cos 𝜃 = sin 𝜃 + cos 𝜃 sin 𝜃 cos 𝜃 cos 𝜃 = sin 𝜃 + 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 = sin 𝜃 𝑐𝑜𝑠𝜃 + sin 𝜃 cos 𝜃 = tan 𝜃 + tan 𝜃 = 2 tan 𝜃
- 23. If tan A = -3 and tan B = 2/3, find tan(A - B). c. -13/9 d. -12/9 a. 11/3 b. -10/9
- 24. If tan A = -3 and tan B = 2/3, find tan(A - B). tan 𝐴 = −3 → tan−1 tan 𝐴 = tan−1 −3 → 𝐴 = tan−1(−3) tan 𝐵 = 2 3 → tan−1 tan 𝐵 = tan−1 2 3 → 𝐵 = tan−1 2 3 tan(𝐴 − 𝐵) = tan tan−1 −3 − tan−1 2 3 = 11 3
- 25. If tan 4x = cot 6y, then c. 4x – 6y = 900 d. 6y – 4x = 900 a. 2x – 3y = 450 b. 2x + 3y = 450
- 26. If tan 4x = cot 6y, then 𝑡𝑎𝑛 4𝑥 = 𝑐𝑜𝑡 6𝑦 𝑆𝑖𝑛𝑐𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑛𝑑 𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑟𝑒 𝑐𝑜𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠, 4𝑥 + 6𝑦 = 900 2𝑥 + 3𝑦 = 450
- 27. The measure of 2.25 revolutions counterclockwise is: c. 835° d. 810° a. – 835° b. – 810°
- 28. The measure of 2.25 revolutions counterclockwise is: 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 3600 2.25 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 3600 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 8100 𝑆𝑖𝑛𝑐𝑒 𝑖𝑡 𝑖𝑠 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒, 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 2.25 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 = +8100
- 29. SUB-TOPIC 2 Trigonometric Identities and Equations
- 30. • To know the solutions to right and oblique triangles. • To analyze the laws of sines, cosines and tangents. • To prove trigonometric identities
- 32. Trigonometric Functions sin Θ = 𝑜 ℎ cos Θ = 𝑎 ℎ tan Θ = 𝑜 𝑎 csc Θ = ℎ 𝑜 sec Θ = ℎ 𝑎 cot Θ = 𝑎 𝑜
- 33. Pythagorean Theorem 𝑎2 + 𝑜2 = ℎ2 “In a right triangle, the sum of the squares of the sides is equal to the square of the hypotenuse” -Pythagoras, 500 BC
- 34. sin2 𝜃 + cos2 𝜃 = 1 1 + cot2 𝜃 = csc2 𝜃 tan2 𝜃 + 1 = sec2 𝜃 sin 𝐴 ± 𝐵 = sin 𝐴 cos 𝐵 + cos 𝐵 sin 𝐴 cos 𝐴 ± 𝐵 = cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵 tan 𝐴 ± 𝐵 = tan 𝐴 ± tan 𝐵 1 ∓ tan 𝐴 tan 𝐵
- 35. Double Angle: sin 2𝐴 = 2 sin 𝐴 cos 𝐴 cos 2𝐴 = cos2 𝐴 − sin2 𝐴 tan 2𝐴 = 2 tan 𝐴 1 − tan2 𝐴 Half Angle: sin 𝐴 2 = 1 − cos 𝐴 2 cos 𝐴 2 = 1 + cos 𝐴 2 tan 𝐴 2 = sin 𝐴 1 + cos 𝐴
- 36. Polar to Rectangular: 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃 Where 𝜃 is the reference angle from x-axis Rectangular to Polar: 𝑟2 = 𝑥2 + 𝑦2 𝑟 = 𝑥2 + 𝑦2 tan 𝜃 = 𝑥 𝑦 𝜃 = tan−1 𝑥 𝑦
- 37. 𝑧 = 𝑎 + 𝑏𝑖 Where 𝑎 = 𝑟 cos 𝜃 and 𝑏 = 𝑟 sin 𝜃 𝑧 = 𝑟 cos 𝜃 + 𝑖 𝑟 sin 𝜃 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) 𝑧 = 𝑟𝑐𝑖𝑠𝜃
- 39. Law of Cosines 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴 𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵 𝑐2 = 𝑏2 + 𝑎2 − 2𝑎𝑏 cos 𝐶 A B C c b a
- 40. If Arccos( x – 2 ) = π/3, find x c. 1/2 d. 1/5 a. 2/5 b. 5/2
- 41. If Arccos( x – 2 ) = π/3, find x For inverse trigonometric functions cos 𝐴 = 𝑏 𝐴 = cos−1 𝑏 cos−1(𝑥 − 2) = 𝜋 3 Take the cosine of both sides, cos cos−1 𝑥 − 2 = cos 𝜋 3 𝑥 − 2 = cos 𝜋/3 Since 𝜋 3 is equivalent to 600, 𝑥 − 2 = cos 600 𝑥 − 2 = 1 2 𝑥 = 1 2 + 2 𝑥 = 5 2
- 42. If sin x = ¼ , find the value of 4sin(x/2)cos(x/2) c. 1/2 d. 1/5 a. 2/5 b. 5/2
- 43. If sin x = ¼ , find the value of 4sin(x/2)cos(x/2) Let 𝐴 = 𝑥 2 Replace all 𝑥 2 by A, 4 sin 𝑥 2 cos 𝑥 2 = 4 sin 𝐴 cos 𝐴 = 2 2 sin 𝐴 cos 𝐴 From double angle, sin 2𝐴 = 2 sin 𝐴 cos 𝐴 2 2 sin 𝐴 cos 𝐴 = 2(sin 2𝐴) Since we assume A = x/2, 2 sin 2𝐴 = 2 sin 2 𝑥 2 = 2 sin 𝑥 If sin x = 1/4, 2 sin 𝑥 = 2 1 4 = 1 2
- 44. The trigonometric expression ( 1 - tan²x ) / ( 1 + tan²x ) is equal to c. sin 2x d. cos 2x a. csc 2x b. tan 2x
- 45. The trigonometric expression ( 1 - tan²x ) / ( 1 + tan²x ) is equal to Since sec2 𝑥 = 1 + tan2 𝑥, 1 − tan2 𝑥 1 + tan2 𝑥 = 1 − tan2 𝑥 sec2 𝑥 = 1 sec2 𝑥 − tan2 𝑥 sec2 𝑥 = 1 1 cos2 𝑥 − sin2 𝑥 cos2 𝑥 1 cos2 𝑥 cos2 𝑥 − sin2 𝑥 From double angles, cos 2𝐴 = cos2 𝐴 − sin2 𝐴 cos2 𝑥 − sin2 𝑥 = cos 2𝑥
- 46. If tanθ = √3, θ in quadrant I, find the value of (1 + cosθ) / (1 – cosθ). c. 3 d. 1/6 a. 1 b. 1/3
- 47. If tanθ = √3, θ in quadrant I, find the value of (1 + cosθ) / (1 – cosθ). tan 𝜃 = 3 𝜃 = tan−1 3 𝜃 = 600 Since angle is less than 900, it is in quadrant 1 1 + cos 𝜃 1 − cos 𝜃 = 1 + cos 600 1 − cos 600 = 1 + 1 2 1 − 1 2 3 2 1 2 = 3 2 2 1 = 3
- 48. If cscθ = 2 and cosθ < 0, then ( secθ + tanθ ) / ( secθ – tanθ ) = c. 3 d. 1/6 a. 1 b. 1/3
- 49. If cscθ = 2 and cosθ < 0, then ( secθ + tanθ ) / ( secθ – tanθ ) = csc 𝜃 = 1 sin 𝜃 = 1 𝑟 𝑦 = 𝑦 𝑟 csc 𝜃 = 𝑦 𝑟 = 2 1 , ∴ 𝑟 = 2 ; 𝑦 = 1 𝑥 = 𝑟2 − 𝑦2 = 22 − 12 = ± 3 But since cos 𝜃 < 0, 𝑥 is negative 𝑥 = − 3 sec 𝜃 = 1 cos 𝜃 𝑎𝑛𝑑 tan 𝜃 = 𝑦 𝑥 sec 𝜃 + tan 𝜃 sec 𝜃 − tan 𝜃 = 𝑟 𝑥 + 𝑦 𝑥 𝑟 𝑥 − 𝑦 𝑥 = 𝑟 + 𝑦 𝑥 𝑟 − 𝑦 𝑥 = 𝑟 + 𝑦 𝑟 − 𝑦 = 2 + 1 2 − 1 = 3
- 50. If sin A + sin B = 1 and sin A – sin B = 1, find A. c. 300 d. 600 a. 450 b. 900
- 51. If sin A + sin B = 1 and sin A – sin B = 1, find A. sin 𝐴 + sin 𝐵 = 1 𝐸𝑞. 1 sin 𝐴 − sin 𝐵 = 1 𝐸𝑞. 2 Equating Eq. 1 and 2, sin 𝐴 + sin 𝐵 = sin 𝐴 − sin 𝐵 sin 𝐵 + sin 𝐵 = sin 𝐴 − sin 𝐴 2 sin 𝐵 = 0 sin 𝐵 = 0 𝐸𝑞. 3 Substitute Eq. 3 to Eq. 1 sin 𝐴 + 0 = 1 sin 𝐴 = 1 𝐴 = sin−1 1 𝐴 = 900
- 52. Simplify ( sec A + csc A ) / ( 1 + tan A ) c. sec A d. csc A a. cos A b. sin A
- 53. Simplify ( sec A + csc A ) / ( 1 + tan A ) sec 𝐴 + csc 𝐴 = 1 cos 𝐴 + 1 sin 𝐴 = sin 𝐴 + cos 𝐴 cos 𝐴 sin 𝐴 1 + tan 𝐴 = 1 + sin 𝐴 cos 𝐴 = cos 𝐴 + sin 𝐴 cos 𝐴 sec 𝐴 + csc 𝐴 1 + tan 𝐴 = sin 𝐴 + cos 𝐴 cos 𝐴 sin 𝐴 cos 𝐴 + sin 𝐴 cos 𝐴 = sin 𝐴 + cos 𝐴 cos 𝐴 sin 𝐴 cos 𝐴 𝑐𝑜𝑠𝐴 + sin 𝐴 sin 𝐴 + cos 𝐴 (cos 𝐴 + sin 𝐴) cos 𝐴 cos 𝐴 sin 𝐴 = 1 sin 𝐴 = csc A
- 54. Three times the sine of an angle is equal to twice the square of the cosine of the same angle. Find the angle. c. 450 d. 900 a. 300 b. 600
- 55. Three times the sine of an angle is equal to twice the square of the cosine of the same angle. Find the angle. 3 sin 𝐴 = 2 cos2 𝐴 But sin2 𝐴 + cos2 𝐴 = 1 → cos2 𝐴 = 1 − sin2 𝐴 → 3 sin 𝐴 = 2 1 − sin2 𝐴 → 3 sin 𝐴 = 2 − 2 sin2 𝐴 → 2 sin2 𝐴 + 3 sin 𝐴 − 2 = 0 Let 𝑥 = sin 𝐴 → 2 sin 𝐴 2 + 3 sin 𝐴 − 2 = 0 → 2𝑥2 + 3𝑥 − 2 = 0 By quadratic formula, 𝑥 = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 Where 𝑎 = 2 , 𝑏 = 3 , 𝑐 = −2 𝑥 = 1 2 𝑜𝑟 − 2 Using x = 1/2 and going back to 𝑥 = sin 𝐴 𝑥 = 1 2 = sin 𝐴 𝐴 = sin−1 1 2 = 300
- 56. Simplify sin(4x) / cos(2x) c. 4sin(x)cos(x) d. 2sin(x)cos(x) a. 4sin(2x) b. 2cos(4x)
- 57. Simplify sin(4x) / cos(2x) sin 4𝑥 = sin[2(2𝑥)] Let 𝐴 = 2𝑥 sin 2 2𝑥 = sin(2𝐴) From double angles, sin(2𝐴) = 2 sin 𝐴 cos 𝐴 Going back to 𝐴 = 2𝑥, 2 sin 𝐴 cos 𝐴 = 2 sin 2𝑥 cos(2𝑥) Therefore, sin 4𝑥 = 2 sin 2𝑥 cos(2𝑥) sin(4𝑥) cos(2𝑥) = 2 sin 2𝑥 cos(2𝑥) cos(2𝑥) = 2 sin(2𝑥) For double angles, sin 2𝑥 = 2 sin(𝑥) cos(𝑥) = 2 sin(2𝑥) = 2[2 sin(𝑥) cos(𝑥)] = 4 sin 𝑥 cos(𝑥)
- 58. Find the product of (4cis1200)(2cis300) in rectangular form. c. 4(√3 + i) d. 4(√3 – i) a. -4(√3 + i) b. -4(√3 – i)
- 59. Find the product of (4cis1200)(2cis300) in rectangular form. The general term for polar coordinates, 𝑧 = 𝑟 𝑐𝑖𝑠𝜃 𝑧 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 Therefore, 4 𝑐𝑖𝑠 1200 = 4 cos 1200 + 𝑖 sin 1200 = 4 − 1 2 + 𝑖 3 2 = −2 + 2𝑖 3 2 𝑐𝑖𝑠 1300 = 2(cos 300 + 𝑖 sin 300) = 2 3 2 + i 1 2 = 3 + i 4 𝑐𝑖𝑠 1200 2 𝑐𝑖𝑠 300 = −2 + 2𝑖 3 3 + i = −2 3 − 2𝑖 + 2𝑖 3 2 + 2𝑖2 3 = −2 3 − 2𝑖 + 6𝑖 + 2 −1 2 ( 3) = −2 3 + 4𝑖 − 2 3 = −4 3 + 4𝑖 = −4( 3 − 𝑖)
- 60. SUB-TOPIC 3 Triangles and Spherical Trigonometry
- 61. • To know the solutions to right and oblique triangles. • To solve angle of elevation and depression. • To solve problems involving spherical triangles
- 62. Trigonometric Functions sin Θ = 𝑜 ℎ cos Θ = 𝑎 ℎ tan Θ = 𝑜 𝑎 csc Θ = ℎ 𝑜 sec Θ = ℎ 𝑎 cot Θ = 𝑎 𝑜
- 63. Pythagorean Theorem 𝑎2 + 𝑜2 = ℎ2 “In a right triangle, the sum of the squares of the sides is equal to the square of the hypotenuse” -Pythagoras, 500 BC
- 65. Law of Cosines 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴 𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵 𝑐2 = 𝑏2 + 𝑎2 − 2𝑎𝑏 cos 𝐶 A B C c b a
- 67. Given: Base & Altitude 𝐴 = 1 2 𝑏ℎ h b
- 68. Given: 2 sides & angle 𝐴 = 1 2 𝑎𝑏 sin 𝜃 a b ϴ
- 69. Given: 3 sides 𝐴 = 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) a b c where: 𝑠 = 𝑎 + 𝑏 + 𝑐 2
- 70. Triangle Inscribed in a circle 𝐴 = 𝑎𝑏𝑐 4𝑟
- 71. Triangle circumscribing a circle 𝐴 = 𝑟𝑠 where: 𝑠 = 𝑎 + 𝑏 + 𝑐 2
- 72. From the top of a 200-meter-high building, the angle of depression to the bottom of a second building is 20 degrees. From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the second building. c. 209.9 m d. 255.8 m a. 302.5 m b. 296.9 m
- 73. From the top of a 200-meter-high building, the angle of depression to the bottom of a second building is 20 degrees. From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the second building. B1 B2 200 100 200 m 200 m y Let 𝑥 = ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔𝑠 𝑦 = 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡𝑤𝑜 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔𝑠 200 + 𝑦 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑒𝑐𝑜𝑛𝑑 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 tan 100 = 𝑦 𝑥 → 𝑥 = 𝑦 tan 100 𝐸𝑞. 1 tan 200 = 200 𝑥 𝐸𝑞. 2 Substitute Eq. 1 to Eq. 2 tan 200 = 200 𝑦 tan 100 → tan 200 = 200 tan 100 𝑦 x x 𝑦 = 200 tan 100 tan 200 𝑦 = 96.9 𝑚 200 + 𝑦 = 200 + 96.9 200 + 𝑦 = 296.9 𝑚 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑒𝑐𝑜𝑛𝑑 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 = 296.9 𝑚
- 74. A 20-meter-high mast is placed on the top of the cliff whose height above sea level is unknown. An observer at sea sees the top mast at an elevation of 46 degrees 42 mins, the foot at 38 degrees 23 mins. The height of the cliff is closest to: c. 54 m d. 59 m a. 57 m b. 51 m
- 75. A 20-meter-high mast is placed on the top of the cliff whose height above sea level is unknown. An observer at sea sees the top mast at an elevation of 46 degrees 42 mins, the foot at 38 degrees 23 mins. The height of the cliff is closest to: O P 46042’ 38023’ 20 m y x Let 𝑥 = ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑙𝑖𝑓𝑓 𝑎𝑛𝑑 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑦 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑙𝑖𝑓𝑓 tan 38023′ = 𝑦 𝑥 → 𝑥 = 𝑦 tan 38023′ 𝐸𝑞. 1 tan 46042′ = 𝑦 + 20 𝑥 𝐸𝑞. 2 Substitute Eq. 1 to Eq. 2 tan 460 42′ = 𝑦 + 20 𝑦 tan 38023′ → tan 460 42′ = (𝑦 + 20) tan 38023′ 𝑦 𝑦 tan 460 42′ = 𝑦 tan 380 23′ + 20(tan 380 23′ ) 𝑦 = 58.88 𝑚 𝑦 = 59 𝑚
- 76. The hypotenuse of a right triangle is 34 cm. Find the length of the shortest leg if it is 14 cm shorter than the other leg. c. 17 cm d. 18 cm a. 15 cm b. 16 cm
- 77. The hypotenuse of a right triangle is 34 cm. Find the length of the shortest leg if it is 14 cm shorter than the other leg. 34 cm x y Let 𝑥 = 𝑠ℎ𝑜𝑟𝑡𝑒𝑠𝑡 𝑙𝑒𝑔 𝑦 = 𝑜𝑡ℎ𝑒𝑟 𝑙𝑒𝑔 The shortest leg is 14 cm shorter than the other leg, 𝑥 = 𝑦 − 14 𝑥 + 14 = 𝑦 𝐸𝑞. 1 By Pythagorean theorem, 𝑥2 + 𝑦2 = 342 𝐸𝑞. 2 Substitute Eq. 1 to Eq. 2 𝑥2 + 𝑥 + 14 2 = 342 𝑥2 + 𝑥2 + 28𝑥 + 142 = 1156 2𝑥2 + 28𝑥 + 196 = 1156 2𝑥2 + 28𝑥 − 960 = 0 𝑥2 + 14𝑥 − 480 = 0 Using quadratic equation, 𝑥 = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 Where 𝑎 = 1 , 𝑏 = 14 , 𝑐 = −480 𝑥 = 16 𝑜𝑟 − 30 Since there is no negative dimension, 𝑥 = 16 𝑐𝑚 𝑠ℎ𝑜𝑟𝑡𝑒𝑠𝑡 𝑠𝑖𝑑𝑒 = 16 𝑐𝑚
- 78. Find the supplement of an angle whose complement is 62° c. 152° d. 162° a. 28° b. 118°
- 79. Find the supplement of an angle whose complement is 62° Let 𝐴 = 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝐵 = 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 620 Since B and 620 are complementary angles, 𝐵 + 620 = 900 𝐵 = 280 Since A and B are supplementary angles, 𝐴 + 𝐵 = 1800 𝐴 + 280 = 1800 𝐴 = 1800 − 280 𝐴 = 1520
- 80. A pole cast a shadow of 15 meters long when the angle of elevation of the sun is 61°If the pole has leaned 15° from the vertical directly toward the sun, what is the length of the pole? c. 53.25 m d. 53.24 m a. 52.43 m b. 54.23 m
- 81. A pole cast a shadow of 15 meters long when the angle of elevation of the sun is 61°If the pole has leaned 15° from the vertical directly toward the sun, what is the length of the pole? Let 𝑥 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑙𝑒 Solve for the unknown angle 𝜃. The sum of interior angles of triangle is 1800, 𝜃 + 610 + 900 + 150 = 1800 𝜃 = 140 Using Law of Sines, 𝑎 sin 𝐴 = 𝑏 sin 𝐵 𝑥 sin 610 = 15 sin 140 𝑥 = 15 sin 140 sin 610 𝑥 = 54.23 𝑚 15 m 150 610 x θ
- 82. Two sides of a triangle measures 6 cm. and 8 cm. and their included angle is 400. Find the third side. c. 4.256 cm d. 5.645 cm a. 5.144 cm b. 5.236 cm
- 83. Two sides of a triangle measures 6 cm. and 8 cm. and their included angle is 400. Find the third side. 6 cm 8 cm 400 Let 𝑥 = 𝑡ℎ𝑖𝑟𝑑 𝑠𝑖𝑑𝑒 Given two known sides and an included angle, we can use Cosine Law, 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos𝐴 𝑥2 = 62 + 82 − 2 6 8 cos 400 𝑥2 = 36 + 64 − 96 cos 400 𝑥2 = 26.46 𝑥 = 26.46 𝑥 = 5.144 𝑐𝑚 x
- 84. Given a triangle: C = 100°, a = 15, b = 20. Find c: c. 43 d. 35 a. 34 b. 27
- 85. Given a triangle: C = 100°, a = 15, b = 20. Find c: 15 20 C = 1000 Let c = 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑠𝑖𝑑𝑒 Given two known sides and an included angle, we can use Cosine Law, 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶 𝑐2 = 152 + 202 − 2 15 20 cos 1000 𝑐2 = 225 + 400 − 600 cos 1000 𝑐2 = 729.19 𝑐 = 729.19 𝑥 = 27.00 c
- 86. Points A and B 1000 m apart are plotted on a straight highway running from East and West. From A, the bearing of a tower C is 32 deg W of N and from B the bearing of C is 26 deg N of E. Approximate the shortest distance of tower C to the highway. c. 384 d. 394 a. 364 b. 374
- 87. Points A and B 1000 m apart are plotted on a straight highway running from East and West. From A, the bearing of a tower C is 32 deg W of N and from B the bearing of C is 26 deg N of E. Approximate the shortest distance of tower C to the highway. A B C 320 260 d α β Let d = 𝑠ℎ𝑜𝑟𝑡𝑒𝑠𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡𝑜𝑤𝑒𝑟 𝐶 𝑡𝑜 𝑡ℎ𝑒 ℎ𝑖𝑔ℎ𝑤𝑎𝑦 Solve for 𝛽, 𝛽 = 900 − 320 = 580 Solve for 𝛼, 𝛼 = 1800 − 260 + 580 = 960 Using Sine Law to solve for BC, 𝐴𝐵 sin 𝛼 = 𝐵𝐶 sin 𝛽 1000 sin 960 = 𝐵𝐶 sin 580 𝐵𝐶 = 852.719 1000 sin 260 = 𝑑 𝐵𝐶 sin 260 = 𝑑 852.719 𝑑 = 852.719 sin 260 𝑑 = 373.81 𝑚 𝑑 = 374 𝑚
- 88. A ship started sailing S 42°35’ W at the rate of 5 kph. After two hours, ship B started at the same port going N 46°20’ W at the rate of 7 kph. After how many hours will the second ship be exactly north of ship A? c. 5.12 d. 4.83 a. 3.68 b. 4.03
- 89. A ship started sailing S 42°35’ W at the rate of 5 kph. After two hours, ship B started at the same port going N 46°20’ W at the rate of 7 kph. After how many hours will the second ship be exactly north of ship A? 46020’ 42035’ A B Starting point 𝑑 = 𝑣𝑡 During the first 2 hours, the distance travelled by ship A: 𝑑 = 5 2 = 10 Let 7𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑠ℎ𝑖𝑝 𝐵 10 + 5𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑙𝑙𝑒𝑑 𝑏𝑦 𝑠ℎ𝑖𝑝 𝐴 By Sine Law: 7𝑡 sin 42035′ = 10 + 5𝑡 sin 46020′ 7𝑡 = 10 + 5𝑡 sin 420 35′ sin 46020′ 7𝑡 = 9.354 + 4.677𝑡 2.323𝑡 = 9.354 𝑡 = 4.03 ℎ𝑟𝑠
- 90. An aerolift airplane can fly at an airspeed of 300 mph, If there is a wind blowing towards the cast at 50 mph, what should be the plane’s compass heading in order for its course to be 30°? c. 21.7° d. 22.3° a. 19.7° b. 20.1°
- 91. An aerolift airplane can fly at an airspeed of 300 mph, If there is a wind blowing towards the cast at 50 mph, what should be the plane’s compass heading in order for its course to be 30°? 600 50 300 α β V By Sine Law, 50 sin 𝛽 = 300 sin 600 𝛽 = 8.30 𝛼 = 300 − 8.30 𝛼 = 21.70
- 92. An aerolift airplane can fly at an airspeed of 300 mph, If there is a wind blowing towards the cast at 50 mph, If its course is to be 30°? What will be the plane’s ground speed if it flies in this course? c. 321.8 mph d. 319.2 mph a. 307.4 mph b. 309.4 mph
- 93. An aerolift airplane can fly at an airspeed of 300 mph, If there is a wind blowing towards the cast at 50 mph, If its course is to be 30°? What will be the plane’s ground speed if it flies in this course? 600 50 300 α β V By Sine Law, 50 sin 𝛽 = 300 sin 600 𝛽 = 8.30 𝛼 = 300 − 8.30 𝛼 = 21.70 𝛿 + 600 + 𝛽 = 1800 𝛿 + 600 + 8.30 = 1800 𝛿 = 111.70 By Sine Law, 𝑉 sin 111.70 = 50 sin 8.30 𝑉 = 321.8 𝑚𝑝ℎ 𝛿
- 94. The sides of a triangle are 195, 157, and 210, respectively. What is the area of the triangle? c. 14586 d. 11260 a. 73250 b. 10250
- 95. The sides of a triangle are 195, 157, and 210, respectively. What is the area of the triangle? 𝑠 = 𝑎 + 𝑏 + 𝑐 2 = 195 + 157 + 210 2 = 281 𝐴 = 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) 𝐴 = 281(281 − 195)(281 − 157)(281 − 210) 𝐴 = 281(86)(124)(71) 𝐴 = 14586
- 96. The sides of a triangular lot are 130 m, 180 m, and 190 m. The lot is to be divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line. c. 125 d. 128 a. 120 b. 130
- 97. The sides of a triangular lot are 130 m, 180 m, and 190 m. The lot is to be divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line. A B C x a = 130 c/2 = 95 c/2 b = 180 By cosine law (∆𝐴𝐵𝐶) 𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵 1802 = 1302 + 1902 − 2 130 190 cos B 𝐵 = 65.350 By cosine law (∆𝐵𝐶𝐷) 𝑥2 = 𝑎2 + 𝑐 2 2 − 2𝑎 𝑐 2 cos 𝐵 𝑥2 = 1302 + 952 − 2 130 95 cos 65.350 𝑥 = 125 a = 190 D
- 98. Plane & Spherical Trigonometry SPHERICAL TRIANGLES
- 103. NOTES for Terrestrial Sphere 1. Radius of Earth = 3659 miles 2. 1 minute of the great circle area on the surface of the earth = 1 NM 3. 1 NM (nautical mile) = 6080 feet 4. 1 statute mile = 5280 feet
- 104. If Greenwich Mean Time (GMT) is 6 AM, what is the time at a place located 30 degrees east longitude? c. 9 AM d. 4 AM a. 7 AM b. 8 AM
- 105. If Greenwich Mean Time (GMT) is 6 AM, what is the time at a place located 30 degrees east longitude? 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑡𝑖𝑚𝑒 24 = 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 3600 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑡𝑖𝑚𝑒 24 = 300 − 00 3600 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑡𝑖𝑚𝑒 = 2 ℎ𝑜𝑢𝑟𝑠 The time in the place is 2 hours ahead of GMT because the place is at the East. Time is 8AM
- 106. If the longitude of Tokyo is 139 E and that of Manila is 121 E, what is the time difference between Tokyo and Manila? c. 1 hr 8 mins d. 1 hr 10 mins a. 1 hr 12 mins b. 1 hr 5 mins
- 107. If the longitude of Tokyo is 139 E and that of Manila is 121 E, what is the time difference between Tokyo and Manila? 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑡𝑖𝑚𝑒 24 = 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 3600 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑡𝑖𝑚𝑒 24 = 1390 − 1210 3600 𝑑𝑖𝑓𝑓. 𝑖𝑛 𝑡𝑖𝑚𝑒 = 1.2 ℎ𝑜𝑢𝑟𝑠 = 1 ℎ𝑜𝑢𝑟𝑠 12 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
- 108. One degree on the equator of the earth is equivalent to c. 30 mins d. 1 hour a. 1 min b. 4 min
- 109. One degree on the equator of the earth is equivalent to 360 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 = 24 ℎ𝑜𝑢𝑟𝑠 𝑡𝑖𝑚𝑒 = 1 𝑑𝑒𝑔𝑟𝑒𝑒 24 ℎ𝑜𝑢𝑟𝑠 360 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑡𝑖𝑚𝑒 = 0.0667 ℎ𝑟 × 60 𝑚𝑖𝑛 ℎ𝑟 𝑡𝑖𝑚𝑒 = 4 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
- 110. A spherical triangle ABC has an angle C = 90 degrees, and sides a = 50 degrees, and c = 80 degrees. Find the value of “b” in degrees. c. 75.44° d. 76.55° a. 73.22° b. 74.33°
- 111. A spherical triangle ABC has an angle C = 90 degrees, and sides a = 50 degrees, and c = 80 degrees. Find the value of “b” in degrees. Using SIN-CO-OP rule, sin(𝑐𝑜 − 𝑐) = cos 𝑎 cos 𝑏 cos 𝑐 = cos 𝑎 cos 𝑏 cos 𝑏 = cos 𝑐 cos 𝑎 cos 𝑏 = cos 800 cos 500 𝑏 = 74.330
- 112. Solve for the remaining side of the right spherical triangle whose given parts A = B = 80 degrees and a = b = 89 degrees. c. 168° 31’ d. 172° 12’ a. 158° 12’ b. 162° 21’
- 113. Solve for the remaining side of the right spherical triangle whose given parts A = B = 80 degrees and a = b = 89 degrees. Using SIN-TA-AD rule, sin(𝑐𝑜 − 𝐴) = tan 𝑐𝑜 − 𝑏 tan 𝑐 2 cos 𝐴 = 1 tan b tan 𝑐 2 tan 𝑐 2 = cos 𝐴 tan 𝑏 tan 𝑐 2 = cos 80 tan 890 𝑐 = 1860 31′
- 114. Determine the spherical excess of a spherical triangle whose angles are all right angles? c. 60° d. 30° a. 45° b. 90°
- 115. Determine the spherical excess of a spherical triangle whose angles are all right angles? 𝐸 = 𝐴 + 𝐵 + 𝐶 − 1800 𝐸 = 900 + 900 + 900 − 1800 𝐸 = 900
- 116. The area of spherical triangle ABC whose parts are A = 93°40’, B = 64°12’, C = 116°51’ and the radius of the sphere is 100 m is: c. 18645 sq. m d. 25612 sq. m a. 15613 sq. m b. 16531 sq. m
- 117. The area of spherical triangle ABC whose parts are A = 93°40’, B = 64°12’, C = 116°51’ and the radius of the sphere is 100 m is: 𝐴 = 𝜋𝑅2 𝐸 1800 𝐸 = 𝐴 + 𝐵 + 𝐶 − 1800 𝐸 = 930 40′ + 640 12′ + 1160 51′ − 1800 𝐸 = 94043′ 𝐴 = 𝜋 100 2 (940 43′ ) 1800 𝐴 = 16531 𝑠𝑞. 𝑚
- 118. DO YOU HAVE ANY QUESTIONS?
- 119. SEE YOU NEXT MEETING!