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Zolock

  • one year ago

4. The diameter of circle C is 18 cm. Chord AD is the same length as a radius. Use this information, the diagram, and your experiences in geometry this semester to answer these questions. (Use pie = 3.14 .) a.) What are the measures of angles BDA and DBA? b.) What is the sine of angle DBA? c.) Find the length of arc AB to the nearest hundredth. d.) How does the length of arc AD compare to the length of arc AB? e.) If EF = 1/2 BD, then what is the degree measure of arc EF? Answer:

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  1. Zolock
    • one year ago
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  2. geerky42
    • one year ago
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    |dw:1433867445131:dw|

  3. geerky42
    • one year ago
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    |dw:1433867464274:dw| So \(\triangle ACD\) is equilateral triangle, right?

  4. geerky42
    • one year ago
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    So can you tell me what \(m\angle BDA\) is?

  5. Zolock
    • one year ago
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    I can not tell

  6. geerky42
    • one year ago
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    You do agree that \(\triangle ACD\) is equilateral triangle, right? So all angles in equilateral triangle are always ...?

  7. Zolock
    • one year ago
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    yes it looks that way

  8. Zolock
    • one year ago
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    they are alway equal

  9. geerky42
    • one year ago
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    yeah. So since \(m\angle BDA\) is IN △ACD, what is angle?

  10. Zolock
    • one year ago
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    BDA = 60 degress

  11. geerky42
    • one year ago
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    |dw:1433867843613:dw|

  12. geerky42
    • one year ago
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    Yeah, so there you go,

  13. Zolock
    • one year ago
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    BDA is 60

  14. geerky42
    • one year ago
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    |dw:1433867910800:dw|

  15. geerky42
    • one year ago
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    Yeah. BDA is 60. Now what is DBA?

  16. Zolock
    • one year ago
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    30

  17. geerky42
    • one year ago
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    Correct. so you got answers for part A.

  18. geerky42
    • one year ago
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    Now for part B, you have \(\sin(30^\text o)\)

  19. Zolock
    • one year ago
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    so the sin for angle DBA is 30

  20. geerky42
    • one year ago
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    Not quite. \(\sin(30^\text o)\neq 30\)

  21. geerky42
    • one year ago
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    Do you have unit circle to check?

  22. Zolock
    • one year ago
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    I did not get that

  23. geerky42
    • one year ago
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    Ok, you know 30 60 90 triangle? |dw:1433868385520:dw| So since \(\sin\theta\) is ratio of opposite side to hypotenuse side. In other words, \(\sin\theta = \dfrac{Opp}{Hyp}\), right?

  24. geerky42
    • one year ago
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    We have Opp = a and Hyp = 2a. What is Opp/Hyp?

  25. Zolock
    • one year ago
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    2

  26. geerky42
    • one year ago
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    Not quite. Opp / Hyp = a / 2a So it's 1/2

  27. Zolock
    • one year ago
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    oh

  28. geerky42
    • one year ago
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    \(\sin(30^\text o) = \dfrac{Opp}{Hyp} = \dfrac{a}{2a} = \dfrac{1}{2}\) IS that clear?

  29. Zolock
    • one year ago
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    yes

  30. Zolock
    • one year ago
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    so sin(30 degrees) = 1/2

  31. geerky42
    • one year ago
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    Yeah

  32. geerky42
    • one year ago
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    Can you do part C?

  33. Zolock
    • one year ago
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    nope I still need help with part C and part D I think I might me able to do part E on my own.

  34. geerky42
    • one year ago
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    Ok. Whole circle is \(360^\text o\), right?|dw:1433869406908:dw| Here, to find arc AB, we would need to find angle BCA. |dw:1433869460152:dw|

  35. Zolock
    • one year ago
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    120

  36. geerky42
    • one year ago
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    Then we can find length of arc by using formula for circumference, kinda. \(Arc = \dfrac{m\angle BCA}{360^\text o} ~~2\pi r\)

  37. geerky42
    • one year ago
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    Here we have \(m\angle BCA = 120^\text o\)

  38. geerky42
    • one year ago
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    So arc length BA is \(\dfrac{120^\text o}{360^\text o}~~~2\pi r = \dfrac{2\pi r}{3}\) Make sense?

  39. geerky42
    • one year ago
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    You were given that diameter is 18. So clearly radius is 9.

  40. Zolock
    • one year ago
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    so the length of AB is 2πr/3

  41. geerky42
    • one year ago
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    Yeah. now plug in r = 9

  42. Zolock
    • one year ago
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    so AB is 2π^9/3

  43. Zolock
    • one year ago
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    @geerky42 Can you finish helping me with part D and E when you get back online please I have to turn this in tomorrow.

  44. geerky42
    • one year ago
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    π^9? It's just \(\dfrac{2\pi 9}{3} = 2\pi3 = \boxed{6\pi}\)

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