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arindameducationusc

  • one year ago

This question was asked by m friend Sopie. I am not able to solve it. Can anyone help? See attachment

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  1. arindameducationusc
    • one year ago
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  2. triciaal
    • one year ago
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    not in order and may be incomplete ABC =180 -60 = 120 ACB = 180- (25 + 120) = 180 - 145 = 35 AC = BD BDC =25 given AD = 65 DA is perpendicular to AE 65 + 25 = 90 = DAB

  3. arindameducationusc
    • one year ago
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    @triciaal why AC=BD?

  4. triciaal
    • one year ago
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    BC parallel to AD Chord BC similar figures ABC and DCB

  5. arindameducationusc
    • one year ago
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    How BC paralll to AC? impossible

  6. arindameducationusc
    • one year ago
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    @triciaal

  7. Jack1
    • one year ago
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    do u have a screenshot? i cant open the document... :-/

  8. phi
    • one year ago
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    there are a few ways to find angle ADC. one way: i) <CBE is an exterior angle of triangle ABC, so it equals the sum of the "opposite" angles: <CBE= <BAC + <BCA 60 = 25 + <BCA so <BCA= 60-25= 35 ii) inscribed angle <ACD = 1/2 arc BC + 1/2 arc AB from part i) we can find arc BC and arc AB, and so we can find angle ACD

  9. phi
    • one year ago
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    for part (b), angle ADB is 1/2 arc AB which we know from part (a) for part (c), angle CAB= 65+25= 90 i.e. a right angle. if the inscribed angle is 90 degrees, then the chord is a diameter.

  10. phi
    • one year ago
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    in general, use the the idea that inscribed angles that subtend the same arc are equal |dw:1439469998213:dw|

  11. arindameducationusc
    • one year ago
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    i) inscribed angle <ACD = 1/2 arc BC + 1/2 arc AB from part i) we can find arc BC and arc AB, and so we can find angle ACD //This part I didnot understand

  12. arindameducationusc
    • one year ago
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    @phi

  13. arindameducationusc
    • one year ago
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    for part (c), angle CAB= 65+25= 90 i.e. a right angle. if the inscribed angle is 90 degrees, then the chord is a diameter. //How CAB is 90 when it is 25

  14. phi
    • one year ago
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    oops, I meant angle BAD is 90

  15. phi
    • one year ago
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    from the picture, can you find the length (in degrees) of arc BC ?

  16. arindameducationusc
    • one year ago
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    okay BAD, yes

  17. phi
    • one year ago
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    |dw:1439470438536:dw|

  18. arindameducationusc
    • one year ago
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    n I cant find arc BC, let me analyse...hmmm....

  19. phi
    • one year ago
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    look for an inscribed angle that "opens" on arc BC

  20. arindameducationusc
    • one year ago
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    angle BDC?

  21. phi
    • one year ago
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    yes, but not useful. there is another one

  22. arindameducationusc
    • one year ago
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    angle BAC?

  23. phi
    • one year ago
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    yes, BAC also opens on arc BC. now we can figure out what arc BC is (it is 2 times angle BAC)

  24. arindameducationusc
    • one year ago
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    means 50? is there some theorem or axiom. can you tel me?

  25. arindameducationusc
    • one year ago
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    arc BC is 50?

  26. phi
    • one year ago
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    yes

  27. phi
    • one year ago
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    next, we can find angle BCA . do you follow the reasoning?

  28. arindameducationusc
    • one year ago
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    BCA, can we find by 180-25-120=35.? cauz angle ABC is 120

  29. phi
    • one year ago
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    yes.

  30. arindameducationusc
    • one year ago
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    I didn't get about the arc and angle relation. I know one thing theta=L/R But never tried this one

  31. phi
    • one year ago
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    ok, but meanwhile you now know arc AB is what ?

  32. arindameducationusc
    • one year ago
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    2*35=70

  33. arindameducationusc
    • one year ago
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    right?

  34. phi
    • one year ago
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    yes. now we know arc AC 70+50= 120

  35. phi
    • one year ago
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    and finally, we can find inscribed angle ADC (which opens on arc AC)

  36. arindameducationusc
    • one year ago
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    okay arc AC...

  37. arindameducationusc
    • one year ago
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    60?

  38. phi
    • one year ago
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    yes

  39. arindameducationusc
    • one year ago
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    wow..!!

  40. arindameducationusc
    • one year ago
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    But please let me know the theorem. Send me some video link or something.... awesome....

  41. arindameducationusc
    • one year ago
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    Thank you @phi

  42. arindameducationusc
    • one year ago
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    ADB is 35? @phi

  43. phi
    • one year ago
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    yes, both <ADB and <ACB are inscribed angles that subtend arc AB both are 1/2 of 70 degrees i.e. 35 deg.

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