Posers and Puzzles

Posers and Puzzles

  1. Joined
    29 Feb '04
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    22
    06 Jun '04 15:09
    An isosceles triangle ABC has a point D on AB and a point E on AC.

    Given:
    Angle BAC = 20 degrees
    Angle EBC = 50 degrees
    Angle DCB = 60 degrees.
    AB = AC

    Find angle CDE.
  2. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
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    20443
    06 Jun '04 17:58
    All 2 dimensional triangles have 180 degrees as sum of their three angles. You're figure ABC has 130 as sum, and therefore cannot be a triangle.
  3. Joined
    29 Feb '04
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    22
    06 Jun '04 19:13
    Originally posted by TheMaster37
    All 2 dimensional triangles have 180 degrees as sum of their three angles. You're figure ABC has 130 as sum, and therefore cannot be a triangle.
    No, it doesn't.

    I haven't mentioned angles ABC or ACB.
  4. Joined
    26 Apr '03
    Moves
    26597
    07 Jun '04 22:25
    Originally posted by THUDandBLUNDER
    An isosceles triangle ABC has a point D on AB and a point E on AC.

    Given:
    Angle BAC = 20 degrees
    Angle EBC = 50 degrees
    Angle DCB = 60 degrees.
    AB = AC

    Find angle CDE.
    I'm having trouble with this. Is it soluble with the information given?
  5. Joined
    19 Oct '03
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    69376
    07 Jun '04 23:361 edit
    m<cde= 50 degrees...o yeh im only 14...just finished geometry
  6. Joined
    19 Oct '03
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    69376
    08 Jun '04 03:34
    im sorry the answer is 40 degree i had a slite mathmatical error
  7. Joined
    29 Feb '04
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    22
    08 Jun '04 04:312 edits
    Originally posted by TDR1
    im sorry the answer is 40 degree i had a slite mathmatical error
    You needn't be sorry that the angle is 40 degrees, 'cos it isn't. 😛

    Anyway, what method are you using?

    .
  8. Joined
    29 Feb '04
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    22
    08 Jun '04 04:33
    Originally posted by iamatiger
    I'm having trouble with this. Is it soluble with the information given?
    Yes, it is. If you draw a diagram it is easy to see that the required angle is uniquely defined by the given information.

    .
  9. Standard memberNemesio
    Ursulakantor
    Pittsburgh, PA
    Joined
    05 Mar '02
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    32455
    08 Jun '04 06:171 edit
    Best I can tell, it is not solvable, because I have been able to plug in any angle to cde and the resulting triangles on the diagram all equal 180. If I am wrong, message me why (don't spoil it).

    This is what I found: Where X is the intersection between BE and CD, you can figure out all of other angles from several of the resulting triangles with no stress (Triangles ACD, AED, and XBC). But, plug in any number for angle CDE (some number between 1 and 109, since angle DXE is 70), and the resulting angles (BED, BEA, and ADE) are forced into place, and all angles work equally well (resulting in triangles with 180 degrees).
  10. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
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    20443
    08 Jun '04 18:13
    i leave out degrees, and with Z(ABC) i mean angle ABC, with T(ABC) i mean triangle ABC.

    I'm still doubting if it is solvable. I've gotten this far:

    70 < Z(CDE) =< 110
    0 =< Z(BED) < 40
    30 =< Z(ADE) <70
    90 < Z(AED) =< 130
    Z(CDE) + Z(BED) = 110
    Z(ADE) + Z(AED) = 160
    Z(ADE) + Z(CDE) = 140
    Z(AED) + Z(BED) = 130

    For wich i used that AB=AC, CB=CE, DA=DC (follows from equal angles at the bases of T(CEB) and T(DCA)).

    This is as far as i can get, though i already know that, for example
    Z(CDE) cannot be 110, for then Z(BED) would be 0 and thus D=B.

    Furthermore i've tried something with the projection of D on AC, for is exatly halfway AC...
  11. Joined
    19 Oct '03
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    69376
    09 Jun '04 00:48
    does the figure look like wat i think it does???.........
    it is a triangle with a trapazoid EDBC in it. This trapazoid has 2 intersecting lines. lines EB and DC...am i correct?
  12. Guelph Ontario
    Joined
    31 May '04
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    4496
    09 Jun '04 01:062 edits
    The inclusive angle of CDE is 30 degrees.
  13. Standard memberNemesio
    Ursulakantor
    Pittsburgh, PA
    Joined
    05 Mar '02
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    32455
    09 Jun '04 01:291 edit
    I've taken the tack that you were trying, Master, (bringing 90 angles down all over the place), but I can't make it such that any figure is missing only one angle; there are always two angles missing. I can narrow the range of Z(CDE) but that's all.

    Nemesio
  14. Australia
    Joined
    16 Jan '04
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    7971
    09 Jun '04 02:05
    The answer is 10 degrees I believe??????????
  15. Joined
    19 Oct '03
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    69376
    09 Jun '04 02:28
    i think this is true.....opposite angles in a trapazoid are supplementary.....im not 100% sure bout this....but with this you can conclude that the answer is 60....im not taking random guesses...that may be wat it seems like
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