1. Joined
    29 Feb '04
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    02 May '05 18:281 edit
    The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees.

    The point D lies on AC such that angle DBC = 60 degrees.
    The point E lies on AB such that angle ECB = 50 degrees.

    Find angle CED.
  2. Standard memberBowmann
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    02 May '05 18:511 edit
    Do you mean an isosceles triangle?


    0 degrees?
  3. Joined
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    02 May '05 19:15
    Originally posted by Bowmann
    Do you mean an isosceles triangle?


    0 degrees?
    Yeah, thanks for the correction.
    Have amended the wording.
  4. Joined
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    02 May '05 19:35
    Originally posted by THUDandBLUNDER
    The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees.

    The point D lies on AC such that angle DBC = 60 degrees.
    The point E lies on AB such that angle ECB = 50 degrees.

    Find angle CED.
    I get that angle CED must be 80:

    Without going through every step, I solved the following set of equations simultaneously. The equations follow from the Law of Sines and from the problem statement:

    AD/[sin(130 - CED)] = AE/[sin(30 + CED)];
    AE/sin(30) = AC/sin(130);
    AD/sin(20) = AB/sin(140);
    AB = AC;

    Solving this set of equations, I get CED = 80.
  5. Joined
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    02 May '05 20:27
    Originally posted by davegage
    I get that angle CED must be 80:
    Nope. 😛
  6. Joined
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    02 May '05 21:04
    Originally posted by THUDandBLUNDER
    Nope. 😛
    hmmm....got me then. everytime i work i it out, i get that it must be 80.

    i must be missing something.
  7. Joined
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    02 May '05 21:10
    Originally posted by davegage
    hmmm....got me then. everytime i work i it out, i get that it must be 80.

    i must be missing something.
    Oops, sorry I meant to ask for angle EDB. 😳
  8. Joined
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    02 May '05 21:12
    Originally posted by THUDandBLUNDER
    Oops, sorry I meant to ask for angle EDB. 😳
    ah...then i would get 30
  9. Joined
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    02 May '05 21:33
    Originally posted by davegage
    ah...then i would get 30
    Correct!

    Too easy?

    As a generalization, consider the case where BAC = 2a, DBC = b, and ECB = c. Find EDB in terms of a, b, and c.


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