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Measure that angle (Posted on 2002-06-18) Difficulty: 5 of 5
Given that:
  • ABC is an isosceles triangle in which
        AB = AC
  • The lengths of the following segmets are equal:
        AD
        DE
        EC
        BC
    Find the measure of angle A.
  • See The Solution Submitted by vohonam    
    Rating: 4.2000 (20 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Some Thoughts First thoughts | Comment 2 of 59 |
    A few of things occur to me off the top pf my head. I'm not sure how many will be useful in solving the problem.

    Andle DEA = Angle DAE
    Angle BDE = 2 * Angle BAC

    Law of Sines:
    sin (BAC)/(BC) = sin (ABC)/(AC) = sin (ACB)/(AC)
    and
    sin(DAE)/(DE) = sin(ADE)/(AE) = sin(AED)/(AE)

    Law of Cosines:
          (BC)^2= (AB)^2 + (AC)^2 -2 (AB)(AC)(cos[BAC])
          (AC)^2= (AB)^2 + (BC)^2 -2 (AB)(BC)(cos[ABC])
          (BA)^2= (CB)^2 + (AC)^2 -2 (CB)(AC)(cos[BCA])
    and
          (DE)^2= (AD)^2 + (AE)^2 -2 (AD)(AE)(cos[DAE])
          (AD)^2= (AE)^2 + (DE)^2 -2 (AE)(DE)(cos[AED])
          (AE)^2= (AD)^2 + (DE)^2 -2 (AD)(DE)(cos[ADE])

    Since we are looking for Angle BAC or Angle DAE, which is the same angle) then solving for cos (BAC) seems to be the most promising route, especially when we can substitute using:
    (AE) + (EC) = AE and
    (AD) = (DE) = (EC)
      Posted by TomM on 2002-06-18 02:20:23
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