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geometry

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obtuse triangle ABC with obtuse angle B. Point D on side AC such that angle BDC is also obtuse. Angle DAB is 2/3 of angle ABD. Angle BCD is 1/5 of angle CBD is 4 more than angle BAD. Find the measure of angle ABC.

  • geometry -

    The wording of " Angle BCD is 1/5 of angle CBD is 4 more than angle BAD." is rather ambiguous.

    Did you mean
    " Angle BCD is 1/5 of angle CBD, AND IT is 4 more than angle BAD." ?

  • geometry -

    yes

  • geometry -

    Ok, let me use < for angle

    Let <ABD = 3x (I can avoid fractions this way)
    then <DAB = 2x
    by exterior angle theorem <BDC = 5x

    <BCD = 2x + 4 (it said so)
    <CBD is 5 times <BCD (it said so)
    so <CBD = 5(2x+4) = 10x + 20

    so by exterior angle theorem :
    < BDC = 12x + 24

    But CDA is a straight line, so
    5x + 12x+24 = 180
    x = 216/7 or 9.1765

    then <ABC = 10x+20 + 3x
    = 13x + 20 = 139.294°

    let's check the rest
    <ABD = 27.529
    <BAD = 18.353
    and 18.535/27.529 = .6666678 or 2/3
    < BDA = 134.118
    for a total sum of 134.118+27.529+18.353 = 180

    In triangle BDC
    <C = 4 more than <BAD = 22.353
    which is 1/5 of <CBD making
    <CBD = 111.765
    Also we knew that <BDC = 5x = 45.882

    check: is CDA a straight line?
    is 17x+24 = 180 ?
    17x + 24 = 17(9.1765) + 24 = 180.0005 , OK

    what about the sum of angles in that triangle?
    what is 111.765+45.882+22.353 = 180

    So again
    <ABC = 10x+20 + 3x
    = 13x + 20 = 139.294°

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