Friday, February 10, 2012

Problem 727: Triangle, Isosceles, Congruence, Angles, Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 727 details.

Online Geometry Problem 727: Triangle, Isosceles, Congruence, Angles, Measurement.

7 comments:

  1. http://img197.imageshack.us/img197/8941/problem727.png

    Let DD’ is the perpendicular bisector of AC and DD’ cut BC at M
    AM is the angle bisector of angle BAC ( see picture)
    Draw BK//AC and locate point L on BK such that BL=BA=BD
    Note that m(BAM)=m(MAC)=44
    Triangle ABL is isosceles => m(BLA)=1/2*(180-92)=44
    So m(BAM)=m(BAL)=44 => A, M ,L are collinear
    Triangle BKM congruence to LKM ( case ASA)
    so BK=KL => triangle BDL is equilateral
    and m(DBL)=60 => x=60-44=16

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  2. Nice proof, Peter. I applied Ceva's Trigonometric Theorem to Tr. ABC and landed up with: sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)
    Can anyone tell how this is solved the answer, of course, being x=16 deg.?
    Ajit

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  3. A trigonometric Proof:
    B = 48°
    ∠ABD = 48° - x
    ΔABD is isosceles
    AD = 2 AB sin(∠ABD/2)= 2 c sin(24° - x/2)
    ∠BAD = ∠BDA = 66° + x/2
    ∠DAC = ∠DCA = 22° - x/2
    ΔABD is isosceles
    b = AC = 2 AD cos (22° - x/2)
    b = 4c sin(24° - x/2) cos(22° - x/2)
    sin B = 4 sin C sin(24° - x/2) cos(22° - x/2)
    sin 48° = 4 sin 44° sin(24° - x/2) cos(22° - x/2)
    sin 48° = (2 sin 44°)[2 sin(24° - x/2) cos(22° - x/2)]
    sin 48° = (2 sin 44°)[sin(46° - x) + sin 2°]
    cos 42° = 2 cos 46° sin(46° - x) + 2 sin 44°sin 2°
    cos 42° = 2 cos 46° sin(46° - x) + cos 42° - cos 46°
    sin(46° - x) = 1/2
    46° - x = 30°
    x = 16°

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  4. Dear Ajit:
    sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)
    =2sin(24-x/2)*cos(24-x/2)*sin(22+x/2)
    Observe sin(66+x/2)and cos(24-x/2)get cancelled
    since the angles are complementary
    We now have
    sin(x)= 2sin(24-x/2)*sin(22+x/2)= cos(2-x)-cos(46)
    cos(46)=cos(2-x)-sin(x)=sin(88+x)-sin(x)=2cos(44+x)sin(44)
    Again cos(46)and sin(44)get cancelled
    since the angles are complimentary
    So we are left with
    1=2cos(44+x)
    cos(44+x)= cos(60)
    x=16 deg
    Pravin

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  5. Thanks, Pravin.
    My email: ajitathle@gmail.com
    Need ur email address.
    Ajit

    ReplyDelete
  6. Dear Ajit,

    sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)

    sin(x)*cos(24-x/2)=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2), hence cos(24-x/2) get cancelled.

    sin(x)=2sin(24-x/2)*sin(22+x/2). Using product to sum formula,

    cos(90-x)=cos(2-x)-cos46

    cos(2-x)-cos(90-x)=cos46. After using sum to product formula,

    -2sin(46-x)*sin(-44)=sin44

    -2sin(46-x)*[-sin44]=2sin30*sin44

    2sin44*sin(46-x)=2sin30*sin44, hence 2sin44 get cancelled.

    Finally sin(46-x)=sin30, then x=16.

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  7. Problem 727
    Forms the isosceles trapezoid ABEC (the point Eis the right of point B and above the point C).
    (ΒΕ//ΑC,<BAC=88=<ECA). Then AB=BD=EC=BE and BC is bisector <ACE.So ABEC is cyclic. But triangle ABD=triangle CED (AB=BD=CE, AD=DC,<BAD=<ECD).Then BD=DE=BE, so triangle BED is equilateral .Therefore x=<DBC=<DBE-<CBE=60-44=16.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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