Monday, May 19, 2008

Elearn Geometry Problem 60



See complete Problem 60
Isosceles triangle and angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:36 PM
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10 comments:

  1. venkat sriramOctober 22, 2011 at 9:38 AM

    use sine rule for both the triangles and u may get answer

    ReplyDelete
  2. Eder Contreras OrdenesNovember 29, 2011 at 8:00 PM

    Let P be a point such that triangle BPC is equilateral, then triangles CAB and DBP are congruent by LAL, because this, triangle DPC is an 80-80-20.

    Therefore, x=20°

    Quod Erat Demonstrandum.

    ReplyDelete
  3. AnonymousJanuary 6, 2013 at 3:03 PM

    Or, we can let E be the point on the opposite side of AC as B such that triangle ACE is congruent to triangle BDC, where angle CAE=10. Then BAE is isoceles, so angle CBE=10. Also, CD=CE, so B must lie on the circumcircle of triangle BDE as angle DBC and angle CBE are equal and intercept equal arcs of this circle. Thus angle DCE=160, so
    160+(70-x)+(170-x)=360, so x=20.

    ReplyDelete
  4. Sumith PeirisJuly 26, 2015 at 3:34 AM

    Let E be such that EB = AB and < EBA = 20

    Then Tr. EBC is equilateral and Tr. ABE is isoceles Tr.s ACE & Tr, BDC Tr. are congruent and x = 20

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Stan FulgerJanuary 1, 2016 at 11:57 AM

    Let E inside the triangle such that ACE be an equilateral triangle; see that triangles BCE and CBD are congruent, s.a.s., so <BCD=<CBE=20 degs.

    A Happy New Year 2016!!

    ReplyDelete
  6. Geek37November 7, 2018 at 11:17 AM

    https://youtu.be/9mtWyND8kQ8

    ReplyDelete
  7. rv.littlemanAugust 26, 2020 at 12:16 AM

    See the drawing : Drawing
    - Define F in AC with AF ⊥FB
    - (ABF) ̂ = (FBC) ̂ = 40°/2=20°

    - Define I such as ΔAIC is equilateral
    - I is on BF => (ABI) ̂ = 20°

    - ΔIAB is congruent to ΔDBC (SAS)
    - Therefore <ABI = <BCD = 20°

    ReplyDelete
  8. rv.littlemanFebruary 16, 2022 at 7:14 AM

    See the drawing

    ∠ABC = 30°+10°=40°
    Define F on AC and on the angle bisector of ∠ABF=∠FBC = 40°/2=20°
    Define I such as ΔAIC is equilateral => ∠IAC=60°
    ΔABC isosceles in B => ∠ABI=∠CBI = 20°

    ∠ABC = 40° and ΔABC isosceles in B => ∠BAC=70°
    ∠BAI = ∠BAC - ∠IAC=70° - 60° =10°
    AC=BD (given) and AC=AI=> ΔIAB is congruent to ΔDBC (SAS)
    Therefore ∠ ABI = ∠ BCD = 20°

    ReplyDelete
  9. Sailendra TMarch 7, 2022 at 1:21 PM

    E be point on right of BC such that AE = AB = BC and m(CAE)=10 degrees. CAE is congruent to DBC and ABE is an equilateral triangle. Since BC=BE, BCE is an isosceles triangle with m(BCE)=m(BEC)=80. As m(BEA)=60 => m(CEA)=m(DCB)=20

    ReplyDelete
  10. MarcoDecember 6, 2022 at 6:46 AM

    <BAC=<BCA=70
    Consider triangle ABC
    sin70/BC=sin40/AC
    AC/BC=sin40/sin70
    =sin40/cos20
    =2sin20---------(1)

    Consider triangle BDC
    sinx/BD=sin(170-x)/BC
    BD/BC=sinx/sin(170-x)----------(2)

    Since AC=BD, by equating (1) & (2)
    2sin20=sinx/sin(170-x)
    2sin20sin(170-x)=sinx
    cos(150-x)-cos(190-x)=sinx
    cos150cosx+sin150sinx-cos190cosx-sin190sinx=sinx
    (sin150-sin190-1)+(cos150-cos190)cotx=0
    cotx=(1-sin150+sin190)/(cos150-cos190)
    With the help of calculator, x=20

    ReplyDelete

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