Sunday, August 15, 2010

Problem 504: Equilateral Triangle, Angles, 45, 60 Degrees

Geometry Problem
Click the figure below to see the complete problem 504 about Equilateral Triangle, Angles, 45, 60 Degrees.

Problem 504. Equilateral Triangle, Angles, 45, 60 Degrees


See also:
Complete Problem 504

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:37 PM
Labels: , , , ,

11 comments:

  1. AnonymousAugust 16, 2010 at 5:50 AM

    in triangle ABD:<A=45,<B=105,<D=30
    with the sine law
    BDsin30=ABsin45;BD=ABsqr(2)
    in triangle BDC
    with the cosine law
    CD²=BC²+BD²-2BC.BD.cos45=AB²
    CD=AC
    triangle ACD is isoscele
    x=15
    .-.

    ReplyDelete
  2. AnonymousAugust 16, 2010 at 1:39 PM

    let t=AB, it is known that BD=t\sqrt{2} (because tr ABD is 45-30-105).
    Consecuently CD=t (by tracing CX, altitude of tr BCD and using Pithagorean Theorem).
    So tr ACD is isosceles and finally <CDA= < CAD=15.

    by mathreyes

    ReplyDelete
  3. c .t . e. oAugust 16, 2010 at 11:56 PM

    draw DG perpendicular to AD, G on extensin of AC
    ABDG cyclic AC = CB = R => C midpoint => AC = CD

    ReplyDelete
  4. c .t . e. oAugust 18, 2010 at 8:41 AM

    arc BG = 2∙A = 120° => arc AB = 60° => C midpoint

    ReplyDelete
  5. kadir altıntaşJuly 3, 2011 at 4:29 PM

    C origin, A,B and D circular
    IBCI=ICDI=r
    so trianle BCD isosceles and x=15

    kadir Altintaş/emirdağ

    ReplyDelete
  6. AnonymousJune 21, 2015 at 2:10 AM

    Draw perpendiculars BP,CQ to AD,BD respectively. Then Tr ABP congruent to Tr BQC SAA. So BP =QC=BQ=QD since < BDP=30. So Tr BQC= Tr DQC. Hence <QDC=45. x=15


    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisJune 21, 2015 at 3:25 AM

    Draw perpendiculars BP and CQ to AD and BD respectively. Then these 2 peroenduculars are easily seen to be equal from congruent Tr.s. Also BP=CQ=BQ=1/2 BD (90 60 30 Tr) = QD, so Tr BQC =Tr BQD. Hence<CDQ = 45,'X=15

    ReplyDelete
  8. APOSTOLIS MANOLOUDISJune 21, 2016 at 1:25 AM

    Problem 504
    Is <ABD=60+45=105 ,so <BDA=30=60/2=(<BCA)/2. Then the point C is the circumcenter of triangle ABD.So AC=CD.Therefore <ADC=<DAC=60-45=15.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  9. Sumith PeirisNovember 28, 2017 at 9:31 AM

    Solution2

    Let the bisector of < ACB meet AD at E.
    Then <BDE <BCE = 30, hence BECD is concyclic.
    So x = <CBE = < CAE =15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  10. Sumith PeirisNovember 30, 2017 at 8:58 PM

    Solution 3

    Let X be the foot of the perpendicular from B to AD

    If BD = 2p then since <BDX = 30, BX = AX = p and so AB =BC = sqrt2.a

    So in Triangle BCD, <B = 45, BC = sqrt2.p and BD = 2p, hence the same is easily seen to be right angled at C.

    Hence BDCX is concyclic and x = <XBC = 15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  11. MarcoDecember 21, 2023 at 11:05 PM

    <DAC=15
    sin15/CD=sinx/AC
    AC/CD=sinx/sin15------(1)

    <BDA=30
    sin(30+x)/BC=sin45/CD
    BC/CD=sin(30+x)/sin45-------(2)

    Since AC=BC, (1)=(2)
    sinx/sin15=sin(30+x)/sin45
    sin45/sin15=sinx/sin(30+x)
    x/(30+x)=1/3
    3x=30+x
    2x=30
    x=15

    ReplyDelete

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