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

8 comments:

  1. 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
    .-.

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  2. 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

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

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

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

    kadir Altintaş/emirdağ

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  6. 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

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  7. 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

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  8. 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

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