Saturday, October 10, 2009

Problem 363: Right triangle, Congruence, Angles

Proposed Problem
Click the figure below to see the complete problem 363 about Right triangle, Congruence, and Angles.

Problem 363. Right triangle, Congruence, Angles.
See more:
Complete Geometry Problem 363
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:26 AM
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5 comments:

  1. By considering isos triangle, alpha=45
    Then, by considering tan(beta+theta),
    tan(beta+theta)=(tan(beta)+tan(theta))/(1+tan(beta)tan(theta))
    =(1/2+1/3)/(1-1/2*1/3)
    =1
    So beta+theta=45,
    alpha+beta+theta=90

    ReplyDelete

  2. How do you solve the problem without useing trigonometric approch???
    Just useing pure Geometry!!!!
    Is there anybody who has any idea???

    ReplyDelete

  3. http://geometri-problemleri.blogspot.com/2009/11/problem-43-ve-cozumu.html

    ReplyDelete
  4. bae deok rak(bdr@korea.com)October 2, 2010 5:43 PM

    Let D_1 be a point with ABDD_1 is a square and let D_2 be a point with DD_1 =DD_2 and BD is a perpendcular to D_1D_2.
    Then we have D_2A=D_2C and ang(D_2CD)=ang(AEB)=beta, that is, the triangle D_2AC is a right isoscele triangle. Therfore ang(D_2CA)=45=beta +gamma and alpha=45. We get
    ang(ADB)+ang(AEB)+ang(ACB)=alpha+beta+gamma=90 degrees.




    sum of the angles ADB, AEB, and ACB is 90 degrees.

    BDer

    ReplyDelete

  5. 1.in tri ABD, ang.ADB=45 ...(iso. tri thm)

    2.hence alpha=45degree

    3.tri. DEA similar to tri CAD ...(sss test of simlarity)

    4.hence ang.AED=ang.DAC ...(c.a.s.t)

    5.hence ang.DAC=Beta ...(From 4)

    6.in triangle ADC,
    ang.DAC+ang.ACD=ang.ADB...(exterior ang. thm)

    7.hence beta+theta=45 degree...(from 5,6 & given)

    8.hence alpha+beta+theta=45+45=90 degree ...(from 2 & 7)

    Q.E.D

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