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ashwin_gowda Just gettin' started!
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thumpin_termis Rising GMAT Star
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Posted: Thu Jun 07, 2007 8:27 am Post subject: |
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First,
Angle ADB = 180 - Angle ADC = 180 - 60 = 120
Angle ADB is an exterior angle of triangle ABD, which means it is the sum of two non-adjacent angles. This means:
Angle ADB = Angle CDA + Angle CAD. so,
120 = 60 + Angle CAD => Angle CAD = 120 - 60 = 60
So finally,
x = 180 - Angle CDA - Angle CAD = 180 - 60 - 60 = 60
Answer choice B:
Anyone see it differently?
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tejpreet Just gettin' started!
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Posted: Fri Jun 08, 2007 2:17 am Post subject: My Answer is 90 degree |
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E
Angle ADB + Angle ADC =180
==> Angle ADB = 180 - Angle ADC
= 180 - 60
= 120
Angle BAD = 180 - 120 - 45
= 180 -165
= 15
Angle DAC = 30
As side opposite Angle BAD which is BD is 1 unit and Side opposite Angle DAC which is DC is 2 units so the respective angle DAC is twice of Angle BAD and hence it is equal to 30 degrees
Angle ACD = Angle x = 180- 60 - 30
= 90
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800GMAT Really wants to Beat The GMAT!
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Posted: Fri Jun 08, 2007 3:24 am Post subject: |
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| I agree with x=90
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thumpin_termis Rising GMAT Star
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Posted: Fri Jun 08, 2007 8:13 am Post subject: |
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| Yup. I take my answer back - I see where I made the mistake (Angle ADB is not CDA + CAD; rather, it shoud've been ACD + CAD, which doesn't help it to get the answer)
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jayhawk2001 Moderator

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Posted: Fri Jun 08, 2007 4:22 pm Post subject: Re: My Answer is 90 degree |
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| tejpreet wrote: |
As side opposite Angle BAD which is BD is 1 unit and Side opposite Angle DAC which is DC is 2 units so the respective angle DAC is twice of Angle BAD and hence it is equal to 30 degrees
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I initially got 90 degrees using the same approach but after a bit more
thought, I think it is incorrect to assume that the angles will be in the
same ratio of the sides.
See attached image. One triangle is formed with the red line and another
one with the blue line. We can see that the sides are not in proportion.
Using the angle bisector property, all we can get is AC/CD = AB/BD
or AC = 2*AB. But knowing this alone will not help us in determining x.
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800GMAT Really wants to Beat The GMAT!
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Posted: Fri Jun 08, 2007 7:04 pm Post subject: |
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jayhawk, u can apply the angle bisector property only when a line bisects an angle..
How did u assume that the angles originating from A are equal or in other words, how did u assume that the line originating from A bisects A.........can u pls explain how did u arrive at the above conclusion...thnkx
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jayhawk2001 Moderator

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Posted: Fri Jun 08, 2007 8:36 pm Post subject: |
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| 800GMAT wrote: | jayhawk, u can apply the angle bisector property only when a line bisects an angle..
How did u assume that the angles originating from A are equal or in other words, how did u assume that the line originating from A bisects A.........can u pls explain how did u arrive at the above conclusion...thnkx |
Oh yes. I meant to point out the fact that even a bisector can divide the
opposite sites non equally. So, we cannot assume that the angles
will be in the ratio of 2:1 if the line is divided in the ratio of 2:1.
Statement came out the wrong way in the context of the question
Well, essentially I haven't found the correct way to solve this.
I'm sure there's some property of triangles that we can use...just isn't
strikingly clear what that is.
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800GMAT Really wants to Beat The GMAT!
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Posted: Sat Jun 09, 2007 5:16 pm Post subject: |
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jayhawk,,u r right...
we cannot assume that the angles will be in the ratio of 2:1 if the line is divided in the ratio of 2:1.
so answer cannot be 90 based on this approach
I too am unable to find an approach to solve this problem
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ashwin_gowda Just gettin' started!
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jayhawk2001 Moderator

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Posted: Mon Jun 11, 2007 6:57 am Post subject: |
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| Interesting solution.
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800GMAT Really wants to Beat The GMAT!
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Posted: Mon Jun 11, 2007 11:29 am Post subject: |
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Thanks a lot for posting the solution.....
I agree...pretty interesting...
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