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A difficult problem

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by thumpin_termis » Thu Jun 07, 2007 8:27 am
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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My Answer is 90 degree

by tejpreet » Fri Jun 08, 2007 2:17 am
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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by 800GMAT » Fri Jun 08, 2007 3:24 am
I agree with x=90
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by thumpin_termis » Fri Jun 08, 2007 8:13 am
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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Re: My Answer is 90 degree

by jayhawk2001 » Fri Jun 08, 2007 4:22 pm
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
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.
Attachments
triangle_prop.GIF
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by 800GMAT » Fri Jun 08, 2007 7:04 pm
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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by jayhawk2001 » Fri Jun 08, 2007 8:36 pm
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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by 800GMAT » Sat Jun 09, 2007 5:16 pm
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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solution to the Difficult Problem

by ashwin_gowda » Mon Jun 11, 2007 2:43 am
Hi All the solution to the problem is available @ this link.

https://www.manhattangmat.com/ChallProbLastWk.cfm
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by jayhawk2001 » Mon Jun 11, 2007 6:57 am
Interesting solution.
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by 800GMAT » Mon Jun 11, 2007 11:29 am
Thanks a lot for posting the solution.....


I agree...pretty interesting...
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