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DS: Triangle

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DS: Triangle [#permalink] New post 07 May 2005, 04:07
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
I don't understand OA on this one.
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Director
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 [#permalink] New post 07 May 2005, 04:45
I am getting B as answer...let me know if that is right i would then explain
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 [#permalink] New post 07 May 2005, 06:25
gmat2me2 wrote:
I am getting B as answer...let me know if that is right i would then explain



B = 90 degree

Angle ACB=60 degree, and angle CAB=30 degree,

Hence, the triangle must not be isosceles triangle.

A seems insufficient
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 [#permalink] New post 07 May 2005, 06:25
gmat2me2 wrote:
I am getting B as answer...let me know if that is right i would then explain



Go for B

angle B = 90 degree

Angle ACB=60 degree, and angle CAB=30 degree,

Hence, the triangle must not be isosceles triangle.

(1) seems insufficient
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 [#permalink] New post 07 May 2005, 06:29
B will disporve its not an isoceles triangle

so answer = B
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 [#permalink] New post 07 May 2005, 10:25
It has to be B because the second stmt shows that the triangle abc is not isoceles triangle.
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 [#permalink] New post 07 May 2005, 10:56
Another vote for B.

In A , the angles of triangle ABC are not known to be equal or not.
In B, the angles are known and are 30,60
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Re: DS: Triangle [#permalink] New post 07 May 2005, 18:47
mirhaque wrote:
I don't understand OA on this one.


1 is suff too.

acb=2adc.....1
acb=cad+adc.....2 (external angle equal two other internal angle)

combining 1 & 2 adc=cad.
that means cd=ad.

This goes to prove that triangle abc is not isosceles
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Re: DS: Triangle [#permalink] New post 07 May 2005, 20:57
mirhaque wrote:
mirhaque wrote:
I don't understand OA on this one.


1 is suff too.

acb=2adc.....1
acb=cad+adc.....2 (external angle equal two other internal angle)

combining 1 & 2 adc=cad.
that means cd=ad.

This goes to prove that triangle abc is not isosceles


I dont think you can use that theorem here simply because angle A is shared between the triangles .

So your (2) in your explanation is a a doubt for me.
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Re: DS: Triangle [#permalink] New post 08 May 2005, 05:47
mirhaque wrote:
mirhaque wrote:
I don't understand OA on this one.


1 is suff too.

acb=2adc.....1
acb=cad+adc.....2 (external angle equal two other internal angle)

combining 1 & 2 adc=cad.
that means cd=ad.

This goes to prove that triangle abc is not isosceles


I get your point now mirhaque. the external angle of ADC is angle B and that should be suffiecient enough to say that A is also sufficient.....
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Re: DS: Triangle [#permalink] New post 08 May 2005, 07:06
gmat2me2 wrote:
mirhaque wrote:
mirhaque wrote:
I don't understand OA on this one.


1 is suff too.

acb=2adc.....1
acb=cad+adc.....2 (external angle equal two other internal angle)

combining 1 & 2 adc=cad.
that means cd=ad.

This goes to prove that triangle abc is not isosceles


I get your point now mirhaque. the external angle of ADC is angle B and that should be suffiecient enough to say that A is also sufficient.....


A is not sufficient.
I don't understand how you guys got A sufficient. B should be the answer.
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Re: DS: Triangle [#permalink] New post 08 May 2005, 08:22
ashkg wrote:
gmat2me2 wrote:
mirhaque wrote:
mirhaque wrote:
I don't understand OA on this one.


1 is suff too.

acb=2adc.....1
acb=cad+adc.....2 (external angle equal two other internal angle)

combining 1 & 2 adc=cad.
that means cd=ad.

This goes to prove that triangle abc is not isosceles


I get your point now mirhaque. the external angle of ADC is angle B and that should be suffiecient enough to say that A is also sufficient.....


A is not sufficient.
I don't understand how you guys got A sufficient. B should be the answer.


Well the theorem the external angle of a triangle is equal to the sum of opposite angles is used here.

let acb = 2x while ADC = x

since ACB=2x ==>> ACD=180-2x

In triangle ACD angles A + D +C =180

So we have C =180-2x D =x So angle A for ACD has to be x.

So it is isoceles
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Re: DS: Triangle [#permalink] New post 08 May 2005, 08:54
gmat2me2 wrote:
ashkg wrote:
gmat2me2 wrote:
mirhaque wrote:
mirhaque wrote:
I don't understand OA on this one.


1 is suff too.

acb=2adc.....1
acb=cad+adc.....2 (external angle equal two other internal angle)

combining 1 & 2 adc=cad.
that means cd=ad.

This goes to prove that triangle abc is not isosceles


I get your point now mirhaque. the external angle of ADC is angle B and that should be suffiecient enough to say that A is also sufficient.....


A is not sufficient.
I don't understand how you guys got A sufficient. B should be the answer.


Well the theorem the external angle of a triangle is equal to the sum of opposite angles is used here.

let acb = 2x while ADC = x

since ACB=2x ==>> ACD=180-2x

In triangle ACD angles A + D +C =180

So we have C =180-2x D =x So angle A for ACD has to be x.

So it is isoceles


The question is asking if trianlge ABC isosceles and not ACD.
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 [#permalink] New post 08 May 2005, 09:09
Looks I have drifted to the other triangle :-D

Anyway ...As per the question B would be enough.......Thanks ashkg.....
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 [#permalink] New post 08 May 2005, 19:09
For 1), if you draw a isoceles triangle with B a right angle and BA=BC, then you draw a nother isoceles triangle where AC=AD, then 1) would be satisfied. However you can also draw a right triangle ABC where BA<>BC, and then still draw a triangle where AC=AD to satisfy 1). In other words 1) is not sufficient to determine whether ABC is isoleles.
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  [#permalink] 08 May 2005, 19:09
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