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# In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C

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Intern
Joined: 17 Jul 2014
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In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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Updated on: 02 May 2016, 03:40
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Difficulty:

95% (hard)

Question Stats:

32% (02:43) correct 68% (02:43) wrong based on 136 sessions

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In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C are the vertices of the triangle and a$$^{\circ}$$and b$$^{\circ}$$ denote the angles shown. Is the length of AB greater than that of AC?

(1) a – b = 30

(2) b = 70

Attachment:

100215.jpg [ 9.91 KiB | Viewed 4727 times ]

Originally posted by jyotibrata on 12 Mar 2016, 10:03.
Last edited by ENGRTOMBA2018 on 02 May 2016, 03:40, edited 2 times in total.
Edited the question.
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Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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12 Mar 2016, 10:40
2
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jyotibrata wrote:
In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C are the vertices of the triangle and ao and bo denote the angles shown. Is the length of AB greater than that of AC?

(1) a – b = 30

(2) b = 70

Attachment:
The attachment 100215.jpg is no longer available

We can clearly see that in order for getting an answer for is AB>AC we need to find the relation between the angles opposite to this sides as in any triangle the side opposite to a greater angle is greater.

Also, from the property of quadrilaterals, sum of all the angles of a quad = 360 and when used in conjunction with the fact that sum of angles on a triangle = 180 --->

$$\angle {ACB} = a-90$$ (refer to the attached picture to see the calculations).

Attachment:

2016-03-12_13-35-52.jpg [ 12.77 KiB | Viewed 3920 times ]

Per statement 1, a-b=30 ---> a=30+b ---> c = a-90=30+b-90 = b-60 ---> c<b and as sides opposite to greater angles are longer ---> AC>AB

Per statement 2, b=70. Clearly not sufficient.

A is thus the correct answer.

Hope this helps.
##### General Discussion
Manager
Joined: 29 Nov 2011
Posts: 87
Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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01 May 2016, 00:21
Engr2012 wrote:
jyotibrata wrote:
In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C are the vertices of the triangle and ao and bo denote the angles shown. Is the length of AB greater than that of AC?

(1) a – b = 30

(2) b = 70

Attachment:
100215.jpg

We can clearly see that in order for getting an answer for is AB>AC we need to find the relation between the angles opposite to this sides as in any triangle the side opposite to a greater angle is greater.

Also, from the property of quadrilaterals, sum of all the angles of a quad = 360 and when used in conjunction with the fact that sum of angles on a triangle = 180 --->

$$\angle {ACB} = a-90$$ (refer to the attached picture to see the calculations).

Attachment:
2016-03-12_13-35-52.jpg

Per statement 1, a-b=30 ---> a=30+b ---> c = a-90=30+b-90 = b-60 ---> c<b and as sides opposite to greater angles are longer ---> AC>AB

Per statement 2, b=70. Clearly not sufficient.

A is thus the correct answer.

Hope this helps.

Why are you assuming the angle as 90 degree in the triangle you have shown in your figure.
Manager
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Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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01 May 2016, 00:33
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Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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01 May 2016, 03:08
1

It's not courteous to post another message without letting the person reply. Do give the OP sufficient time in replying to your question.

I did not assume anything. The question says that the triangle ABC is drawn on the coordinate plane and as such the 2 lines can only be the coordinate axes. Hence, they are perpendicular, making that angle 90 degree

Posted from my mobile device
Manager
Joined: 29 Nov 2011
Posts: 87
Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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01 May 2016, 04:09
Engr2012 wrote:

It's not courteous to post another message without letting the person reply. Do give the OP sufficient time in replying to your question.

I did not assume anything. The question says that the triangle ABC is drawn on the coordinate plane and as such the 2 lines can only be the coordinate axes. Hence, they are perpendicular, making that angle 90 degree

Posted from my mobile device

I apologies , I did not mean that way. I came across few posts from experts where it is clearly written that don't be decisive in DS question with coordinate geometer figures. Thanks for explanation I got your point.
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Concentration: Finance, Strategy
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Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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01 May 2016, 04:16
Engr2012 wrote:

It's not courteous to post another message without letting the person reply. Do give the OP sufficient time in replying to your question.

I did not assume anything. The question says that the triangle ABC is drawn on the coordinate plane and as such the 2 lines can only be the coordinate axes. Hence, they are perpendicular, making that angle 90 degree

Posted from my mobile device

I apologies , I did not mean that way. I came across few posts from experts where it is clearly written that don't be decisive in DS question with coordinate geometer figures. Thanks for explanation I got your point.

You do raise a very valid question, something GMAT question will not do. All GMAT questions are unambiguous (or 99.9% of them) and I'm sure that if this would've been an official question, the question stem would've stated that the vertical and the horizontal lines that make the 90 degrees are the coordinate axes to remove any ambiguity.

Hope this helps.
Intern
Joined: 18 Jun 2017
Posts: 1
Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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15 Jan 2018, 05:58
Sorry I don´t understand why this is not a D.

I think that if I only know the value of B i can Solve de problem.

(180-a) +90 + (180-b) + (180-a) = 360

...
...

2a + b = 270

2) b= 70

----> a= 100 and as a + b+ c = 180 then C= 10 so AC > AB

is that right?
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Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C  [#permalink]

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01 Dec 2019, 01:11
Quote:
Also, from the property of quadrilaterals, sum of all the angles of a quad = 360 and when used in conjunction with the fact that sum of angles on a triangle = 180 --->

∠ACB=a−90∠ACB=a−90

Hi can you pls explain how you got a-90 for angle ACB?

I did not get which quadrilateral property you are saying
Re: In triangle ABC (figure shown) drawn on a coordinate plane, A, B and C   [#permalink] 01 Dec 2019, 01:11