If points A, B, and C form a triangle, is angle ABC>90 degre : GMAT Data Sufficiency (DS)
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# If points A, B, and C form a triangle, is angle ABC>90 degre

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If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink]

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16 Nov 2007, 07:59
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If points A, B, and C form a triangle, is angle ABC>90 degrees?

(1) AC = AB + BC − 0.001

(2) AC = AB

M15-24
[Reveal] Spoiler: OA
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Re: C 15.24 degrees of a triangle [#permalink]

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16 Nov 2007, 20:04
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?

1. AC = AB + BC - .001
2. AC = AB

I think the answer is A.

S1 :

AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2

AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001

By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.

PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )

St2 :

As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
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Re: C 15.24 degrees of a triangle [#permalink]

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16 Nov 2007, 20:47
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?

1. AC = AB + BC - .001
2. AC = AB

I think the answer is A.

S1 :

AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2

AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001

By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.

PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )

St2 :

As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.

I find myself inclined to agree with your logic about statement 1.

However, I find statement 2 to be sufficient by itself as well.

If ac=ab, then angle ABC = angle ACB.

Therefore angle ABC cannot be greater than 90.
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Re: C 15.24 degrees of a triangle [#permalink]

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16 Nov 2007, 21:18
jbs wrote:
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?

1. AC = AB + BC - .001
2. AC = AB

I think the answer is A.

S1 :

AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2

AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001

By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.

PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )

St2 :

As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.

I find myself inclined to agree with your logic about statement 1.

However, I find statement 2 to be sufficient by itself as well.

If ac=ab, then angle ABC = angle ACB.

Therefore angle ABC cannot be greater than 90.

Ooops .. I missed that .. I think these are the traps that are set by GMAC to fool us around..

yes, D it is ..

Good question !!
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Re: C 15.24 degrees of a triangle [#permalink]

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16 Nov 2007, 22:20
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?

1. AC = AB + BC - .001
2. AC = AB

I think the answer is A.

S1 :

AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2

AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
--> AC^2 = AB^2 + BC^2 + 2AB.BC - (2AC*.001 + .001^2)

By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.

PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )

St2 :

As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.

Please see the correction in blue above.

I pick B.
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Re: C 15.24 degrees of a triangle [#permalink]

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17 Nov 2007, 00:28
jbs wrote:
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?

1. AC = AB + BC - .001
2. AC = AB

I think the answer is A.

S1 :

AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2

AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001

By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.

PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )

St2 :

As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.

I find myself inclined to agree with your logic about statement 1.

However, I find statement 2 to be sufficient by itself as well.

If ac=ab, then angle ABC = angle ACB.

Therefore angle ABC cannot be greater than 90.

Since AC = AB + BC - .001, what if BC = 0.001? then AC = AB again as in statement 2.
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Re: C 15.24 degrees of a triangle [#permalink]

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17 Nov 2007, 22:39
GMAT TIGER wrote:
jbs wrote:
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?

1. AC = AB + BC - .001
2. AC = AB

I think the answer is A.

S1 :

AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2

AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001

By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.

PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )

St2 :

As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.

I find myself inclined to agree with your logic about statement 1.

However, I find statement 2 to be sufficient by itself as well.

If ac=ab, then angle ABC = angle ACB.

Therefore angle ABC cannot be greater than 90.

Since AC = AB + BC - .001, what if BC = 0.001? then AC = AB again as in statement 2.

when dealing with triangles, i usually look for defined size and shape.

-.001 is a concrete size. however, we dont know whether that is a material size that can change the size of the sides of a triangle. From 1, we cannot infer anything.
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17 Nov 2007, 22:47
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1. AC=AB+BC-0.001

this is the same as AC>AB+BC (common for triangles)
for example,
AC=1000.001, AB=500, BC=500 => ABC~180

AC=0.001, AB=500, BC=500.001 =>ABC~0

insuf.

2. AB=AC

ABC=ACB => 2ABC<180> ABC<90

suf.

B is correct

P.S if one can draw it solution will come easy.
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17 Nov 2007, 22:55
walker wrote:
AC=0.001, AB=500, BC=500.001 =>ABC~0

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12 Apr 2013, 16:27
If Angle ABC is > 90, then AC has to be the hypotenuse.

With Point 1:

If AB is 1, and BC is 1, then AC would be 1.999, making it the hypotenuse

But if AB is .0006, and BC is .0007, then AC would be .0003, making it not the hypotenuse.

Because the .001 gives us no reference, we cannot conclude anything from Point 1 alone.

If AB = AC, then that means that there is no possible way that AC could be the hypotenuse since there is another side of equal length right next to it. Even if BC is infinitely small, it is still >0 and therefore ABC cannot be >90. Therefore, Point 2 is enough for us to disqualify it alone.
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19 Apr 2014, 11:21
HI Bunnel,

Thanks.
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21 Apr 2014, 05:05
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PathFinder007 wrote:
HI Bunnel,

Thanks.

THEORY:

Say the lengths of the sides of a triangle are a, b, and c, where the largest side is c.

For a right triangle: $$a^2 +b^2= c^2$$.
For an acute (a triangle that has all angles less than 90°) triangle: $$a^2 +b^2>c^2$$.
For an obtuse (a triangle that has an angle greater than 90°) triangle: $$a^2 +b^2<c^2$$.

Points A, B and C form a triangle. Is ABC > 90 degrees?

(1) AC = AB + BC - 0.001.

If AC=0.001, AB=0.001 and BC=0.001, then the triangle will be equilateral, thus each of its angles will be 60 degrees.

If AC=10, AB=5 and BC=5.001, then AC^2>AB^2+BC^2, which means that angle ABC will be more than 90 degrees.

Not sufficient.

(2) AC = AB --> triangle ABC is an isosceles triangle --> angles B and C are equal, which means that angle B cannot be greater than 90 degrees. Sufficient.

Similar questions to practice:
are-all-angles-of-triangle-abc-smaller-than-90-degrees-129298.html
if-10-12-and-x-are-sides-of-an-acute-angled-triangle-ho-90462.html

Hope it's clear.
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Re: If points A, B, and C form a triangle... [#permalink]

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24 May 2014, 19:41
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bekerman wrote:
If points A, B, and C form a triangle, is angle ABC>90 degrees?

(1) AC=AB+BC−0.001

(2) AC=AB

M15-24 in GMATClub tests - I am wondering whether the OA is incorrect?

Statement-1:
AC = AB+ BC - .001,
If AB, BC are quite big numbers (greater than .01), then angle ABC would be greater than 90 degrees. But if length of AB, BC are in the same range of .001, then angle ABC could be acute angle also.
So statement 1 is not sufficient.

Statement -2:
AC= AB, it means angle ABC = angle ACB, now in any triangle sum all the angles is 180 degree, thus ABC +ACB+BAC = 180 degree. Now as ABC = ACB -> 2ABC + BAC = 180 -> ABC = 90 - BAC/2. Hence angle ABC is always less than 90 degree.

Statement 2 is sufficient
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink]

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05 Mar 2015, 09:53
Bunuel, can we also claim that when the angle us obtuse c will be greater than a and b?
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink]

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05 Mar 2015, 10:28
Ergenekon wrote:
Bunuel, can we also claim that when the angle us obtuse c will be greater than a and b?

Yes, the greatest side is opposite the greatest angle.
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink]

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Re: If points A, B, and C form a triangle, is angle ABC>90 degre   [#permalink] 25 Dec 2016, 05:58
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