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SVP  Joined: 21 Jan 2007
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If points A, B, and C form a triangle, is angle ABC>90 degre  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58347

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PathFinder007 wrote:
HI Bunnel,

Could you please provide your comments on statement defined in question.

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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Senior Manager  Joined: 09 Aug 2006
Posts: 431
Re: C 15.24 degrees of a triangle  [#permalink]

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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.
Manager  Joined: 25 Jul 2007
Posts: 94
Re: C 15.24 degrees of a triangle  [#permalink]

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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.
Senior Manager  Joined: 09 Aug 2006
Posts: 431
Re: C 15.24 degrees of a triangle  [#permalink]

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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 !!
Director  Joined: 09 Aug 2006
Posts: 608
Re: C 15.24 degrees of a triangle  [#permalink]

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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.
SVP  Joined: 29 Aug 2007
Posts: 1975
Re: C 15.24 degrees of a triangle  [#permalink]

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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.
SVP  Joined: 21 Jan 2007
Posts: 2260
Location: New York City
Re: C 15.24 degrees of a triangle  [#permalink]

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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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1
2
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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walker wrote:
AC=0.001, AB=500, BC=500.001 =>ABC~0

BC=500.002 instead of 500.001
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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.
Manager  B
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HI Bunnel,

Could you please provide your comments on statement defined in question.

Thanks.
Intern  Joined: 13 Dec 2013
Posts: 8
Re: If points A, B, and C form a triangle...  [#permalink]

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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?

IMO Answer is "B"

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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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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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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Bunuel wrote:
PathFinder007 wrote:
HI Bunnel,

Could you please provide your comments on statement defined in question.

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:
http://gmatclub.com/forum/are-all-angle ... 29298.html
http://gmatclub.com/forum/if-10-12-and- ... 90462.html

Hope it's clear.

Hi Bunuel, to find an obtuse angle within the constraints set by 1) I did the following (in bold). Is this approach okay?

1) If AB + BC = 100, then angle ABC will be close to 180. This triangle is allowed because AC<AB+AC. I felt that this triangle allowed easier visualization of the obtuse angle.

And, as you stated if all sides = 0.001, then angle ABC will be 60.

2) Means that the triangle is isosceles and therefore has 2 equal angles.
2x+y=180
2x=180-y

Because y cannot be 0, x must be less than 90. Suff.
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre  [#permalink]

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bmwhype2 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

hi

I don't know whether this is okay, but I tried this problem this way
statement 1 actually says that

AC < AB + BC

now, if we suppose that AC is the largest side, then it is true for any triangle, regardless of whether the triangle is acute or obtuse
so clearly insufficient

statement 2 says that the triangle is an isosceles triangle with its 2 angels equal

now, since the sum of all the angles of a triangle adds up to 180 degrees, angle B cannot even equal to 90 degrees let alone be more than 90 degrees
hence the statement 2 is clearly sufficient

So the answer is B

thanks Senior Manager  G
Joined: 23 Nov 2016
Posts: 295
GMAT 1: 690 Q50 V33 Re: If points A, B, and C form a triangle, is angle ABC>90 degre  [#permalink]

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Experts can someone please explain, Seems like S1 is sufficient. His explanation seems logical .

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.

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If my post anyway helped you,please spare Kudos ! Re: If points A, B, and C form a triangle, is angle ABC>90 degre   [#permalink] 27 Mar 2019, 19:10
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