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In the triangle above(refer attached file), is x > 90?

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In the triangle above(refer attached file), is x > 90? [#permalink]

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New post 02 Sep 2010, 12:57
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In the triangle above(refer attached file), is x > 90?
(1) \(a^2\) + \(b^2\) 4

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Re: Tringle [#permalink]

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New post 02 Sep 2010, 13:29
udaymathapati wrote:
In the triangle above(refer attached file), is x > 90?
(1) \(a^2\) + \(b^2\) < 15
(2) c > 4


(1) \(a^2+b^2<15\), no info about \(c\). Not sufficient.
(2) \(c>4\), no info about \(a\) and \(b\). Not sufficient.

(1)+(2) Now if angle \(x\) were 90 degrees than must be true that \(a^2+b^2=c^2\), but \(a^2+b^2<15<16<c^2\) so \(c\) is more than it would be if \(x\) were 90 degrees, so \(x>90\) degrees (if we increase \(c\) we increase angle \(x\)). Sufficient.

Answer: C.
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Re: Tringle [#permalink]

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New post 07 Sep 2010, 05:58
I cannot download the file..some different format...can you upload again
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Re: Tringle [#permalink]

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In the triangle above, is x > 90? [#permalink]

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In the triangle above, is x > 90?

(1) a^2 + b^2 < 15
(2) c > 4
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Originally posted by enigma123 on 09 Jun 2012, 14:33.
Last edited by Bunuel on 09 Jun 2012, 14:41, edited 1 time in total.
Edited the question.
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Re: In the triangle above, is x > 90? [#permalink]

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New post 09 Jun 2012, 14:44
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In the triangle above, is x > 90?

(1) a^2 + b^2 < 15
(2) c > 4

Each statement alone is clearly insufficient. When taken together:

If angle x were 90 degrees than we would have a^2+b^2=c^2, since a^2+b^2<15<16=c^2 then angle x must be greater than 90 degrees (c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90).

Answer: C.
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Re: In the triangle above, is x > 90? [#permalink]

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New post 09 Jun 2012, 14:53
Hm...

1) This only tells us that a^2 + b^2 < 15 but nothing about c.

Insufficient.
[strike]AD[/strike]
BCE

2) Knowing that c > 4 without knowing anything about a or b is clearly not enough.

Insufficient.

So answer is C or E.

1+2) We should know that if a^2 + b^2 = c^2, then x=90, and if a^2 + b^2 < c^2, then x > 90, and if a^2 + b^2 > c^2, x < 90.

So, we know that c^2 > 16 because c > 4
We also know that a^2 + b^2 < 15

15 < 16 (Although keep in mind that this is the limiting factor. For instance, they could be 4 & 80, but they can't be 17 & 15, respectively.)
:. x > 90

Answer = C
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Re: In the triangle above, is x > 90? [#permalink]

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New post 13 Jun 2012, 06:52
c² = b² + a² - 2ba cosC
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Re: In the triangle above(refer attached file), is x > 90? [#permalink]

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New post 14 Aug 2017, 09:18
udaymathapati wrote:
In the triangle above(refer attached file), is x > 90?
(1) \(a^2\) + \(b^2\) 4


OPEN DISCUSSION OF THIS QUESTION IS HERE: abc-with-a-trainagle-with-three-sides-abc-and-one-of-the-50700.html

--== Message from GMAT Club Team ==--

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the triangle above(refer attached file), is x > 90?   [#permalink] 14 Aug 2017, 09:18
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In the triangle above(refer attached file), is x > 90?

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