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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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02 Sep 2010, 12:57
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In the triangle above(refer attached file), is x > 90?
(1) $$a^2$$ + $$b^2$$ < 15
(2) c > 4
[Reveal] Spoiler: OA

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Triangle.docx [14.03 KiB]

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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.

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07 Sep 2010, 05:58
I cannot download the file..some different format...can you upload again
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07 Sep 2010, 06:02
oldstudent wrote:
I cannot download the file..some different format...can you upload again

Here it is:
Attachment:

untitled.PNG [ 4.96 KiB | Viewed 1445 times ]

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

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09 Jun 2012, 14:33
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In the triangle above, is x > 90?

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

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

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09 Jun 2012, 14:53
Hm...

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

Insufficient.
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

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

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13 Jun 2012, 06:52
c² = b² + a² - 2ba cosC

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

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28 Sep 2015, 05:18
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Re: In the triangle above, is x > 90? [#permalink]

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06 Nov 2016, 12:58
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Re: In the triangle above(refer attached file), is x > 90? [#permalink]

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14 Aug 2017, 09:18
udaymathapati wrote:
In the triangle above(refer attached file), is x > 90?
(1) $$a^2$$ + $$b^2$$ < 15
(2) c > 4

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/abc-with-a-t ... 50700.html
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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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