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# In the triangle above, is x > 90?

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In the triangle above, is x > 90? [#permalink]  09 Jun 2012, 13:33
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Difficulty:

15% (low)

Question Stats:

80% (01:36) correct 20% (00:59) wrong based on 44 sessions
Attachment:

triangle.GIF [ 1.66 KiB | Viewed 1859 times ]
In the triangle above, is x > 90?

(1) a^2 + b^2 < 15
(2) c > 4
[Reveal] Spoiler: OA

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Last edited by Bunuel on 09 Jun 2012, 13:41, edited 1 time in total.
Edited the question.
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Re: In the triangle above, is x > 90? [#permalink]  09 Jun 2012, 13:44
Expert's post
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]  09 Jun 2012, 13:53
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]  13 Jun 2012, 05:52
c² = b² + a² - 2ba cosC
Re: In the triangle above, is x > 90?   [#permalink] 13 Jun 2012, 05:52
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