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Re: In the triangle above(refer attached file), is x > 90? [#permalink]
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oldstudent wrote:
I cannot download the file..some different format...can you upload again


Here it is:
Attachment:
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untitled.PNG [ 4.96 KiB | Viewed 2626 times ]
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Re: In the triangle above(refer attached file), is x > 90? [#permalink]
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In the triangle above, is x > 90?

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

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(refer attached file), is x > 90? [#permalink]
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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(refer attached file), is x > 90? [#permalink]
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(refer attached file), is x > 90? [#permalink]
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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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



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