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Triangle

by student22 » Thu May 27, 2010 7:56 pm
In the triangle above, is x > 90?
1. a^2 + b^2 < 15
2. c > 4

OA:c

Image

I ended up just trying values until I was 75% sure and then guessed. Afterwards, I plugged in more values in Excel and confirmed the answer.

My question, is, what's the theoretical way to solve this? How do I go about proving that this cannot be a right triangle. Thanks!
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by liferocks » Thu May 27, 2010 8:03 pm
I have used trigonometry to solve.

either of the statements not sufficient alone

now combining we get

a^2+b^<c^
or a^2+b^-c^<0...1

again from cosign rule of triangle we have a^2+b^2-2abCos(x)=c^ or a^2+b^2-c^2=2abCos(x)..2

From relation 1 and 2 we have
2abCos(x)<0..since both ab are positive we get Cos(x)<0 ..this is possible only when 90<x<180..hence x>90..sufficient

Ans optionC

But I think trigo is not in GMAT..so there must be some other way of solving this which I might have been missing.
"If you don't know where you are going, any road will get you there."
Lewis Carroll
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by 4GMAT_Mumbai » Thu May 27, 2010 8:09 pm
If c is the largest side of the triangle, then

if c^2 < a^2 + b^2, then it is an acute angled triangle
if c^2 = a^2 + b^2, then it is a right angled triangle
if c^2 > a^2 + b^2, then it is an obtuse angled triangle

In this particular case, when statement 1 and statement 2 are combined, it can be deduced that

c^2 > a^2 + b^2. Hence, it is deduced that it is an obtuse angled triangle.

Hope this helps.
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by student22 » Thu May 27, 2010 8:16 pm
Thanks 4GMAT_Mumbai, I didn't know those rules. But the answer now makes sense, logically. Since c > 4 and it says that a^2 + b^2 < 15, then c^2 is > a^2 + b^2, so it's an obtuse triangle. This is exactly what I was looking for.

liferocks, you're lucky that you still remember how to do trigonometry. It's been years since I've done it, and I remember how powerful and useful it was. But unfortunately, I don't remember how to do any of it.
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