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GMAT SET 16 - 4

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Manager
Joined: 14 Jan 2006
Posts: 96

Kudos [?]: 40 [0], given: 2

Schools: HKUST
GMAT SET 16 - 4 [#permalink]

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01 Oct 2008, 01:48
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Kudos [?]: 40 [0], given: 2

Manager
Joined: 30 Sep 2008
Posts: 111

Kudos [?]: 20 [0], given: 0

Re: GMAT SET 16 - 4 [#permalink]

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01 Oct 2008, 08:38
x> 90 <=> a^2 + b^2 < c^2

(1) insufficient, c is unknown
(2) insufficient, a,b are unknown

(1) & (2)

a^2 + b^2 < 15 < 4^2 < c^2
<=> a^2 + b^2 < c^2

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Senior Manager
Joined: 04 Jan 2006
Posts: 276

Kudos [?]: 43 [0], given: 0

Re: GMAT SET 16 - 4 [#permalink]

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01 Oct 2008, 09:14
This is the trigonometry approach. Some people can totally ignore this solution.

c > 4
c^ > 16 ---------- (1)

a^2 + b^2 < 15
-(a^2) - (b^2) > -15 ---------- (2)

Since; $$c^2 = a^2 + b^2 - 2abcos(x)$$
and c^2 > 16

a^2 + b^2 - 2abcos(x) > 16 ---------- (3)
-2abcos(x) > 16 - 15 ---------- (3) + (2)
-2abcos(x) > 1
cos(x) > -1/(2ab)

cos(x) < 0 if and only if 90 < x < 180

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SVP
Joined: 17 Jun 2008
Posts: 1535

Kudos [?]: 279 [0], given: 0

Re: GMAT SET 16 - 4 [#permalink]

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01 Oct 2008, 12:30
What the question is really asking is whether c^2 is more than (a^2 + b^2). In a right angled triangle, c^2 = (a^2 + b^2). However, if the angle is more than 90 degree, the triangle will open up and in order to cover that, c has to be longer than it was for right angled triangle.

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Re: GMAT SET 16 - 4   [#permalink] 01 Oct 2008, 12:30
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