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One of GRE question

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Intern
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Joined: 07 Jan 2017
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One of GRE question [#permalink] New post 07 Jan 2017, 09:50
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Hi everyone, I couldn't confirm my answer whether it is correct or not, can anyone confirm for me?? Or correct me if I'm wrong??
Thanks.

BC = (10√3)/3
BC = 5.77

angle of C = (cos θ = AC/BC)
since radius of the circle with center C is 5
AC = 5
cos θ = 5/5.77
θ = (sin-1) 5/5.77
angle of C ≈ 30

sin θ = AB/BC
sin 30 = AB/5.77
0.5 = AB/5.77
AB = 2.88

Therefore Quantity B is greater.
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CEO
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Re: One of GRE question [#permalink] New post 07 Jan 2017, 17:45
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Expert's post
Hey,

Your answer is correct. But use Pythagoras theorem.

BC^2 = AC^2 + AB^2

or (\frac{10}{3}\sqrt{3})= 5^2 + AB^2

or AB^2= 100/3 -25 = 25/3 \approx 8.3333

or AB \approx 2.88

Regards
_________________

Sandy
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Intern
Intern
Joined: 07 Jan 2017
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Kudos [?]: 6 [0], given: 3

Re: One of GRE question [#permalink] New post 07 Jan 2017, 20:58
Thanks for reply
Re: One of GRE question   [#permalink] 07 Jan 2017, 20:58
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