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Intern Joined: 07 Jan 2017
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One of GRE question [#permalink] 00:00

Question Stats: 100% (00:05) correct 0% (00:00) wrong based on 1 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.
Attachments question.jpg [ 81.4 KiB | Viewed 2154 times ] GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: One of GRE question [#permalink]
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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 Joined: 07 Jan 2017
Posts: 10
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Kudos [?]: 6 , given: 3

Re: One of GRE question [#permalink] Intern Joined: 26 Feb 2018
Posts: 4
GRE 1: Q162 V148 Followers: 0

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Re: One of GRE question [#permalink]
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darkdevil8z wrote:
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.

Both of you are going a very long way.

Method 1 : darkdevil8z method simplified : Only Mental Mathematic required. Two sides are given and one angle is 90 degree so try diving both the length and see if you recognise the number.
I just divide (10√3)/3 with 5 and remainder is 2/√3.

I don't remember Cos/tan/cosec/cot/sec but I do remember Sin θ i.e (perpendicular / hypotenuse OR P/B)and you too have to.
This is the only requirement for GRE as far as trigonometry application is required.

I remember 1 - 1/2 - 1/√2 - √3/2 - 1 for Sin θ and recall that the above number is familiar. It is reciprocal of (10√3)/3 by 5

Now I know Sin 60 = √3/2 = 5 divided by (10√3)/3 = AC/CB = perpendicular/Hypotenuse
Therefore angle ABC is 60 degree and side in-front will definitely be bigger.

AC is greater

This was simpler method but for those who can't recite Sin θ at that moment, go for Pythagoras theorem.

Method 2 : Pythagoras theorem. method simplified : Only Mental Mathematic required. ABˆ2 = (10/√3) - 5ˆ2 = 100/3 - 25 = 25/3 = Less than 9
AB = Less than 3

AC=5 is greater.

To get more simplified solutions.
Start Follow and Hit Kudos / Like

Regards
Shekhar  Re: One of GRE question   [#permalink] 26 Feb 2018, 02:19
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