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solve the problem

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by Frankenstein » Wed Jun 22, 2011 1:52 am
Hi,
I don't know whether such questions fall into GMAT category, I will explain this
In the triangle ABC, 4/sine B = 5/sine A
So, Sine A/sine B = 5/4 --(1)
Drop perpendiculars from O to AC and BC. Let the foot of perpendiculars be P, Q respectively
In triangle APO, sine A = radius/OA
similarly in triangle BQO, sine B = radius/OB
So, sine A/ sine B = OB/OA
From (1): Sine A/sine B = 5/4 So, OB/OA = 5/4
OA+OB = AB =6
So, OB = 5/9(AB) = (5/9)*6 = 10/3
OA = 8/3
So, OA*OB = (8/3)*(10/3) = 80/9
Cheers!

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by amit2k9 » Wed Jun 22, 2011 3:35 am
hmm good question. Learnt a new concept here.
with a= 5,b=5 and c=6 we have

a/sin A = b/sin B = c/sin C.
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by shavkatmirzaev » Wed Jun 22, 2011 4:33 am
I solved through another way, anyways it is correct. well done
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by casperkamal » Wed Jun 22, 2011 7:40 am
Hi Do you mind sharing the other way that you picked up to solve this ???
Kamal
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