PR and QR are two tangents drawn to a circle with centre O. What is angle PRQ?
(1) AnglePOQ = 120°
(2) PRO and ROQ are isosceles triangles.
OA is C but i dnt agree with it . please help
Geometry
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ritula wrote:PR and QR are two tangents drawn to a circle with centre O. What is angle PRQ?
(1) AnglePOQ = 120°
(2) PRO and ROQ are isosceles triangles.
OA is C but i dnt agree with it . please help
I guess you are assuming that every tangent drawn to the circle makes a 90 degree angle.
One cannot simply assume that, when it is not mentioned.
Statement I
angle formed at the center of the circle is 120.
We dont know anything about the other 2 tangents or their angles.
Insufficient.
Statement II
two opposite triangles are isosceles triangles, we dont know anything about the other angle formed at the center, therefore Insufficient.
Combining I & II
center angle120 , two sides isosceles means both the angles are 60 degree,
120/ 2= 60
In two triangles we 6060 formation, therefore angle PRQ=60
Sufficient.
Hence C is the answer.
Last edited by parallel_chase on Sat Sep 06, 2008 4:40 am, edited 2 times in total.

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Hi parallel_chase,
Would you please explain in detail following points in explanation?
a)center angle120 , two sides isosceles means both the angles are 90 degree,
b)120/ 2= 60
Thanks
Would you please explain in detail following points in explanation?
a)center angle120 , two sides isosceles means both the angles are 90 degree,
b)120/ 2= 60
Thanks

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I actually made a mistake, i have revised my above post.shahdevang87 wrote:Hi parallel_chase,
Would you please explain in detail following points in explanation?
a)center angle120 , two sides isosceles means both the angles are 90 degree,
b)120/ 2= 60
Thanks
"two sides isosceles means both the angles are 90 degree" this means the angle is 60 degrees not 90 thats a typo
120/ 2= 60
this is done to divide the center angle into two equal parts since the two triangles share the same angle at the center.
Hope its clear, Let me know if you have any doubts.

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Angle formed by tangent and radius is always 90 deg
you do nt have to assume anything, in light of this D will be OA
you do nt have to assume anything, in light of this D will be OA

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@ Vignesh.4384
the answer you get in both the statements is different
In st 1 you get PRQ = 60 degrees
In st 2 you get PRQ = 90 degrees
We should be getting an unique answer with both the statements for D to be correct.
I cant understand how st 2 can give an answer of PRQ = 60 degrees
the answer you get in both the statements is different
In st 1 you get PRQ = 60 degrees
In st 2 you get PRQ = 90 degrees
We should be getting an unique answer with both the statements for D to be correct.
I cant understand how st 2 can give an answer of PRQ = 60 degrees
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 Ian Stewart
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The only way both statements can be true together is if R, P and Q are not points on the circle. We can certainly have a tangent line PR that touches the circle at a different point A, say, and QR might touch the circle at B. Once you see this, it quickly becomes clear that neither statement is sufficient on its own. When you consider both statements together, there is a lot of work involved, unless I'm missing a trick. If POR is isosceles, we need to consider three possibilities PO = PR, PO = OR and PR = OR and we need to do the same for the other isosceles triangle. It's far too much work for a GMAT question, and I'm wondering if additional information was provided in the question (a diagram, perhaps?). Also curious about where the question is from.
@ IAn
The source of this problem is www.tcyonline.com
The source of this problem is www.tcyonline.com
Philosophers have interpreted world in various ways, the point is to change it!