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In the attached figure, points P and Q lie on the circle

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Director
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In the attached figure, points P and Q lie on the circle [#permalink] New post 15 Jun 2006, 23:55
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In the attached figure, points P and Q lie on the circle with centre O. What is the value of S?

A. 1/2
B. 1
C. sqrt(2)
D. sqrt(3)
E. sqrt(2)/2

I've solved it, checking if there is a better way of doing it... :roll:
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SVP
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 [#permalink] New post 16 Jun 2006, 01:06
Took me some time because I had the som many ideas

What I finally chose.

If we draw a line from P to Q we get a 90-45-45 triangle.

Find the distance of OP = 2
Hence PQ must be 2rt(2).

Now take individual values of s and calculate value of t for OQ (=2)

1,sqr(3) and -sqr(3),1 satisifes PQ..........

this is a little long but can be a made short
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 [#permalink] New post 16 Jun 2006, 04:44
My method:

if two line are perpendicular, product of slopes = -1

Therefore
((t-0)/(s-0))*((1-0)/(srt(3)-0)) = -1
t=s* sqrt(3) ------ A

Also distance of both points from origin is same

Sqrt(t^2 + s^2) = sqrt (3+1) = 2
t^2 + s^2 = 4 -----B

Solving both A & B

3*s^2 + s^2 = 4
s = +1 or -1 (corresponding to Q1 & Q3)

from diagram choose +1
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 [#permalink] New post 16 Jun 2006, 06:04
First calculate the distance between [-sqr(3), 1] and the origin

= sqt ( -sqr(3)^2 + 1^2) = sqt (3+1) = 2

This distance must be equal to the distance between the (s,t) and the origin since it's a circle.

2 = sqt (s^2 + t^2) 4 = s^2 + t^2

Next calculate the slopes. Slope of [-sqr(3), 1] and origin = 1/-sqt(3)
Perpindicular slope must be sqt (3).
Therefore, formula of the line is t = sqt(3)*s + 0

Plug in t = sqt(3)*s inot the above distance equation:
4 = s^2 + [sqt (3)*s]^2
4 = s^2 (1+3)
1 = s

Checking the formulas and eyeballing the graph confirms the answer.
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 [#permalink] New post 16 Jun 2006, 07:10
OA is B.

Thx all. I used the method by shobhitb and tl372.

jaynayak, could you elaborate on your method plz? i did not quite get it.
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 [#permalink] New post 16 Jun 2006, 08:07
Another approach is to use sin/cos functions.
Given P's coordiantes and radius = 2, you can find out that it makes an angle of 30 degrees with X axis, then Q must make an angle of 60 degrees with X axis, which gives s = 1 (cos 60 = 1/2)
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 [#permalink] New post 16 Jun 2006, 11:10
I solved it in under 1 minute.

OP = OQ = 2
Angle of OP with -ve side of x-axis = 30 (because Sin30 = 1/2). So angle of OQ with +ve side of x axis = 60

So s/OQ = Cos60 i.e s/2 = 1/2 i.e s = 1
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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 [#permalink] New post 26 Jun 2006, 05:28
Intersting Question
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Learn and let others learn

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 [#permalink] New post 26 Jun 2006, 15:39
ps_dahiya, sgrover - great method! thx :)
  [#permalink] 26 Jun 2006, 15:39
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In the attached figure, points P and Q lie on the circle

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