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by Anurag@Gurome » Wed Nov 02, 2011 4:33 am
gmatblood wrote:Please help in solving this one!

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant, so s = 1.
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by [email protected] » Wed Nov 02, 2011 4:38 am
Draw a perpendicular line from point p to x axis (at say point R) this will make a right triangle OPR.

Where Length of OP would be (- Root 3) and length of PR would be 1.

using pythogorus theorem we find out th length of PO which is 2.

As OP is radius of circle and is 2 so OQ will also be 2.

Draw a line between P and Q to make a right triangle POQ where PO=OQ=2

Using pythogorus theorem again PQ would be root 8 OR 2 root 2

since we need to find out S which is part of line PQ we will reduce the left hand side length of that line that is - root 3

s will be
2 root 2 - root3

Please confirm
Ashish

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by [email protected] » Wed Nov 02, 2011 4:44 am
Dear Anurag,

M not clear about both the formula you used.. Can you explain in detail please. and where i did mistake in interepreting while answering
Ashish

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by shankar.ashwin » Wed Nov 02, 2011 4:44 am
A more intuitive way to look at this, would be if you know the sides of a 30-60-90 triangle would be in the ratio 1-Sqrt(3) - 2.

You know the sides satisfy this condition here. Now the x-coordinate of point 'P' is sort(3), which means its opposite to angle 60.

Now in the x-axis, 60 + 90 (POQ-mentioned) , so the angle QOX would be 30. And using the ratio you know side opposite to 30 degrees is 1.
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by [email protected] » Wed Nov 02, 2011 4:55 am
Dear shankar how do we know the sides satisfy the 30 60 90 condition as we only get to know the angle is 90degree

any property involved to identify here that this is 30 60 90 triangle
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by shankar.ashwin » Wed Nov 02, 2011 5:01 am
You know it because the sides are 1,sqrt(3) and 2 (Complete the triangle and calculate the hyp)
We plot point P at (-Sqrt(3),1). THose are values of 'x' and 'y' respectively. Usually when we know a triangle is a 30-60-90, we deduce the sides.
Here the sides are given, inversely we could deduce the angles.
[email protected] wrote:Dear shankar how do we know the sides satisfy the 30 60 90 condition as we only get to know the angle is 90degree

any property involved to identify here that this is 30 60 90 triangle
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