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Value of s - Please help

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by Anurag@Gurome » Wed Feb 22, 2012 4:37 am
rjanardhanan wrote:In the figure shown, point P (-√3, 1) and Q (s, t) lie on the circle with center O.

Image

What is value of s?
Coordinates of O = (0, 0) and that of P are (-√3, 1)
Slope of OP = (1 - 0)/(-√3 - 0) = -1/√3, which implies slope of OQ is √3 since product of slopes of perpendicular lines is -1.
Slope of OQ = (t - 0)/(s - 0) = t/s = √3/1, which clearly implies that t = √3 and [spoiler]s = 1[/spoiler]
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by Lasve » Fri Apr 27, 2012 11:13 am
Image

I Solved it in a different way.
First of all it's obvious that point P and Q creates two triangles with the x and y Axis.
I drew thus triangles and name the new angles respectively a & b.

we know that segment OA is √3 and PA is 1, consequently PO must be 2.
Since PO and QO are both radius of the half circle drew in the picture, they must both be 2.

since that is the classic 1: √3 : 2 right triangle that means that angle pOa must be 30 degree.
Since pOa + pOq + qOb = 180 -> 30 + 90 + qOb =180 -> qOb = 60°

Since It's obvious that qBo = 90

We now know that also riangle QBO is a right 30:60:90 Triangle,and since
qBo = 90° and QO 2
qOb = 60°, THEN Segment QB = √3
oQb = 30, then OB = 1

So the x coordinate of the point Q is s=1
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