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# Coordinate Geometry Problem

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Coordinate Geometry Problem [#permalink]  22 Oct 2008, 16:38
Can someone help me out on this question.
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Re: Coordinate Geometry Problem [#permalink]  22 Oct 2008, 17:20
Since OP = OQ = radius

s^2 + t^2 = 4

Since OP and OQ are perpendicular

Slopes of OP* slope of OQ = -1
t/s = -sqrt(3)

4s^2 = 4
s = 1
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Re: Coordinate Geometry Problem [#permalink]  22 Oct 2008, 18:04
LiveStronger wrote:
Since OP = OQ = radius

s^2 + t^2 = 4

Since OP and OQ are perpendicular

Slopes of OP* slope of OQ = -1
t/s = -sqrt(3)

4s^2 = 4
s = 1

good solution...btw t/s = +sqrt(3)....since (t/s)*(1/-sqrt 3) = -1

i solved by trigo
you get the angles for the triangle on left and thus you get angles of the triangle on right and thus the sides
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Re: Coordinate Geometry Problem [#permalink]  23 Oct 2008, 04:00
LiveStronger wrote:
Since OP = OQ = radius

s^2 + t^2 = 4

Since OP and OQ are perpendicular

Slopes of OP* slope of OQ = -1
t/s = -sqrt(3)

4s^2 = 4
s = 1

Good one

except that t/s = SQRT(3) as pointed by kandyhot
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Re: Coordinate Geometry Problem [#permalink]  25 Oct 2008, 06:54
problem requires no calculation...just know your right triangles and sides (30,60,90 here)
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Re: Coordinate Geometry Problem [#permalink]  25 Oct 2008, 07:01
ChristopherKu wrote:
problem requires no calculation...just know your right triangles and sides (30,60,90 here)

I think its 45-45-90 here ..... The two sides of the triangle are of same length ( radii of the circle)

An isosceles right angle triangle has angles 45-45-90
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Re: Coordinate Geometry Problem [#permalink]  25 Oct 2008, 07:06
sorry...should've been clear

1:sqrt3:2 = 30:60:90

1) gives you the angle between the the radius and X
2) you know the interior angle is 90
3) gives you the last angle and you apply the 30:60:90 sides again accordingly
Re: Coordinate Geometry Problem   [#permalink] 25 Oct 2008, 07:06
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