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geometry Q

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Senior Manager
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Joined: 07 Feb 2008
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geometry Q [#permalink]

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New post 20 May 2009, 06:40
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A
B
C
D
E

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Please don't forget the explanations. Thanks.
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Re: geometry Q [#permalink]

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New post 20 May 2009, 08:38
First find distance between P and O.

Dist OP=2
Therefor PQ=2 *2^1/2

(2 *2^1/2)^2 = (s+3^1/2)^2 + (t-1)^2

Since t=1, Therefore s=1

Hence B.

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Re: geometry Q [#permalink]

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New post 22 May 2009, 21:51
Can you pls. tell why t is 1?

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Re: geometry Q [#permalink]

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New post 22 May 2009, 22:47
t is NOT equal to 1 . Otherwise distance OQ should be (2)^0.5 and that´s out of the question since OP=OQ=2 .

I guess many people incorrectly assume from the drawing that the Y axis bisects the right angle POQ when it is not really the case.

First solution (The long one):

Draw perpendicular segments to the X axis from points P and Q, call the intersection points of these segments with the X axis P´ and Q´ respectively. You will now have two right-angle triangles POP´ and QOQ´

Applying basic trigonometry: Sin(POP´) = |PP´| / |OP| = 1/2 so

angle POP´=arcsin(1/2) = 30° , since angle POP´ + angle QOQ´=90° then angle QOQ´= 60°

s= |OQ| * cos (angle QOQ´) = 2*cos(60°) = 2*0.5 =1

Second Solution (The short one)= Since |OQ|=|OP| and they form a right angle at the origin, points P(x1,y1) and Q(x2,y2) will have the following symmetry(symmetry of sine and cosine values for complementary angles) : y1 = x2=s = 1 (Our answer! ) and |x1|=y2=t

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Re: geometry Q   [#permalink] 22 May 2009, 22:47
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