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AdrianG
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Can anybody please explain this to me?



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Fig
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(B) it is :)

The drawn circle has a radius of 2. It represents 2 times the unit circle.

So, we can imagine the 2 points :
> P'(-sqrt(3)/2 , 1/2) the projected point of P on the unit circle.
> Q'(r/2 , s/2) the projected point of Q on the unit circle.

Also, -sqrt(3)/2 = cos(5*Pi/6) and s/2 = cos(5*Pi/6 - pi/2)

Then,
s = 2*cos(5*Pi/6 - 3*pi/6)
<=> s = 2*cos(2*Pi/6)
<=> s = 2*cos(Pi/3)
<=> s = 2*1/2 = 1
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subhen
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sorry Fig I did not understand how you arrived at s=1. I understood how you arrived at the radius but beyond that I could not follow. Besides cos functions are they within scope of GMAT. I am sure there must be a simler explanation on how to solve these types of problem. Would appreciate if you can elaborate.
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From fig (attached)

AO = sqrt(3) and PO = 1. Use special triangles. Sides opposite 30:60:90 are in ratio 1: sqrt(3) :2 (This is tested regularly by GMAT, required to be known)

Thus radius is 2. And angle (POA) = 30

angle(POY) = 90-30 = 60
angle(YOQ) = 90-60 = 30
angle(QOB) = 90-30 = 60

Again using special triangles: 30:60: 90
we know the hypotinuse is 2. The side opposite 60 degree angle (angle(QOB), has to be 1.
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trivikram
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hsampath
From fig (attached)

AO = sqrt(3) and PO = 1. Use special triangles. Sides opposite 30:60:90 are in ratio 1: sqrt(3) :2 (This is tested regularly by GMAT, required to be known)

Thus radius is 2. And angle (POA) = 30

angle(POY) = 90-30 = 60
angle(YOQ) = 90-60 = 30
angle(QOB) = 90-30 = 60

Again using special triangles: 30:60: 90
we know the hypotinuse is 2. The side opposite 60 degree angle (angle(QOB), has to be 1.
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I think in your diagram PA=1 not PO as the radius is 2
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AdrianG
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Thanks for explanation. Now everything is clear :)
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Andr359
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AdrianG
Can anybody please explain this to me?
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A good shortcut for solving this sort of problem is noting that, for each point P(a,b) on a circumference with center O = the origin of axes x and y, the point Q for which PO and QO are perpendicular can be either (-b,a) or (b,-a). This is from a property of vectors, etc, which is not the scope of Q Gmat but interesting and certainly useful.

In this case: (-sqrt(3), 1) => (1, sqrt(3)) OR (-1, -sqrt(3)). Given that Q is on the 1st Quadrant => s = 1.

B.



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