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Source: — Problem Solving |
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by vinay1983 » Sat Oct 05, 2013 5:08 pm
Java_85 wrote:Image
My guess D
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by ducefun » Sat Oct 05, 2013 5:14 pm
radius^2 = sqrt(3)^2 + 1^2
radius = 2

since the angle PO(-x) is 30 deg (very common triangle)
=> angle SOx is 60 degree.
s = radius*cos(60) = 2*1/2 = 1
Choose B.
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by rakeshd347 » Sat Oct 05, 2013 5:15 pm
Java_85 wrote:Image
It should be C

Whats the OA.
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by Brent@GMATPrepNow » Sat Oct 05, 2013 5:39 pm
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So, s = 1
Answer: B

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by [email protected] » Sat Oct 05, 2013 5:46 pm
Hi Java_85,

Many graphing questions can be made easier to solve IF you use the diagonal lines you're given to draw RIGHT TRIANGLES. In this question, try drawing a line from P down to the X-axis and from Q down to the X-axis. Now you will have two right triangles to work with.

You can figure out the radius from the left-most right triangle, then use the angles to figure out the right-most triangle.

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by Java_85 » Sat Oct 05, 2013 9:12 pm
Thank you all for explanations, as some of you said the OA is B.
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by ganeshrkamath » Sat Oct 05, 2013 11:54 pm
Java_85 wrote:Image
product of slopes of the 2 lines = -1
(1-0)/(-√3 - 0)* (t/s) = -1
-1/√3 * t/s = -1
t/s = √3

PO = √((-√3)^2 + (1)^2) = √(3+1) = 2
QO = √(s^2 + t^2) = 2
(s^2 + t^2) = 4
s^2 + 3s^2 = 4
4s^2 = 4
s^2 = 1
s = 1

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