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30 Jun 2007, 20:23
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Please explain.
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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30 Jun 2007, 21:42
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Slope of line PO: -(1/sqrt(3))
Slope of line SO: t/s
Also, product of slopes of perpendicular lines is always -1.
Hence, -(1/sqrt(3)) * (t/s) = -1
Thus, t = s * sqrt(3) ...I
Also, PO=SO (radii)
s^2 + t^2 = (sqrt(3))^2 + 1 = 4 ...II
From I and II,
s^2 + 3s^2 = 4
4*s^2=4
s^2=1 => s = 1 (s cannot be -1 in the first quadrant).
Hence, B.
But I hope there is some easier way of doing this.
Slope of line SO: t/s
Also, product of slopes of perpendicular lines is always -1.
Hence, -(1/sqrt(3)) * (t/s) = -1
Thus, t = s * sqrt(3) ...I
Also, PO=SO (radii)
s^2 + t^2 = (sqrt(3))^2 + 1 = 4 ...II
From I and II,
s^2 + 3s^2 = 4
4*s^2=4
s^2=1 => s = 1 (s cannot be -1 in the first quadrant).
Hence, B.
But I hope there is some easier way of doing this.
01 Jul 2007, 20:54
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sumande
Nice, I have no idea how I could have figured that out in 2 mins. :wink:
I guess if you know what you are doing from the start, you could do that quickly in 2 mins.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
