Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
05 Oct 2008, 03:37
Kudos
Bookmarks
PFA
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 below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
05 Oct 2008, 08:25
Kudos
Bookmarks
Start with the unit circle, a circle of radius 1 centered around the origin.
\((-sqrt3,1)\) is not on the unit circle, but \(((-sqrt3)/2,1/2)\) is.
This leads me to believe that point P lies on the circle with radius 2 at 150 degrees.
150 - 90 = 60
2*(cos 60) = 1
My answer is B.
\((-sqrt3,1)\) is not on the unit circle, but \(((-sqrt3)/2,1/2)\) is.
This leads me to believe that point P lies on the circle with radius 2 at 150 degrees.
150 - 90 = 60
2*(cos 60) = 1
My answer is B.
05 Oct 2008, 12:45
Kudos
Bookmarks
Since OP and OQ are perpendicular to each other, multiplication of their slopes will be -1.
Or. (t/s)(1/-sqrt3) = -1 or t = (sqrt3)s.
Also, equation of circle with radius 2 is x^2 + y^2 = 4 and since, (s,t) lies on circle, hence s^2 + t^2 = 4 or s^2 + 3s^2 = 4 or s = 1 (as -1 is not the answer choice
)
Or. (t/s)(1/-sqrt3) = -1 or t = (sqrt3)s.
Also, equation of circle with radius 2 is x^2 + y^2 = 4 and since, (s,t) lies on circle, hence s^2 + t^2 = 4 or s^2 + 3s^2 = 4 or s = 1 (as -1 is not the answer choice
) 
