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ritula
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In the figure above, points P 13 Nov 2008, 03:35
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In the figure above, points P 13 Nov 2008, 08:01
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good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)
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In the figure above, points P 13 Nov 2008, 12:06
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prasun84
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)
Show more

it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
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alpha_plus_gamma
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In the figure above, points P 13 Nov 2008, 20:00
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calculate radius of the circle
radius^2= (sqrt(3))^2 + 1^2
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -1/sqrt(3)x [ slope = (1-0)/(-sqrt(3) - 0) and y intercept of the line is 0)

The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
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ritula
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In the figure above, points P 13 Nov 2008, 21:42
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Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
alpha_plus_gamma
calculate radius of the circle
radius^2= (sqrt(3))^2 + 1^2
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -sqrt(3)x [ slope = (-sqrt(3) - 0) / (1-0) and y intercept of the line is 0)
The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
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ritula
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In the figure above, points P 13 Nov 2008, 21:48
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Wow ! wht a shortcut. Applying angle formula never occured to me.
+1 from me
hibloom
prasun84
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)

it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
Show more
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alpha_plus_gamma
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In the figure above, points P 13 Nov 2008, 21:55
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ritula
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
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Yes you are correct. I have edited my post. thanks
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In the figure above, points P 14 Nov 2008, 12:28
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There's a much faster way to do this problem. To find the coordinates of q, all you have to do is switch the x- and y- values of the coordinates of P, then multiply the new y-value (t) by -1.

To understand why, imagine the full circle. Draw any two perpendicular lines through the center. They will intersect the circle at (a, b), (b, -a), (-a, -b), and (-b, a).
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In the figure above, points P 14 Nov 2008, 12:45
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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!