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Please help me on this question: Geometry

by desihokie » Sun Jan 13, 2008 11:20 pm
Got this question on powerprep software. What is the right answer? and why?






Also...I just found this forum looking for some math fundamentals on i'net and i wish i found this earlier. My test is on next saturday, jan 19th at 8 AM and i am a bit stressed since i really need to score >=700 on this. I took two of the practice tests from GMAT official CD and got 640 and 650. Did worse than expected in quant section in both tests. Any advice?
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by simplyjat » Mon Jan 14, 2008 12:55 am
This is a trigonometry question, You have to find the angle PO makes with negative X axis.

sqrt (3) / 1 = tan 60, thus the angle is 60 degrees. Thus the angle OQ with respect to positive X axis is 30 degrees. And thus you come to
sin 60 = sqrt(3)/2, so radius of the circle is 2

s/t = tan 30 = 1/2, length of OQ = 2 ( radius of the circle ) , and using sin 30 you can find out the value of X...

sin 30 = 1/2 = s/r = s/2 => s = 1
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by camitava » Mon Jan 14, 2008 1:03 am
desihokie,
This Qs is already been discussed in this forum. I forgot the link! :cry: But refer the approach below (in the attached document) -
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by sibbineni » Mon Jan 14, 2008 3:50 am
In the fig. side OP = OQ=2...

since 2 sides are equal it is an isoceles triangle
then angle OPQ = OQP = 45 degree ...

since we know length of 2 sides then length of PQ=?

by pythagaros theorem

PQ^2=OP^2+OQ^2
=4+4
=8
PQ=2sqrt(2)

there fore length of PQ=2sqrt(2)

Formula:
Distance between 2 points (x1,y1) and (x2,y2) is sqrt (x2-x1)^2+(y2-y1)^2....

then calculate the distance between O and Q

since O is origin theno(0,0)
Q(s,t)

after calculating the distance then
s^2+t^2=4----(1)

then the distance of P and Q

s^2+t^2+t+4+s sqrt(3)=8----(2)

substiting the 1 value in 2 we have

4+t+4+s sqrt(3)=8
t=- s sqrt(3) ---- (3)

substiting the value of 3 in 1 we have
s^2+3 s^2=4
4s^2=4
s=1

and substitute the value s=1 in eq 1 we have

t=sqrt(3)
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by Stuart@KaplanGMAT » Mon Jan 14, 2008 5:11 am
Let's solve this with reasoning, instead of really complicated math (you definitely don't need to know any trigonometry for test day).

Points P and Q lie on the same circle, which has a centre O. O lies on the origin (i.e. at 0,0).

Since points P and Q are 90 degrees apart, they lie on perpendicular lines. In other words, the slope of line OQ will have the inverse negative slope of the line OP.

Since point O is on the origin, it's very easy to calculate the slope of OP. Rise/Run will simply be 1/-(root3). Therefore, the slope of OQ will be (root 3)/1.

So, to get to point Q, we need to go root3 up and 1 across. Therefore, the x coordinate of point Q is 1 (since we start at (0,0)).

As a general rule, any 2 points separated by 90 degrees on a circle whose centre is the origin will have reversed coordinates (determine sign by the quadrant in which you find the point). So, since P is at (-root3, 1), Q will be at (1, root3).
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by StarDust845 » Mon Jan 14, 2008 8:40 am
Stuart Kovinsky wrote:Let's solve this with reasoning, instead of really complicated math (you definitely don't need to know any trigonometry for test day).

Points P and Q lie on the same circle, which has a centre O. O lies on the origin (i.e. at 0,0).

Since points P and Q are 90 degrees apart, they lie on perpendicular lines. In other words, the slope of line OQ will have the inverse negative slope of the line OP.

Since point O is on the origin, it's very easy to calculate the slope of OP. Rise/Run will simply be 1/-(root3). Therefore, the slope of OQ will be (root 3)/1.

So, to get to point Q, we need to go root3 up and 1 across. Therefore, the x coordinate of point Q is 1 (since we start at (0,0)).

As a general rule, any 2 points separated by 90 degrees on a circle whose centre is the origin will have reversed coordinates (determine sign by the quadrant in which you find the point). So, since P is at (-root3, 1), Q will be at (1, root3).

" Therefore, the slope of OQ will be (root 3)/1. So, to get to point Q, we need to go root3 up and 1 across."

Just because the slope of the line is sqrt(3), doesn't necessarily lead to the conclusion that the point is 1 unit across and root3 up. What if I go 2 units across and 2 * sqrt(3) up? The slope is still root3. I still have to calculate the distance of the point from origin and make sure that is same as that of the earlier point. Correct me if I am wrong.

Thanks,
Calista.
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by Stuart@KaplanGMAT » Mon Jan 14, 2008 9:39 am
StarDust845 wrote:

" Therefore, the slope of OQ will be (root 3)/1. So, to get to point Q, we need to go root3 up and 1 across."

Just because the slope of the line is sqrt(3), doesn't necessarily lead to the conclusion that the point is 1 unit across and root3 up. What if I go 2 units across and 2 * sqrt(3) up? The slope is still root3. I still have to calculate the distance of the point from origin and make sure that is same as that of the earlier point. Correct me if I am wrong.

Thanks,
Calista.
Since the two points are on the same circle (with its centre at the origin), they'll be equidistant from the origin.

Therefore, if point P to the origin is 1 down and root3 across, to get to point Q we go root3 up and 1 across.

If point P to the origin had been 2 down and 2root3 across, the origin to point Q would have been 2root3 up and 2 across.
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by StarDust845 » Mon Jan 14, 2008 10:00 am
Okay, I see it now. Thanks Stuart.
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by rb90 » Thu Oct 21, 2010 10:05 am
Hi stuart,
I still didnt understand. How could srt3 across and 1 up in quadrant 2 , lead to 1 across and sqrt 3 up. That means the line PO is coinciding with the x-axis.
Please help. Im really confused.
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by N:Dure » Thu Oct 21, 2010 3:17 pm
Thank you Stuart for your reply! I thought in the same way (i.e something is opposite) but i only did this in the signs i.e it must be root 3 on the right side instead of -root 3, but apparently I was wrong.
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by pesfunk » Sun Oct 31, 2010 3:24 am
This explanation was really nice stuart.

Agree...we dont need trigonometric formulas to solve this...thanks
Stuart Kovinsky wrote:Let's solve this with reasoning, instead of really complicated math (you definitely don't need to know any trigonometry for test day).

Points P and Q lie on the same circle, which has a centre O. O lies on the origin (i.e. at 0,0).

Since points P and Q are 90 degrees apart, they lie on perpendicular lines. In other words, the slope of line OQ will have the inverse negative slope of the line OP.

Since point O is on the origin, it's very easy to calculate the slope of OP. Rise/Run will simply be 1/-(root3). Therefore, the slope of OQ will be (root 3)/1.

So, to get to point Q, we need to go root3 up and 1 across. Therefore, the x coordinate of point Q is 1 (since we start at (0,0)).

As a general rule, any 2 points separated by 90 degrees on a circle whose centre is the origin will have reversed coordinates (determine sign by the quadrant in which you find the point). So, since P is at (-root3, 1), Q will be at (1, root3).
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