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solve without trig [#permalink]
 19 Jul 2008, 16:26
hi, any suggestions on how to use the following question without trig i.e. without using sines or cosines cheers S
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Re: solve without trig [#permalink]
19 Jul 2008, 16:29
draw the vertical lines from point P and point Q to the x axis. solve by using 30-60-90 degrees triangle values.
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Re: solve without trig [#permalink]
19 Jul 2008, 16:32
using 30/60/90 triangle ratios can allow you to solve it,
its 1/2/sqrt3
30 degrees on the left hand side, 60 on the right hand side (because the middle is 90 degrees, and a semi circle has 180)
so the answer is 1
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Re: solve without trig [#permalink]
19 Jul 2008, 17:22
Or notice the lines are perpendicular, so their slopes are negative reciprocals: -from (-root(3), 1) to (0,0), you go right by root(3) and down by 1. -from the origin, on a perpendicular line, to find a point the same distance away as (-root(3), 1), you can go right by 1 and up by root(3). The point is (1, root(3)).
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Re: solve without trig [#permalink]
19 Jul 2008, 17:35
Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.
20 sec solution:
2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).
Ans is D
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Re: solve without trig [#permalink]
19 Jul 2008, 17:43
rao_1857, There are actually quite a few triangles made with this. I see what you're saying that the radius forms a triangle also, so it would be an iscoceles triangle, but I don't believe that is the easiest method for solving this problem. We know point O is the origin, and then the point on the left is (\sqrt{3},1) If you use the x-axis as one side of that triangle, and then go up to the point, you get another triangle. This one is actually a 30:60:90 triangle. The bottom (or x-axis) has a length of sqrt(3) and the other side of the triangle (the hypotnuse is the same as the radius) is 1, so 1^2 + sqrt{3}^2 = 4 so the radius is 2. We don't even really need to know this to solve the problem. Because the middle angle formed by the two radii (?) is a 90 degree angle, we know that the values of the points for x and y will switch. x = sqrt{3} so t on the other side = sqrt{3} and y = 1, so s =1 on the other side. Actually a very simple problem and no trig needed.
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Re: solve without trig [#permalink]
19 Jul 2008, 18:10
rao_1857 wrote: Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.
20 sec solution:
2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).
Ans is D while you're right that the middle triangle is an isoceles triangle, your conclusion is flawed - the angles are not the same on either side of the middle isoceles triangle and so the distance does not remain the same. Your reasoning would only be correct if the angle on either side of the middle triangle is equal - which is not the case here. So the answer is actually 1 - choice B, and not sqrt 3 - choice D
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Re: solve without trig [#permalink]
19 Jul 2008, 19:45
jallenmorris wrote: rao_1857,
There are actually quite a few triangles made with this. I see what you're saying that the radius forms a triangle also, so it would be an iscoceles triangle, but I don't believe that is the easiest method for solving this problem.
We know point O is the origin, and then the point on the left is (\sqrt{3},1) If you use the x-axis as one side of that triangle, and then go up to the point, you get another triangle. This one is actually a 30:60:90 triangle. The bottom (or x-axis) has a length of sqrt(3) and the other side of the triangle (the hypotnuse is the same as the radius) is 1, so 1^2 + sqrt{3}^2 = 4 so the radius is 2.
We don't even really need to know this to solve the problem. Because the middle angle formed by the two radii (?) is a 90 degree angle, we know that the values of the points for x and y will switch. x = sqrt{3} so t on the other side = sqrt{3} and y = 1, so s =1 on the other side. Actually a very simple problem and no trig needed. Thats right and I see your point but why do we need to consider that triangle if we can easily solve it using the isoceles traigle with 2 radii. I am little confused now....
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Re: solve without trig [#permalink]
19 Jul 2008, 19:48
jasonc wrote: rao_1857 wrote: Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.
20 sec solution:
2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).
Ans is D while you're right that the middle triangle is an isoceles triangle, your conclusion is flawed - the angles are not the same on either side of the middle isoceles triangle and so the distance does not remain the same. Your reasoning would only be correct if the angle on either side of the middle triangle is equal - which is not the case here. So the answer is actually 1 - choice B, and not sqrt 3 - choice DPardon my ignorance. May be I am looking from diff view. But you are that angles are not same on the isocelec traingle..... by definition of isoceles triangle this is not true.
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Re: solve without trig [#permalink]
19 Jul 2008, 22:57
rao_1857 wrote: Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.
20 sec solution:
2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).
Ans is D If Q is any other point on the circle, you'll get an isosceles triangle. If Q is (0,2), or (2,0), or (root(2), root(2)), the distances OP and OQ will be equal. If you aren't somehow using the right angle in the diagram, you aren't doing the problem correctly. The answer is 1, as explained above.
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Re: solve without trig [#permalink]
19 Jul 2008, 23:16
you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer sqrt(3)the 30-60-90 traingle mentioned by JasonC are actually these two traingles. the answer is 1. Attachment: 
circle.JPG [ 9.22 KiB | Viewed 489 times ]
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Re: solve without trig [#permalink]
19 Jul 2008, 23:40
Nice diagram, durgesh, and a good illustration of what's really happening here (but your 30 degree angle is bigger than the 60 degree angle!  ) The angle on the left should be 30, and on the right should be 60, with the given co-ordinates for P and Q.
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Re: solve without trig [#permalink]
19 Jul 2008, 23:42
durgesh79 wrote: you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer sqrt(3)the 30-60-90 traingle mentioned by JasonC are actually these two traingles. the answer is 1. Attachment: circle.JPG nice one, but 60 should be 30 and 30 should be 60.
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Re: solve without trig [#permalink]
19 Jul 2008, 23:48
GMAT TIGER wrote: durgesh79 wrote: you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer sqrt(3)the 30-60-90 traingle mentioned by JasonC are actually these two traingles. the answer is 1. Attachment: circle.JPG nice one, but 60 should be 30 and 30 should be 60. oops 
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Re: solve without trig [#permalink]
19 Jul 2008, 23:58
IanStewart wrote: Nice diagram, durgesh, and a good illustration of what's really happening here (but your 30 degree angle is bigger than the 60 degree angle! ) The angle on the left should be 30, and on the right should be 60, with the given co-ordinates for P and Q. Thanks Ian... it was a typo ... i think I now can distinguish between 60 and 30 angles....  so no need for me to go back to highschool geomatry books ....
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Re: solve without trig [#permalink]
20 Jul 2008, 06:23
Quote: Thats right and I see your point but why do we need to consider that triangle if we can easily solve it using the isoceles traigle with 2 radii. I am little confused now.... Because the isoceles triangle with 2 radii is "tilted" - its hypotenuse is not parallel to the x-axis. If, instead, P was (-2/sqrt(2),2/sqrt(2)) then Q was (2/sqrt(2),2/sqrt(2))
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Re: solve without trig [#permalink]
20 Jul 2008, 07:06
jallenmorris wrote: rao_1857,
There are actually quite a few triangles made with this. I see what you're saying that the radius forms a triangle also, so it would be an iscoceles triangle, but I don't believe that is the easiest method for solving this problem.
We know point O is the origin, and then the point on the left is (\sqrt{3},1) If you use the x-axis as one side of that triangle, and then go up to the point, you get another triangle. This one is actually a 30:60:90 triangle. The bottom (or x-axis) has a length of sqrt(3) and the other side of the triangle (the hypotnuse is the same as the radius) is 1, so 1^2 + sqrt{3}^2 = 4 so the radius is 2.
We don't even really need to know this to solve the problem. Because the middle angle formed by the two radii (?) is a 90 degree angle, we know that the values of the points for x and y will switch. x = sqrt{3} so t on the other side = sqrt{3} and y = 1, so s =1 on the other side. Actually a very simple problem and no trig needed. how do you know "the values of the points for x and y will switch"? by trying values, i can see this, but is there a theorom for this?
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Re: solve without trig [#permalink]
20 Jul 2008, 07:08
so i guess i have to remember my cosine and sine ratios for 30, 45 and 60?
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Re: solve without trig [#permalink]
20 Jul 2008, 07:16
sset009 wrote: so i guess i have to remember my cosine and sine ratios for 30, 45 and 60? if you are talking abt this question, i dont think so.... Slope of line OP = (1-0)/(-sqrt3 - 0) = - 1/sqrt3slope of line OQ = t/s both lines are perpendicular = (-1/sqrt3) * (t/s) = -1 t = sqrt3 * s -------- (1) both P and Q are points on circle t^2 + s^2 = 1^2 + (sqrt3)^2 = 4 --------(2) Solve for s and you'll get s = 1
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Re: solve without trig [#permalink]
20 Jul 2008, 11:18
durgesh79 wrote: you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer sqrt(3)the 30-60-90 traingle mentioned by JasonC are actually these two traingles. the answer is 1. Attachment: circle.JPG aah ... I see what you are saying. the two triangles are only sharing the hypotenus. the bases are exchanged. Thnaks ... PS:- GOD! when am I gonna stop doing these stupid mistakes.
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Re: solve without trig
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20 Jul 2008, 11:18
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