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# Coordinate system

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Intern
Joined: 25 Dec 2008
Posts: 18
Schools: HBS, Stanford

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12 Apr 2009, 09:00
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Please explain - OA is B >> 1

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Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson

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12 Apr 2009, 21:08

This is how:

Draw a perpendicular line from the point P onto X-axis. Lets call the point of intersection as A. Now we have a 30-60-90 Right Triangle POA. PA=1. AO= . PO = 2 = QO. (OP = OQ = radius of circle).

This is 1:2: triangle. Angle POA=30 degrees.

Draw a perpendicular line from the point Q onto X-axis. Lets call that point of intersection as B. QBO is a 30-60-90 Right Triangle as well.

Extend the line PO and QO. Using Vertical Angles theorem (Opposite angles), you know Angle QOB = 60.

We know the ratio of a 30-60-90 Right Triangle to be 1:2:.
We know QO = 2. Thus, OB = 1 = s.

s=1.

Although it seems complicated, once you extend the line segments PO and QO and find out all the angles, the problem becomes simple.
Remembering the ratio of the sides of a 30-60-90 Right Triangle and 45-45-90 Right Triangle is very handy for GMAT.

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Intern
Joined: 19 Oct 2008
Posts: 21

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13 Apr 2009, 12:03
That was perfect explanation. Thank you. We can also make use of Sin and Cos formulae but it makes the process more time consuming and complex

goldeneagle94 wrote:

This is how:

Draw a perpendicular line from the point P onto X-axis. Lets call the point of intersection as A. Now we have a 30-60-90 Right Triangle POA. PA=1. AO= . PO = 2 = QO. (OP = OQ = radius of circle).

This is 1:2: triangle. Angle POA=30 degrees.

Draw a perpendicular line from the point Q onto X-axis. Lets call that point of intersection as B. QBO is a 30-60-90 Right Triangle as well.

Extend the line PO and QO. Using Vertical Angles theorem (Opposite angles), you know Angle QOB = 60.

We know the ratio of a 30-60-90 Right Triangle to be 1:2:.
We know QO = 2. Thus, OB = 1 = s.

s=1.

Although it seems complicated, once you extend the line segments PO and QO and find out all the angles, the problem becomes simple.
Remembering the ratio of the sides of a 30-60-90 Right Triangle and 45-45-90 Right Triangle is very handy for GMAT.
Manager
Joined: 19 Aug 2006
Posts: 232

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13 Apr 2009, 18:41
goldeneagle94 wrote:

This is how:

Draw a perpendicular line from the point P onto X-axis. Lets call the point of intersection as A. Now we have a 30-60-90 Right Triangle POA. PA=1. AO= . PO = 2 = QO. (OP = OQ = radius of circle).

This is 1:2: triangle. Angle POA=30 degrees.

Draw a perpendicular line from the point Q onto X-axis. Lets call that point of intersection as B. QBO is a 30-60-90 Right Triangle as well.

Extend the line PO and QO. Using Vertical Angles theorem (Opposite angles), you know Angle QOB = 60.

We know the ratio of a 30-60-90 Right Triangle to be 1:2:.
We know QO = 2. Thus, OB = 1 = s.

s=1.

Although it seems complicated, once you extend the line segments PO and QO and find out all the angles, the problem becomes simple.
Remembering the ratio of the sides of a 30-60-90 Right Triangle and 45-45-90 Right Triangle is very handy for GMAT.

Why is POA in this explanation a 30-60-90 triangle? Could anyone pls explain?
Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson

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13 Apr 2009, 19:33
The ratio of the sides of a 30-60-90 triangle is always 1:2:

Intern
Joined: 29 Dec 2006
Posts: 31

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15 Apr 2009, 00:37
how did you know that this was a 30-60-90 triangle vs 45-45-90?
Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson

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15 Apr 2009, 06:19
Co-ordinates of point P are (-,1).
Now you know that, The Right Triangle PAO has a height of 1 and a base of .

Whenever you see sides in such a ratio in a right triangle, it is a 30-60-90 triangle.
Manager
Joined: 19 Aug 2006
Posts: 232

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15 Apr 2009, 10:20
goldeneagle94 wrote:
Co-ordinates of point P are (-,1).
Now you know that, The Right Triangle PAO has a height of 1 and a base of .

Whenever you see sides in such a ratio in a right triangle, it is a 30-60-90 triangle.

Excellent, the light bulb turned on. I'm getting more and more comfortable with 30-60-90 triangle. Thanks.
Intern
Joined: 29 Dec 2006
Posts: 31

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15 Apr 2009, 15:19
it all finally clicks! thanks!!

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Re: Coordinate system   [#permalink] 15 Apr 2009, 15:19
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