GMAT Prep XY Coordinate with Half Circle

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dnkcdnguy: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Sat Sep 06, 2008 12:27 pm

by dnkcdnguy » Tue Feb 17, 2009 4:57 pm

This one took me awhile and I kind of guessed that it was "1".

I'm interested to see how others solve this quesiton.

Thanks

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Source: Beat The GMAT — Problem Solving | Tue Feb 17, 2009 4:57 pm

griscomtestprep: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Sat Feb 07, 2009 6:14 pmFollowed by: 1 member; Location: Las Vegas, NV; Thanked: 7 times; GMAT Score:760

Re: GMAT Prep XY Coordinate with Half Circle

by griscomtestprep » Wed Feb 18, 2009 12:44 am

dnkcdnguy wrote:This one took me awhile and I kind of guessed that it was "1".

I'm interested to see how others solve this quesiton.

Thanks

Draw vertical line from P to X axis. Call that point A.
Draw vertical line from Q to X axis. Call that point B.
PAO is a 30-60-90 triangle.
OP=2
OQ=2
<POA=30
<QOB=60
QOB is 30-60-90 triangle.
OB=1

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valdemas: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jan 10, 2009 12:44 pm

by valdemas » Wed Feb 18, 2009 11:27 am

the answer is D
if you drop the line from point Q the value is the square rout of 3

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x2suresh: Master | Next Rank: 500 Posts; Posts: 258; Joined: Thu Aug 07, 2008 5:32 am; Thanked: 16 times

Re: GMAT Prep XY Coordinate with Half Circle

by x2suresh » Wed Feb 18, 2009 12:04 pm

griscomtestprep wrote:
dnkcdnguy wrote:This one took me awhile and I kind of guessed that it was "1".

I'm interested to see how others solve this quesiton.

Thanks
Draw vertical line from P to X axis. Call that point A.
Draw vertical line from Q to X axis. Call that point B.
PAO is a 30-60-90 triangle.
OP=2
OQ=2
<POA=30
<QOB=60
QOB is 30-60-90 triangle.
OB=1

good approach..!! without solving.. we can answer the question

I did it in another approach

slope of the line OP = 1-0/-SQRT(3)-0 = - 1/SQRT(3)

slope of the line OQ, which is perpendicular to OP = sqrt(3) = t-0/s-0 = t/s
radius=op =
{(-sqrt(3))}^{2} + 1^2=4 =OQ= 4=
{s*sqrt(3)}^2+ s^2
--> s=1

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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Fri Apr 10, 2009 1:22 pm

<POA=30
<QOB=60
QOB is 30-60-90 triangle.

How did you figure out that they were 30 60 90 triangles? Couldn't they be 45-45-90 triangles as well?

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pakaskwa: Master | Next Rank: 500 Posts; Posts: 107; Joined: Wed Mar 04, 2009 4:39 am; Location: Vancouver; Thanked: 9 times; GMAT Score:750

by pakaskwa » Fri Apr 10, 2009 5:07 pm

For any right-angled triangle,
- if 3 internal angles are 30-60-90, the ratio of triangle's 3 sides is 1:sqrt3:2; and vice versa
- if 3 internal angles are 45-45-90, the ratio of triangle's 3 sides is 1:1:sqrt2; and vice versa

In the original question, if you draw a vertical like from P to axis X, and intersect X at point A, then triangle PAO is 30-60-90.

Since it's given that angle POQ is a right angle, if you draw a vertical like from Q to axis X and interset at point B, QBO is 30-60-90 triangle.

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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Fri Apr 10, 2009 9:10 pm

pakaskwa wrote:For any right-angled triangle,
- if 3 internal angles are 30-60-90, the ratio of triangle's 3 sides is 1:sqrt3:2; and vice versa
- if 3 internal angles are 45-45-90, the ratio of triangle's 3 sides is 1:1:sqrt2; and vice versa

In the original question, if you draw a vertical like from P to axis X, and intersect X at point A, then triangle PAO is 30-60-90.

Since it's given that angle POQ is a right angle, if you draw a vertical like from Q to axis X and interset at point B, QBO is 30-60-90 triangle.

You have not answered his question. Why can Q B O not be 45 is his question. The only facts you know according the problem is that the other two angles along the x axis sum to 90 and the two radii are each 2. How do these facts force QBO to be 60?.

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de_sandip: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Wed Mar 25, 2009 10:02 am

by de_sandip » Tue Apr 14, 2009 2:34 am

How can you determine (or rather assume) that the triangle you are referring to, is a 30-60-90 trinagle?

According to me , the question does not provide enough data to support the arguement that whether the triangle(as referred earlier) will be 30-60-90 or 45-90-45.

So you need to make an assumption here.

From the figure, one can see that PO and QO are equially distanced from Y axis as the Y axis bisects the 90 degree angle symbol.

So, I assume that s = negative of (-sqrt(3)), i.e. sqrt(3).

Therefore, D seems to be the right answer.

Whats the OA?

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sacx: Master | Next Rank: 500 Posts; Posts: 139; Joined: Wed Oct 22, 2008 4:36 am; Thanked: 17 times

by sacx » Tue Apr 14, 2009 4:45 am

https://www.beatthegmat.com/what-is-the- ... 27181.html

SACX

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andes1: Senior | Next Rank: 100 Posts; Posts: 48; Joined: Sun Nov 25, 2007 3:27 pm; Location: Colombia Southamerica; Thanked: 1 times

by andes1 » Tue Apr 14, 2009 5:36 pm

The small square in the angle means right angle… meaning 90 degrees.

So only use logic.

Imaging a point (0,1) the radius is 1 (circle)
Add the fact that is a 90 degrees figure

The other point will be the opposite (1,0)

In this case the initial point is – sqr3 , 1 so the opposite is 1, sqr3

Answer 1

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dumb.doofus: Master | Next Rank: 500 Posts; Posts: 435; Joined: Sat Sep 27, 2008 2:02 pmFollowed by: 1 member; Location: San Jose, CA; Thanked: 43 times; GMAT Score:720

by dumb.doofus » Tue Apr 14, 2009 10:30 pm

Explanation for why it isnt 45 45 90 triangle
I am not sure, if it is ok to give an explanation wrt trigonometry as to how much it is used in GMAT land. But it is the simplest thing to put here.

In triangle PAO, where A is the point on x-axis directly below P, and just for the sake of finding the angle, let's forget that we are in second quadrant. So let's take point P as {root(3), 1}
OP = 2
PA = 1
AO = root(3)

Sin (x) = PA/OP = 1/2
We know that Sin(30) is 1/2. That implies Angle POA is 30 degrees.

Hope this helps.

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