BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT Prep XY Coordinate with Half Circle

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Sat Feb 07, 2009 6:14 pm
Location: Las Vegas, NV
Thanked: 7 times
Followed by:1 members
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
Free Hour of Live Online GMAT Training
Griscom Test Prep
www.griscomtestprep.net
[email protected]
702-406-7688
Top
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
Top
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
Top
User avatar
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?
Top
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.
Top
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?.
Top
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?
Top
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
Top
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
LEARNING ENGLIS H
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Sat Sep 27, 2008 2:02 pm
Location: San Jose, CA
Thanked: 43 times
Followed by:1 members
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.
One love, one blood, one life. You got to do what you should.
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
Top
Post new topic Post Reply
• Page 1 of 1