GMAT PREP GEOMETRY QUES???

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 446
Joined: Thu Jul 26, 2007 1:07 pm
Thanked: 6 times

GMAT PREP GEOMETRY QUES???

by dferm » Mon Mar 10, 2008 12:57 pm
PLEASE HELP
Attachments
Doc45.doc
(58 KiB) Downloaded 123 times

User avatar
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Sun Mar 02, 2008 6:23 pm
Location: NY

by 2008nv » Mon Mar 10, 2008 1:46 pm
***
Last edited by 2008nv on Tue Mar 11, 2008 8:50 am, edited 1 time in total.

Junior | Next Rank: 30 Posts
Posts: 17
Joined: Wed Feb 27, 2008 9:32 pm
Location: Mauritius
Thanked: 2 times

by mnjoosub » Tue Mar 11, 2008 6:41 am
My ans is B, i.e 1 :wink:

Junior | Next Rank: 30 Posts
Posts: 17
Joined: Wed Feb 27, 2008 9:32 pm
Location: Mauritius
Thanked: 2 times

by mnjoosub » Tue Mar 11, 2008 6:55 am
More detailed explanation.

From properties of triangle : r=2, and angle PO with horizontal = 30 deg.
Therefore angle QO with horizontal = 60 deg.

=> from properties of triangle once more s=1 and t=root(3).

That' it.

:wink: :wink:

Master | Next Rank: 500 Posts
Posts: 446
Joined: Thu Jul 26, 2007 1:07 pm
Thanked: 6 times

GMAT QUES GEOMETRY

by dferm » Tue Mar 11, 2008 7:40 am
STILL HAVING TROUBLE UNDERSTANDING THIS QUEST. I KNOW THE PROPERTIES OF A TRIANGLE BUT IT STILL DOESN'T MAKE SENSE.

Junior | Next Rank: 30 Posts
Posts: 17
Joined: Wed Feb 27, 2008 9:32 pm
Location: Mauritius
Thanked: 2 times

by mnjoosub » Tue Mar 11, 2008 7:49 am
What is the OA?

Senior | Next Rank: 100 Posts
Posts: 72
Joined: Mon Mar 10, 2008 10:29 am
Thanked: 25 times

by tmmyc » Tue Mar 11, 2008 7:51 am
This was the solution I posted on the manhattangmat forums.
tmmyc wrote:First, see that after dropping perpendicular lines, we have two right triangles.

Image

Detailed Explanation:
Let's begin with the triangle on the left.

We know the sides are 1 and (sqrt 3) from point P.
If you know your special right triangles, you will quickly see that this is a 30-60-90 right triangle.

The angle opposite '1' is 30 degrees.



Let's move on to the triangle on the right.

We know that a straight line has 180 degrees.

Since we know the lower angle of the triangle on the left is 30 degrees, and we also know the angle between the two line segments is 90 degrees, the lower angle of the triangle on the right must be 60 degrees in order to sum to 180 degrees. (30 + 90 + x = 180; x = 60)

This means the triangle on the right is also a 30-60-90 triangle. The hypotenuse of this triangle is the same as the other triangle's (which is '2' by the Pythagorean Theorem), since both are radii of the same circle.

Using the same properties of a 30-60-90 triangle, you can find the side lengths and finally the point (s,t) which gives the value for s.

Master | Next Rank: 500 Posts
Posts: 446
Joined: Thu Jul 26, 2007 1:07 pm
Thanked: 6 times

GMAT GEOMETRY QUES

by dferm » Tue Mar 11, 2008 9:19 am
THANKS .

Senior | Next Rank: 100 Posts
Posts: 62
Joined: Wed Mar 05, 2008 8:57 am
Location: mumbai, India
Thanked: 5 times

by ikant » Tue Mar 11, 2008 9:23 am
I dunno but I have a doubt... what led u people to consider O as the origin.

The question can be solved without making that assumption.

Since the angle changed is 90, we need to change the x component to keep the other point on the circle.

Done.
"To do is to be"