Co-ordinate - gmatprep

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 127
Joined: Tue Oct 02, 2007 9:35 am
Thanked: 1 times

Co-ordinate - gmatprep

by arocks » Fri Oct 12, 2007 8:49 am
Please explain. Thanks.

OA - B
Attachments
qu 10.JPG

Senior | Next Rank: 100 Posts
Posts: 50
Joined: Thu Oct 04, 2007 12:26 pm
Location: RTP, NC
Thanked: 26 times
Followed by:3 members

by achandwa » Fri Oct 12, 2007 11:32 am
Say A is a point of intersection of a perpendicular to X axis from point P.
Therefore, triangle PAO is right triangle.
Apply Pythagorean theorem: PA^2 + AO^2 = PO^2
That is, 1^2+Root3^2= PO^2
PO^2 = 4 => PO = 2

Alternatively you should have realized the ratio of the sides PA:AO = 1:Root 3 and realized it was a 30:60:90 triangle right away.
So measure of angle POA is 30 degrees.
Therefore angle QOx is 180-(angle POQ + angle POA) = 180-90-30=60 degrees.
Say B is a point of intersection of a perpendicular to X axis from point Q.
Therefore, triangle QOB is 30:60:90 triangle with angle OQB = 30 degrees.
Length of the smallest side of a 30:60:90 triangle is half that of the largest side. Thus s = 1.