GMAT Prep / Geometry

[Blast]

xxx

[lunarpower]

here's an awesome way to approach this.

see that 90 degree angle there? ok. that means that this question is really asking where the point (-√3, 1) would go if the paper were rotated clockwise by 90 degrees.

so...

draw that point on your paper, and then physically rotate the paper by 90 degrees.

originally the x coordinate was negative √3 (to the left). when you rotate the paper, this is now upward, so it's positive √3 in the y direction.
originally the y coordinate was positive 1 (upward). when you rotate the paper, this is now to the right, so it becomes positive 1 in the x direction.

ergo, new coordinates = (1, √3)

sweet

[orel]

Another way of solving this problem:

We can find the Radius of the semicircle using the ccordinates of P:
Radius= sqrt(3+1)=2

So, the R=2, which is a hypotenuse of a right angled triangle with the catets: sqrt(3) and 1. If the hypotenuse is twice as long as one catet, it means that the angle in front of that catet is 30 degrees. OPQ=30, which meach that PQO is 60. Here, Hypotenuse is 2, and the side s, in front of 30 degree angle, is half as long as hypotenuse = 1