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
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GMAT Prep / Geometry
Source: Beat The GMAT — Problem Solving |
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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
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
