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GMAT PREP GEOMETRY QUES???
Source: Beat The GMAT — Problem Solving |
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.
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.
STILL HAVING TROUBLE UNDERSTANDING THIS QUEST. I KNOW THE PROPERTIES OF A TRIANGLE BUT IT STILL DOESN'T MAKE SENSE.
This was the solution I posted on the manhattangmat forums.
tmmyc wrote:First, see that after dropping perpendicular lines, we have two right triangles.
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.
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.
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"
