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Circle..I can't figure this one out

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Circle..I can't figure this one out

by mikebarmy » Thu Aug 04, 2011 8:57 pm
I can't figure this one out...Can someone please explain how to arrive at the answer?

A semicircle is on an xy plane with center O at the origin. Points A and B are on the circle. Point A (in the second quadrant) is at point (-radical(3),1). Point B (in the first quadrant) is at point (s, t). Lines drawn from point A to the origin and point B to the origin form a 90 degree angle. What is the value of S?




The answer is 1 <-

I would appreciate any explanation of this. Thanks!
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by gmatboost » Thu Aug 04, 2011 9:33 pm
Let's add a few more points:
Draw vertical lines down to the x-axis from A and from B.
Call these points C and D.
C is at (-sqrt(3), 0) and D is at (s, 0).
Also, put a point Y somewhere along the y-axis, let's say it's at (0, 5) but this is just for reference.

Triangle ACO is a right triangle with legs of length sqrt(3) and 1. You should recognize this as a 30-60-90 right triangle based on the side ratios. The hypotenuse of a 30-60-90 is twice the shorter leg, so AO has length 2.

Angle COA is 30, because it is opposite the side of length 1. This means angle AOY is 60, since angle COY is 90.

Now, we know angle AOB is 90. This means angle YOB is 30. This in turn means that angle BOD is 60.

That means that triangle BOD is also a 30-60-90 right triangle. Since AO and BO are both radii of the semi-circle, they have the same length, so BO, the hypotenuse of triangle BOD, has length of 2 also.

Since angle BOD is 60, the side opposite that, BD, has a length of sqrt(3) because of the 30-60-90 side ratios. Similarly, since angle OBD is 30, the side opposite that, OD, has a length of 1 because of the 30-60-90 side ratios.

That means point B is [spoiler]at (1, sqrt(3)).[/spoiler]
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