This is a hard question b/c of how the picture is drawn, but I think 1 is the right answer.
If you draw a line perpendicular to the x axis down from P, you form a right triangle with height of 1 and base of sqrt(3) (known by P's coordinates). This is the ratio of the sides of a 30-60-90 right triangle, so the angle formed in this triangle around the origin is 30 degrees since it is opposite the shorter side. The angle already drawn is known to be 90 degrees (angle POQ), which leaves 60 degrees for the measure of the angle b/w the line from O to Q and the x-axis.
So, if we draw a perpendicular line from Q to the x axis, we again have a 30-60-90 right triangle. The side opposite the 60 degree angle (line from Q to axis) is of length sqrt(3) and side opposite the 30 degree angle is of length 1. Therefore, we know the coordinates of point Q are (1,sqrt(3)).
prep question
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t^2 + s^2 = 3 + 1 = 4 (radius of circle)
Line connecting (-sqrt-3, 1) to origin is of the form y = -1/sqrt-3 * x
Line perpendicular to this is of the form y = sqrt-3 * x
Since s,t is a point on this line, we have t = sqrt-3*s
substituting in first eqn, we have 4*s^2 = 4 or s = 1 (we know it is in the
first quadrant and hence positive)
Hence s = 1
Line connecting (-sqrt-3, 1) to origin is of the form y = -1/sqrt-3 * x
Line perpendicular to this is of the form y = sqrt-3 * x
Since s,t is a point on this line, we have t = sqrt-3*s
substituting in first eqn, we have 4*s^2 = 4 or s = 1 (we know it is in the
first quadrant and hence positive)
Hence s = 1
