There are a few different approaches we can take to this question.
First, we can know a very esoteric rule about circles with centre O:
If you draw a right angle triangle with the right angle at the origin and with the other two vertices on the circuference of the triangle, the x/y co-ordinates of the two points will be reversed (with signs appropriate for the respective quadrant).
For example:
If one point were at (1,0), the other point would be at either (0,1) or (0,-1).
If one point were at (1/2, 1/3), the other point would be at (-1/3, 1/2) or (1/3, -1/2).
In this question, since we have one point at (-root3, 1), and the point we want is in quadrant I, the other point is at (1, root3) (and so the value of s is 1).
If you're really bored one day, you can use either the pythagorean formula or trig to prove this rule.
Second, we can recognize that we can make a special right triangle as half of the right triangle linking the two points.
Going from (-root3, 1) to the origin (0,0) and then to (0,1), we have a 1/root3/2 triangle. If we know that this is a 30/60/90 triangle, we realize that the angle formed near the origin is 60 degrees (since it's opposite the root3 side of the triangle).
Now we know that the portion of the right triangle on the other side of the y-axis is 30 degrees (90- the 60 we used on the left side). We can draw another right triangle from the origin (0,0) to (s,t) to (0,t), which is also a 30/60/90. In this triangle, the side connecting points (0,t) and (s,t) is opposite the 30 degree angle, so that side has a length of 1.
Therefore, the value of s is 1.
This long-winded explanation is a great lesson in why it's vital to redraw diagrams on your scratch paper - this question is MUCH easier to explain (and solve) while referring directly to a picture.
