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For this problem, you can search, but the used property is the 30-60-90 rule (1:root3:2, isnt the triangle on the left following this? )

If you drop a perpendicular from the point (-sqrt3, 1) to the x axis you get a 30:60:90 triangle, with the legs equal to sqrt3 and 1 and the hypotenuse = 2.

That makes the angle to the left of the origin along the x axis 30 degrees. Since we also have a 90 degree angle as marked, the angle to the right of the origin along the x-axis must be 60 degrees. We have another 30:60:90 triangle. Because the figure is inscribed in a semicircle, the line from the origin to (-sqrt3, 1)and the line from the origin to (s,t) must both be radii and therefore equal.

Again, dropping a line from (s,t) perpendicular to the x axis we know that with a hypotenuse of 2 in a 30:60:90 triangle our t value (representing the length of the side opposite the 60 degree angle is sqrt3, and the s value is 1.