TITLE: Finding points on the circumference of an arc
QUESTION [5 upvotes]: Sorry, it may be a simple task but it has been 30 years since I finished high school. :)
I have an initial course to a center of a circle (x0=100 and y0=100), which is, say, 150 degrees. The radius of the circle is 10. I also have an angle at which I draw an arc around the circumference of the circle, 30 degrees.
I need to find four things:
1) Left top corner of the bounding square for the circle. This is easy. x3=90 and x4=90.
2) First point of the original line entering the arc (x1 and y1).
3) Second point of the final line leaving the arc (x2 and y2).
4) Angle between right corner of the circle and the first point of the original line entering the arc (x1 and y1).

Could someone please remind of trigonometry?
Thanks a lot!
EDIT:
Since this is needed for Java code, these are some of the specific conditions:
1) Original coordinates (0,0) are located in the top left corner of the screen;
2) Original angle (0º) is at the rightmost point of the circle.
3) Positive is a counterclockwise rotation starting at 0º.

REPLY [3 votes]: $$
(x_1, y_1) = (100 + 10 \cos 150^\circ, 100 - 10 \sin 150^\circ)
$$
because $(x_1, y_1)$ is 150 degrees around from the rightmost point of the circle, and the circle-center is at $(100, 100)$, and the radius is 10, and $y$ increases down instead of up, which makes for the "-" in the second term.
$$
(x_2, y_2) = (100 + 10 \cos 180^\circ, 100 - 10 \sin 180^\circ)
$$
because that point is $30$ degrees further around from a point at angle 150 degrees. 
The angle you're looking at (the red arc) is $150^\circ$. (That also means that your picture is wrong, because the black arrow in the lower left should be headed about west-north-west, not west-south-west. )