I'm standing in a circle (Point P), radius of 2000 with a center point 0/0. Then, I'm looking straight to the edge of the circle on a random point (B1 or B2) of the edge. How can I get the distance (f1 or f2) to the point when my rotation/angle is given?

Advanced: What happens if teta is not given, is it still possible to solve it?

Image of Problem

My first tries: Depending on the angle, I'm using sin, cos, tan of my rotation * (the distance to the center point + radius). I tested some numbers and it looks like it's working. Does that makes sense and is there a better solution to solve my problem? It looks like I need several formulas depending on the angle teta and the Position P.

  • $\begingroup$ making sure here: you don't wish to find the point on the circle that's closest to you, but the one you're looking at in a particular direction? $\endgroup$ – Dan Uznanski Sep 1 '16 at 4:43
  • $\begingroup$ What do you mean by "my rotation is given"? $\endgroup$ – gambler101 Sep 1 '16 at 4:57
  • $\begingroup$ I recommend drawing a picture. Geogebra is a good tool. It is much easier to understand what you are asking. $\endgroup$ – Ross Millikan Sep 1 '16 at 5:05
  • $\begingroup$ No, this is not a dupe, I updated my question to make the problem more understandable <3. $\endgroup$ – The Hero Sep 1 '16 at 10:24
  • $\begingroup$ You are overcomplicating. Extend your line d so that it forms a diameter of the circle. Your picture will then be the same as that in the answer below, and the answer below will give you either f1 or f2. $\endgroup$ – daniel Sep 1 '16 at 12:21

If I understand you, the situation is shown below. $g=2000$ is the radius of the circle, $r$ is your distance to the center, $\theta$ is the angle from the radius you are on to the direction you are looking. You want to find $f$. First use the law of sines: $\frac g{\sin \theta}=\frac r{\sin \angle ADC}$ to get $\angle ADC$, then $\angle CAD=\pi-\theta-\angle ADC$ and now use the law of cosines $f^2=r^2+g^2-2rg\cos \angle CAD$

enter image description here

  • $\begingroup$ Hey, thanks for the answer <3. I added more information about the problem, I think this is a part of the solution <3. $\endgroup$ – The Hero Sep 1 '16 at 10:21
  • $\begingroup$ @TheHero: I think this is the entire solution and that you have overcomplicated the question. $\endgroup$ – daniel Sep 1 '16 at 12:12
  • $\begingroup$ So basically it is sqrt(square(dc)+square(r)-2*dcrcos(3.14-rot-(sin(rot)/r)*dc)) where dc: distance to center, r: radius and rot: Rotation of myself/angle $\endgroup$ – The Hero Sep 1 '16 at 12:53
  • $\begingroup$ Also, is it sind or sinr, degree or radian? I'm using this to compute a fog in a game, somehow it's not working, I'll share some screens later <3. $\endgroup$ – The Hero Sep 1 '16 at 13:10
  • $\begingroup$ Actually it's degree, my other problem seems to be my calculation of teta. $\endgroup$ – The Hero Sep 1 '16 at 13:41

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