|written by Kynreeve on May 18, 2011 16:47|
Figure 1: Here I need to find out the angle X, knowing that side AB (radius) = 4pt, side BC = 20 pt, and angle BA, AC is right.
I guess for starters I need to find out sin X (strangely enough, Anynowhere refuses greek letters...), which is AB/BC, which is 4/20 = 0.2. The rest is not so clear to me. Maybe, the next step is to calculate AC?
Figure 2: When the angle X on the first figure is known, I see there is two choices to find out X for each of the triangles (1, 2 and 3):
1) Do calculations of X for each of them, knowing their sides BC are 20pt, 36pt, 52 pt and so on. Straightforward approach.
2) Or, more interesting, find out and use the relation between the sequence 20, 36, 52 ... and the sequence of angles X of each of the triangles. This must be a propotion of some kind, I just don't know. Help is much appreciated.
|written by E_net4 on May 18, 2011 17:10|
|For figure 1, we do have a right triangle over there, right?|
sin x = 4 / (20+4)
x = arcsin (4/24) ~ 9.594 degrees
As for the explanation... there's a right triangle that connects the "origin", the centre of the circle you're taking and the point in the circumference that's tangent to that line. You know the radius of the circle, and the distance of the origin to the surface of the circle. So now you have 2 measures.
|written by Kynreeve on May 18, 2011 17:22|
|That was fast I'll test this solution. Thanks!|
What do you think about the relation on fig.2?
|written by E_net4 on May 18, 2011 17:26|
|Ah, there is! Assume you have an independent variable x, which holds this distances of possible values 20pt, 36pt, 52pt, etc...|
Therefore, the angle alpha can be defined by:
alpha = arcsin(4/(x+4))
You can put arcsin(4/(x+4)) in WolframAlpha to see how this function develops. Presto.
And yes, I R edit samurai.
|I believe the given distances are up to the center of each circle, so no adding +4... But other than that, I think E_net4 nailed it: you have a right triangle, so each angle is arcsin(4/distance). As for AC, again, since it's a right triangle, AC = sqrt(BC^2 - AB^2) = sqrt(distance^2 - 16).|
|written by Neuzd on May 18, 2011 22:29|
|I think Barebones is right.|
Also in the first figure the arrows indicate that the hypotenuse of the right triangle (BC) is 20, you probably got fooled by the blue and orange lines, E_net4.
|written by Kynreeve on May 18, 2011 22:59|
|I've got 11.54°, 6.379° and 4.412° tangent angles for each circle and they fit perfectly. If any propotions will change I can calculate these angles easily now. Thanks E_net4, Barebones, Neuzd!|
|written by E_net4 on May 18, 2011 23:13|
|you probably got fooled by the blue and orange lines, E_net4.|