[Golist] OT trig

[Cadete B7]

reposting the solution, which i'm embarrassed not to have realized  
myself, was pulled from a math forum response...  i'm using this to  
create a series of tweens at initialization (so i can't use sceneX and  
the like) which will orient planes back to a 1:1 aspect ratio and  
screen center while being nested within other DisplayObject3D's that  
have an axis rotation and z-depth...

Let us call the given central angle as theta.
Let us call the given radius of the circle as r.
Let call the upper "?" as x, and the lower "?" as y.
Let us call the hypotenuse of the legs x and y as z.
So the whole vertical line from the center of the circle up to the  
bottom of y is (r+z).

In the large right triangle, whose hypotenuse is (r+z):
cos(theta) = r / (r+z)
So,
(r+z)cos(theta) = r
(r+z) = r / cos(theta) = r*sec(theta)
z = r*sec(theta) -r
z = r*[sec(theta) -1] -------**

In the smaller right triangle, whose hypotenuse is z:
Since r and y are parallel ...they are both perpendicular to the same  
green line...then the angle between y and z is equal to theta.
Then,

sin(theta) = x / z
So,
x = z*sin(theta)
x = r[sec(theta) -1]*sin(theta) ------------answer.

cos(theta) = y / z
y = z*cos(theta)
y = r[sec(theta) -1]*cos(theta) --------answer.

------------------
EDIT:

Or, x = r[1 /cos(theta) -1] sin(theta) = r[sin(theta)/cos(theta) - 
sin(theta)] = r[tan(theta) -sin(theta)]

And, y = r[cos(theta) /cos(theta) -cos(theta)] = r[1 -cos(theta)].

- Wilbur

On Jul 26, 2008, at 10:59 PM, cadete at bigote7.com wrote:

> i was wondering if anyone could lend me a little trig help...  given  
> the radius and theta, what would the two lengths indicated by  
> question marks be?  exsecant?  thanks.
>
> link to diagram...
> http://wilbcorp.com/links/trig.jpg
>
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