What is the Trigonometry Behind Finding x in Two Triangles?

[Jerry1989]

Hi Guys,

This is my second post relating to my problem, But I've boiled it down to more simple trig.
sadly, I still can't figure it out. See the sketch below; what I want to obtain is x as a function of R.
the angle in red, θ is the same for the two corners.

What is known:
1) two corners are θ, so te third corner is π-2θ
2) the triangle is isosceles, so R=x+C (but mind you, R≠D)
What I want to obtain:
1) a function R(x)=... independent of θ.


left triangle: B=x tan θ
right: B=R sin (π-2θ) = R sin (2θ)
also known: C= R-x= -R cos (2θ)

However, when I try to substitute these two formulas, I do this:
x tan θ = R sin (2θ) → X = R * 2cost(t)^2
R-x= -R cos (2θ) → X = R (1+cos(2θ))

and these are the exact same formula :(


My other approach was this
C= R-x= -R cos (2θ) → θ = 0.5* arccos (X/R - 1)
then substituting this into the others
but also didn't quite work out.

can you guys help me out?

https://www.physicsforums.com/attachment.php?attachmentid=70550&d=1402581368

 

[Adithyan]

Find the third side using sine rule.

 

[Jerry1989]

Thanks, I tried that, however it didn't work out that well.
if D = X /cos (t)
and D = sin(2t)*(R/sin(t))
then the same relation X = 2Rcos(t)^2 pops out.
just like the other ones.

I can't get t (Theta) to 'drop out'

 

[haruspex]

That link doesn't work for me. I get a 404 error.

 

[adjacent]

That link doesn't work for me. I get a 404 error.

I can see it.
https://www.physicsforums.com/attachment.php?attachmentid=70550&d=1402581368