- #1
Jerry1989
- 4
- 0
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
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