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Trigonometry of two triangles

  1. Jun 12, 2014 #1
    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 aproach 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
     
  2. jcsd
  3. Jun 12, 2014 #2
    Find the third side using sine rule.
     
  4. Jun 13, 2014 #3
    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'
     
  5. Jun 13, 2014 #4

    haruspex

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    That link doesn't work for me. I get a 404 error.
     
  6. Jun 13, 2014 #5

    adjacent

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    I can see it.
    https://www.physicsforums.com/attachment.php?attachmentid=70550&d=1402581368
     
    Last edited: Jun 13, 2014
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