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    If only there was a circle around the triangle, ez 150/2 :laugh:
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    (Original post by morgan8002)
    I'm pretty confused. My last line of working was \cos \frac{\theta}{2} = 0, which I wrote down the wrong solution \theta = 90^o. There was something else that could have been 0 instead which I'm working on now, but I don't see any reason why I would accidentally come up with the right answer.
    :dontknow: if it's stupid but works, then it ain't stupid!
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    *applies to all fields of life except mathematics :getmecoat:
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    (Original post by morgan8002)
    I'm pretty confused. My last line of working was \cos \frac{\theta}{2} = 0, which I wrote down the wrong solution \theta = 90^o. There was something else that could have been 0 instead which I'm working on now, but I don't see any reason why I would accidentally come up with the right answer.
    what did you get as the lengths of triangle?
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    Let φ=CBA=BCA
    AD=1 and we take a parallel line to BC from D which intersects CA at E
    DE= x
    cosφ = x/2
    sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
    sin^2φ+cos^2φ=1 which gives x thus φ thus θ
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    (Original post by Vesniep)
    Let φ=CBA=BCA
    AD=1 and we take a parallel line to BC from D which intersects CA at E
    DE= x
    cosφ = x/2
    sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
    sin^2φ+cos^2φ=1 which gives x thus φ thus θ
    :coma:
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    (Original post by Student403)
    :coma:
    I know
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    (Original post by Vesniep)
    I know
    Very humble :rofl:
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    (Original post by Vesniep)
    Let φ=CBA=BCA
    AD=1 and we take a parallel line to BC from D which intersects CA at E
    DE= x
    cosφ = x/2
    sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
    sin^2φ+cos^2φ=1 which gives x thus φ thus θ
    I think I did this and ended up with a quartic. Is that what you got?
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    (Original post by notnek)
    I think I did this and ended up with a quartic. Is that what you got?
    yes it is a quartic , then you ask help from geogebra and the work is done .
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    So it looks this can be solved by getting help to solve a quartic, which is what I thought.

    The challenge is to find a nicer method that gives the answer directly. It could be that there isn't one.
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    (Original post by notnek)
    So it looks this can be solved by getting help to solve a quartic, which is what I thought.

    The challenge is to find a nicer method that gives the answer directly. It could be that there isn't one.
    Did you find a solution ?
    I found two roots for that awful equation .
    0<φ < 90 so I eliminated the negative root and the positive was x=1.5
    cosφ=0.75 thus θ=7.18.
    Perhaps I did something wrong of course.
    The thing is the roots are only two not four so we either have two imaginary roots or they are both double roots which means that the quartic was not that awful.
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    (Original post by Student403)
    Wha?

    I just checked using the sine and cosine rules and 90 was fine
    90 works? Now I'm confused.

    I'll have another look later when I have time.
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    (Original post by notnek)
    90 works? Now I'm confused.

    I'll have another look later when I have time.
    Unless I'm being stupid, which isn't an unreasonable possibility
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    absolutely vile
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    I had a go and got:
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    Name:  Possible Z-shape triangle solution.png
Views: 52
Size:  15.3 KB
 
 
 
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