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Math Help - Finding theta of right triangle

  1. #1
    Junior Member
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    Jun 2008
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    Finding theta of right triangle

    Hello,

    I have a right triangle with hypotenuse x, adjacent side 4, and opposite side sqrt(x^(2) - 16) with angle theta.

    I'm having trouble remembering trigonometric approaches to finding theta.

    Previous parts of this problem involve integration using trigonometric substitution and I'm building the triangle to convert back to the original variable, x. It is very well possible that I have erred in doing that part of the problem, but this is where I got to, but couldn't proceed. I have the solution in terms of theta being, theta/16 + cosine(theta)sine(theta) + C.

    Might I express theta as an inverse trig function, or is there a better way to do it?

    If anyone has any pointers or tips, I'd greatly appreciate it.

    Thanks for your consideration,

    Austin Martin
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  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    i dont know if this would help you.. but the mnemonics we use is SOH-CAH-TOA

    for a given right triangle with \theta as an interior (acute) angle, then

    \sin \theta = \frac{Opposite}{Hypotenuse}
    \cos \theta = \frac{Adjacent}{Hypotenuse}
    \tan \theta = \frac{Opposite}{Adjacent}

    therefore, you can find \theta using the inverse trigo function..
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