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Solving for unknown in arccos equation help?

  1. Aug 10, 2011 #1
    I need help with the following, is it possible to get the below equation into a form like:

    THETA = 30a

    or somehow get the 'a' on the left hand side?

    Equation:

    THETA = arccos(1 - 0.005a)


    Those aren't the exact numbers/arrangement I'm working with but any help would be greatly appreciated.
     
  2. jcsd
  3. Aug 10, 2011 #2

    eumyang

    User Avatar
    Homework Helper

    It would be better if you posted the original problem instead of using numbers and/or an arrangement that isn't exact. Are you trying to solve for a? If so, as long as you're mindful of the domain/range restrictions, why can't you take the cosine of both sides?
     
  4. Aug 10, 2011 #3
    Sorry it is a bit hard to post the entire problem as this is only part of quite a large engineering question.

    Photo 10-08-11 10 15 49 PM (HDR).jpeg

    From the above the unknown is 'x' which is what I want to solve for (again I have skipped out a lot of the working that is irrelevant to you).

    If I substitute in THETA to the area 'A' equation then put that into the equation on the bottom I get a huge awful looking equation, only the one unknown but I just don't know how to solve for it. Any guidance please?
     
  5. Aug 10, 2011 #4

    Mark44

    Staff: Mentor

    You're starting with cos([itex]\theta/2[/itex]) = (150 - .77x)/150.
    You solved for [itex]\theta/2[/itex], but didn't solve for x.

    To isolate x (i.e., solve for x), multiply both sides of the original equation by 150. Then add -150 to both sides. Finally, divide both sides by -.77 to get x all by itself.
     
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