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Need help with trigonometric problems

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img135.imageshack.us/img135/4150/trighelp.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution
    http://img18.imageshack.us/img18/8480/workz.jpg [Broken]

    Please help I'm totally confused, if anyone can help I can offer a reward of monetary value :).
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 4, 2009 #2

    Mark44

    Staff: Mentor

    There are 3 triangles: the big one, the little one, and the middle one (the one with the 12 deg. angle). You don't need to do anything with the middle one -- the one that you spent most of your efforts on. Your value for the angle that the bottom of the beam makes with the floor is close, but it's slightly more than 2 degrees. Don't round anything until the very end.

    What you want is the altitude of the little triangle. Since the angle is given in degrees, make sure that your calculator is in degree mode. To get that, get the angle (alpha) of the big triangle by using the tangent function. tan(alpha) = 10/40 = 1/4, or alpha = arctan(1/4).

    The angle, beta, of the small triangle is alpha - 12. Store this value in memory. tan(beta) = h/40, where h is the altitude of the small triangle. Solve for h in this equation.

    The nefarious trig student has to be able to fit into a rectangular hole 16" high by 24" wide. Use the angle beta again -- that's why you stored it -- to calculate the altitude of a slightly smaller triangle that is h2 in height and 38' along its base.

    Be sure to include units, since you're working with both feet and inches. If you leave off the units, you be more likely to forget what they are, and won't know what you've ended up with.

    Good luck!
     
  4. Mar 4, 2009 #3
    For the new angle of the smallest triangle why do i have to solve for the altitude? Isn't the altitude = .089 feet? It says to solve for the new angle so that the thief can't pass underneath the beam. So why wouldn't I solve for theta = (1.333 feet )/(38 feet) which equals = 2.009 degrees?
     
  5. Mar 4, 2009 #4

    Mark44

    Staff: Mentor

    I notice that you cross-posted this problem in the Precalc forum as well. That's a no-no, since people will waste time giving you two explanations.

    The smaller, small triangle has the same acute angle as the one with the longer base. I didn't say you needed to solve for it. Angle beta (what I called it, not theta) is larger than what you show. The relationship is tan(beta) = h2/38, not beta = h2/38, and no, h2 is not .089 feet, which is only about an inch.

    DON'T ROUND YOUR INTERMEDIATE RESULTS!
     
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