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Homework Help: NEED HELP Finding the length of an angle on a right triangle

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data


    ^This is a right triangle (not drawn to scale). The left side is 18,944, then top side is 30,556, and the hypotenuse is 35,952. The right angel is between the left side and the top side. Now, how do I find the measurement of the angle between the hypotenuse and the left side?

    2. Relevant equations

    sin, cos, and tan?

    3. The attempt at a solution

    Since the angle is between the hypotenuse and the adjacent, I assume that I do cos, but I don't know what numbers to use.
  2. jcsd
  3. Nov 5, 2007 #2


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    For the angle from vertical (left side) and hypotenuse, 18,944 = 35,952 cos [itex]\theta[/itex].
  4. Nov 5, 2007 #3
    theres a relationship between the size of a leg and the vertex angle (the angle opposite of the leg)
  5. Nov 5, 2007 #4
    You can use any of them to find the angle since you know all three lengths, you just have to setup the right ratios.
  6. Nov 5, 2007 #5
    You can use any of the trig functions that you have listed to establish the proper relationship. But, for now we can stick to cos.

    What ratio does cos represent? [tex]\cos\theta=\frac{?}{?}[/tex]

  7. Nov 5, 2007 #6
    cos of angle = adjacent/hypotenuse

    So, cos x = 18,944/35,952 right?
  8. Nov 5, 2007 #7
    Use the law of cosine: http://en.wikipedia.org/wiki/Law_of_cosines

    Let A = 18944, B = 30556, C = 35952

    B^2 = A^2 + C^2 - 2*A*C*cos(angleB)
    cos(angleB) = (A^2 + C^2 - B^2 ) / (-2*A*C)

    Solve using inverse cosine, and you get B.
    Last edited: Nov 5, 2007
  9. Nov 5, 2007 #8
    Thank you guys I got it.
  10. Nov 6, 2007 #9
    Why? I mean you can but it's a complete waste of time in this situation :-/
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