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Hard trig question

  1. Nov 15, 2007 #1
    hard trig question :(

    1. The problem statement, all variables and given/known data
    A,B, and C are three vertices of a cube, as shown in the diagra,. Find the measure of <ACB, to the nearest tenth of a degree.

    this is the diagram of the cube

    with no given side and angle how do u solve this problem? the answer in the book states 35.3 degrees. but HOW:confused:
  2. jcsd
  3. Nov 15, 2007 #2


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    Find teh distance between B and C, say, in terms of the length of one edge (call this "a" or whatever you want). That should be easy. Now, consider the triangle formed by the line going from C to B, then to A and then back to C. Can you visualize that triangle? You know the length of two sides, so you should be able to find all the angles.
  4. Nov 15, 2007 #3
    yes I can visualize the triangle, but what do you mean i now the length of two sides? theres no given lengths.
  5. Nov 15, 2007 #4
    The angle is independent of the size of the cube- e.g. if you double all lengths then the angle remains the same.

    I would solve this problem using the dot product of two vectors.
  6. Nov 15, 2007 #5


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    Like I said, simply give a name to the length of one edge. call this lenght "a", let's say. Then you may find the distance BC in terms of a. Then, when you find the angle, you will see that the calculation will involve taking the ratio of two lengths so the "a" will cancel out and you can get a number (because when you use a trig function, for example tan theta = opposite/adjacent, there is always a ratio of two lenghts involved). It's not surprising that teh answer does not depend on the size of the cube for the reason that christianjb explained.
  7. Nov 15, 2007 #6
    im sorry but im not understanding your method :(. I labled Length BC "a", but Im not sure how you can get an number angle from variables.
  8. Nov 15, 2007 #7
    If it's simpler- just set the length=1.
  9. Nov 16, 2007 #8


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    Assume each side of the cube has length "1". That is, side AB has length 1.
    Can you find the length of BC?

    Once you know those, tan(ACB)= "opposite side divided by near side".

    If you used a general "a" you would find that the "a" cancels out when you do that division.
  10. Nov 26, 2011 #9
    Re: hard trig question :(

    set AC as x, set BC as x(sqrt 2) because it is a square.
    tan(ACB) = x/x(sqrt2)
    = 1/sqrt2
    ACB = tan^-1 (sqrt2/2)
  11. Feb 15, 2012 #10
    Re: hard trig question :(

    Let s be the length of the segment from A to B.

    The length of the diagonal from A to C =s√3
    The length of the segment from B to C = s√2

    These three segments form a right triangle ABC, with hypotenuse AC.

    We can use trigonometry to find angle ACB. Let angle ACB = x

    cosx = s√2 / s √3;
    cosx = √2/√3 = √6/3;
    x = acos(√6/3) = 35.264.
  12. Feb 15, 2012 #11
    Re: hard trig question :(

    This thread is more than 4 years old...
  13. Feb 15, 2012 #12


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    Re: hard trig question :(

    A double necro lol
  14. Feb 16, 2012 #13
    Re: hard trig question :(

    lol :d
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