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Cosine law

  1. Aug 20, 2008 #1
    hello , How are you all?
    look at this question
    what the magnitude of the angle that between two ribs from the heads of the cube to its center ?
    I attempted by using cosine law [look at the pecture in attachment]
    but, Hwo to apply this law on the cube (3D)?
    any one help me >>and thank you anyway
     

    Attached Files:

  2. jcsd
  3. Aug 20, 2008 #2

    HallsofIvy

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    I am afraid that I don't understand what you mean by "ribs" or "heads" of a cube. Are the "ribs" diagonals and the "heads" vertices? In that case, the angle between two diagonals is a right angle. You don't need the cosine law to see that.
     
  4. Aug 20, 2008 #3
    yes ,,
    look at attachment the picture is clear that explanation
    I calculat it by using simple way but How to use cosine law to giveangle theta in pecture
    an thanks Mr.HallsofIvy
     

    Attached Files:

  5. Aug 20, 2008 #4
    Hmm in the drawing is theta the same as angle BOA? Well either way it's BOA or half of BOA due to symmetry so you can find BOA easily and divide if you have to.

    Angle BOA has two [tex]\frac{\sqrt{3}r}{2}[/tex] sides that make up Angle BOA. The law of cosines tells us that [tex](AB)^2 = (OA)^2 + (OB)^2 + 2(OA)(OB)cos(Angle BOA)[/tex]. Then just solve for Angle BOA.

    Although really you shouldn't need the law of cosines.
     
  6. Aug 20, 2008 #5
    Hmm in the drawing is theta the same as angle BOA? Well either way it's BOA or half of BOA due to symmetry so you can find BOA easily and divide if you have to.

    sorry , my explanation not clearly anyway


    now I understand

    used cosine law directly and I don't need analysis all angles because I can find it from symmetry considerations .
    thank you very much
     
  7. Aug 20, 2008 #6
    Err, the formula is [tex](AB)^2 = (OA)^2 + (OB)^2 - 2(OA)(OB)cos(Angle BOA)[/tex].
     
  8. Oct 27, 2008 #7
    thanks Mr. BoundByAxioms for your correcting :>
    I hope the best for all
     
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