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Soh Cah Toa?

  1. Oct 15, 2003 #1
    I understand that Sin is Opposite over the Hypotenuse, Cosine is Adjacent over Hypotenue, and Tangent is Opposite over the Adjacent. But I don't know how to use it. Sin has its own equation, as does Tangent...I am just confused. What do I do to find a specific angle? What about a certain side?

    Thanks!

    P.S I have done Soh Cah Toa during previous school years, but I never understood!
     
  2. jcsd
  3. Oct 15, 2003 #2

    HallsofIvy

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    Then all I can recommend is that you draw a right triangle, label the sides and angles, stare at it for a while and THINK!

    It might help if you specified on your triangle, which angle you are talking about "sin is opposite over hypotenuse" is non-sense until you specify the angle! Never say "sin". It is always sin OF A PARTICULAR ANGLE.
     
  4. Oct 15, 2003 #3
    Ok....here could be an example:

    One angle of a right triangle is 20 degress. The length of the hypotenuse is 6 cm. How would I get the other two sides?

    For the opposite side, would it be something like Sin20 is x/6?

    It just doesn't make sense...What about the whole Sin is the square root of a plus b times a and b or something like that?

    Thanks!
     
  5. Oct 16, 2003 #4

    Integral

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    So you have

    x= 6* Sin20 for the opposite side,

    the adjacent side would be

    y = 6* Cos20

    I am not sure what your last comment is about, are you refering to the Pythagorean Theorm.

    a2+ b2= c2?
     
  6. Oct 16, 2003 #5
    Thanks for the reply.

    No, it wasn't the pathagorem theorum. I don't know what it was.

    Thanks again!
     
  7. Oct 16, 2003 #6

    HallsofIvy

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    "What about the whole Sin is the square root of a plus b times a and b or something like that? "

    Now THAT'S the part that doesn't make sense! I recommend you erase that from your mind immediately!
     
  8. Oct 16, 2003 #7
    Ok, I will. If I do find out what I was thinking of, I'll post it!
     
  9. Dec 9, 2006 #8
    pay attention in class

    your thinkin of 30 60 90 triangles. thats where the side across from the right angle ( hypotenuse) equals 2x. the side across from the 30 degree angle equals x and the side across from the 60 degree angle is x times the squareroot of 3
     
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