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Thread: Triginomotry problem please. (really easy one)

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    Default Triginomotry problem please. (really easy one)

    Ok, So there's a right triangle,

    angle of it is 64 degrees
    Adjacent leg is X
    opposite leg is Y
    Hypotunuse is 18

    Can anyone help me solve this?

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    The only thing I remember is a2+b2=c2

    So just uhm....find the square root of that
    and yeah. i don't remember anything else

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    the Sq root of the hypotenuse is 4.242 if that helps

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    You guys are using trig identities, the hard way.

    a^2 + b^2 = c^2

    That's for losers.

    SOH-CAH-TOA is where it's at.

    Oh and for the above post, yes the triangle does have a and b. Each leg corresponds.

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    Quote Originally Posted by Mr. First Name Basis View Post
    You guys are using trig identities, the hard way.

    a^2 + b^2 = c^2

    That's for losers.

    SOH-CAH-TOA is where it's at.

    Oh and for the above post, yes the triangle does have a and b. Each leg corresponds.
    He doesn't have the values for a and b though.

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    Quote Originally Posted by Rendascus View Post
    The only thing I remember is a2+b2=c2

    So just uhm....find the square root of that
    and yeah. i don't remember anything else
    The problem with that is he doesn't have a OR b, just c. So that won't work.

    I wish I could be more helpful, I was really good at math last year, Almost made a 100 in Algebra II. But our math class is so damn laid back this year I feel like I have forgotten just about everything I used to know. We aren't really pressured into trying and I don't give a shit about math enough to actually take any initiative myself so I just kind of cruise by with a C now.

    Useless post is useless... sorry I can't help.

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    Can you use the trigonometry table with the sin cos and tg values?
    If you can it's easy...
    All you need to do is sin64 = y/hipotenuse
    It is 0,898794 = y/18
    y = 0,898794 x 18
    y = 16,178292

    And to find x you just do cos64 = x/hipotenuse
    0,438371 = x/18
    x = 0,438371 x 18
    x = 7,890678

    If you can't use the trigonometry table then I have to think more...

    Hope I helped you...
    (:


    Post above...
    To do a cosine law you usually need 2 sides and 1 angle..
    But it's not neccessary at all..
    It's a triangle with a 90 degrees angle...

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    I hope you've been taught this accronym.

    SOH CAH TOA

    Sine - Opposite over Hypotenuse
    Cosine - Adjacent over Hypotenuse
    Tangent - Opposite over Adjacent

    Excess Information:And to tell the truth, well you could set this up as a proportion because each angle corresponds with it's opposite side. Anyways...

    Sin64 = opposite leg/18
    So 18sin64 = opposite leg

    If you can't figure it out from there, may God help you.

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    Quote Originally Posted by Mr. First Name Basis View Post
    I hope you've been taught this accronym.

    SOH CAH TOA

    Sine - Opposite over Hypotenuse
    Cosine - Adjacent over Hypotenuse
    Tangent - Opposite over Adjacent

    Excess Information:And to tell the truth, well you could set this up as a proportion because each angle corresponds with it's opposite side. Anyways...

    Sin64 = opposite leg/18
    So 18sin64 = opposite leg

    If you can't figure it out from there, may God help you.
    what this guy said.

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    Isn't this cosine law because he has all 3 angles.

    See, 90+64-180= 26

    So thats the other side.

    And with 3 angles you do cosine, or sine law. I forgot which one.

    Doesn't matter.. it's easy from there.

    If you STILL can't figure it out from there that's sad, and I guess I'll just answer it for you...

    Actually I drew it, and it looks like you can do simple sohcahtoa....

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    I remember Soh Cah Toa, remember it too good, because my friend was saying it like it was desu.

    But how? I only know Sine, Cosine, and tangent. I don't want to get too far, I don't even understand how to find the sides with those three things, or what they do.
    Edit: OH OH OH OHO HOHOHOHO I GOT IT THANKS GUyZ LOL

    Side y is about 16 right?

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