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Math Help - Cos and Sin in a triangle that has no right angle

  1. #1
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    Thumbs up Cos and Sin in a triangle that has no right angle

    Hey MHF!
    Here something I found.
    I'm still learning geometry again, so this may seem easy for you guys.

    The sides of \Delta ABC are AB = 6 BC=4 and CA=5

    \cos <A =
    (The < means the angle)
    But!
    AB is not the Hypotenuse, since 16+25=41
    So, I figure this is not a Triangle Rectangle, but something else.
    No right angle on this one, right?

    I'm still kind of amateur in Geometry, so go easy! hahaha
    Thanks
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  2. #2
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    Quote Originally Posted by Zellator View Post
    Hey MHF!
    Here something I found.
    I'm still learning geometry again, so this may seem easy for you guys.

    The sides of \Delta ABC are AB = 6 BC=4 and CA=5

    \cos <A =
    (The < means the angle)
    You need to know the Law of Cosines.
    a^2=b^2+c^2-2bc\cos(\angle A)
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  3. #3
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    Yes! This makes sense!
    Thanks so much Plato!
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