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Math Help - General question about a triangle's height

  1. #1
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    Thumbs up General question about a triangle's height

    Hi Forum

    I studied parallelograms and triangles and I got to the formula of the area of both of them.
    Namely, 1absinA/2 or absinA.
    Naturally, an area of a triangle is also 1bh/2.

    So, if we mash them up, considering that sinA=c/b
    1/2 x bh = 1/2 x ab x c/b

    Could this mean that bh=ac?

    But this formulas work for both right angled triangles and non-right angled triangles, right?
    So we can't consider sinA=c/b at all times, since the angle is not always right.
    Does this has -something- to do with the Law of Sines or Cosines?

    Sorry if I haven't made myself clear.
    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Zellator View Post
    Hi Forum

    I studied parallelograms and triangles and I got to the formula of the area of both of them.
    Namely, 1absinA/2 or absinA.
    Naturally, an area of a triangle is also 1bh/2.

    So, if we mash them up, considering that sinA=c/b
    1/2 x bh = 1/2 x ab x c/b

    Could this mean that bh=ac?

    But this formulas work for both right angled triangles and non-right angled triangles, right?
    So we can't consider sinA=c/b at all times, since the angle is not always right.
    Does this has -something- to do with the Law of Sines or Cosines?

    Sorry if I haven't made myself clear.
    Thanks.
    You have got everything confused. Even in a right triangle sin(A) !=c/b, assuming you have labelled the sides and angles in the conventional manner.

    CB
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  3. #3
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    Yes, that is what I meant.
    I know that in a right triangle sin(A) is possible.
    I'm considering b as the base of the triangle.

    Yes I got confused on this one, now I get it.
    bh=absin(a)
    so sin(a)=h/a

    Second opinions get me back to reality.
    That's why they are important.

    Thanks CaptainBlack
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