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Trig proof

  1. Mar 23, 2005 #1
    I have to prove that [itex]a/\sin\alpha =b/\sin\beta[/itex]. A triangle has sides "a", "b", "c" and angles [itex]\alpha[/itex] and [itex]\beta[/itex] (opposite of the sides "a" and "b" respectively).

    This is what I did:

    Draw a line "h" as the height of the triangle on the side "c".

    [tex]\sin\alpha = h/b[/tex]. Multiplying by "b" gives [tex]b\sin\alpha = h[/tex]

    [tex]\sin\beta = h/a[/tex]. Multiplying by "a" gives [tex]a\sin\beta = h[/tex]

    From this we see that [tex]b\sin\alpha = a\sin\beta[/tex] and from this


    - Kamataata
  2. jcsd
  3. Mar 23, 2005 #2


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    It looks okay to me...:smile:

  4. Mar 23, 2005 #3
    Thanks for the quick help!
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