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Math Help - Proving the Law of Sines

  1. #1
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    Proving the Law of Sines

    Hey Everyone,

    I have a question here that asks:

    Prove the Law of Sines for triangle ABC:

    sinA/a = sinB/b = sinC/c

    We have looked at all the side and angle axioms and up to euclidean geometry, any help here would be greatly appreciated. Thanks guys.
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  2. #2
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    Quote Originally Posted by GreenDay14 View Post
    Hey Everyone,

    I have a question here that asks:

    Prove the Law of Sines for triangle ABC:

    sinA/a = sinB/b = sinC/c

    We have looked at all the side and angle axioms and up to euclidean geometry, any help here would be greatly appreciated. Thanks guys.
    Hi GreenDay14.

    If we do not have the triangle altitude on a particular base,
    we can simply split the triangle into back to back right-angled triangles,
    using any of the 3 sides as base.

    Hence, there are 3 ways to write the triangle area..

    A=\frac{1}{2}abSinC=\frac{1}{2}acSinB=\frac{1}{2}b  cSinA

    Hence

    bSinC=cSinB

    aSinC=cSinA

    aSinB=bSinA

    Hence

    \frac{b}{SinB}=\frac{c}{SinC}\ or\ \frac{SinB}{b}=\frac{SinC}{c} etc
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  3. #3
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    You can prove the Law of Sines by starting with a triangle consisting of vectors A,B, and C. This means means that A+B+C=0. Then take the cross product of both sides with vector A to get the first part of the relation. Do the same thing with vector B to get the second part of the relation.
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  4. #4
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    thanks a lot for the help guys.
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  5. #5
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    Hello, GreenDay14!

    Prove the Law of Sines for triangle ABC: . \frac{\sin A}{a} \:=\: \frac{\sin B}{b} \:=\: \frac{\sin C}{c}
    Here is the classic textbook proof.
    Code:
                  C
                  *
                 *| *
                * |   *
             b *  |     *  a
              *   |h      *
             *    |         *
            *     |           *
           *      |             *
        A * - - - * - - - - - - - * B
                  D

    Draw altitude CD to side AB.
    . . Call it h.

    In right triangle CDA\!:\;\;\sin A \:=\:\frac{h}{b} \quad\Rightarrow\quad h \:=\:b\sin A .[1]

    In right triangle CDB\!:\;\;\sin B \:=\:\frac{h}{a} \quad\Rightarrow\quad h \:=\:a\sin B .[2]


    Equate [1] and [2]: . b\sin A \:=\:a\sin B  \quad\Rightarrow\quad \boxed{\frac{\sin A}{a} \:=\:\frac{\sin B}{b}}


    In a similar fashion, we can prove that: . \frac{\sin B}{b} \:=\:\frac{\sin C}{c}

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