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Math Help - trigonometry ABC triangle

  1. #1
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    trigonometry ABC triangle

    of a ABC triangle
    a=5 , b=4 and cos(A-B)=31/32

    show that c=6
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  2. #2
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    Hello, stud_02!

    I found a solution . . . but there must be a better way!


    In \Delta ABC\!:\;\;a\:=\:5,\;b\:=\:4,\;\cos(A-B)\:=\:\tfrac{31}{32}
    Show that: . c\:=\:6
    Code:
                  C
                  *
                 *| *
                * |   *
           b=4 *  |     * a=5
              *   |y      *
             *    |         *
            *     |           *
           *      |             *
        A * - - - * - - - - - - - * B
          :   x   D      c-x      :
          : - - - - - c - - - - - :

    In right triangle CDA\!:\;\;x^2+y^2\:=\:16 \quad\Rightarrow\quad y^2 \:=\:16-x^2 .[1]

    In right triangle CDB\!:\;\;(c-x)^2 + y^2 \:=\:25 \quad\Rightarrow\quad y^2 \:=\:25-(c-x)^2 .[2]

    Equate [1] and [2]: . 16-x^2 \:=\:25 - (c-x)^2 \quad\Rightarrow\quad x \:=\:\frac{c^2-9}{2c} .[3]

    Substitute into [1]: . y^2 \:=\:16 - \left(\frac{c^2-9}{2c}\right)^2 \quad\Rightarrow\quad y^2\:=\:\frac{(c^2-1)(81-c^2)}{4c^2} .[4]


    We are given: . \cos(A-B) \:=\:\tfrac{31}{32} \quad\Rightarrow\quad \cos A\cos B + \sin A\sin B \:=\:\tfrac{31}{32}

    . . . . \frac{x}{4}\cdot\frac{c-x}{5} + \frac{y}{4}\cdot\frac{y}{5} \:=\:\frac{31}{32} \quad\Rightarrow\quad x(c-x) + y^2 \:=\:\frac{155}{8}

    Substitute [3] and [4]: . \left(\frac{c^2-9}{2c}\right)\left(c - \frac{c^2-9}{2c}\right) + \frac{(c^2-1)(81-c^2)}{4c^2} \:=\:\frac{155}{8}

    . . which simplifies to: . 9c^2 \:=\:324 \quad\Rightarrow\quad c^2 \:=\:36 \quad\Rightarrow\quad\boxed{c \:=\:6}

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