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Math Help - law of cosines (distance)

  1. #1
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    law of cosines (distance)

    a 100 ft verticle tower is to be erected on a side of a hill that makes a 6 degree angle rising up. find the length of each of the guide wires that will be anchored 75 ft uphill and hown hill from the base of the tower
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  2. #2
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    So you have two triangles with sides of 100 and 75 feet. The angle between those sides for the uphill cable is 90- 6= 94 degrees. The angle for the downhill cable is 90+ 6= 96 degrees. Now, use the cosine law.
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  3. #3
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    Hello, kenzie103109!

    A 100-ft vertical tower is to be erected on a side
    of a hill that makes a 6 angle with the horizontal.
    Find the length of each of the guide wires that will be anchored 75 ft
    uphill and downhill from the base of the tower.
    Code:
                            D
                            *
                            |
                          * |  *
                            |
                        *   |     *  L2
                    L1      |100
                      *     |        *
                            |                 * P
                    *       |           o
                        96 | 84 *     B
                  *         o  96
               A      * 84 |C
                o 6        |
        Q * - - - - - - - - * - - - - - - - - * R

    The slope of the hill is PQ\!:\;\angle PQR = 6^o
    The tower is CD = 100.

    The guide wires are anchord at A and B.
    . . AC \,=\,CB\,=\,75.

    The guide wires are: . L_1 = DA,\;L_2 = DB.

    We find that: . \angle DCA = 96^o,\;\angle DCB = 84^o.


    In \Delta DCA\!:\;\;L_1^2 \:=\:25^2 + 100^2 - 2(25)(100)\cos96^o \:=\:11,\!147,64232

    . . Therefore: . L_1 \;\approx\;105.6\text{ ft}


    In \Delta DCB\!:\;\;L_2^2 \;=\;25^2 + 100^2 - 2(25)(100)\cos84^o \;=\;10,\!727.35768

    . . Therefore: . L_2 \;\approx\;103.6\text{ ft}

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