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Math Help - Trig. Help...please!

  1. #1
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    Trig. Help...please!

    Town Y is 9 km due north of town Z. Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

    A) Draw triangle XYZ and find angle YZX

    B) During an earthquake , town X moves due south until it is due west of Z. Find how far it has moved.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by <3<3 View Post
    Town Y is 9 km due north of town Z. Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

    A) Draw triangle XYZ and find angle YZX

    B) During an earthquake , town X moves due south until it is due west of Z. Find how far it has moved.
    A) Use law of cosines:

    z^2=x^2+y^2-2xy \cos Z

    8^2=9^2+5^2-2(9)(5) \cos Z

    etc....
    Last edited by masters; September 9th 2008 at 12:07 PM.
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  3. #3
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    Hello, <3<3!

    I assume you made sketches . . .


    Town Y is 9 km due north of town Z.
    Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

    A) Draw \Delta XYZ and find \angle YZX
    Code:
                      * Y
                    * |
               8  *   |
                *     |
              *       | 9
            *         |
        X *           |
             *        |
              5 *   θ |
                   *  |
                      * Z

    Let \theta = \angle YZX

    Law of Cosines: . \cos\theta \:=\:\frac{5^2 + 9^2 - 8^2}{2(5)(9)} \:=\:0.46666...

    Therefore: . \theta \:\approx\:62.2^o




    B) During an earthquake, town X moves due south until it is due west of Z.
    Find how far it has moved.
    Code:
                      * Y
                    * |
               8  *   |
                *     |
              *       | 9
            *         |
        X *    5      |
          :  *   62.2|
          :     *     |
          :  27.8 *  |
        W * - - - - - * Z

    Town X has moved directly south to W.

    Since \angle YZX = 62.2^o, then: . \angle XZW = 27.8^o

    In right triangle XWZ\!:\;\;\sin27.8^o \:=\:\frac{XW}{5} \quad\Rightarrow\quad XW \:=\:5\sin27.8^o \:=\:2.3319...


    Therefore, town X moved about 2.3 km.

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  4. #4
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    Quote Originally Posted by <3<3 View Post
    Town Y is 9 km due north of town Z. Town X is 8km from Y, 5km from Z and somewhere to the west of the line YZ.

    A) Draw triangle XYZ and find angle YZX

    B) During an earthquake , town X moves due south until it is due west of Z. Find how far it has moved.
     \left( a \right){\text{ In }}\Delta {\text{XYZ, use the cosine law to find }}\angle {\text{YZX (}} = \angle {\text{Z):}} \hfill \\

     \cos Z = \frac{{x^2  + y^2  - z^2 }}<br />
{{2xy}} \hfill \\

    \cos Z = \frac{{\left( 9 \right)^2  + \left( 5 \right)^2  - \left( 8 \right)^2 }}<br />
{{2 \times 9 \times 5}} \hfill \\

     Z = 62.2^\circ  \hfill \\

      \left( b \right){\text{ Town X has moved south, let us say, to point P, which is west of Z}}{\text{.}} \hfill \\

     {\text{In }}\Delta {\text{XPZ, }}\angle XZP = 90^\circ  - 62.2^\circ  = 27.8^\circ  \hfill \\

    {\text{So, the distance moved by town X is }} = XP \hfill \\

     {\text{In }}\Delta {\text{XPZ, }}\frac{{XP}}<br />
{{XZ}} = \sin 27.8^\circ  \hfill \\

      \frac{{XP}}<br />
{5} = 0.46638664 \hfill \\

      XP = 2.3{\text{  km}}{\text{.}} \hfill \\

    {\text{Distance moved by town X is 2}}{\text{.3 km}}{\text{.}} \hfill \\ <br />
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  5. #5
    A riddle wrapped in an enigma
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    Am I seeing double?!
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