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Thread: 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
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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:

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

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

    etc....
    Last edited by masters; Sep 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 $\displaystyle Y$ is 9 km due north of town $\displaystyle Z.$
    Town $\displaystyle X$ is 8km from $\displaystyle Y$, 5km from $\displaystyle Z$ and somewhere to the west of the line $\displaystyle YZ.$

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

    Let $\displaystyle \theta = \angle YZX$

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

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




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

    Town $\displaystyle X$ has moved directly south to $\displaystyle W.$

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

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


    Therefore, town $\displaystyle X$ 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.
    $\displaystyle \left( a \right){\text{ In }}\Delta {\text{XYZ, use the cosine law to find }}\angle {\text{YZX (}} = \angle {\text{Z):}} \hfill \\$

    $\displaystyle \cos Z = \frac{{x^2 + y^2 - z^2 }}
    {{2xy}} \hfill \\$

    $\displaystyle \cos Z = \frac{{\left( 9 \right)^2 + \left( 5 \right)^2 - \left( 8 \right)^2 }}
    {{2 \times 9 \times 5}} \hfill \\$

    $\displaystyle Z = 62.2^\circ \hfill \\$

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

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

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

    $\displaystyle {\text{In }}\Delta {\text{XPZ, }}\frac{{XP}}
    {{XZ}} = \sin 27.8^\circ \hfill \\$

    $\displaystyle \frac{{XP}}
    {5} = 0.46638664 \hfill \\$

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

    $\displaystyle {\text{Distance moved by town X is 2}}{\text{.3 km}}{\text{.}} \hfill \\
    $
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  5. #5
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