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Math Help - Bearing/trig problem pleaasee help !! T.t

  1. #1
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    Unhappy Bearing/trig problem pleaasee help !! T.t

    A man walking due north along a level road observes a church spire in the direction 342degrees True north. After walking 1500m he observes it in the direction 337 degress true north. How far is the church from the road?

    Any help will be GREATLY APPRECIATED !
    thanks ! =D
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  2. #2
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    Quote Originally Posted by iiharthero View Post
    A man walking due north along a level road observes a church spire in the direction 342degrees True north. After walking 1500m he observes it in the direction 337 degress true north. How far is the church from the road?

    Any help will be GREATLY APPRECIATED !
    thanks ! =D

    The church spire will be west of the road.

    The angle at the first point (call that point A)
    from the road at point A to the spire is 342 degrees or
    the acute angle is 360 - 342 = 18 degrees.

    The angle at the second point (call that point B)
    from the road at point B to the spire is 337 degrees or
    the acute angle is 360 - 337 = 23 degrees.

    You want to know the distance the spire is from the road.
    The shortest distance or the perpendicular distance.
    Let's call that distance "h"
    Let's call the distance walked "d" which we know is 1500m.

    The distance from point A to the perpendicular point is
    1500 + unknown distance along the road.
    (let's call the unknown distance "x".)

    The distance from point B to the perpendicular point is
    the unknown distance along the road.
    Let's call the unknown distance "x"

    for a right triangle:
    the height of the triangle is the base times the tangent of the angle from the base to the apex of the triangle.

    Triangle 1:
    (d+x) \times \tan (18deg) = h ----> EQUATION 1

    Triangle 2:
     x \times \tan (23deg) = h ---->EQUATION 2

    Since both equations are equal to h, they are equal to each other.
    (d+x) \times \tan (18deg) = x \times \tan (23deg)

    expanding the LHS
     d \times \tan (18deg) + x \times \tan (18deg) = x \times \tan (23deg)

    subtracting
     d \times \tan (18deg) = x \times \tan (23deg) - x \times \tan (18deg)

    simplify the RHS
     d \times \tan (18deg) = x \times \left ( \tan (23deg) - \tan (18deg) \right )

    & divide


     \frac {d \times \tan (18deg)}{ \tan (23deg) - \tan (18deg)} = x

    Then plug X back into equation 2 to determine the perpendicular distance the spire is west of the road -- the "h" in both equations.
    or
    plug it back into equation 1.
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  3. #3
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    thank you SO much !!!!!!!!
    ur completely aweSUM. =DDDD, and genius too XD
    thanx again !
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  4. #4
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    Hello, iiharthero!

    A slight variation of Aidan's excellent approach . . .


    A man walking due north observes a church spire with bearing 342.
    After walking 1500m, he observes it with bearing 337.
    How far is the church from the road?

    I'll use Aidan's diagram . . .

    Code:
                     h
        C * - - - - - - - - - * D
            * *               |
              *   *           | x
                *     *   23 |
                  *       *   |
                    *         * B
                      *       |
                        * 18 | 1500
                          *   |
                            * |
                              * A

    The church is at C. .The man walks along the road AD.

    He makes his first observation at A\!:\;\angle A = 18^o

    He walks 1500 m to B \!:\;AB = 1500
    . . And \angle CBD = 23^o.

    The distance from the church to the road is: h = CD.
    . . Let x = BD.


    In right triangle CDB\!:\;\;\tan23^o \:=\:\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan23^o} .[1]

    In right triangle CDA\!:\;\;\tan18^o \:=\:\frac{h}{x+1500} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan18^o} - 1500 .[2]


    Equate [2] and [1]: . \frac{h}{\tan18^o}-1500 \:=\:\frac{h}{\tan23^o}


    Multiply by \tan18^o\tan23^o\!:\quad h\tan23^o - 1500\tan18^o\tan23^o \:=\:h\tan18^o

    . . and we have: . h\tan23^o - h\tan18^o \:=\:1500\tan18^o\tan23^o


    Factor: . h(\tan23^o -\tan18^o) \:=\:1500\tan18^o\tan23^o


    Therefore: . h \;=\;\frac{1500\tan18^o\tan23^o}{\tan23^o-\tan18^o} \;\approx\;2078\text{ m}

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  5. #5
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    thanx soroban ! realli appreciate your help !!!! =]]
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