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Math Help - Help Needed

  1. #1
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    Help Needed

    Hey im having trouble finishing the following question:

    The Width of the river between A and B is to be measured. It is known that [ABD = 90 degrees and survey shows that [ADB = 38 degrees [ACB = 28 degrees and distance between C and D is 40 meters. Find out the Width (w)
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  2. #2
    Super Member
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    Quote Originally Posted by mikel03 View Post
    Hey im having trouble finishing the following question:
    The Width of the river between A and B is to be measured. It is known that [ABD = 90 degrees and survey shows that [ADB = 38 degrees [ACB = 28 degrees and distance between C and D is 40 meters. Find out the Width (w)
    As stated there are some assumptions:
    1) points C and A are on the opposite side of the river from where points B and D are.
    2) the angle BAC is 90 degrees.
    3) the banks or edges of the river are parallel.

    There are several answers:
    1) if you are at point B looking at point A, then you need to know if D is 90 degrees to your left or 90 degrees to you right.
    2) if you are at point A looking at point B, then you also need to know if C is 90 degrees to your left or 90 degrees to you right.
    You have 4 possible solutions.
    No wonder you are having trouble with it.
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  3. #3
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    Thanks Aidan,

    Ive attached the question, thanks
    Attached Thumbnails Attached Thumbnails Help Needed-q6.jpg  
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  4. #4
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    Hello, mikel03!

    The width of the river between A and B is to be measured.
    It is known that \angle ABD = 90^o,\;\angle ADB = 38^o,\;\angle ACB = 28^o,\;CD = 40 meters.
    Find the width (w).
    Code:
        A *
          | * *
          |   *   *
        w |     *     *
          |       *       *
          |     38 *     28 *
          * - - - - - * - - - - - *
          B     x     D    40     C

    We have: . AB = w,\;\angle ADB = 38^o,\;\angle ACB = 28^o, CD = 40
    Let BD = x.

    In right triangle ABD\!:\;\;\tan38^o \:=\:\frac{w}{x} \quad\Rightarrow\quad x \:=\:\frac{w}{\tan37^o} .[1]

    In right triangle ABC\!:\;\;\tan28^o \:=\:\frac{w}{x+40}\quad\Rightarrow\quad x \:=\:\frac{w}{\tan28^o} - 40 .[2]


    Equate [2] and [1]: . \frac{w}{\tan28^o} - 40 \:=\:\frac{w}{\tan38^o} \quad\Rightarrow\quad \frac{w}{\tan28^o} - \frac{w}{\tan38^o} \:=\:40


    Multiply by \tan28^o\tan38^o\!:\;\;w\tan38^o - w\tan28^o \:=\:40\tan28^o\tan38^o

    Factor: . w(\tan38^o - \tan28^o) \:=\:40\tan28^o\tan38^o \quad\Rightarrow\quad w \:=\:\frac{40\tan28^o\tan38^o}{\tan38^o-\tan28^o}


    Therefore: . w \;=\;66.5795769 \;\approx\;66.6 m.

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  5. #5
    MHF Contributor
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    South Coast of England
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    Trigonometry

    Hello mikel03
    Quote Originally Posted by mikel03 View Post
    Hey im having trouble finishing the following question:

    The Width of the river between A and B is to be measured. It is known that [ABD = 90 degrees and survey shows that [ADB = 38 degrees [ACB = 28 degrees and distance between C and D is 40 meters. Find out the Width (w)
    ... or, if you want the easy method:

    \angle DAC = 10^o (Exterior angle of triangle = sum of interior opposite angles)

    \Rightarrow \frac{AD}{\sin 28^o}= \frac{40}{\sin 10^o} (Sine Rule on \triangle ADC)

    \Rightarrow AB = AD \sin 38^o = \frac{40\sin 38^o\sin 28^o}{\sin 10^o} \approx 66.6

    Grandad
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