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Math Help - angles problem

  1. #1
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    angles problem

    A man can row at 4 km/h, and run at 3 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 5 km, and the water is 4 km wide. He starts rowing with an angle between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of .
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  2. #2
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    Hello, samtheman17!

    A man can row at 4 km/h, and run at 3 km/h.
    He needs to get from a point A, on the south bank of a stretch of still water,
    to point B on the north bank of the water.
    The direct distance from A to B is 5 km, and the water is 4 km wide.
    He starts rowing with an angle θ between North and the direction in which he rows.

    Find an expression for the time he will take to get from A to B, in terms of θ.
    Code:
              3
        C * - - - o B
          |     *
        4 |   * 5
          | *
        A o

    The width of the water is: AC = 4 km.
    The distance from A to B is: AB = 5 km.
    Pythagorus tells us that: CB = 3 km.


    He will start at A, row to P, then run to B.

    Code:
              x   P   3-x
        C * - - - * - - - - * B
          |      /
          |     /
          |    / y
        4 |   /
          |  /
          |θ/
          |/
        A *

    Let \theta \,=\, \angle CAP
    Let x \,=\, CP, then: 3-x \:=\:PB.
    Let y \,=\,AP

    In right triangle ACP\!:\;\;y^2 \:=\:x^2 + 4^2 \quad\Rightarrow\quad y \:=\:\sqrt{x^2+16}

    He rows \sqrt{x^2+16} km at 4 km/hr.
    . . This will take him: . \frac{\sqrt{x^2+16}}{4} hours.

    Then he runs 3-x km at 3 km/hr.
    .This will take him: . \frac{3-x}{3}\:=\:1-\tfrac{1}{3}x hours.

    Hence, his total time is: . T \;=\;\tfrac{1}{4}\sqrt{x^2+16} + 1 -\tfrac{1}{3}x .[1]


    We want this answer in terms of θ.

    In right triangle ACP\!:\;\tan\theta \:=\:\frac{x}{4} \quad\Rightarrow\quad x \:=\:4\tan\theta .[2]

    Note that: . x^2+16 \;=\;(4\tan\theta)^2 + 16 \;=\;16\tan^2\!\theta + 16 \;=\;16(\tan^2\!\theta+1) \;=\;16\sec^2\!\theta

    . . Hence: . \sqrt{x^2+16} \;=\;4\sec\theta .[3]


    Subnstitute [2] and [3] into [1]: . T \;=\;\tfrac{1}{4}(4\sec\theta) + 1 - \tfrac{1}{3}(4\tan\theta)


    Therefore: . T \;=\;\sec\theta + 1 - \tfrac{4}{3}\tan\theta

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  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
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    thank you so much!

    i was trying to do something like that but kept messing something up as there are multiple steps....

    it is so clear now, thanks again
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