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Math Help - Trig angle problem

  1. #1
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    Ornskoldsvik
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    Exclamation Trig angle problem

    A woman on an island wishes to reach point R on a straight shore on the mainland from a point P on the island. The point P is 9km from the shore and 15km from the point R. If the woman rows a boat at a rate of 3 km/h to a point Q on land, then walks the rest of the way at a rate of 5km/h, express the total time it takes the woman to reach point R as a function of the indicated angle  \Theta


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  2. #2
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    So this is what I've gotten so far

    sin(theta) = 9/PQ so PQ = 9/sin(theta).


    Call the point at the right angle S.


    RS = sqrt(15^2 + 9^2) = sqrt(225 + 81) = sqrt(306) = 17.5


    9/QS = tan(theta)


    QS = 9/tan(theta)


    QR = 17.5 - 9/tan(theta)
    PQ = 9/sin(theta)


    she travels 3km/h on PQ and 5 on QR. So it's PQ/3 + QR/5.

    Is there a way to find out the lengths of PQ or QR or is what I have the answer?
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  3. #3
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    use s=vt to find t.find time taken to travel from P to Q and then add it to the time taken to travel from Q to R. You will get the answer in terms of theta. The lengths PQ and QR can only be found if the angle is given.
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  4. #4
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    Hello, OVechkin8!

    A woman on an island wishes to reach point R on a straight shore
    on the mainland from a point P on the island.
    The point P is 9 km from the shore and 15k m from the point R.
    If the woman rows a boat at a rate of 3 km/h to a point Q on land,
    then walks the rest of the way at a rate of 5 km/h,
    express the total time it takes to reach point R as a function of the indicated angle \theta.
    Code:
          : - - - - - 12  - - - - :
          S   x   Q      12-x     R
          o - - - o - - - - - - - o
          |      *             *
          |     *           *
          |    *         *
        9 |   *       * 15
          |  *     *
          |θ*   *
          |* *
          o
          P
    Let \theta = \angle SPQ

    In right triangle PSR\!:\;PS = 9,\;PR = 15
    Pythagorus says: . SR = 12

    Let: x = SQ, then QR = 12-x


    In right triangle PSQ\!:\;\;\cos\theta \,=\,\frac{9}{PQ} \quad\Rightarrow\quad PQ \,=\,\frac{9}{\cos\theta} \,=\,9\sec\theta

    She rows 9\sec\theta km at 3 km/hr.

    . . This will take her: . \frac{9\sec\theta}{3}\:=\:3\sec\theta hours.


    In right triangle PSQ\!:\;\;\tan\theta \,=\,\frac{x}{9} \quad\Rightarrow\quad x \,=\,9\tan\theta
    Hence: . QR \:=\:12-x \:=\:12 - 9\tan\theta

    She walks (12 - 9\tan\theta) km at 5 km/hr.

    . . This will take her: . \frac{12-9\tan\theta}{5} hours.


    Therefore, her total time is: . T \;=\;3\sec\theta + \frac{12-9\tan\theta}{5} hours.

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