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Math Help - Exam Review - Geometric Vectors

  1. #1
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    Exam Review - Geometric Vectors

    A light plane is travelling at 175km/h on a heading of N8E in a 40km/h wind from N80E. Determine the plane's ground velocity using geometric vectors.

     |r| = \sqrt {175^2 + 40^2 - 2(175)(40)cos144}

     |r| = 208.69km/h

     \frac {sin144}{208.69} =   \frac {sinx}{40}

    x = 6.47


    Therefore, the plane's ground velocity is 208.69km/h on a bearing of N6.47E.



    Textbook Answer: 167km/h, N5W
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  2. #2
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    Hello, Macleef!

    A light plane is travelling at 175 km/h on a heading of N8E
    in a 40 km/h wind from N80E.
    Determine the plane's ground velocity using geometric vectors.

     |r| \:=\: \sqrt {175^2 + 40^2 - 2(175)(40)\cos{\color{red}144}} . ??
    Code:
                            o B
                        * *
               40   * 72*
                *       *
       C    *          *
        o             *
         *     :     * 175
          *    :    *
        r  *   :   *
            *  :8*
             * : *
              *:*
               o
               A

    We find that: . \angle B \:=\:72^o

    Then: . r^2 \:=\:40^2 + 175^2 - 2(40)(175)\cos72^o \:=\:27,898.76208

    Therefore: .  r \:=\:167.0292252 \:\approx\:\boxed{167} km/hr



    \cos A \:=\:\frac{167^2 + 175^2 - 40^2}{2(167)(175)} \:=\:0.973721129

    Hence: . A \:=\:13.16426633 \:\approx\:13^o

    Therefore, the heading is: . \boxed{N\,5^o\,W}

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