Results 1 to 2 of 2

Math Help - Trig/Vector help!

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    2

    Trig/Vector help!

    I tried this question and managed to get an answer of 56.5km/h using the cosine law, however I can still not get a heading angle and am not sure if I used the formular correctly.

    A bird flies at a speed of 22km/h at a bearing of 265degrees. If the wind is blowing at 50km/h in a bearing direction of 10degrees, determine the resultant velocity (side x + direction) to a nearest tenth and degree, respectively.

    Need some help...thanks =]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    692
    Hello, rebeccasnarfer!

    \text{A bird flies at a speed of 22 km/hr at a bearing of 265}^o.
    \text{If the wind is blowing at 50 km/hr in a bearing direction of 10}^o,
    \text{determine the resultant speed to the nearest tenth of a kilometer,}
    \text{and bearing to the nearest degree.}
    Code:
                    C         N
                    *         : 
          P        *  *       :
          :       *     *     :
          :      *        *   :
          :  50 *           * : 
          :    *              * A
          :10d*           *85d:
          :  *        *       :
          : * 75d * 22        : 
          :*  *               :
        B *                   :
                              S

    The bird flies from \,A to \,B at 22 km/hr: . ABV = 22.

    Major angle NAB = 265^o \quad\Rightarrow\quad \angle BAS = 85^o.

    The wind blows from \,B to \,C at 50 km/hr: . BC = 50.

    \angle PBC = 10^o \quad\Rightarrow\quad \angle CBA = 75^o


    In \Delta ABC, use the Law of Cosines:

    . . AC^2 \;=\;50^2 + 22^2 - 2(50(22)\cos75^o \:=\:2414.598101

    Hence: . AC \:\approx\:49.1\text{ km/hr.}



    Let A \,=\,\angle CAB.

    Law of Cosines: . \cos A \;=\;\dfrac{22^2 + 49.1^2 - 50^2}{2(22)(49.1)} \;=\;0.182748565

    . . Hence: . \angle A \;=\;79.47010295^o \;\approx\;79^o

    Then: . \angle CAN \;=\;180^o - 79^o - 85^o \;=\;16^o


    The bird is flying at 49.1 km/hr at a bearing of 344^o.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: January 11th 2011, 10:17 AM
  2. Replies: 2
    Last Post: January 27th 2010, 08:23 PM
  3. Replies: 11
    Last Post: December 23rd 2009, 01:30 AM
  4. Replies: 2
    Last Post: October 5th 2009, 03:25 PM
  5. trig help for vector calc class
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 30th 2008, 02:31 AM

Search Tags


/mathhelpforum @mathhelpforum