Results 1 to 2 of 2

Thread: 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
    12,028
    Thanks
    848
    Hello, rebeccasnarfer!

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

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

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

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

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


    In $\displaystyle \Delta ABC$, use the Law of Cosines:

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

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



    Let $\displaystyle A \,=\,\angle CAB.$

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

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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

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

Search Tags


/mathhelpforum @mathhelpforum