Results 1 to 2 of 2

Thread: i need help qiet urgently

  1. #1
    Banned
    Joined
    Feb 2008
    Posts
    3

    i need help qiet urgently

    AN aircraft flies 400km from a point "o" on a bearing of 025degrees and then 700km from a bearing of 80 degrees to arrive at b.
    a)how far north of o is b?
    b)how far east of o is b?
    c)find the distance and bearing of b from c?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, kasak!

    I'll get you started . . .


    An aircraft flies 400km from a point $\displaystyle O$ on a bearing of 25 to $\displaystyle A$
    and then 700km on a bearing of 80 to arrive at $\displaystyle B$.

    a) How far north of $\displaystyle O$ is $\displaystyle B$?
    b) How far east of $\displaystyle O$ is $\displaystyle B$?
    c) Find the distance and bearing of $\displaystyle B$ from $\displaystyle O$?
    Code:
          P     Q                 o B
          :     :              *  |
          :     :           *     |
          :     :        *        |
          :     : 80 *           |
          :    A:  * 10          |
          :     o - - - - - - - - + C
          :    *|                 |
          :25* |                 |
          :  *  |                 |
          : *   |                 |
          :*65 |                 |
          o - - + - - - - - - - - *
          O     D                 E

    The plane starts at $\displaystyle O$ on a bearing of 25: .$\displaystyle \angle POA = 25^o \quad\Rightarrow\quad \angle AOD = 65^o $
    . . The distance is 400 km: .$\displaystyle OA = 400$

    Then it is on a bearing of 80: .$\displaystyle \angle QOB = 80^o\quad\Rightarrow\quad \angle BOC = 10^o$
    . . The distance is 700 km: .$\displaystyle AB = 700$


    (a) We want the distance $\displaystyle BE$

    In right triangle ADO: .$\displaystyle \sin65^o \:= \:\frac{AD}{400}\quad\Rightarrow\quad AD \:= \:400\sin65^o$
    . . Since $\displaystyle CE = AD\!:\;\;CE \:=\:400\sin65^o$

    In right triangle BCA: .$\displaystyle \sin10^o \:=\:\frac{BC}{700}\quad\Rightarrow\quad BC \:=\:700\sin10^o$

    Therefore: .$\displaystyle BE \;=\;BC + CE \;=\;700\sin10^o + 400\sin65^o \;\approx\;484$ km.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need Help Urgently...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 25th 2009, 05:02 AM
  2. Help urgently!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 19th 2008, 03:17 AM
  3. need help urgently plz
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jun 1st 2008, 11:25 AM
  4. [SOLVED] Need Help Urgently
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 4th 2007, 05:03 AM
  5. Need help urgently
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Jun 13th 2006, 09:04 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum