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Thread: i need help qiet urgently

  1. #1
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    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?
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  2. #2
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    Hello, kasak!

    I'll get you started . . .


    An aircraft flies 400km from a point O on a bearing of 25 to A
    and then 700km on a bearing of 80 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 O?
    Code:
          P     Q                 o B
          :     :              *  |
          :     :           *     |
          :     :        *        |
          :     : 80 *           |
          :    A:  * 10          |
          :     o - - - - - - - - + C
          :    *|                 |
          :25* |                 |
          :  *  |                 |
          : *   |                 |
          :*65 |                 |
          o - - + - - - - - - - - *
          O     D                 E

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

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


    (a) We want the distance BE

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

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

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

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