i need help qiet urgently

**kasak** · February 26th 2008, 09:41 AM

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?

**Soroban** · February 26th 2008, 01:28 PM

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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