# Thread: i need help qiet urgently

1. ## 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?

2. 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.

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# a motorist starts from O and moves in a direction 030 for 36km

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