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.