Hello, dolly kohli!

Did you make a sketch?

I'll get you started on this one . . .

An airplane sets off from $\displaystyle G$ on a bearing of 024 degrees

towards $\displaystyle H$ a point 250 km away. At $\displaystyle H$ it heads towards $\displaystyle J$

on a bearing of 055 degrees and a distance of 180 km away. Code:

Q o J
: 180 * :
: * :
:55°* :
: * 35° :
P H o - - - - + C
: /: :
: 240/ : :
: / : :
:24°/ : :
: / : :
: / : :
:/ 66° : :
G o - - - + - - - - +
A B

$\displaystyle \angle PGH = 24^o \quad\Rightarrow\quad \angle HGA = 66^o \quad\Rightarrow\quad \angle GHA = 24^o$

$\displaystyle \angle QHJ = 55^o \quad\Rightarrow\quad \angle JHC = 35^o$

$\displaystyle GH = 240,\; HJ = 180$

1)How far is $\displaystyle H$ to the north of $\displaystyle G$ ? We want $\displaystyle HA.$

In $\displaystyle \Delta HAG\!:\;\;\sin66^o = \frac{HA}{240} \quad\Rightarrow\quad HA \:=\:240\sin66^o \:\approx\: 219.25$

2)How far is $\displaystyle H$ to the east of $\displaystyle G$ ? We want GA.

In $\displaystyle \Delta HAG\!:\;\;\cos66^o = \frac{GA}{240} \quad\Rightarrow\quad GA \:=\:240\cos66^o \:\approx\:97.62$

3) How far is $\displaystyle J$ to the north of $\displaystyle H$ ? We want $\displaystyle JC.$

In $\displaystyle \Delta JCH\!:\;\;\sin35^o = \frac{JHC}{180} \quad\Rightarrow\quad JC \:=\:180\sin35^o \:\approx\:103.24$

Are you getting the idea?