Math Help - trignometry

1. trignometry

please solve this sum for me---
an aeroplane sets off from G on a bearing of 024 degrees towards H a point 250 km away .At H it changes course and heads towards J on a bearing of 055 degrees and a distance of 180 km away.
1)How far is H to the north of G.
2)How far is H to the east of G.
3)How far is J to the north of H.
4)How far is J to the east of H.
5)What is the shortest distance between G and J.
6)What is the bearing of G from J.

2. Hello, dolly kohli!

Did you make a sketch?
I'll get you started on this one . . .

An airplane sets off from $G$ on a bearing of 024 degrees
towards $H$ a point 250 km away. At $H$ it heads towards $J$
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

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

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

$GH = 240,\; HJ = 180$

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

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

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

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

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

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

Are you getting the idea?

3. thanku for ur help but i wanted the last three answers because first 3 even i got correct

4. Originally Posted by dolly kohli
thanku for ur help but i wanted the last three answers because first 3 even i got correct
It would have saved a lot of time if you'd said that in your original post and given the answers