# trignometry

• Sep 7th 2008, 02:42 AM
dolly kohli
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.
• Sep 7th 2008, 07:30 PM
Soroban
Hello, dolly kohli!

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

Quote:

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

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

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

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

Quote:

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

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

Quote:

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

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

Quote:

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?

• Sep 8th 2008, 02:06 AM
dolly kohli
thanku for ur help but i wanted the last three answers because first 3 even i got correct
• Sep 8th 2008, 03:43 AM
mr fantastic
Quote:

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 (Headbang)
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