1. ## Bearings Help

Hi i have a scanned copy of what i need help with. if you could post and email me, that would be AWSOME! but i do need this soon!

Well i just found out, the file is too big, so, if you could email me, i will send you a copy of the sheet. thnx.

but yes, i am going to type the problem, but the Page is more help....

Just had a break through! i just croped the scan to the diagram only, so here is what it says....

A ship is travelling on a straight course at a constant speed, during fog, to avoid detection.

The ship's journey is to be logged by two government run observatory posts on a stretch of shoreline running north/south. (see diagram [this is were you might need the diagram.]). The observatories are 10Km apart.

At midnight observatory B detects a sighting of a ship bearing 045(degrees). At the same time observatory T detects the ship at a bearing of 129(degrees)

You must draw a scaled diagram to represent the situation and calculate the psotion of the ship from both observatory B and T at midnight.

Email : fire_monkey4me@hotmail.com

thnx!

YES I KNOW the image is huge, that is not my fault. sorry![/

2. Hello, firemonkey!

A ship is travelling on a straight course at a constant speed.
Two observation posts $\displaystyle B$ and $\displaystyle T$ are 10 miles apart on a north-south shoreline.
$\displaystyle T$ is north of $\displaystyle B.$

$\displaystyle B$ detects the ship at bearing 045°.
At the same time $\displaystyle T$ detects the ship at bearing of 129°.

Draw a diagram to represent the situation.
Calculate the postion of the ship from both $\displaystyle B$ and $\displaystyle T.$
Code:
      N
|
|
T o 129°
| *
|51°*
|     * y
|       *
|         *
10 |       84° o S
|         *
|       *
|     * x
|45°*
| *
B o
The ship is at $\displaystyle S.$

$\displaystyle \angle TBS = 45^o,\;\angle NTS = 129^o\quad\Rightarrow\quad\angle STB = 51^o$

$\displaystyle \angle S \:=\:180^o - 51^o - 45^o \:=\:84^o$

We want: .$\displaystyle x \,=\,SB,\;y \,=\,ST$

Law of Sines: .$\displaystyle \frac{x}{\sin51^o} \:=\:\frac{10}{\sin84^o}\quad\Rightarrow\quad x \:=\:\frac{10\sin51^o}{\sin84^o} \:=\:7.814266987$

. . Therefore, the ship is about 7.81 miles from $\displaystyle B.$

Law of Sines: .$\displaystyle \frac{y}{\sin45^o} \:=\:\frac{10}{\sin84^o}\quad\Rightarrow\quad y \:=\:\frac{10\sin45^o}{\sin84^o} \:=\:7.11001723$

. . Therefore, the ship is about 7.11 miles from $\displaystyle T.$