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 northsouth 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 $\displaystyle T.$