Hello, qbkr21!

A ship is sailing due north.

At a certain point the bearing of a lighthouse is $\displaystyle N\,48.6\,E$ and the distance is 12.5 (miles?)

After a while, the captain notices that the bearing of the lighthouse is now $\displaystyle S\,42.9\,E$.

How far did the ship travel between the two observations of the lighthouse? Code:

B *
| *
| *
|42.9°*
| *
c| *
| 88.5°* C
| *
|48.6°* b=12.5
| *
A *

At $\displaystyle A$ the ship takes a bearing of the lighthouse at $\displaystyle C$

. . and finds angle $\displaystyle A = 48.6^o$ and $\displaystyle b = AC = 12.5$ miles.

Later at point $\displaystyle B$, they find that angle $\displaystyle B = 42.9^o$.

. . Hence, angle $\displaystyle C\:=\:180^o - 48.6^o - 42.9^o \:=\:88.5^o$

To find side $\displaystyle c$, use the Law of Sines: .$\displaystyle \frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}$

So we have: .$\displaystyle c \;= \;\frac{12.5\sin88.5^o}{\sin42.9^o} \;=\;18.35659392$

Therefore, the distance is about $\displaystyle 18.4$ miles.