Hello, Brndo4u!

$\displaystyle \text{The bearing of a lighthouse from a ship was found to be }N\,37^o\,E.$

$\displaystyle \text{After the ship sailed 2.5 mi due south, the new bearing was }N\,25^o\,E.$

$\displaystyle \text{Find the distance between the ship and the lighthouse at each location.}$

Code:

N ♠ L
: * *
: *12o*
: * *
: 37o * *
: * *
A ♥ 143o *
: *
: *
2.5 : 25o *
: *
: *
B ♥

The lighthouse is at $\displaystyle \,L.$

When the shp is at $\displaystyle \,A,\:\angle N\!AL = 37^o.$

The ship sails 2.5 miles to $\displaystyle \,B\!:\:AB = 2.5;\;\angle NBL = 25^o$

We see that $\displaystyle \angle LAB = 143^o,\,\text{ hence: }\:\angle ALB = 12^o$

In $\displaystyle \Delta ALB$, use the Law of Sines:

. . $\displaystyle \displaystyle \frac{AL}{\sin25^o} \:=\:\frac{2.5}{\sin12^o} \quad\Rightarrow\quad AL \:\approx\:5.08\text{ miles.}$

. . $\displaystyle \displaystyle \frac{BL}{\sin143^o} \:=\:\frac{2.5}{\sin12^o} \quad\Rightarrow\quad BL \:\approx\:7.24\text{ miles.}$