Hello, mitchoboy!
Bearings are measured clockwise from North.
And a good diagram is essential.
A ship leaves at port $\displaystyle A$ and travels for 30 km on bearing of 120°.
It then changes course and travels for 50 km on bearing of 080° arriving at port $\displaystyle B.$
Calculate distance $\displaystyle AB$ and bearing of $\displaystyle {\color{blue}A}$ from $\displaystyle {\color{blue}B}.$ . Is this correct? Code:
N


A o R
 * * 
60°* * 
 * Q o B
 30 *  *
S *60°80°* 50
*  *
o
P
The ship starts at A and sails 30 km to point $\displaystyle P$:
. . $\displaystyle AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$
Then it turns and sails 50 km to point $\displaystyle B$:
. . $\displaystyle PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$
In $\displaystyle \Delta APB$, use the Law of Cosines:
. . $\displaystyle AB^2 \:=\:AP^2 + PB^2  2(AP)(BP)\cos(\angle APB)$
. . $\displaystyle AB^2 \:=\:30^2+50^2  2(30)(50)\cos140^o \:=\:5698.133329$
Therefore: .$\displaystyle \boxed{AB \;\approx\;75.5\text{ km}}$
In $\displaystyle \Delta APB$, use the Law of Cosines.
. . $\displaystyle \cos A \:=\:\frac{75.5^2 + 30^2  50^2}{2(75,5)(30)} \:=\:0.90413245$
Hence: .$\displaystyle \angle A \;\approx\;25.2^o$
Then: .$\displaystyle \angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$
Therefore, the bearing of $\displaystyle A$ from $\displaystyle B$ is: .$\displaystyle 360^o  85.2^o \:=\:\boxed{274.8^o} $