Hello, aeroflix!
A fisherman leaves his homeport and heads in the direction $\displaystyle N\,70^o\,W$.
He sails 30 miles and reaches Egg Island.
The next day he sails $\displaystyle N\,10^o\,E$ for 50 miles to Forest Island.
(b) Find the bearing from Forest Island back to his homeport.
In my diagram, I omitted the distances (they wouldn't fit).
I hope you read it.
Code:
F
♥
Q *:
: * :*
:10* :
: * : * P
:* S :
E ♥100 * :
: * :
:70 * * :
: * 70:
R * *:
* :
♥ H
The homeport is $\displaystyle H.$
He sails 30 miles to $\displaystyle E.$
. . $\displaystyle \angle PHE = \angle HER = 70^o,\;HE = 30$
Then he sails 50 miles to $\displaystyle F.$
. . $\displaystyle \angle QEF = 10^o \quad\Rightarrow\quad \angle FEH = 100^o,\;\; EF = 50$
I assume that part (a) asked for the distance from $\displaystyle F$ to $\displaystyle H.$
. . $\displaystyle FH^2 \;=\;30^2 + 50^2 - 2(30)(50)\cos100^o \;=\;3920.944533 $
Hence: .$\displaystyle FH \;=\;62.61744592 \;\approx\;62.62$
Now we determine $\displaystyle \angle FHE.$
. . $\displaystyle \cos(\angle FHE) \;=\;\dfrac{30^2 + 62.62^2 - 50^2}{2(30)(62.62)} \;=\; 0.61781763$
Hence: .$\displaystyle \angle FHE \;=\;52.84305918^o \;\approx\;51.8^o$
Then:. . $\displaystyle \angle PHF \;=\;70^o - 51.8^o \;=\;18.2^o$
Note that: .$\displaystyle \angle SFH = \angle PHF$
Therefore, the bearing is: .$\displaystyle S\,18.2^o\,E$