Hello, musicbeat!

pickslides is absolutely correct!

A fisherman leaves his home port and heads in the direction N 70° W.

He travels 30 mi and reaches Egg Island.

he next day he sails N 10° E for 50 mi, reaching Forrest Island.

a) Find the distance between the fisherman's home port and Forrest Island.

b) Find the bearing from Forrest Island back to his home port. Code:

F
N o
| *
| 10° * Q
| * 50 |
| * |
E o 100° |
| * 30 |
|70° * 70°|
| * |
| o
S P

The fisherman's port is $\displaystyle P.$

He sails 30 miles to Egg Island $\displaystyle (E)\!:\;\angle QPE = 70^o,\;PE = 30$

Then he sails 50 miles to Forrest Island $\displaystyle (F)\!:\;\angle NEF = 10^o,\;EF = 50$

Draw line $\displaystyle FP.$

Since $\displaystyle NS \parallel QP\!:\;\;\angle SEP = \angle QPE = 70^o$

. . Hence: .$\displaystyle \angle FEP \,=\,100^o$

Using the Law of Cosines on $\displaystyle \Delta FEP$, we have:

.$\displaystyle d^2 \:=\:F\!P^2 \:=\:50^2 + 30^2 - 2(50)(30)\cos100^o \;=\;3920.944533$

Therefore: .$\displaystyle d \;=\;62.61744592 \;\approx\;63\text{ miles}$

Again, using the Law of Cosines:

. . $\displaystyle \cos(\angle FPE) \:=\:\frac{30^2 + 63^2 - 50^2}{2(30)(63)} \;=\;0.626719577$

Hence: .$\displaystyle \angle FPE \;=\;51.13148859^o \;\approx\;51^o$

Then: .$\displaystyle \angle FPQ \;=\;70^o - 51^o \;=\;19^o$

The heading from $\displaystyle F$ to $\displaystyle P$ is: .$\displaystyle S\,19^o\,E$