Answer
(a) $62.6 mi$
(b) South $18.2^{\circ} East$
Work Step by Step
(a) Use the figure given in the Exercise, in the triangle of interest, two sides are given as 30mi and 50mi, the angel between the two sides is given by $180-70-10=100^{\circ}$ using the geometry. The distance (d) between the fisherman’s home port and Forrest Island, based on the Law of Cosines, $d^2=30^2+50^2-2\times30\times50\times cos100^{\circ}$ which gives $d\approx62.6 mi$
(b) The bearing angle from Forrest Island back to his home port is the angle formed between the connection (of Forest Island to home port) and the vertical direction (N). This angle is the top angle $\theta$ of the triangle less $10^{\circ}$ as shown in the figure. Use the Law of Sines $\frac{sin\theta}{30}=\frac{sin100^{\circ}}{62.6}$ which gives $\theta\approx28.2^{\circ}$ so the bearing is South $18.2^{\circ} East$
