A fisherman leaves his home port and heads in the direction N 70 ° W. He travels d1 = 40 mi and reaches Egg Island. The next day he sails N 10 ° E for d2 = 65 mi, reaching Forrest Island.
(a) Find the distance between the fisherman's home port and Forrest Island. (Round your answer to two decimal places.)
(b) Find the bearing from Forrest Island back to his home port. (Round your answer to one decimal place.)
S ° E

From the image attached, we can see the angles and distance depicted as given in the question. Using parallel angles, we have been able to establish that the internal angle at egg island is 100°.

A) Thus, we can find the distance between the home port and forrest island using law of cosines which is that;

a² = b² + c² - 2bc Cos A

Thus, let the distance between the home port and forrest island be x.

So,

x² = 40² + 65² - 2(40 × 65)cos 100

x² = 1600 + 4225 - (2 × 2600 × -0.1736)

x² = 6727.72

x = √6727.72

x = 82.02 mi

B) To find the bearing from Forrest Island back to his home port, we will make use of law of sines which is that;

A/sinA = b/sinB = c/sinC

82.02/sin 100 = 40/sinθ

Cross multiply to get;

sinθ = (40 × sin 100)/82.02

sin θ = 0.4803

θ = sin^(-1) 0.4803

θ = 28.7°

From the diagram we can see that from parallel angles, 10° is part of the total angle θ.

Thus, the bearing from Forrest Island back to his home port is;

28.7 - 10 = 18.7° SE