Question

a ship leaves at noon heads due west at 30knots(nautical miles)(6080 ft) per hour . At 3pm the ship changes course to N58degreesW. FInd the ships bearing and distance from the part of departure at 4 pm

Answer 1

Using law of cosine \[c^2 = a^2+b^2-2ab*\cos(C)\] where c^2 is equal to the total travelled distance (hypothenuse) gives the following equation.
Given numbers:
a = 3*30 = 90 naut miles (Calling this b)
b = 1*30 = 30 naut miles (Calling this a)
Angle: 180-58 = 122
Solving for c and Alpha (Bearing)\[c = \sqrt{30^2+90^2-30*90*\cos(122)}\]
c = 108 Naut miles
Alpha (Bearing) = adj/hypothenuse = 90/108 = 0.826, inverse cosine of that is 34°

Answer 2

c^2 is not equal to total travelled distance, sqrt(c^2) is. My bad.