A ship leaves port and travels 30 miles at a bearing of N42°E. The ship then changes direction to a bearing of S50°E and continues for another 50 miles.

Question

find the bearing from the ship back to the home port , and find the distance to the port.

Answer 1

Draw and label a diagram.

The ship makes a right turn of 92° (42° + 50°)

Use Law of Cosines to determine distance to port.

Use Law of Sines to determine angle of turn and bearing to port.

Answer 2

Sketch from the origin (port) a diagonal in quadrant I at 42° from North, making it 48° above the x-axis. From the end of this vector, drop a dashed line straight down and draw the second vector angling 50° off to the right as it drops. Consider x and y component lengths of each vector, in the first case from the origin and in the second case from where the ship turns: x₁ = 30 cos 48° = 20.07, y₁ = 30 sin 48° = 22.29, x₂ = 50 sin 50° = 38.30, y₂ = - 50 cos 50° = -32.14. Final bearing back to port comes from the angle θ of the final position of the ship below the positive x-axis, tan θ = y/x = -9.85 / 58.37 from x = x₁ + x₂ and y = y₁ + y₂. This gives θ = tan^-1 (-0.1686) = -9.6° for the direction of the ship from port, but the problem asks for the bearing of the port from the ship which is 170.4° from x-axis and 80.4 from y-axis giving bearing = N80.4W.

Distance from port d² = x² + y², d = 59.2 miles.