A ship leaves port and travels 40 miles at a bearing of N50°W. The ship then changes direction to a bearing of S20°W and travels another 100 miles.

Question

Find the ships distance from home to port and find the bearing back to the home port.

Answer 1

Sketch this out: in quadrant II the first leg of length 40 is 50° to left of positive y-axis, giving a reference angle of 40°. Δx₁ = -40 cos 40 = -30.64, and Δy₁ = +40 sin 40 = +25.71. The second leg of length 100 goes down from here and somewhat to the left at an angle of 20° from the negative y-axis.

I made a right triangle by extending a vertical line Δy₂ down from point 1, with a horizontal line Δx₂ connecting this to the tip of the second vector (point 2). Now Δx₂ = opposite side and Δy₂ = adjacent side of 20° angle; Δx₂ = -100 sin 20 = -34.20 and Δy₂ = -100 cos 20 = -93.97. Add the x and y components separately to get total Δx = -64.84 and Δy = -68.26. The distance formula gives 94.15 mi for the distance from port.

Finally, find an angle θ from tan θ = (-68.26) / (-64.84) yielding θ = 46.5° below the negative x axis, for a bearing 43.5° left of negative y-axis, S43.5W.