Anonymous

Let t hours be the sailing time. Add the northing components of the two courses. 18(2)cos(130°) + 18tcos(0°) = 0 2cos(130°) + t(1) = 0 t = -2cos(130°) t ≈ 1.286 1.286 hr ≈ 1hr 17 min

Anonymous

Let x nautical miles = distance sailing due north x = 36sin40 nautical miles Time = 36sin40/18 = 1.2856 hours Answer 1 hour 17 minutes

Anonymous

2 hrs at 18 knots = 36 nm 36sin(130°) = 27.578 nm east 36cos(130°) = 23.140 nm south ^^ where the ship is after the first 2 hours So it needs to sail 23.14 nm north at a rate of 18 knots. This will take 23.14/18 = 1.286 hours. (1 hour, 17 min, 10 sec)