2.5 @ W45N = -1.77i + 1.77j
4.7 @ E60S = 2.35i - 4.07j
do the others likewise.
Then add them all up and convert back to a distance and bearing
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines:
2.5 km 45° north of west; then
4.70 km 60° south of east; then
5.1 km straight east; then
7.2 km 55° south of west; and finally
2.8 km 10° north of east.
What is his final position relative to the island?
2 answers
All angles are measured CCW from +x-axis.
Disp. = 2.5[135o]+4.7[300]+5.1+
7.2[235]+2.8[10o].
Disp. = -1.77+1.77i + 2.35-4.07i + 5.1 + 4.13-5.90i + 2.76+0.486i
Disp. = 12.6 - 7.71i = 14.8km[ -31.5o] = 14.8km[31.5o] S. of E.
Disp. = 2.5[135o]+4.7[300]+5.1+
7.2[235]+2.8[10o].
Disp. = -1.77+1.77i + 2.35-4.07i + 5.1 + 4.13-5.90i + 2.76+0.486i
Disp. = 12.6 - 7.71i = 14.8km[ -31.5o] = 14.8km[31.5o] S. of E.