A ship leaves a port and travels along a bearing of 165 degrees for 270km. To avoid an incoming storm, it then changes course to a bearing of 203 degrees for another 130km and anchors. Find its bearing and distance from port
+36915 165 degrees magnitude 270 firnd the x and y components
x = 270 cos 165 = - 260.8
y = 270 sin 165 = 69.9
203 degrees magnitude 130 find the x and y components
x = 130 cos 203 = -119.7 (edited)
y= 130 sin 203 = -50.8 (edited)
Add the x components and the y components
x =- 260.8+-119.7 = -380.5
y = 69.9 - 50.8 = 19.1
d = sqrt (x^2 + y^2) = 381 km from origin bearing = arctan (y/x) = 177.1 degrees
