A ship travels 84 km on a bearing of 10 degrees, and then travels on a bearing of 100 degrees for 146 km. Find the distance of the trip from the starting point, to the nearest kilometer.
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|D| = √[(84cos10+146cos100)² + (84sin10+146sin100)²] ≈ 168 km
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When doing navigation type problems it is convienent to use a coordinate system with x North and y East.
and use heading (the angle from North that is measured positive toward the east)
a vector v of magnitude vm is then < vm cos(heading), vm sin(heading)>
bearing and heading are the same
vaMH = 84 km, Heading =10 deg.
va = Calc Vector from vaMH = < 82.72385, 14.58645> km
vbMH = 146 km, Heading =100 deg.
vb = Calc Vector from vbMH = < -25.35263, 143.78193> km
km
vc = va + vb = < 57.37122, 158.36838> km
start = < 0, 0>
vd = start - vc = < -57.37122, -158.36838>
vcMH = Calc Heading & Mag of vd = 168.43990 , Heading =250.08635 deg.
The distance of the trip is 84+146 = 230 km
The distance from the starting point is 168.4 km
The heading back to the starting point is 250 deg.
and use heading (the angle from North that is measured positive toward the east)
a vector v of magnitude vm is then < vm cos(heading), vm sin(heading)>
bearing and heading are the same
vaMH = 84 km, Heading =10 deg.
va = Calc Vector from vaMH = < 82.72385, 14.58645> km
vbMH = 146 km, Heading =100 deg.
vb = Calc Vector from vbMH = < -25.35263, 143.78193> km
km
vc = va + vb = < 57.37122, 158.36838> km
start = < 0, 0>
vd = start - vc = < -57.37122, -158.36838>
vcMH = Calc Heading & Mag of vd = 168.43990 , Heading =250.08635 deg.
The distance of the trip is 84+146 = 230 km
The distance from the starting point is 168.4 km
The heading back to the starting point is 250 deg.