math  bearings
posted by anonymous
Been working on this question for a while. Have drawn a diagram (looks like a tilted trapezium, correct me if i'm wrong), but I still can't find any right angles to calculate the distance. Help would be appreciated, thanks.
Question:
A plane flies on a true bearing of 320° for 450 km. It then flies on a true bearing of 350° for 130 km and finally on a true bearing of 050° for 330 km. How far north of its starting point is the plane?

Damon
First of all you do not fly on a "bearing". You fly, or sail, on a "heading". A "bearing" is the direction a lighthouse or whatever is from you. Math texts are totally ignorant of navigation.
Anyway, let's do it in conventional xy axes:
320 is 50 degrees above x axis
so
in x y coordinates
x1 = 450 cos 50
y1 = +450 sin 50
350 is 80 degrees above x axis
so
in x y coordinates
x2 = 130 cos 80
y2 = +130 sin 80
50 is 40 degrees above +x axis
so
in x y coordinates
x3 = 330 cos 40
y3 = 330 sin 40
NOW ADD those orthogonal (perpendicular to each other) vector components
x = x1 + x2 + x3
y = y1 + y2 + y3
north dist = y
no need to even calculate sqrt(x^2 + y^2) :) 
anonymous
Wow, you are really good at this, and also BIG thanks for the extremely quick reply and easytounderstand solution <3

Damon
You are welcome :)
( I have spent some time at sea :) 
Henry
When using bearings, multiplying the distance by the Cos of the angle gives the vertical component instead of the usual hor. component. We need the vertical component only.
Displacement = 450km[320o] + 130[350o] + 330[50o].
Y = 450*Cos320 + 130*Cos350 + 330*Cos50 = 344.72 + 128.03 + 212.12 = 685 km, North.
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